Answer:
B. the molecules in liquid are loosely packed and scattered thus, they cannot be arranged
Give the IUPAC name for the following compound: Multiple Choice (1R,3R)-1-ethyl-3-methylcyclohexane (1S,3S)-1-ethyl-3-methylcyclohexane (1R,3S)-1-ethyl-3-methylcyclohexane (1S,3R)-1-ethyl-3-methylcyclohexane
Answer:
(1R,3R)-1-ethyl-3-methylcyclohexane.
Explanation:
NOTE: The question is not complete since we do not have the diagram to the chemical structure in the question. Kindly check the attached picture for the diagram of the chemical structure.
So, in order to name Enantiomers or chemical structure through the use of the R,S system requires series of rules and regulations to follow for the proper naming.
There is an ethyl attached to the compound as the first substituents and methyl at the third which are the secondary prefix.
=> The longest chain is 6, thus the compound has hexane as the root compound.
=> It is (1R,3R) because when we draw from the highest substituents to the lower substituents, this is done in a clockwise direction.
CAN SOMEONE HELP ME!!
Solutions of each of the hypothetical acids in the following table are prepared with an initial concentration of 0.100 M. Which of the four solutions will have the lowest pH and be most acidic? Explain please.
Acid pKa
HA 4.00
HB 7.00
HC 10.00
HD 11.00
a. HA
b. HB
c. HC
d. HD
e. All will have the same pH because the concentrations are the same.
Answer: HA has lowest pH and it is the most acidic as compared to the rest of given acids.
Explanation:
We know that relation between [tex]pK_a[/tex] and [tex]K_a[/tex] is as follows.
[tex]pK_a = -log K_a[/tex]
This means that more is the value of [tex]K_a[/tex], smaller will be the [tex]pK_a[/tex]. Also, more is the value of [tex]K_a[/tex] smaller will be the pH of a solution.
As, larger is the value of [tex]K_a[/tex] more negative will be the [tex]pK_a[/tex] value. Hence, stronger will be the acid.
In the given options, HA has the smallest [tex]pK_a[/tex] value.
Therefore, we can conclude that HA has lowest pH and it is the most acidic as compared to the rest of given acids.
A sample of helium has a volume of 325 mL and a pressure of 655 mmHg. What will be the pressure, in mmHg, if the sample of helium is compressed to 125 mL (T, n constant)? (Show calculations.)
Answer:
1703 mmHg
Explanation:
Volume and pressure are presumed to be inversely proportional. Hence a change in volume by a factor of 125/325 = 5/13 is expected to change the pressure by a factor of 13/5:
(13/5)(655 mmHg) = 1703 mmHg
Which process absorbs the greatest amount of heat?
a. the cooling of 10 g of liquid water from 100°C to 0°C.
b. the heating of 10 g of liquid water from 0°C to 100°C.
c. the freezing of 10 g of liquid water the melting of 10 g of ice.
d. the condensation of 10 g of gaseous water.
Answer:
b. the heating of 10 g of liquid water from 0°C to 100°C.
Explanation:
Hello,
In this case, we must notice a., c. and d. processes are not actually absorbing heat but releasing it since cooling, freezing and condensation are processes with negative heat sign since matter changes from a state of more energy to a state of less energy. We can prove this by realizing that freezing enthalpy of water is -6.00 kJ/mol, condensation enthalpy of eater is -40.8 kJ/mol and a change of temperature from 100 °C to 0 °C is negative.
In such a way, the only process absorbing heat is b. the heating of 10 g of liquid water from 0°C to 100°C since energy must be added to the system, or absorbed by it in order to attain the heating.
Regards.
The process having the greatest amount of heat is:
b. the heating of 10 g of liquid water from 0°C to 100°C.
Looking at all the options:The options a., c. and d. processes are not actually absorbing heat but releasing it since cooling, freezing and condensation are processes with negative heat sign since matter changes from a state of more energy to a state of less energy.
The freezing enthalpy of water is -6.00 kJ/mol, condensation enthalpy of eater is -40.8 kJ/mol and a change of temperature from 100 °C to 0 °C is negative.
So out of all the options, only process at b is a heating process thus it will absorb greatest amount of heat.
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The major source of aluminum in the world this bauxite (mostly aluminum oxide). It’s thermal decomposition can be represented by:
Al2 O3 (s) —> 2 Al (s) + 3/2 O2 (g)
ΔH rxn = 1676
If aluminum is produced this way, how many grams of aluminum can conform when 1.000×10^3 kJ of heat is transferred?
Answer:
The correct answer is 32.2 grams.
Explanation:
Based on the given information, the enthalpy of formation for aluminum oxide is 1676 kJ/mol. It signifies towards the energy that is required to generate aluminum and oxygen, and both of these exhibit zero enthalpy of formation. Therefore, the ΔHreaction is the required energy to generate 2 moles of aluminum. Thus, the energy needed for the formation of single mole of aluminum is,
ΔHrxn = 1676/2 = 838 kJ/mol
Q or the energy input mentioned in the given case is 1000 kJ. Therefore, the number of moles of Al generated is,
(1000 kJ) / (838 kJ/Al mole) = 1.19 moles of Aluminum
The grams of aluminum produced can be obtained by using the formula,
mass = moles * molecular mass
= 1.19 * 26.98
= 32.2 grams.
In the thermal decomposition of aluminum oxide, the transference of 1.000 × 10³ kJ of heat can produce 32.19 g of Al.
What is a thermochemical equation?A thermochemical equation is a balanced stoichiometric chemical equation that includes the enthalpy change.
Step 1: Write the thermochemical equation.Al₂O₃(s) ⇒ 2 Al(s) + 3/2 O₂(g) ΔH rxn = 1676 kJ
Step 2: Calculate the moles of Al formed when 1.000 × 10³ kJ of heat is transferred.According to the thermochemical equation, 2 moles of Al are formed when 1676 kJ of heat is transferred.
1.000 × 10³ kJ × (2 mol Al/1676 kJ) = 1.193 mol Al
Step 3: Calculate the mass corresponding to 1.193 moles of AlThe molar mass of Al is 26.98 g/mol.
1.193 mol × 26.98 g/mol = 32.19 g
In the thermal decomposition of aluminum oxide, the transference of 1.000 × 10³ kJ of heat can produce 32.19 g of Al.
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Mass is:
measured in kilograms
measured using a scale
affected by gravity
all of the above
differentiate between sol,aerosol and solid soluti
Answer:
Sol is a colloidal suspension with solid particles in a liquid. Foam is formed when many gas particles are trapped in a liquid or solid. Aerosol contains small particles of liquid or solid dispersed in a gas. While solid solution contain solid as solute in either solid, liquid or gas.
(a) Titration curve for the titration of 5.00 mL 0.010 M NaOH(aq) with 0.005 M HCl(aq), indicating the pH of the initial and final solutions and the pH at the stoichiometric point.
What volume of HCl has been added at
(b) the stoichiometric point
(c) the halfway point of the titration?
Answer:
AT STOICHIOMETRIC POINT, THE VOLUME OF ACID ADDED IS 0.01 L
AT HALF-WAY POINT, THE VOLUME OF ACID IS 0.0050 L
Explanation:
In solving titration problems, you must remember this formula;
MaVa = MbVb
Since M a= 0.005 M
Mb = 0.010 M
Vb = 5 mL = 5 /1000 = 0.005 L
Va = unknown.
Solving for Va, we have:
Va = MbVb / Ma
Va = 0.010 * 0.005 / 0.005
Va = 0.01 L
So therefore, the volume of acid added at:
1. the stoichiometric point is 0.01 L
2. half-way point of titration is 0.01 /2 = 0.0050 L
For the pH:
Since HCl is a strong acid, it dissociate into {H30}+ ion.
First calculate the number of moles of hydronium ion
number of mole = concentration of hydronium ion {H30}+ * Volume
n = 0.005 * 0.01 = 0.00005 moles
A. At initial point of the titration, the volume of base added is 0 L
{H30]+ = n(H+)/ V = 0.00005 / 0.01 = 0.005 M
pH = - log {0.005}
pH = 2.3
B. At the final point, since the volumes and concentrations of acid and base are the same, the pH is equal to 7.
n(H+) = n(OH^-)
pH = 7
Using the following balanced chemical equation 8 H2 + S8à 8 H2S. Determine the mass of the product (molar mass = 34.08g/mol) if you start with 1.35 g of hydrogen and 6.86 g of S8 (Molar mass = 256.5 g/mole).
Answer: 7.29 g of [tex]H_2S[/tex] will be produced from the given masses of both reactants.
Explanation:
To calculate the moles :
[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]
[tex]{\text{Moles of} H_2}=\frac{1.35g}{2.01g/mol}=0.672moles[/tex]
[tex]\text{Moles of} S_8=\frac{6.86g}{256.5g/mol}=0.0267moles[/tex]
[tex]8H_2+S_8\rightarrow 8H_2S[/tex]
According to stoichiometry :
1 mole of [tex]S_8[/tex] require = 8 moles of [tex]H_2[/tex]
Thus 0.0267 moles of [tex]S_8[/tex] will require=[tex]\frac{8}{1}\times 0.0267=0.214moles[/tex] of [tex]H_2[/tex]
Thus [tex]S_8[/tex] is the limiting reagent as it limits the formation of product and [tex]H_2[/tex] is the excess reagent.
As 1 mole of [tex]S_8[/tex] give = 8 moles of [tex]H_2S[/tex]
Thus 0.0267 moles of [tex]S_8[/tex] give =[tex]\frac{8}{1}\times 0.0267=0.214moles[/tex] of [tex]H_2S[/tex]
Mass of [tex]H_2S=moles\times {\text {Molar mass}}=0.214moles\times 34.08g/mol=7.29g[/tex]
Thus 7.29 g of [tex]H_2S[/tex] will be produced from the given masses of both reactants.
A 25.0 mL solution of quinine was titrated with 1.00 M hydrochloric acid, HCl. It was found that the solution contained 0.125 moles of quinine. What was the pH of the solution after 50.00 mL of the HCl solution were added
Answer:
pH = 9.08
Explanation:
Quinine, C₂₀H₂₄O₂N₂, Q, is a weak base that, in water, has as equilibrium:
Q + H₂O ⇄ QH⁺ + OH⁻
Where pKb is 5.10
Using H-H equation for weak bases:
pOH = pKb + log₁₀ [QH⁺] / [Q]
The reaction of quinine with HCl is:
Q + HCl → QH⁺ + Cl⁻
Initial moles of quinine are 0.125 moles and moles added of HCl are:
0.05000L × (1.00mol / L) = 0.05000moles.
That means after the addition of 50.00mL of the HCl solution, moles of Q and QH⁺ are:
Q = 0.125mol - 0.050mol = 0.075 moles
QH⁺ = 0.050 moles
Replacing in H-H equation:
pOH = 5.10 + log₁₀ [0.050] / [0.075]
pOH = 4.92
As pH = 14 - pOJ
pH = 9.08Le Chatelier's Principle. For the reaction below, if the equilibrium concentrations were NH3 = 2 x 10-4, H3O+ = 2 x 10-4M and NH4+ = 18.0M, what is the equilibrium constant for the reaction and what would happen if you were to add some acid to this reaction? NH3 + H3O+ --> NH4+ + H2O
Answer:
Explanation:
NH3 + H3O+ --> NH4+ + H2O
equllibrium constant =K = [ H2O] [NH4+] / [NH3] [H3O+ ]
=
by inserting thier respecive values can you calcaulte, by the way coniseder [ H2O] =1 ,
From the unbalanced reaction: B2H6 + O2 ---> HBO2 + H2O
How many grams of O2 (32g/mol) will be needed to burn 36.1 g of B2H6 (Molar mass = 27.67g/mol)? ______g
Include the correct number of significant figures in your final answer
Answer: 125 g
Explanation:
To calculate the moles :
[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]
[tex]\text{Moles of} B_2H_6=\frac{36.1g}{17}=1.30moles[/tex]
The balanced reaction is:
[tex]B_2H_6+3O_2\rightarrow 2HBO_2+2H_2O[/tex]
According to stoichiometry :
1 mole of [tex]B_2H_6[/tex] require = 3 moles of [tex]O_2[/tex]
Thus 1.30 moles of [tex]B_2H_6[/tex] will require=[tex]\frac{3}{1}\times 1.30=3.90moles[/tex] of [tex]O_2[/tex]
Mass of [tex]O_2=moles\times {\text {Molar mass}}=3.90moles\times 32g/mol=125g[/tex]
Thus 125 g of [tex]O_2[/tex] will be needed to burn 36.1 g of [tex]B_2H_6[/tex]
A student takes a measured volume of 3.00 M HCl to prepare a 50.0 mL sample of 1.80 M HCI. What volume of 3.00 M HCI
did the student use to make the sample?
Use M,V;-MV
3.70 mL
16.7 ml
30.0 mL
83.3 mL
Mark this and return
Save and Exit
Next
Submit
Answer:
30 mL VOLUME OF 3.0 M HCl SHOULD BE USED BY THE STUDENT TO MAKE A 1.80 M IN 50 mL OF HCl.
Explanation:
M1 = 3.00 M
M2 = 1.80 M
V2 = 50 .0 mL = 50 /1000 L = 0.05 L
V1 = unknown
In solving this question, we know that number of moles of a solution is equal to the molar concentration multiplied by the volume. To compare two samples, we equate both number of moles and substitute for the required component.
So we use the equation:
M1 V1 = M2 V2
V1 = M2 V2 / M1
V2 = 1.80 * 0.05 / 3.0
V2 = 0.09 /3.0
V2 = 0.03 L or 30 mL
To prepare the sample of 1.80 M HCl in 50.0 mL from a 3.0 M HCl, 30 mL volume should be used.
Answer:
C on edg 2021Explanation:
i dont like reading
what is the sign of Mercury
Answer:
The answer is Hg.
Explanation:
Symbol for Mercury is Hg.
A geochemist in the field takes a 46.0 mL sample of water from a rock pool lined with crystals of a certain mineral compound X. He notes the temperature of the pool, 21°C, and caps the sample carefully. Back in the lab, the geochemist filters the sample and then evaporates all the water under vacuum. Crystals of X are left behind. The researcher washes, dries and weighs the crystals. They weigh 0.87 g.
Required:
Using only the information above, can you calculate the solubility of X in water at 21°C? If yes, calculate it.
Answer: The solubility of X in water is [tex]1.891 \times 10^{-2}[/tex] g/ml.
Explanation:
The given data is as follows.
Volume of sample water = 46 ml
Temperature = [tex]21^oC[/tex]
After vaporization, washes and then drying the weight of mineral X = 0.87 g
This means that 46.0 ml of water contains 0.87 g of X. Therefore, grams present in 1 ml of water will be calculated as follows.
1 ml of water = [tex]\frac{0.87 g}{46.0 ml}[/tex]
= [tex]1.891 \times 10^{-2}[/tex] g/ml
Therefore, we can conclude that solubility of X in water is [tex]1.891 \times 10^{-2}[/tex] g/ml.
Given that S is the central atom, draw a Lewis structure of OSF4 in which the formal charges of all atoms are zero. Draw the molecule by placing atoms on the grid and connecting them with bonds. Include all lone pairs of electrons.
Answer:
Here's what I get
Explanation:
A Lewis structure shows the valence electrons surrounding the atoms.
Your structure has two problems:
It shows too many valence electrons It violates the octet rule for O — there are 10 electrons around the O atom.Here's one way to draw a Lewis structure.
1. Draw a trial structure
Make F and O terminal atoms and give each one an octet (Fig. 1).
2. Count the valence electrons in the trial structure
5 BP + 15 LP = 10 + 30 = 40 electrons
3. Check the number of valence electrons available
1 S = 1 × 6 = 6 electrons
1 O = 1 × 6 = 6
4 F = 4 × 7 = 28
TOTAL = 40 electrons
The trial structure has the correct number of electrons.
4. Determine the formal charge on each atom.
To get the formal charges, we cut the covalent bonds in half.
Each atom gets the electrons on its side of the cut.
Formal charge = valence electrons in isolated atom - electrons on bonded atom
FC = VE - BE
(a) On S
VE = 6
BE = 5 bonding electrons = 5
FC = 6 - 5 = +1
(b) On O:
VE = 6
BE = 3 LP(six electrons) + 1 bonding electron = 7
FC = 6 - 7 = -1
(c) On F:
VE = 6
BE = 3 lone pairs(6 electrons) + 1 bonding electron = 6 + 1 =7
FC = 7 - 7 = 0
5. Minimize the formal charges
We must rearrange the valence electrons so that S gets one more and O gets one fewer.
Move a lone pair from the O to make an S=O double bond (Fig. 2).
6. Recalculate the formal charges
(a) On S
VE = 6
BE = (3 bonding electrons) = 6
FC = 6 - 6 = 0
(b) On O:
VE = 6
BE = 2 LP(four electrons) + 2 bonding electrons = 6
FC = 6 - 6 = 0
Fig. 2 shows the Lewis structure in which all atoms have a formal charge of zero.
The formal charge of the atoms can be concluded zero with the bond formation between the sulfur and oxygen atom.
The lewis structure can be defined as the dot structure of the valence bond with the bonded atoms. The formal charge can be calculated with the difference in the valence electrons and the bonding electrons.
The formal charge of an atom can be zero when the valence electrons and the bonding electrons are equal. In the structure of [tex]\rm OSF_4[/tex], the formal charge has been assigned zero with the bond formation resulting in the valence electrons and bonding electrons being equal.
The lewis structure with the central S atom has been attached.
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Pyruvate is the end product of glycolysis. Its further metabolism depends on the organism and on the presence or absence of oxygen. Draw the structure of the product from each reaction as it would exist at pH 7. Include the appropriate hydrogen atoms. Reaction A: aerobic conditions in humans or yeast
The given question is incomplete. The image present in the question for Reaction A is attached below along with the answer.
Explanation:
Pyruvate molecule reacts with Coenzyme A in the presence of oxygen and it results in the formation of acetyl Coenzyme A and carbon dioxide.
The enzyme pyruvae dehydrogenase helps in catalyzing this reaction. As in this biochemical reaction [tex]NAD^{+}[/tex] gets converted into NADH.
This reaction is shown in the image attached below.
The solubility of N2 in water at a particular temperature and at a N2 pressure of 1 atm is 6.8 × 10–4 mol L–1. Calculate the concentration of dissolved N2 in water under normal atmospheric conditions where the partial pressure of N2 is 0.78 atm.
Answer:
The correct answer is 5.30 * 10^-4 mol per L.
Explanation:
Based on Henry's law, in a solution solubility of the gas is directly proportional to the pressure, that is, C is directly proportional to P. Here P is the pressure and C is the concentration of the dissolved gases.
Therefore, it can be written as,
C2/C1 = P2/P1
Here, C1 is 6.8 * 10^-4 mol/L, P1 is 1 atm and P2 is 0.78 atm, then the value of C2 obtained by putting the values in the equation,
C2/(6.8*10^-4) = 0.78/1
C2 = 0.78 * 6.8*10^-4
C2 = 5.30 * 10^-4 mol per L.
Hence, the concentration of dissolved nitrogen at 0.78 atm is 5.30*10^-4 mol/L.
9. Predict the major products formed when: (a) Toluene is sulfonated. (c) Nitrobenzene is brominated. (b) Benzoic acid is nitrated. (d) Isopropylbenzene reacts with acetyl chloride and AlCl3. If the major products would be a mixture of ortho and para isomers, you should so state.
Answer:
a) ortho-para isomers predominates
b) 3-nitrobenzoic acid ( meta isomer predominates)
c) 3-bromo nitrobenzene ( meta isomer predominates)
d) the ortho- para isomers predominates
Explanation:
a) Toluene contains -CH3 which is an ortho- para- director hence the major product of the sulphonation of toluene should be the ortho- para isomers.
b) The major product of the nitration of benzoic acid is 3-nitrobenzoic acid. This is an electrophilic substitution in which the meta isomer predominates.
c) The meta isomer predominates giving 3-bromo nitrobenzene as the major product.
d) The isopropyl group is an ortho- para director hence the ortho- para isomers predominates .
Who proposed the plum pudding model and what does it say about the structure of the atom
Answer:
J. J. Thomson
Explanation:
First proposed by J. J. Thomson in 1904 soon after the discovery of the electron, but before the discovery of the atomic nucleus, the model tried to explain two properties of atoms then known: that electrons are negatively-charged particles and that atoms have no net electric charge.
The fluoride ion is the conjugate base of the weak acid hydrofluoric acid. The value of Kb for F-, is 1.39×10-11. Write the equation for the reaction that goes with this equilibrium constant.
Answer:
F⁻(aq) + H₂O(l) ⇄ HF(aq) + OH⁻(aq)
Explanation:
According to Brönsted-Lowry acid-base theory, an acid is a substance that donates H⁺ ions. In this sense, hydrofluoric acid is an acid according to the following equation.
HF(aq) + H₂O(l) ⇄ F⁻(aq) + H₃O⁺(aq)
According to Brönsted-Lowry acid-base theory, a base is a substance that accepts H⁺ ions. In this sense, the fluoride ion is a base according to the following equation.
F⁻(aq) + H₂O(l) ⇄ HF(aq) + OH⁻(aq)
The equilibrium constant for this reaction is Kb = 1.39 × 10⁻¹¹.
Liquid octane CH3CH26CH3 will react with gaseous oxygen O2 to produce gaseous carbon dioxide CO2 and gaseous water H2O. Suppose 3.4 g of octane is mixed with 15.6 g of oxygen. Calculate the maximum mass of carbon dioxide that could be produced by the chemical reaction. Round your answer to 2 significant digits.
Answer:
10 g of CO2
Explanation:
Equation of the reaction:
CH3(CH2)6CH3 + 17O2 ----> 18H2O + 8CO2
Fom the above balanced equation,
1 mole of Octane gas reacts with 17 moles of oxygen gas to produce 8 moles of CO2
Molar mass of Octane = 114 g/mol
Molar mass of oxygen gas = 32 g/mol
Molar mass of CO2 = 44 g/mol
Therefore, 114 g of Octane reacts completely with 17 * 32g (= 544 g) of oxygen to produce 8 * 44 g(=352g) of CO2.
From the given mass of reactants;
3.4 g of Octane will react with (544 * 3.4)/114 g of oxygen = 16.22g of oxygen.
Therefore oxygen is the limiting reactant.
15.6 g of oxygen will react with (114 * 15.6)/544 g of CO2 = 3.27 g of octane.
Mass of CO2 produced will be
(352 * 15.6)/544 = 10 g of CO2
calculate the moles of 25.2 g Na2S2O8
Answer:
To calculate the moles we must first find the molar mass M
M (Na2S2O8) = (23*2) + (32*2) + (16*8)
= 46 + 64 + 168
= 278g/mol
Molar mass = mass/moles
moles =mass / molar mass
= 25.2/278
= 0.0906mol
Hope this helps.
Which is true regarding a water molecule?
Answer:
Has many answers, but one is that it consists of small polar v shaped molecules with a molecular formula H20.
Explanation:
Water molecules consists of 2 hydrogen atoms bonded with on oxygen atom. Each molecule is electrically neutral but polar, with the center of positive and negative charges located in different places.
Each hydrogen atom has a nucleus consisting of a single positively-charged proton surrounded by a 'cloud' of a single negatively-charged electron and the oxygen atom has a nucleus consisting of eight positively-charged protons and eight uncharged neutrons surrounded by a 'cloud' of eight negatively-charged electrons.
Hoped this helped!
When a sample of Mg(s) reacts completely with O2(g), the Mg(s) loses 5.0 moles of electrons. How many moles of electrons are gained by the O2(g)? *
Answer:
if magnesium looses five moles of electrons, oxygen will also gain five moles of electrons.!
Explanation:
Oxidation refers to the loss of electrons. Any specie that looses electrons in a redox reaction is said to be the reducing agent. Hence the reducing agent participates in the oxidation half equation. In this case, magnesium is the reducing agent.
Reduction has to do with the gain of electrons. The oxidizing agent participates in the reduction half equation. Hence the oxidizing agent is reduced in the redid reaction. The reducing agent in this case is the oxygen molecule.
Oxidation half equation;
Mg(s)-----> Mg^2+(aq) + 2e
Reduction half equation;
O2(g) + 2e ------> 2O^2-(aq)
From the balanced reaction equation, two moles of electrons is transferred.
Hence if magnesium looses five moles of electrons, oxygen will also gain five moles of electrons.
Round to 3 significant figures.
1.4593
Answer : The correct answer is 1.46
Explanation :
The following rules are used to round off a number to the required number of significant figures:
(1) If the rightmost digit to be removed is more than 5, the preceding number is increased by one.
(2) If the rightmost digit to be removed is less than 5, the preceding number is not changed.
(3) If the rightmost digit to be removed is 5, then the preceding number is not changed if it is an even number but it is increased by one if it is an odd number.
(4) The same procedure is follow for decimal values.
As we are given, 1.4593
In the given answer, there are 5 significant figures. Now we have to convert it into 3 significant figures.
According to the rules, round off the given measurement in three significant figures as 1.46
Therefore, the correct answer is 1.46
An equilibrium mixture of the three gases in a 1.00 L flask at 350 K contains 5.35×10-2 M CH2Cl2, 0.173 M CH4 and 0.173 M CCl4. What will be the concentrations of the three gases once equilibrium has been reestablished, if 0.155 mol of CH4(g) is added to the flask?
Answer:
[CH₂Cl₂] = 7.07x10⁻² M
[CH₄] = 0.319 M
[CCl₄] = 0.164 M
Explanation:
The equilibrium reaction is the following:
2CH₂Cl₂(g) ⇄ CH₄(g) + CCl₄(g)
The equilibrium constant of the above reaction is:
[tex] K = \frac{[CH_{4}][CCl_{4}]}{[CH_{2}Cl_{2}]^{2}} = \frac{0.173 M*0.173 M}{(5.35 \cdot 10^{-2} M)^{2}} = 10.5 [/tex]
When 0.155 mol of CH₄(g) is added to the flask we have the following concentration of CH₄:
[tex] C = \frac{\eta}{V} = \frac{0.155 mol}{1.00 L} = 0.155 M [/tex]
[tex]C_{CH_{4}} = 0.328 M[/tex]
Now, the concentrations at the equilibrium are:
2CH₂Cl₂(g) ⇄ CH₄(g) + CCl₄(g)
5.35x10⁻² - 2x 0.328 + x 0.173 + x
[tex]K = \frac{[CH_{4}][CCl_{4}]}{[CH_{2}Cl_{2}]^{2}} = \frac{(0.328 + x)(0.173 + x)}{(5.35 \cdot 10^{-2} - 2x)^{2}}[/tex]
[tex]10.5*(5.35 \cdot 10^{-2} - 2x)^{2} - (0.328 + x)*(0.173 + x) = 0[/tex]
Solving the above equation for x:
x₁ = 0.076 and x₂ = -0.0086
Hence, the concentration of the three gases once equilibrium has been reestablished is:
[CH₂Cl₂] = 5.35x10⁻² - 2(-0.0086) = 7.07x10⁻² M
[CH₄] = 0.328 + (-0.0086) = 0.319 M
[CCl₄] = 0.173 + (-0.0086) = 0.164 M
We took x₂ value because the x₁ value gives a negative CH₂Cl₂ concentration.
I hope it helps you!
ultraviolet photon (λ = 58.4nm) from a helium gas discharge tube is absorbed by a hydrogen molecule which is at rest. Since momentum is conserved, what is the velocity of the hydrogen molecule after absorbing the photon? What is the translational energy of the hydrogen molecule in Jmol-1.
[h = 6.626 x 10-34 Js; NA = 6.022 x 1023 mol-1]
Answer:
Translation energy of 1 mole of H2 molecules = KE x Avogadros number
[tex]= 1.923 * 10^{-26} * 6.022 * 10^{23}\\\\= 0.0116 J \\\\= 1.16 * 10^{-2} \ J[/tex]
Explanation:
Photon wavelength [tex]= 58.4 nm = 58.4 * 10^{-9} m[/tex]
Photon momentum = h/wavelength
[tex]= (6.626 * 10^{-34})/(58.4 * 10^{-9})\\\\ = 1.1346 * 10^{-26} \ kg.m/s[/tex]
Mass of H2 molecule m = molar mass/Avogadros number
[tex]= (2.016)/(6.022 * 10^{23})\\\\= 3.3477 * 10^{-24} \ g = 3.3477 * 10^{-27} \ kg[/tex]
Since momentum is conserved:
Photon momentum = H2 molecule momentum = mass x velocity of H2
[tex]1.1346 * 10^{-26} = 3.3477 * 10^{-27} * v[/tex]
velocity [tex]v = 3.389 m/s = 3.39 m/s[/tex]
Translation energy of 1 H2 molecule = kinectic energy (KE) = (1/2)mv^2
[tex]= 1/2 * 3.3477 * 10^{-27} * 3.389^2\\\\= 1.923 * 10^{-26} J[/tex]
Translation energy of 1 mole of H2 molecules = KE x Avogadros number
[tex]= 1.923 * 10^{-26} * 6.022 * 10^{23}\\\\= 0.0116 J \\\\= 1.16 * 10^{-2} \ J[/tex]
In the compound Fe2O3, iron's oxidation number is +3, and oxygen's oxidation
number is
Answer here
Answer: The oxygen's oxidation number is -2.
Explanation:
For formation of a neutral ionic compound, the charges on cation and anion must be balanced. The cation is formed by loss of electrons by metals and anions are formed by gain of electrons by non metals.
In [tex]Fe_2O_3[/tex], Fe is having an oxidation state of +3 called as cation and oxygen is an anion with oxidation state of -2. Thus they combine and their oxidation states are exchanged and written in simplest whole number ratios to give neutral [tex]Fe_2O_3[/tex]
The cations and anions being oppositely charged attract each other through strong coloumbic forces and form an ionic bond.
A growing concern in agricultural and food chemistry is the presence of residues in food. We use many forms of organic chemicals in agriculture and food chemistry and there is growing concern as to how safe these materials are. Choose an organic chemical used in agricultural of food chemistry and report on the functional groups contained in your compound, the uses of the compound, and the safety of that compound.
Answer:
Monosodium Glutamate (MSG) is a chemical which is used in agricultural of food chemistry.
Explanation:
Monosodium Glutamate (MSG) is a chemical that is used in types of different food as food additives. The functional group that is present in Glutamate are carboxylic acid and amine. This chemical is used in different types of foods which is responsible for enhancing the taste of the food. Monosodium Glutamate is safe if it is used in moderate dose but adversely affected when it is used in large amount.