Answer:0.30 m KI ---- A. Highest boiling point
0.19 m Mg(CH3COO)2 ---- B. Second highest boiling point
0.53 m Glucose(nonelectrolyte) ---- Third highest boiling point-C
0.13 m FeCl3---- Lowest boiling point-D
Explanation:
Using the boilng point elevation formula
ΔTb=m* kb *i
where m= molality
kb= elevated boiling point constant( here kb values will be same for all soluton)
i= vant hoff factor = number of ions present in a solution
Using the number of ions and molarity present in a solution as a collagative property, since kb is constant, we can determine which of the species has the highest boiling point.
1.) 0.13 m FeCl3= Fe³⁻ + Cl⁻
i=4
ΔTb=m* kb* i= molarity x number of ionsx Kb= 0.13 x 4= 0.52kb
2) 0.19 m Mg(CH3COO)2 = Mg²⁺ + CH₃COO⁻
i= 3
ΔTb=m* kb* i= molarity x number of ions= 0.19 x 3= 0.57kb
3. 0.30 m KI = K⁺ + I⁻
i= 2
ΔTb=m *kb *= imolarity x number of ions xKb= 0.30x 2= 0.60kb
4. 0.53 m Glucose(nonelectrolyte) =
i= 1 for nonelectroytes
ΔTb=m* kb* i = molarity x number of ionsx Kb= 0.53 x 1= 0.53Kb
therefore,
0.30 m KI ---- A. Highest boiling point
0.19 m Mg(CH3COO)2 ---- B. Second highest boiling point
0.53 m Glucose(nonelectrolyte) ---- Third highest boiling point
0.13 m FeCl3---- Lowest boiling point
Phosphofructokinase is a four‑subunit protein with four active sites. Phosphofructokinase catalyzes step 3 of glycolysis, converting fructose‑6‑phosphate to fructose‑1,6‑bisphosphate. Phosphoenolpyruvate (PEP) is the product of step 9 of glycolysis. The PEP concentration in the cell affects phosphofructokinase activity.Select the true statements about PEP regulation of phosphofructokinase.
1. PEP is a feedback inhibitor of phosphofructokinase.
2. The apparent affinity of phosphofructokinase for its substrate increases when PEP binds.
3. PEP is a positive effector of phosphofructokinase.
4. PEP inhibition of phosphofructokinase yields a sigmoidal velocity versus substrate curve.
5. PEP competes with fructose-6-phosphate for the active site of phosphofructokinase.
6. The binding of PEP to one phosphofructokinase subunit causes a conformation change that affects the ability of the substrate to bind to the other subunits.
Answer:
1. PEP is a feedback inhibitor of phosphofructokinase.
4. PEP inhibition of phosphofructokinase yields a sigmoidal velocity versus substrate curve.
6. The binding of PEP to one phosphofructokinase subunit causes a conformation change that affects the ability of the substrate to bind to the other subunits.
Explanation:
Phosphofructokinase-1, PFK-1, is an allosteric enzymes composed of four protein subunits.
Allosteric enzymes are enzymes that function through non-covalent binding of allosteric modulators which may be activators or inhibitors. They produce a characteristic velocity versus substrate sigmoidal curve. PFK-1 has a separate binding site for its substrate, fructose-6-phosphate and it's allosteric modulators: ATP, ADP or phosphoenolpyruvate, PEP.
The enzyme can exist in two conformations, the T-state (tense) or the R-state (resting). Binding of substrate causes a conformational change from T-state to R-state, whereas binding of allosteric inhibitors returns it to the T-state.
PEP, the product of step 9 in glycolysis, is an allosteric inhibitor of PFK-1. When it binds to the the allosteric site, it leads to conformational changes in PFK-1 from the R-state to the T-state which reduces the enzymes ability to bind the substrate. These changes are responsible for the sigmoidal velocity/substrate curve in allosteric enzymes.
Therefore, the true statements from the options above are 1, 4, 6.
Options 2,3 and 5 are wrong because PEP is a negative effector of PFK-1, thus its binding reduces the affinity of PFK-1 for its substrate. Also, PFK-1 being an allosteric enzyme has separate binding sites for its substrate and its modulators. Thus, there is no competition for active site binding by substrate and modulators.
At that volume is measured to be 755 mm of Hg. If the lungs are compressed to a newA healthy male adult has a lung capacity around 6.00 liters. The pressure in the lungs volume of 3.81 liters, what would be the new pressure in the lungs? What would happen to the air in the lungs?
Answer:
1188.976 mmHg
Explanation:
Initial pressure P1= 755 mmHg
Initial volume V1 = 6.00 litres
Final volume V2 = 3.81 litres
Final pressure P2= the unknown
Now applying Boyle's Law,we have;
P1V1 = P2V2
Since P2 is the unknown then it has to be made the subject of the formula.
P2=P1V1/ V2
P2= 755 × 6.00/ 3.81
P2= 1188.976 mmHg
Therefore, the new pressure is; 1188.976 mmHg
ions always have the same electronic structure as elements in which group of the periodic table?
Answer:
In 0 group of the periodic table
Explanation:
So they will not react with other atoms because they have a full outer shell of electrons and an overall charge of 0.
Hope it helps.
Which correctly lists the three processes that are affected by freeze and thaw cycles?
creep, landslide, and deposition
deposition, creep, and weathering
landslide, slump, and deposition
O slump. weathering, and creep
Answer:
slump. weathering, and creep
Explanation:
Freezing and thawing cycle in geology is the process in which water gets in between soil space or rock cracks, freeze in a cold season, and then melt in a warmer season, exerting a force on the soil or rock around it. This force is due to the expansion and contraction of water when it changes from ice to liquid water.
The three geological processes slump, weathering and creep all depend on thawing and freezing cycle among other factors.
Slump: Slump is a type of geological process that occurs when coherent mass of loosely consolidated materials or a rock layer moves a short distance down a slope. The movement of a slump is characterized by sliding along a concave-upward or planar surface. Causes includes earthquake shocks, thorough wetting, freezing and thawing, undercutting, and loading of a slope.Weathering: This is a geological process that results in the gradual disintegration of rocks into smaller sizes. It is one of the most important soil formation process, and is different from erosion by the degree of movement of the soil formed. Weathering does not move the soil from its origin. Thawing and freezing cycle plays a major role in weathering by helping crack up the rocks and by also tearing the rock apart. plays a major role.Creep: This is the slow, often imperceptible downslope movement of soil or other debris. The effects of creep is often seen in the presence of physical characteristics like bent trees, tilted fences, and cracked walls. Creep is caused by multiple factors, of which heaving is likely the most important process. Heaving involves the expansion and contraction of rock fragments, and occurs during cycles of wetting and drying, as well as freezing and thawing.Answer:
It is slump, weathering and creep
Explanation:
Took the test on edg
Arrange the following compounds in order of increasing solubility in
water and explain your sequence.
C7H15OH C6H13OH C6H6 C2H5OH
Answer:
C6H6<C7H15OH<C6H13OH<C2H5OH
Explanation:
Organic substances are ordinarily nonpolar. This means that they do not dissolve in water. However, certain homologous series of organic compounds actually dissolve in water because they possess certain functional groups that effectively interact with water via hydrogen bonding.
A typical example of this is alcohol family. All members of this homologous series contain the -OH functional group. This group can effectively interact with water via hydrogen bonding, leading to the dissolution of low molecular weight alcohols in water.
Low molecular weight alcohols are miscible with water in all proportions. This implies that they are highly soluble in water. However, as the size of the alkyl moiety in the alcohol increases, the solubility of the alcohol in water decreases due to less effective interaction of the -OH group with water via hydrogen bonding. This explains the fact that C2H5OH is the most soluble alcohol in the list.
C6H6 is insoluble in water since it is purely a hydrocarbon with no -OH group capable of interaction with water via hydrogen bonding.
A vegetable soup recipe requires one teaspoonful of salt. A chef accidentally puts in one tablespoonful. Now the soup is much too salty.
a) What can the chef do to reduce the salty taste of the soup?
b) What effects would your suggestion in a) have on the soup?
Answer:
a. Put a piece of fresh sliced yam with a bore into it into the soup.
Explanation:
b. Osmosis may occur
The chef can put a slice of yam in the soup with a hole in it as it will absorb excess of salt by process of diffusion.
What is diffusion?
Diffusion is defined as the process of movement of molecules which takes place under concentration gradient. It helps in movement of substances in and out from the cell.The molecules move from lower concentration region to a higher concentration region till the concentration becomes equal.
There are 2 main types of diffusion:
1) simple diffusion-process in which substances move through a semi-permeable membrane without the aid of transport proteins.
2) facilitated diffusion- It is a passive movement of molecules across cell membrane from higher concentration region to lower concentration.
There are 2 types of facilitated diffusion one is osmosis and dialysis.
Learn more about diffusion,here:
https://brainly.com/question/14852229
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A laser is used in eye surgery to weld a detached retina back into place. The wavelength of the laser beam is 503 nm, while the power is 1.4 W. During surgery, the laser beam is turned on for 0.070 s. During this time, how many photons are emitted by the laser?
Answer:
Number of proton emmitted by laser=[tex]2.48*10^17proton[/tex]
Explanation:
Energy is the ability to cause change; power is directly proportional to energy and its the rate energy is utilized.
Power=energy/time.
First we need to calculate the total energy used which is equal to the total power utilized.
E(total)= P( total) = 1.4W × 0.070 s =[tex]0.098J[/tex]
CHECK THE ATTACHMENT FOR THE REMAINING DETAILED CALCULATION
The acetate ion is the conjugate base of the weak acid acetic acid. The value of Kb for CH3COO-, is 5.56×10-10. Write the equation for the reaction that goes with this equilibrium constant.
Answer: The equation for the reaction that goes with this equilibrium constant is [tex]5.56\times 10^{-10}=\frac{[CH_3COOH]}{[CH_3COO^-]\times [H^+]}[/tex]
Explanation:
[tex]CH_3COOH\rightarrow CH_3COO^-+H^+[/tex]
Here [tex]CH_3COOH[/tex] donates a proton and thus behaves as an acid and forms [tex]CH_3COO^-[/tex] which is called as the conjugate base of [tex]CH_3COOH[/tex]
The dissociation constant of acids is given by the term [tex]K_a[/tex] and the dissociation constant of bases is given by the term [tex]K_b[/tex] and is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios.
[tex]K_a[/tex] for [tex]CH_3COOH[/tex] :
[tex]K_a=\frac{[CH_3COO^-]\times [H^+]}{[CH_3COOH]}[/tex]
[tex]CH_3COO^-+H^+\rightarrow CH_3COOH[/tex]
[tex]K_b=\frac{[CH_3COOH]}{[CH_3COO^-]\times [H^+]}[/tex]
[tex]5.56\times 10^{-10}=\frac{[CH_3COOH]}{[CH_3COO^-]\times [H^+]}[/tex]
The equation for the reaction that goes with this equilibrium constant is [tex]K_b=\frac{[CH_3COOH]}{[CH_3COO^-]\times [H^+]}[/tex]
Which of the following is an example of a mechanical wave?
O A. A light ray
B. A seismic wave
C. A radio wave
D. An X-ray
Answer:
A seismic wave
Explanation:
It requires a medium for its propagation.
Consider each pair of compounds listed below and determine whether a fractional distillation would be necessary to separate them or if a simple distillation would be sufficient.
a. Ethyl acetate and hexane
b. Diethyl Ether and 1-butanol
c. Bromobenzene and 1,2-dibromobenzene
Consider the following reaction where Kc = 2.90×10-2 at 1150 K: 2 SO3 (g) 2 SO2 (g) + O2 (g) A reaction mixture was found to contain 4.71×10-2 moles of SO3 (g), 5.00×10-2 moles of SO2 (g), and 4.53×10-2 moles of O2 (g), in a 1.00 liter container.
Answer:
The reaction is not in equilibrium and must move in a backward manner i.e towards the reactant so that it will attain equilibrium
Explanation:
The complete question is as follows;
Consider the following reaction where Kc = 2.90×10-2 at 1150 K: 2 SO3 (g) 2 SO2 (g) + O2 (g) A reaction mixture was found to contain 4.71×10-2 moles of SO3 (g), 5.00×10-2 moles of SO2 (g), and 4.53×10-2 moles of O2 (g), in a 1.00 liter container.
Is the reaction at equilibrium? If not, what direction must it run in order to reach equilibrium? The reaction quotient, Qc, equals . The reaction A. must run in the forward direction to reach equilibrium. B. must run in the reverse direction to reach equilibrium. C. is at equilibrium.
Solution
The first thing to do here is to calculate the pressure of each of the gases. This would be useful in the equilibrium calculations. We calculate this by dividing the respective number of moles by the volume of the container.
Now, since the volume of the container is 1L, then the number of moles will be equal to the pressure of the gaseous substances, although units will be different.
So, [SO3] = 4.71 * 10^-2 mol/L
[SO2] = 5.00 * 10^-2 mol/L
[O2] = 4.53 * 10^-2 mol/L
The equation of the reaction is as follows;
[tex]2SO_{3(g)}[/tex] ⇆ [tex]2SO_{2(g)}[/tex] + [tex]O_{2(g)}[/tex]
We proceed to calculate the reaction quotient Qc
Mathematically Qc for this reaction = [[tex]SO_{2}[/tex]]^2 × [[tex]O_{2}[/tex]]/ [[tex]SO_{3}[/tex]]^2
Qc = {(5 * 10^-2)^2 * (4.53 * 10^-2)}/ (4.71 * 10^-2)^2 = 5.11 × 10^-2 mol/L
Now, we are given that the value of Kc = 2.9 * 10^-2 which is less than Qc
Since Kc < Qc, the backward reaction is favored.
Now to the question;
The reaction is not in equilibrium and must move in a backward manner i.e towards the reactant so that it will attain equilibrium
A gas company in Massachusetts charges $2.80 for 15.0 ft3 of natural gas (CH4) measured at 20.0°C and 1.00 atm. Calculate the cost of heating 2.00 × 102 mL of water (enough to make a cup of coffee or tea) from 20.0°C to 100.0°C. Assume that only 50.0% of the heat generated by the combustion is used to heat the water; the rest of the heat is lost to the surroundings. Assume that the products of the combustion of methane are CO2(g) and H2O(l).
Answer:
$0.0238
Explanation:
The energy you need to increase the temperature of water from 20°C to 100°C is obtained from:
Q = C×m×ΔT
Where Q is the energy, C is specific heat of water (4.184J/g°C), m is mass of water (2.00x10²g - Density of water 1g/mL), ΔT is change in temperature (100.0°C - 20.0°C)
Replacing:
Q = 4.184J/g°C × 2.00x10²g × 80.0°C
Q = 66944J = 66.944kJ
As you are assuming the energy of combustion will be just 50.0% to heat the water the energy you need is 66.944kJ × 2 = 133.888kJ
The combustion of methane is:
CH4(g) + 2O2(g) ⟶ CO2(g) + 2H2O(l) ΔH = −890.8kJ
That means 1 mole of methane produce 890.8kJ. As you need 133.888kJ, moles of methane are:
133.888kJ × (1 mol CH₄ / 890.8kJ) = 0.150 moles of CH₄.
Using PV = nRT, moles of 15.0ft³ (424.8L) at 20.0°C (293.15K) and 1.00atm:
1.00atmₓ424.8L = moles CH₄ₓ0.082atmL/molKₓ293.15K
17.67 = moles CH₄
As 17.67 moles of CH₄ cost $2.80, the cost of 0.150 moles of CH₄ is:
0.150 moles CH₄ ₓ ($2.80 / 17.67 moles) =
$0.0238A balanced equation has
Answer:
A balanced equation is an equation for a chemical reaction in which the number of atoms for each element in the reaction and the total charge is the same for both the reactants and the products.In other words, the mass and the charge are balanced on both sides of the reaction.
Explanation:
During chemical reaction 7.55gKI and 9.06g were allowed to react. How many grams of excess reagent are left over after the reaction is complete. Reaction: Pb(NO3)2(s) + 2KCI(s) > 2KNO3(s) + PbI(s)
Answer: 7.45 g of [tex]Pb(NO_3)_2[/tex] excess reagent are left over after the reaction is complete.
Explanation:
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
a) [tex]{\text{Number of moles of} KI}=\frac{7.55g}{166g/mol}=0.045moles[/tex]
b) [tex]{\text{Number of moles of} Pb(NO_3)_2}=\frac{9.06g}{331.2g/mol}=0.027moles[/tex]
The balanced chemical reaction is :
[tex]Pb(NO_3)_2(s)+2KI(s)\rightarrow 2KNO_3(s)+PbI(s)[/tex]
According to stoichiometry :
2 moles of [tex]KI[/tex] require = 1 mole of [tex]Pb(NO_3)_2[/tex]
Thus 0.045 moles of [tex]KI[/tex] will require=[tex]\frac{1}{2}\times 0.045=0.0225moles[/tex] of [tex]Pb(NO_3)_2[/tex]
Thus [tex]KI[/tex] is the limiting reagent as it limits the formation of product and [tex]Pb(NO_3)_2[/tex] is the excess reagent as (0.045-0.0225) = 0.0225 moles are left
Mass of [tex]Pb(NO_3)_2=moles\times {\text {Molar mass}}=0.0225moles\times 331.2g/mol=7.45g[/tex]
Thus 7.45 g of [tex]Pb(NO_3)_2[/tex] of excess reagent are left over after the reaction is complete.
Pick the odd one out?
Ethanol
Hexane
Oil
Carbon tetrachloride
Answer: Ethanol is the odd one out.
Explanation:
A polar compound is defined as the compound which is formed when there is a difference of electronegativities between the atoms. It is also defined as the bond which is formed due to the unequal sharing of electrons between the atoms.
Non-polar compound is defined as the compound which is formed when there is no difference of electronegativities between the atoms or the polarities cancel out.
Hexane [tex](C_6H_{14}), Oil (mixture of hydrocarbons) and carbon tetrachloride [tex](CCl_4)[/tex] all are non polar whereas ethanol is polar due to electronegative difference between hydrogen and oxygen.
Draw every stereoisomer for 1-bromo-2-chloro-1,2-difluorocyclopentane. Use wedge-and-dash bonds for the substituent groups, and be sure that they are drawn on the outside of the ring, adjacent to each other. The skeletal structure of one molecule is included to indicate the proper format.
Answer:
Explanation:
The objective here is mainly drawing the diagrams of every stereoisomer for 1-bromo-2-chloro-1,2-difluorocyclopentane.
Stereoisomerism is the difference of the spatial arrangement of atoms in a molecule or a compound with the same molecular formula.
For 1-bromo-2-chloro-1,2-difluorocyclopentane.
We have the stereoisomers as follows:
(1R,2S)-1-bromo-2-chloro-1,2-difluorocyclopentane.
(1S,2R)-1-bromo-2-chloro-1,2-difluorocyclopentane.
(1S,1S)-1-bromo-2-chloro-1,2-difluorocyclopentane.
(1R,1R)-1-bromo-2-chloro-1,2-difluorocyclopentane.
Their diagrams are drawn and shown in the attached file below in the order with which they are listed above.
molar mass of A1C1 3
Answer:
Gold(III) chloride
Identify the person who made the correct statement.
Mike said petrified fossils are hard and heavy like rock.
Bobby said that petrified fossils have the same appearance as when they were alive.
Neither Mike nor Bobby is correct.
Mike is correct.
Bobby is correct.
Both Mike and Bobby are correct.
Answer: Both Mike and Bobby are correct.
Explanation:
Petrifcation can be defined as the process in which the organic material of the dead living being becomes fossil by the replacement of mineral deposition in the bony, hard material.
Thus although the body components gets decomposed wiped out due to this process. The body shape of the dead organism remains the same as that was in living.
Thus the statements made by Mike and Bobby both are correct. The fossils are hard and have the same appearance as when they were alive.
A tank at is filled with of dinitrogen monoxide gas and of boron trifluoride gas. You can assume both gases behave as ideal gases under these conditions. Calculate the mole fraction and partial pressure of each gas, and the total pressure in the tank. Round each of your answers to significant digits.
Answer:
(1). Mole fraction = 0.152 for sulfur tetrafluoride gas.
Mole fraction = 0.848 For dinitrogen monoxide gas.
(2). Partial Pressure for dinitrogen monoxide gas = 187 kPa
Partial Pressure for sulfur tetrafluoride gas = 33.4 kpa.
(3). Total Partial Pressure = 220.4 kpa.
Explanation:
So, we are given the following data or parameters or information in the question above;
• Volume of the tank = 5.00L per tank;
• Temperature of the tank = 7.03°C;
• The mass of the content in the tank =
17.7g of dinitrogen monoxide gas and
7.77g of sulfur tetrafluoride gas.
So, we will be making use of the formulae below to calculate the MOLE FRACTION:
Moles, n= mass/molar mass and mole fraction = n(1)/ n(1) + n(2) per each constituents.
Moles, n1 = 17.7g of dinitrogen monoxide gas/ 44 grams per mole. =0.4023 moles.
Moles, n2 = 7.77g of sulfur tetrafluoride gas/ 108.1 grams per mole. = 0.07188 moles.
Total numbers of moles = n1 + n2 = 0.47415 moles
Mole fraction =0.4023 / 0.47415 = 0.848 of dinitrogen monoxide gas.
Mole fraction = 0.07188/0.47415 = 0.152 of sulfur tetrafluoride gas.
PART TWO: CALCULATE THE PARTIAL PRESSURE AND TOTAL PRESSURE BY USING THE FORMULA BELOW;
pressure × volume = number of moles × gas constant, R × temperature.
Pressure = n × R × T/ V.
For dinitrogen monoxide gas. ;
Partial Pressure = 0.4023 × 8.314 × 280.03 / 5 × 10^-3 = 187 kPa.
For sulfur tetrafluoride gas
Partial Pressure = 0.07188 × 8.314 ( × 280.03 / 5 × 10^-3. = 33.4 kpa.
(3). Total pressure = (187 + 33.4)kpa = 220.4 kpa
Propane (C3H8) is widely used in liquid form as a fuel for barbecue grills and camp stoves. For 67.7 g of propane, determine the following.(a) Calculate the moles of compound.mol(b) Calculate the grams of carbon.g
Answer:
A. 1.54 mole.
B. 55.39g of carbon
Explanation:
A. Determination of the number of mole in 67.7g of C3H8.
Mass of C3H8 = 67.7g
Molar mass of C3H8 = (3x12) + (8x1) = 44g/mol
Number of mole of C3H8 =..?
Number of mole = Mass/Molar Mass
Number of mole of C3H8 = 67.7/44
Number of mole of C3H8 = 1.54 mole
B. Determination of the mass of carbon in the compound.
This is illustrated below:
The mass of C in compound can be obtained as follow:
=> 3C/C3H8 x 67. 7
=> 3x12 / 44 x 67.7
=> 36/44 x 67.7
=> 55.39g
Therefore, 55.39g of carbon is present in the compound.
A boy with pneumonia has lungs with a volume of 1.7 L that fill with 0.070 mol of air when he inhales. When he exhales, his lung volume decreases to 1.3 L. Enter the number of moles of gas that remain in his lungs after he exhales. Assume constant temperature and pressure.
Answer:
0.053moles
Explanation:
Hello,
To calculate the number of moles of gas remaining in his after he exhale, we'll have to use Avogadro's law which states that the volume of a given mass of gas is directly proportional to its number of moles provided that temperature and pressure are kept constant. Mathematically,
V = kN, k = V / N
V1 / N1 = V2 / N2= V3 / N3 = Vx / Nx
V1 = 1.7L
N1 = 0.070mol
V2 = 1.3L
N2 = ?
From the above equation,
V1 / N1 = V2 / N2
Make N2 the subject of formula
N2 = (N1 × V2) / V1
N2 = (0.07 × 1.3) / 1.7
N2 = 0.053mol
The number of moles of gas in his lungs when he exhale is 0.053 moles
In Chapter 4, we will learn that single bonds experience free rotation at room temperature, while double bonds do not. Consider the two C-N bonds in the structure. One of these bonds exhibits free rotation, as expected for a single bond, but the other C-N bond exhibits restricted rotation. Identify the C-N bond with restricted rotation, and justify your answer by drawing resonance structures.
Answer:
Explanation:
The main objective here is to draw a diagram of an heterocyclic compound containing two C-N bonds in the structure. One with free rotation, as expected for a single bond, but the other C-N bond exhibits restricted rotation. After that ; we will identify the C-N bond with restricted rotation, and also justify our answer by drawing resonance structures.
So; the first image below shows the structure of the heterocyclic compound containing two C-N bonds in the structure with One with free rotation, as expected for a single bond, but the other C-N bond exhibits restricted rotation. From the first diagram. the squared area indicates the C-N bond that exhibits restricted rotation.
The amide bonds in the C-N bonds offers the resonance characteristics and thus exhibits restricted rotation. The resonance is shown in the second image below
A 8.00g of a certain Compound X, known to be made of carbon, hydrogen and perhaps oxygen, and to have a molecular molar mass of 44./gmol, is burned completely in excess oxygen, and the mass of the products carefully measured: product mass carbon dioxide 24.01g water 13.10g Use this information to find the molecular formula of X.
Answer:
C3H6.
Explanation:
Data obtained from the question:
Mass of the compound = 8g
Mass of CO2 = 24.01g
Mass of H2O = 13.10g
Next, we shall determine the mass of C, H and O present in the compound. This is illustrated below:
Molar Mass of CO2 = 12 + (2x16) = 44g/mol
Molar Mass of H2O = (2x1) + 16 = 18g/mol
Mass of C in compound = Mass of C/Molar Mass of CO2 x 24.01
=> 12/44 x 24.01 = 6.5g
Mass of H in the compound = Mass of H/Molar Mass of H2O x 13.1
=> 2x1/18 x 13.1 = 1.5g
Mass of O in the compound = Mass of compound – (mass of C + Mass of H)
=> 8 – (6.5 + 1.5) = 0
Next, we shall determine the empirical formula of the compound. This is illustrated below:
C = 6.5g
H = 1.
Divide by their molar mass
C = 6.5/12 = 0.54
H = 1.4/1 = 1.
Divide by the smallest
C = 0.54/0.54 = 1
H = 1/0.54 = 2
Therefore, the empirical formula is CH2
Finally, we shall determine the molecular formula as follow:
The molecular formula of a compound is a multiple of the empirical formula.
Molecular formula = [CH2]n
[CH2]n = 44
[12 + (2x1)]n = 44
14n = 44
Divide both side by 14
n = 44/14
n = 3
Molecular formula = [CH2]n = [CH2]3 = C3H6
Therefore, the molecular formula of the compound is C3H6
Balance the following chemical equation:
NH4NO3
N20+
H2O
Answer:
NH4NO3 = N2O + 2(H2O)
Explanation:
there are 2 N, 4 H, 3 O
Answer:
NH4NO3=N2O+2H2O
Explanation:
N-2,O-3,H-4
How many moles of h2 can be formed if a 3.25g sample of Mg reacts with excess HCl
Answer:
0.134 moles of H₂ can be formed if a 3.25g sample of Mg reacts with excess HCl
Explanation:
The balanced reaction is:
Mg + 2 HCl → MgCl₂ + H₂
By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles react:
Mg: 1 moleHCl: 2 molesMgCl₂: 1 moleH₂: 1 moleBeing:
Mg: 24. 31 g/moleH: 1 g/moleCl: 35.45 g/molethe molar mass of the compounds participating in the reaction is:
Mg: 24.31 g/moleHCl: 1 g/mole + 35.45 g/mole= 36.45 g/moleMgCl₂: 24.31 g/mole + 2*35.45 g/mole= 95.21 g/moleH₂: 2*1 g/mole= 2 g/moleThen, by stoichiometry of the reaction, the following quantities of mass participate in the reaction:
Mg: 1 mole* 24.31 g/mole= 24.31 gHCl: 2 moles* 36.45 g/mole= 72.9 gMgCl₂: 1 mole* 95.21 g/mole= 95.21 gH₂: 1 mole* 2 g/mole= 2 gThen you can apply the following rule of three: if by stoichiometry 24.31 grams of Mg form 1 mole of H₂, 3.25 grams of Mg how many moles of H₂ will they form?
[tex]moles of H_{2} =\frac{3.25 grams of Mg*1 mole of H_{2} }{24.31 grams of Mg}[/tex]
moles of H₂= 0.134
0.134 moles of H₂ can be formed if a 3.25g sample of Mg reacts with excess HCl
0.134 moles of H₂ are formed by the reaction of 3.25 g of Mg with excess HCl.
Let's consider the balanced equation between Mg and HCl.
Mg + 2 HCl ⇒ MgCl₂ + H₂
The molar mass of Mg is 24.3 g/mol. The moles corresponding to 3.25 g of Mg are:
[tex]3.25 g \times \frac{1mol}{24.3g} = 0.134 mol[/tex]
The molar ratio of Mg to H₂ is 1:1. The moles of H₂ formed by 0.134 moles of Mg are:
[tex]0.134 mol Mg \times \frac{1molH_2}{1molMg} = 0.134molH_2[/tex]
0.134 moles of H₂ are formed by the reaction of 3.25 g of Mg with excess HCl.
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An aqueous KNO3 solution is made using 72.5 g of KNO3 diluted to a total solution volume of 2.00 L. Calculate the molarity, molality, and mass percent of the solution. (Assume the density of 1.05 g/mL for the solution.)
Answer:
The molarity is 0.359[tex]\frac{moles}{L}[/tex]
The molality is 0.354 [tex]\frac{moles}{kg}[/tex]
The mass percent of the solution es 3.45%
Explanation:
Molarity is a unit of concentration that indicates the amount of moles of solute that appear dissolved in each liter of the mixture. It is determined by:
[tex]Molarity (M)=\frac{number of moles of solute}{volume}[/tex]
Being:
K: 39 g/moleN: 14 g/moleO: 16 g/moleThe molar mass of KNO₃ is:
KNO₃= 39 g/mole + 14 g/mole + 3*16 g/mole= 101 g/mole
You can apply the following rule of three: if 101 grams of KNO₃ are present in 1 mole, 72.5 grams in how many moles are present?
[tex]moles of KNO_{3}=\frac{72.5 grams*1 mole}{101 grams}[/tex]
moles of KNO₃= 0.718
So you have:
moles of KNO₃= 0.718volume= 2 LApplying this quantity in the definition of molarity:
[tex]molarity=\frac{0.718 moles}{2 L}[/tex]
Molarity= 0.359[tex]\frac{moles}{L}[/tex]
The molarity is 0.359[tex]\frac{moles}{L}[/tex]
Molality is a way of measuring the concentration of solute in solvent and indicates the amount of moles of solute in each kilogram of solvent.
Then the molality is calculated by:
[tex]Molality=\frac{moles of solute}{mass of solvent in kilograms}[/tex]
Density is defined as the property that matter, whether solid, liquid or gas, has to compress in a given space. That is, it is the amount of mass per unit volume. So, if the density of 1.05 g / mL for the solution indicates that in 1 mL of solution there are 1.05 grams of solution, in 2000 mL (where 2L = 2000 mL, because 1 L = 1000mL) how much mass is there?
[tex]mass=\frac{2000 mL*1.05 grams}{1 mL}[/tex]
mass= 2100 grams
Since mass solution = mass water + mass KNO₃
then mass water = mass solution - mass KNO₃
Being mass solution 2100 grams and mass KNO₃ 72.5 grams, and replacing you get: mass water= 2100 grams - 72.5 grams
mass water= 2,027.5 grams
Then, being:
moles of KNO₃= 0.718mass of solvent in kilograms= 2.0275 kg (being 2,027.5 grams= 2.0275 kilograms because 1,000 grams= 1 kilogram)Replacing in the definition of molality:
[tex]molality=\frac{0.718 moles}{2.0275 kg}[/tex]
molality= 0.354 [tex]\frac{moles}{kg}[/tex]
The molality is 0.354 [tex]\frac{moles}{kg}[/tex]
The mass percent of a solution is the number of grams of solute per 100 grams of solution. Then the mass percent is the mass of the element or solute divided by the mass of the compound or solute and the result of which is multiplied by 100 to give a percentage.
[tex]mass percent=\frac{mass of solute}{mass of solution} *100[/tex]
So, in this case:
[tex]mass percent=\frac{72.5 grams}{2100 grams} *100[/tex]
mass percent= 3.45 % KNO₃ by mass
The mass percent of the solution es 3.45%
The molarity of the solution is 0.36 mol/L. The molality of the solution is 0.34 m. The mass percent of the solution is 3.33%.
Number of moles of KNO3 = mass/molar mass = 72.5 g/101 g/mol = 0.72 moles
Molarity = Number of moles / volume = 0.72 moles/ 2.00 L = 0.36 mol/L
The molality = Number of moles of solute/Mass of solution in kilograms
mass of solution = 1.05 g/mL × 2000 mL = 21000 g or 2.1Kg
Molality of solution = 0.72 moles/2.1 Kg = 0.34 m
Mass percent of solution = mass of solute/mass of solution × 100/1
Mass percent of solution = 72.5 g/ (72.5 g + 2100 g) × 100/1
= 3.33%
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what type of bonds do compounds formed from non metal consist of?
Compounds formed from non-metals consist of molecules. The atoms in a molecule are joined together by covalent bonds. These bonds form when atoms share pairs of electrons.
A sample of carbon dioxide gas at a pressure of 879 mm Hg and a temperature of 65°C, occupies a volume of 14.2 liters. Of the gas is cooled at constant pressure to a temperature of 23°C, the volume of the gas sample will be
Answer:
The correct answer is 12.43 Liters.
Explanation:
Based on the given question, the volume V₁ occupied by the sample of carbon dioxide gas is 14.2 liters at temperature (T₁) 65 degree C or 65+273 K = 338 K.
The gas is cooled at a temperature (T₂) 23 degree C or 273+23 K = 296 K
The volume of the gas (V₂) after cooling can be determined by using the formula,
V₁/T₁ = V₂/T₂
14.2/338 = V₂/296
0.0420 = V₂/296
V₂ = 0.0420 * 296
V₂ = 12.43 Liters.
Consider the following reaction where Kc = 1.80×10-2 at 698 K:
2HI(g) → H2(g) + I2(g)
A reaction mixture was found to contain 0.280 moles of HI (g), 2.09×10^-2 moles of H2 (g), and 4.14×10^-2 moles of I2 (g), in a 1.00 liter container.
Required:
a. Is the reaction at equilibrium?
b. What direction must it run in order to reach equilibrium?
c. The reaction
1. must run in the forward direction to reach equilibrium.
2. must run in the reverse direction to reach equilibrium.
3. is at equilibrium.
Answer:
The system is not in equilibrium and the reaction must run in the forward direction to reach equilibrium.
Explanation:
The reaction quotient Qc is a measure of the relative amount of products and reagents present in a reaction at any given time, which is calculated in a reaction that may not yet have reached equilibrium.
For the reversible reaction aA + bB⇔ cC + dD, where a, b, c and d are the stoichiometric coefficients of the balanced equation, Qc is calculated by:
[tex]Qc=\frac{[C]^{c}*[D]^{d} } {[A]^{a}*[B]^{b}}[/tex]
In this case:
[tex]Qc=\frac{[H_{2} ]*[I_{2} ] } {[HI]^{2}}[/tex]
Since molarity is the concentration of a solution expressed in the number of moles dissolved per liter of solution, you have:
[tex][H_{2} ]=\frac{2.09*10^{-2} moles}{1 Liter}[/tex]=2.09*10⁻² [tex]\frac{moles}{liter}[/tex][tex][I_{2} ]=\frac{4.14*10^{-2} moles}{1 Liter}[/tex]=4.14*10⁻² [tex]\frac{moles}{liter}[/tex][tex][I_{2} ]=\frac{0.280 moles}{1 Liter}[/tex]= 0.280 [tex]\frac{moles}{liter}[/tex]So,
[tex]Qc=\frac{2.09*10^{-2} *4.14*10^{-2} } {0.280^{2} }[/tex]
Qc= 0.011
Comparing Qc with Kc allows to find out the status and evolution of the system:
If the reaction quotient is equal to the equilibrium constant, Qc = Kc, the system has reached chemical equilibrium.
If the reaction quotient is greater than the equilibrium constant, Qc> Kc, the system is not in equilibrium. In this case the direct reaction predominates and there will be more product present than what is obtained at equilibrium. Therefore, this product is used to promote the reverse reaction and reach equilibrium. The system will then evolve to the left to increase the reagent concentration.
If the reaction quotient is less than the equilibrium constant, Qc <Kc, the system is not in equilibrium. The concentration of the reagents is higher than it would be at equilibrium, so the direct reaction predominates. Thus, the system will evolve to the right to increase the concentration of products.
Being Qc=0.011 and Kc=1.80⁻²=0.018, then Qc<Kc. The system is not in equilibrium and the reaction must run in the forward direction to reach equilibrium.
Question 8
1 pts
A closed flask contains a 0.25 moles of O2 which exerts a pressure of
0.50 atm. If 0.75 moles of CO, is added to the container what is the
total pressure in the flask?
Answer:
\large \boxed{\text{2.0 atm}}
Explanation:
We can use Dalton's Law of Partial Pressures:
Each gas in a mixture of gases exerts its pressure separately from the other gases.
0.25 mol of O₂ exerts 0.50 atm.
If you add 0.75 mol of CO, the total amount of gas is
0.25 mol + 0.75 mol = 1.00 mol
[tex]p_{\text{total}} = \text{1.00 mol} \times \dfrac{\text{0.50 atm}}{\text{0.25 mol}}= \textbf{2.0 atm}\\\\\text{The total pressure in the flask is $\large \boxed{\textbf{2.0 atm}}$}[/tex]
The pressure of the closed flask after the addition of 0.75 moles of CO has been 2 atm.
Partial pressure can be defined as the pressure exerted by each gas in a given solution.
The total moles of gas in the container by the addition of CO has been:
Total moles = moles of oxygen + moles of CO
Total moles = 0.25 + 0.75
Total moles = 1 mol.
By using Dalton's law of partial pressure:
Total pressure = total moles [tex]\rm \times\;\dfrac{pressure\;of\;oxygen}{moles\;of\;oxygen}[/tex]
Total pressure = 1 [tex]\rm \times\;\dfrac{0.50}{0.25}[/tex]
Total pressure = 2 atm.
The pressure of the closed flask after the addition of 0.75 moles of CO has been 2 atm.
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