Answer:
218.3 g/mol
Explanation:
Boiling point elevation occurs when a solute is added to a solvent increasing the boiling point of the solution with regard to the pure solvent.
The law is:
ΔT = Kb×m×i
Where ΔT is change in temperature (3.67°C), Kb is the boiling point constant of the solvent (4.95°C/m), m is molality of the solution and i is Van't Hoff factor (1 for a non-electrolyte).
3.67°C = 4.95°C/m×m×i
0.7414m = molality of the solution (Moles solute / kg solvent).
As the mass of the solvent is 50.0g = 0.0500kg:
0.7414m = Moles solute / 0.0500kg
0.0371 = moles of solute
As the mass of the solute is 8.1g, molar mass of the solute (Ratio between mass in g and moles) is:
8.1g / 0.0371mol =
218.3 g/molQuestion 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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A 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:
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
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
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
How many moles of solute are contained in the following solution: 15.25 mL of a 2.10 M CaCl₂
Answer:
0.032moles
Explanation:
2.10moles in 1000ml what about 15.25ml
(15.25×2.10)÷1000
0.032moles
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 system at equilibrium: P(aq)+Q(aq)⇌3R(aq) Classify each of the following actions by whether it causes a leftward shift, a rightward shift, or no shift in the direction of the net reaction. Drag the appropriate items to their respective bins.
Items:1) Increase [P]2) Increase [Q]3) Increase [R]4) Decrease [P]5) Decrease [Q]6) Decrease [R]7) Triple [P] and reduce [Q] to one third8) Triple both [Q] and [R]
Explanation:
P(aq)+Q(aq)⇌3R(aq)
This problem involves applying LeChatelier's principle.
LeChatelier's principle states that whenever a system in equilibrium is disturbed, the equilibrium position would change in order to annul that change.
1) Increase [P]
This would cause the equilibrium position to shift to the right. This is because more reactions have been added, to annul that change more products have to be formed.
2) Increase [Q]
This would cause the equilibrium position to shift to the right. This is because more reactions have been added, to annul that change more products have to be formed.
3) Increase [R]
This would cause the equlibrium position to shift to the left. This is because more products have been formed, to annul that change more reactants have to be formed.
4) Decrease [P]
This would cause the equlibrium position to shift to the left. This is because there are now less reactants, to annul that change more reactants have to be formed.
5) Decrease [Q]
This would cause the equilibrium position to shift to the left. This is because there are now less reactants, to annul that change more reactants have to be formed.
6) Decrease [R]
This would cause the equilibrium position to shift to the right. This is because there are now less products, to annul that change more products have to be formed.
7) Triple [P] and reduce [Q] to one third
No shift in the direction of the net reaction because both changes cancels each other.
8) Triple both [Q] and [R]
No shift in the direction of the net reaction because both changes cancels each other.
Draw the Lewis structure of H2O. Include any nonbonding electron pairs. Draw the molecule by placing atoms on the grid and connecting them with bonds. Include all lone pairs of electrons. - CHONSPFBrClIXMore Request Answer Part B What is the electron geometry of H2O
Answer:
Concepts and reason
Lewis structure is a structure that explains the bonding between atoms of a molecule and lone pair of electrons that is present in the molecule is called a Lewis structure.
With the help of Lewis structure the electronic geometry of a molecule can be determine.
Fundamentals
According to Lewis structure, every atom and their position in the structure of a molecule by using its chemical symbol.
Lines connecting the atoms that are bonded to them are drawn. Lone pairs are expressed by pairs of dots and are located beside the atoms.
Lewis structure of [tex]H_{2}O[/tex] is, the total number of valence electrons is eight in [tex]H_{2}O[/tex].
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
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.
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]
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.
Three different students determined the density of a metal object. Here are their results: 15.12 g/mL, 15.09 g/mL, and 15.12 g/mL. The actual density of the object was 14.41 g/mL. Calculate the percent error. Make sure to include units with your answer, units are %.
Answer:
The correct answers are 4.93 %, 4.72 % and 4.93 %.
Explanation:
Based on the given question, 14.41 g per ml is the actual density of the object. However, the density determined by three different students of the object is 15.12 g per ml, 15.09 g per ml, and 15.12 g per ml. The percent error can be calculated by using the formula,
% error = (actual value - calculated value) / actual value * 100
By 1st student, the calculated value is 15.12 g per ml, the percent error will be,
% error = (14.41 - 15.12) / 14.41 * 100
= 0.71/14.41 * 100
= 4.93 %
By 2nd student, the calculated value is 15.09 g per ml, the percent error will be,
% error = (14.41-15.09)/14.41 * 100
= 0.68/14.41 * 100
= 4.72 %
By 3rd student, the calculated value is 15.12 g per ml, the percent error will be,
% error = (14.41-15.12)/14.41 * 100
= 0.71/14.41 * 100
= 4.93 %
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) =
$0.0238Which 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.
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.
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.
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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The transfer of surface water into the ground to become groundwater is known as
and it can replenish an aquifer.
Answer: Recharge
Explanation:
To solve this we must be knowing each and every concept related to groundwater recharge. Therefore, the transfer of surface water into the ground to become groundwater is known as groundwater recharge.
What is groundwater recharge?The water that is added to the aquifer and through unsaturated zone after percolation (or infiltration) following any storm rainfall event is known as groundwater recharge.
In the natural world, rivers, lakes, streams, rain, and snowmelt all contribute to groundwater recharge. Other surface water trickles and through soil, eventually connecting with a source of water underneath the surface, while other surface water has evaporated or enters another watershed.
Therefore, the transfer of surface water into the ground to become groundwater is known as groundwater recharge.
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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
Which of the following is named using the unmodified element name and adding the word "ion"? Select the correct answer below:
a. simple cations (monatomic cations of elements of only one possible charge)
b. simple anions (monatomic anions of elements of only one possible charge)
c. simple protons
d. simple neutrons
Answer:
simple cations (monatomic cations of elements of only one possible charge)
Explanation:
Simple cations (monatomic cations of elements of only one possible charge) are named using the unmodified element name and adding the word "ion"
For example, the Na+ is named the sodium ion.
An atom or molecule with a net electric charge as a result of the loss or gain of one or more electrons is known as an ion.
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.
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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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:
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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.
Calculate the Kc for the following reaction if an initial reaction mixture of 0.500 mole of CO and 1.500 mole of H2 in a 5.00 liter container forms an equilibrium mixture containing 0.198 mole of H2O and corresponding amounts of CO, H2, and CH4.
Answer:
4.41
Explanation:
Step 1: Write the balanced equation
CO(g) + 3 H₂(g) = CH₄(g) + H₂O(g)
Step 2: Calculate the respective concentrations
[tex][CO]_i = \frac{0.500mol}{5.00L} = 0.100M[/tex]
[tex][H_2]_i = \frac{1.500mol}{5.00L} = 0.300M[/tex]
[tex][H_2O]_{eq} = \frac{0.198mol}{5.00L} = 0.0396M[/tex]
Step 3: Make an ICE chart
CO(g) + 3 H₂(g) = CH₄(g) + H₂O(g)
I 0.100 0.300 0 0
C -x -3x +x +x
E 0.100-x 0.300-3x x x
Step 4: Find the value of x
Since the concentration at equilibrium of water is 0.0396 M, x = 0.0396
Step 5: Find the concentrations at equilibrium
[CO] = 0.100-x = 0.100-0.0396 = 0.060 M
[H₂] = 0.300-3x = 0.300-3(0.0396) = 0.181 M
[CH₄] = x = 0.0396 M
[H₂O] = x = 0.0396 M
Step 6: Calculate the equilibrium constant (Kc)
[tex]Kc = \frac{[CH_4] \times [H_2O] }{[CO] \times [H_2]^{3} } = \frac{0.0396 \times 0.0396 }{0.060 \times 0.181^{3} } = 4.41[/tex]
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.
molar mass of A1C1 3
Answer:
Gold(III) chloride
¿What are the units that make up the 3 quantities (mass, volume of a substance and density)?
Answer:
Grams , centimeters cubed, and grams per centimeter
Explanation: