CHECK COMPLETE QUESTION BELOW
At a given temperature the vapor pressures of benzene and toluene are 183 mm Hg and 59.2 mm Hg, respectively. Calculate the total vapor pressure over a solution of benzene and toluene with Xbenzene = 0.580.
Answer:
The total vapor pressure is [tex]81.3 mmHg[/tex]
Explanation:
We will be making use of Dalton and Raoults equation in order to calculate the total pressure,
Which is [tex]PT= (PA × XA) +(PB ×XB)[/tex]
PT= total vapor pressure
From the question
Benene's Mole fraction = 0.580
then to get Mole fraction of toluene we will substract the one of benzene from 1. because total mole fraction is always 1.
= (1 - 0.580) = 0.420
Vapor pressure of benzene given = 183 mmHg
Vapor pressure of toluene given= 59.2 mmHg
If we substitute those value into above equation, we have
PT=(183×0.580)+(59.2×0.420)
=81.3mmHg
Therefore,, the total vapor pressure of the solution is 81.3 mmHg
What is the edge length of a face-centered cubic unit cell that is made of of atoms, each with a radius of 154 pm
Answer:
The edge length of a face-centered cubic unit cell is 435.6 pm.
Explanation:
In a face-centered cubic unit cell, each of the eight corners is occupied by one atom and each of the six faces is occupied by a single atom.
Hence, the number of atoms in an FCC unit cell is:
[tex] 8*\frac{1}{8} + 6*\frac{1}{2} = 4 atoms [/tex]
In a face-centered cubic unit cell, to find the edge length we need to use Pythagorean Theorem:
[tex] a^{2} + a^{2} = (4R)^{2} [/tex] (1)
Where:
a: is the edge length
R: is the radius of each atom = 154 pm
By solving equation (1) for "a" we have:
[tex] a = 2R\sqrt{2} = 2*154 pm*\sqrt{2} = 435.6 pm [/tex]
Therefore, the edge length of a face-centered cubic unit cell is 435.6 pm.
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I 100 mL o a pharmaceutical preparation contains 20 mL o a 50% w/v solution o benzalkonium chloride, what is the percent strength o that agent in the solution
The percentage strength of the agent is 10 %
Since 100 mL of a pharmaceutical preparation contains 20 mL of a 50% w/v solution of benzalkonium chloride, we desire to find the percentage strength of that agent in the 100 mL solution.
Let C₁ = percentage strength of 20 mL solution = 50 %, V₁ = initial volume = 20 mL, C₂ = percentage strength of agent in 100 mL solution and V₂ = final volume = 100 mL.
Since the initial quantity of agent in 20 mL solution equals the final quantity of agent in 100 mL solution, we have that
C₁V₁ = C₂V₂
So, making C₂ = percentage strength of agent in 100 mL solution the subject of the formula, we have
C₂ = C₁V₁/V₂
Substituting the values of the variables into the equation, we have
C₂ = C₁V₁/V₂
C₂ = 50 % × 20 mL/100 mL
C₂ = 50 % × 0.2
C₂ = 10 %
So, the percentage strength of the agent is 10 %.
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"How much NH4Cl, when present in 2.00 liters of 0.200 M ammonia, will give a solution with pH = 8.20? For NH3, Kb = 1.8 x 10-5"
Answer:
245.66g of NH₄Cl is the mass we need to add to obtain the desire pH
Explanation:
The mixture of NH3/NH4Cl produce a buffer. We can find the pH of a buffer using H-H equation:
pH = pKa + log [A⁻] / [HA]
Where [A⁻] is the molar concentration of the base, NH₃, and [HA] molar concentration of the acid, NH₄⁺. This molar concentration can be taken as the moles of each chemical
First, we need to find pKa of NH₃ using Kb. Then, the moles of NH₃ and finally replace these values in H-H equation to solve moles of NH₄Cl we need to obtain the desire pH.
pKa NH₃/NH₄⁺pKb = - log Kb
pKb = -log 1.8x10⁻⁵ = 4.74
pKa = 14 - pKb
pKa = 14 - 4.74
pKa = 9.26
Moles NH₃2.00L ₓ (0.200mol NH₃ / L) = 0.400 moles NH₃
H-H equation:pH = pKa + log [NH₃] / [NH₄Cl]
8.20 = 9.26 + log [0.400 moles] / [NH₄Cl]
-1.06 = log [0.400 moles] / [NH₄Cl]
0.0087 = [0.400 moles] / [NH₄Cl]
[NH₄Cl] = 0.400 moles / 0.0087
[NH₄Cl] = 4.59 moles of NH₄Cl we need to add to original solution to obtain a pH of 8.20. In grams (Using molar mass NH₄Cl=53.491g/mol):
4.59 moles NH₄Cl ₓ (53.491g / mol) =
245.66g of NH₄Cl is the mass we need to add to obtain the desire pH
Why can gasses change volume?
A. The forces holding the gas particles together are
stronger than gravity.
B. The gas particles have no mass, so they can change volume.
C. Gravity has no effect on gas particles, so they can float away.
O D. There are no forces holding the gas particles together.
Answer:
There are no forces holding the gas particles together.
Explanation:
If the heat of combustion for a specific compound is −1320.0 kJ/mol and its molar mass is 30.55 g/mol, how many grams of this compound must you burn to release 617.30 kJ of heat?
Answer:
14.297 g
Explanation:
From the question;
1 mo of the compound requires 1320.0 kJ
From the molar mass;
1 ml of the compound weighs 30.55g
How many grams requires 617.30kJ?
1 ml = 1320
x mol = 617.30
x = 617.30 / 1320
x = 0.468 mol
But 1 mol = 30.55
0.468 mol = x
x = 14.297 g
when the temperature of an ideal gas is increased from 27C to 927C then kinetic energy increases by
Answer:
The rms speed of its molecules becomes. (T) has become four times. Therefore, v_(rms) will become two times,...
When the equation MnO₄⁻ + I⁻ + H₂O → MnO₂ + IO₃⁻ is balanced in basic solution, what is the smallest whole-number coefficient for OH⁻?
Answer:
The smallest whole-number coefficient for OH⁻ is 2
Explanation:
Step 1: The equation redox reaction is divided into two half equations
Reduction half equation: MnO₄⁻ ----> MnO₂
Oxidation half-equation: I⁻ ---> IO₃⁻
Step 2: Next the atoms are balanced by adding OH⁻ ions and H₂O molecules to the appropriate side of each half equation;
MnO₄⁻ + 2H₂O ----> MnO₂ + 4OH⁻
I⁻ + 6OH⁻ ---> IO₃⁻ + 3H₂O
Step 3 : The charges are then balanced by adding electrons to the appropriate sides of each half equation
MnO₄⁻ + 2H₂O + 3e⁻ ----> MnO₂ + 4OH⁻
I⁻ + 6OH⁻ ---> IO₃⁻ + 3H₂O + 6e⁻
Step 4: Oxidation half equation is multiplied by 2 while reduction half equation is multiplied by 1 to balance the number of electrons gained and lost for the reaction
2MnO₄⁻ + 4H₂O + 6e⁻ ----> 2MnO₂ + 8OH⁻
I⁻ + 6OH⁻ ---> IO₃⁻ + 3H₂O + 6e⁻
Step 5 : addition of the two half equations to yield a net ionic equation
2MnO₄⁻ + I⁻ + H₂O ----> 2MnO₂ + IO₃⁻ + 2OH⁻
The smallest whole number coefficient for OH⁻ is 2
A redox reaction is divided into two half equations which are shown below:
Reduction half equation: MnO₄⁻ ----> MnO₂
Oxidation half-equation: I⁻ ---> IO₃⁻
Atoms are balanced by adding OH⁻ ions and H₂O molecules to the appropriate side of each half equation to make the equation complete ;
MnO₄⁻ + 2H₂O ----> MnO₂ + 4OH⁻
I⁻ + 6OH⁻ ---> IO₃⁻ + 3H₂O
The charges needs to be balanced and this is done by adding electrons to the appropriate sides of each half equation
MnO₄⁻ + 2H₂O + 3e⁻ ----> MnO₂ + 4OH⁻
I⁻ + 6OH⁻ ---> IO₃⁻ + 3H₂O + 6e⁻
The equation needs to be balanced by multiplying the oxidation half equation by 2 while reduction half equation is multiplied by 1 to balance the number of electrons on both sides of the equations.
2MnO₄⁻ + 4H₂O + 6e⁻ ----> 2MnO₂ + 8OH⁻
I⁻ + 6OH⁻ ---> IO₃⁻ + 3H₂O + 6e⁻
The two half equations are then added and written together to form a net ionic equation
2MnO₄⁻ + I⁻ + H₂O ----> 2MnO₂ + IO₃⁻ + 2OH⁻
The smallest whole-number coefficient for OH⁻ is therefore 2.
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The following reaction, catalyzed by iridium, is endothermic at 700 K: CaO(s) + CH4(g) + 2H2O (g) → CaCO3 (s) + 4H2 (g) For the reaction mixture above at equilibrium at 700 K, how would the following changes affect the total quantity of CaCO3 in the reaction mixture once equilibrium is re-established?
a. Increasing the temperature
b. Adding calcium oxide (CaO)
c. Removing methane (CH4)
d. Increasing the total volume
e. Adding iridium
Answer:
A. Increasing the temperature will favor forward reaction and more CaCo3 formed.
B. More CaCo3 will be formed.
C. CaCo3 will decrease and more react ants formed.
D. Less CaCo3 will be formed.
E. Iridium is a catalyst so there is no effect
Explanation:
A. Temperature will increase because it's an endothermic reaction.
B. Adding Cao will favor forward reaction and more CaCo3 formed.
C. Removing methane, more react ants are formed and CaCo3 decreases.
D. Irridi is a catalyst so it has no effect on the CaCo3 but only speeds its rate of reaction.
When 5.58g H2 react by the following balanced equation, 32.8g H2O are formed. What is the percent yield of the reaction? 2H2(g)+O2(g)⟶2H2O(l)
A) 11.7%
B) 17.0%
C) 38.9%
D) 65.7%
Answer:
D) 65.7%
Explanation:
Based on the reaction:
2H2(g)+O2(g)⟶2H2O(l)
2 moles of hydrogen produce 2 moles of water assuming an excess of oxygen.
To find percent yield of the reaction we need to find theoretical yield (The yield assuming all hydrogen reacts producing water). With theoretical yield and actual yield (32.8g H₂O) we can determine percent yield as 100 times the ratio between actual yield and theoretical yield.
Theoretical yield:
Moles of 5.58g H₂:
5.58g H₂ ₓ (1 mol / 2.016g) = 2.768 moles H₂
As 2 moles of H₂ produce 2 moles of H₂O, if all hydrogen reacts will produce 2.768 moles H₂O. In grams:
2.768 moles H₂O ₓ (18.015g / mol) =
49.86g H₂O is theoretical yield
Percent yield:
Percent yield = Actual yield / Theoretical yield ₓ 100
32.8g H₂O / 49.86g ₓ 100 =
65.7% is percent yield of the reaction
D) 65.7%Plz help!!!! Solve this by using factor labeling
Answer:
the answer is 2,000 nickels.
Explanation:
we multiplied 100 by 100, because there are 100 cents in a dollar, and we divided 10,000 by 5, because there are 5 cents in a nickel.
"Aqueous solutions of lead nitrate and ammonium chloride are mixed" together. Which statement is correct
Answer:
PbCl₂ will precipitate from solution.
Explanation:
Statements are:
Insufficient information is given.
Both NH4NO3 and PbCl2 precipitate from solution.
No precipitate forms.
PbCl2 will precipitate from solution.
NH4NO3 will precipitate from solution.
The reaction of ammonium chloride (NH₄Cl) with lead nitrate (Pb(NO₃)₂) is:
Pb(NO₃)₂ + 2NH₄Cl → PbCl₂ + 2 NH₄NO₃
Talking of rules of solubility, all nitrates are soluble in water, that means NH₄NO₃ is soluble and no precipitate is formed.
In the same way, all chlorides are soluble except silver chloride and lead chloride. That means:
PbCl₂ (Lead chloride) will precipitate from solution.What electron configuration represents Nitrogen? A. 2,2 B. 2,8,4 C. 2,4 D. 2,5
Answer:
D. 2:5
Explanation:
It has 5 valency electrons
[tex].[/tex]
In the given question, the electronic configuration of nitrogen is 2,5. The correct answer is option D.
The electronic configuration of an atom describes the arrangement of its electrons in different energy levels or orbitals.
Nitrogen has an atomic number of 7, which means it has 7 electrons.
The first energy level or shell can hold up to 2 electrons, and the second can hold up to 8 electrons. Nitrogen has 2 electrons in its first energy level and 5 electrons in its second energy level.Therefore, the electronic configuration of nitrogen is 2,5, which means it has 2 electrons in its first energy level and 5 electrons in its second energy level. Option D is the correct answer.
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Find the number of ibuprofen molecules in a tablet containing 210.0 mg of ibuprofen (C13H18O2).
Answer:
the answer is 5.83x1020 molecules
Explanation:
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The gas with an initial volume of 24.0 L at a pressure of 565 torr is compressed until the volume is 16.0 L. What is the final pressure of the gas, assuming the temperature and amount of gas does not change
Answer:
848 torr
Explanation:
The only variables are the pressure and the volume, so we can use Boyle's Law.
p₁V₁ = p₂V₂
Data:
p₁ = 565 torr; V₁ = 24.0 L
p₂ = ?; V₂ = 16.0 L
Calculations:
[tex]\begin{array}{rcl}p_{1}V_{1} & = & p_{2}V_{2}\\\text{565 torr} \times \text{24.0 L} & = & p_{2} \times \text{16.0 L}\\\text{13 560 torr} & = & 16.0p_{2}\\p_{2} & = & \dfrac{\text{13 560 torr}}{16.0}\\\\& = &\textbf{848 torr}\\\end{array}\\\text{The final pressure of the gas is $\large \boxed{\textbf{848 torr}}$}[/tex]
1. Explain the test for unsaturation.
2. Write down the balanced chemical equations for the complete and incomplete
combustion of octene
3. Explain how propanol, an alcohol, is formed from propene..
4. How is margarine formed?
Answer:
1)In organic chemistry, the bromine test is a qualitative test for the presence of unsaturation (carbon-to-carbon double or triple bonds), phenols and anilines. ... The more unsaturated an unknown is, the more bromine it reacts with, and the less coloured the solution will appear.
2)The equation for incomplete combustion of propane is: 2 C3H8 + 9 O2 → 4 CO2 + 2 CO + 8 H2O + Heat. If not enough oxygen is present for complete combustion, incomplete combustion occurs. The result of incomplete combustion is, once again, water vapour, carbon dioxide and heat. But it also produces carbon monoxide.
Explanation:
3)Propene, also known as propylene, is an unsaturated organic compound with the chemical formula {\displaystyle {\ce {CH3CH=CH2}}}. It has one double bond, and is the second simplest member of the alkene class of hydrocarbons. It is a colorless gas with a faint petroleum-like odor.
Formula: C3H6
IUPAC ID: Propene
4)Margarines are chemically created during hydrogenation which, until January 1, 2006, relied upon trans fats to solidify their vegetable oils. Food companies have been exploring options for replacing trans fat in partially hydrogenated margarine.
Which response includes all the following processes that are accompanied by an increase in entropy? 1) 2SO 2(g) + O 2(g) → SO 3(g) 2) H 2O(l) → H 2O(s) 3) Br 2(l) → Br 2(g) 4) H 2O 2(l) → H 2O(l) + 1/ 2O 2(g)
Answer: Reaction (1) , (3) and (4) are accompanied by an increase in entropy.
Explanation:
Entropy is the measure of randomness or disorder of a system. If a system moves from an ordered arrangement to a disordered arrangement, the entropy is said to decrease and vice versa.
(1) [tex]2SO_2(g)+O_2(g)\rightarrow SO_3(g)[/tex]
3 moles of reactant are changing to 1 mole of product , thus the randomness is increasing. Thus the entropy also increases.
2) [tex]H_2O(l)\rightarrow H_2O(s)[/tex]
1 mole of Liquid reactant is changing to 1 mole of solid product , thus the randomness is decreasing. Thus the entropy also decreases.
3) [tex]Br_2(l)\rightarrow Br_2(g)[/tex]
1 mole of Liquid reactant is changing to 1 mole of gaseous product , thus the randomness is increasing. Thus the entropy also increases.
4) [tex]H_2O_2(l)\rightarrow H_2O(l)+\frac{1}{2}O_2(g)[/tex]
1 mole of Liquid reactant is changing to half mole of gaseous product and 1 mole of liquid product, thus the randomness is increasing. Thus the entropy also increases.
A 3.140 molal solution of NaCl is prepared. How many grams of NaCl are present in a sample containing 2.692 kg of water
Answer:
494.49 g of NaCl.
Explanation:
Data obtained from the question include the following:
Molality of NaCl = 3.140 m
Mass of water = 2.692 kg
Mass of NaCl =.?
Next, we shall determine the number of mole of NaCl in the solution.
Molality is simply defined as the mole of solute per unit kilogram of solvent. Mathematically, it is expressed as
Molality = mole of solute /Kg of solvent
With the above formula, we can obtain the number of mole NaCl in the solution as follow:
Molality of NaCl = 3.140 m
Mass of water = 2.692 kg
Mole of NaCl =..?
Molality = mole of solute /Kg of solvent
3.140 = mole of NaCl /2.692
Cross multiply
Mole of NaCl = 3.140 x 2.692
Mole of NaCl = 8.45288 moles
Finally, we shall covert 8.45288 moles of NaCl to grams. This can be obtained as follow:
Mole of NaCl = 8.45288 moles
Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol
Mass of NaCl =.?
Mole = mass /Molar mass
8.45288 = mass of NaCl /58.5
Cross multiply
Mass of NaCl = 8.45288 × 58.5
Mass of NaCl = 494.49 g.
Therefore, 494.49 g of NaCl are present in the solution.
The mass of NaCl in 3.140 molal NaCl solution has been 494.493 grams.
Molality can be defined as the mass of solute present in 1000 grams of solvent.
Molality = [tex]\rm \dfrac{moles\;of\;solute}{mass\;of\;solvent\;(kg)}[/tex]
The moles of NaCl has been calculated as;
3.140 = [tex]\rm \dfrac{moles\;of\;NaCl}{mass\;of\;water\;(kg)}[/tex]
3.140 = [tex]\rm \dfrac{moles\;of\;NaCl}{2.692\;kg}[/tex]
Moles of NaCl = 3.140 [tex]\times[/tex] 2.692
Moles of NaCl = 8.45288 mol
The moles can be expressed as;
Moles = [tex]\rm \dfrac{weight}{molecular\;weight}[/tex]
Molecular weight of NaCl = 58.5 g/mol
The mass of NaCl can be calculated as:
8.45288 mol = [tex]\rm \dfrac{mass\;of\;NaCl}{58.5\;g/mol}[/tex]
Mass of NaCl = 58.5 g/mol [tex]\times[/tex] 8.45288 mol
Mass of NaCl = 494.493 grams.
The mass of NaCl in 3.140 molal NaCl solution has been 494.493 grams.
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Which molecule is NOT hypervalent?
Select the correct answer below:
SF
PBr3
PBr5
XeFo
Answer:
PBr3 is NOT hypervalent
Explanation:
The molecule that is not hypervalent is PBr3
A molecule can be defined as the smallest part of a substance that can exist independently.
It is formed by the chemical combination of two or more atoms.
A molecule is said to be hypervalent when more than four pairs of electrons are around the central atom.
A molecule is said to be hypovalent when less than four pairs of electrons are around the central atom.
From the question, the molecule that is hypovalent is PBr3
This is because, phosphorus can make hypervalent compounds, but in this specific example it is sharing three bonds and has one lone pair, so it has simply a full octet.
Therefore, the molecule that is not hypervalent is PBr3.
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a) Define typical polyfunctional acid ?
b) Show the equations of dissociation mechanism of phosphoric acid as an example.
c) Write the equation for calculating the [H3O*].
a) A polyfunctional acid is an acid that has more than one functional group.
b) The equations of dissociation of phosphoric acid are:
H₃PO₄ + H₂O ⇄ H₂PO₄⁻ + H₃O⁺ H₂PO₄⁻ + H₂O ⇄ HPO₄²⁻ + H₃O⁺ HPO₄²⁻ + H₂O ⇄ PO₄³⁻ + H₃O⁺c) The equation for calculating the concentration of H₃O⁺ is [tex] [H_{3}O^{+}] = (\frac{K_{1}K_{2}K_{3}[H_{3}PO_{4}]}{[PO_{4}^{-3}]})^{1/3} [/tex]
a) A polyfunctional acid can be defined as an acid that has more than one functional group. Phosphoric acid (H₃PO₄) is an example of polyfunctional acid since it is composed of three hydroxyl groups joined to a phosphorus atom, which is also joined to an oxygen atom by a double bound. In that structure, the three hydrogen atoms of the hydroxyl groups give the acidic behavior to this compound.
b) Phosphoric acid has three equations of dissociation:
H₃PO₄ + H₂O ⇄ H₂PO₄⁻ + H₃O⁺ (1)H₂PO₄⁻ + H₂O ⇄ HPO₄²⁻ + H₃O⁺ (2)HPO₄²⁻ + H₂O ⇄ PO₄³⁻ + H₃O⁺ (3)The dissociation constants for the three above equations are:
[tex] K_{1} = \frac{[H_{2}PO_{4}^{-}][H_{3}O^{+}]}{[H_{3}PO_{4}]} [/tex] (4)
[tex] K_{2} = \frac{[HPO_{4}^{2-}][H_{3}O^{+}]}{[H_{2}PO_{4}^{-}]} [/tex] (5)
[tex] K_{3} = \frac{[PO_{4}^{3-}][H_{3}O^{+}]}{[HPO_{4}^{2-}]} [/tex] (6)
c) We can calculate the concentration of H₃O⁺ for each equilibrium with the equations (4), (5), and (6).
The general reaction of dissociation of phosphoric acid is given by the sum of equations (1), (2), and (3):
H₃PO₄ + 3H₂O ⇄ PO₄³⁻ + 3H₃O⁺ (7)
The concentration of H₃O⁺ for the total dissociation reaction (eq 7) can be found as follows:
[tex] K_{t} = \frac{[PO_{4}^{-3}][H_{3}O^{+}]^{3}}{[H_{3}PO_{4}]} [/tex] (8)
Where:
[tex] K_{t} = K_{1}*K_{2}*K_{3} [/tex]
Hence, by knowing the dissociation constants K₁, K₂ and K₃, and the concentrations of PO₄³⁻ and H₃PO₄, the [H₃O⁺] is:
[tex][H_{3}O^{+}] = (\frac{K_{1}K_{2}K_{3}[H_{3}PO_{4}]}{[PO_{4}^{-3}]})^{1/3}[/tex]
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How many moles of bromine will react with 0.0500 mole of C 2H 2 in the reaction C 2H 4 + Br 2 → C 2H 4Br 2?
Answer:
0.05 mole of Br2.
Explanation:
We'll begin by writing the balanced equation for the reaction. This is given below:
C2H4 + Br2 —› C2H4Br2
From the balanced equation above,
1 mole of C2H4 reacted with 1 mole of Br2 to produce 1 mole of C2H4 Br2.
Finally, we shall determine the number of moles bromine that will react with 0.05 mole of C2H2.
The number of mole of Br2 needed for the reaction can be obtained as follow:
From the balanced equation above,
1 mole of C2H4 reacted with 1 mole of Br2.
Therefore, 0.05 mole of C2H4 will also react with 0.05 mole of Br2.
Therefore, 0.05 mole of Br2 is needed for the reaction.
To find the pH of a solution of NaNO2, one would have to construct an ICE chart using:
a. the Kb of NO−2 to find the hydroxide concentration.
b. the Kb of HNO2 to find the hydronium concentration.
c. the Kb of NO-2, to find the hydronium concentration.
d. the Kb of HNO2, to find the hydroxide concentration.
Answer:
a. the Kb of NO₂⁻ to find the hydroxide concentration.
Explanation:
When sodium nitrite is dissolved in water, it dissociates in sodium cation and nitrite anion according to the following equation.
NaNO₂(s) ⇒ Na⁺(aq) + NO₂⁻(aq)
Na⁺ comes from NaOH (strong base) so it doesn't react with water.
NO₂⁻ comes from HNO₂ (weak acid) so it reacts with water according to the following equation.
NO₂⁻(aq) + H₂O(l) ⇄ HNO₂(aq) + OH⁻(aq)
This is the basic reaction of nitrite ion, so we need the Kb of NO₂⁻ to find the hydroxide concentration.
What type of bonding is occuring in the compound below?
A. Covalent polar
B. Metallic
C. Ionic
D. Covalent nonpolar
Answer:
(B). it's metallic bonding
Consider the reaction: C(s) + O2(g)CO2(g) Write the equilibrium constant for this reaction in terms of the equilibrium constants, Ka and Kb, for reactions a and b below: a.) C(s) + 1/2 O2(g) CO(g) Ka b.) CO(g) + 1/2 O2(g) CO2(g) Kb
Answer:
A. Ka = [CO2] / [C] [O2]^1/2
B. Kb = [CO2] / [CO] [O2]^1/2
Explanation:
Equilibrium constant is simply defined as the ratio of the concentration of the products raised to their coefficient to the concentration of the reactants raised to their coefficient.
Now, we shall obtain the expression for the equilibrium constant for the reaction as follow:
A. Determination of the expression for equilibrium constant Ka.
This is illustrated below:
C(s) + 1/2 O2(g) <==> CO(g)
Ka = [CO2] / [C] [O2]^1/2
B. Determination of the expression for equilibrium constant Kb.
This is illustrated below:
CO(g) + 1/2 O2(g) <==> CO2(g)
Kb = [CO2] / [CO] [O2]^1/2
Which of the following is an example of a physical, rather than a chemical, change?
Answer: The question is not clear
Explanation:
What is pH of a buffer made by combining 45.0mL of 0.150M nitrous acid and 20.0mL of 0.175M sodium nitrate
Answer:
3.11
Explanation:
Any buffer system can be described with the reaction:
[tex]HA~->~H^+~+~A^-[/tex]
Where is the acid and is the base. Additionally, the calculation of the pH of any buffer system can be made with the Henderson-Hasselbach equation:
[tex]pH~=~pKa~+~Log(\frac{ [A^-]}{[HA]})[/tex]
With all this in mind, we can write the reaction for our buffer system:
-) Nitrous acid: [tex]HNO_2[/tex]
-) Sodium nitrate: [tex]NaNO_2[/tex]
[tex]HNO_2~->~H^+~+~NO_2^-[/tex]
In this case, the acid is [tex]HNO_2[/tex] with a concentration of 0.150 M and a volume of 45.0 mL (0.045 L). The base is [tex]NO_2^-[/tex] with a concentration of 0.175 M and a volume of 20.0 mL (0.020 L).
We can calculate the moles of each compound is we take into account the molarity equation ([tex]M=\frac{mol}{L}[/tex]). So:
-) moles of [tex]HNO_2[/tex]:
[tex]mol=0.150~M*0.045~L=0.00657[/tex]
-) moles of [tex]NO_2^-[/tex]:
[tex]mol=0.175~M*0.020~L=0.0035[/tex]
The total volume would be:
0.020 L + 0.045 L = 0.065 L
With this in mind, we can calculate the molarity of each compound:
-) Concentration of [tex]HNO_2[/tex]
[tex]M=\frac{0.00657~mol}{0.065~L}=0.101~M[/tex]
-) Concentration of [tex]NO_2^-[/tex]
[tex]M=\frac{0.0035~mol}{0.065~L}=0.0538~M[/tex]
The pKa reported is 3.39, therefore we can plug the values into the Henderson-Hasselbach equation:
[tex]pH~=~3.39~+~Log(\frac{[0.0538~M]}{[0.101~M]})~=~3.11[/tex]
The final pH value would be 3.11
I hope it helps!
The pH of a buffer made by combining 45.0 mL of 0.150M nitrous acid and 20.0mL of 0.175M sodium nitrate is 2.87.
We have a buffer made by combining 45.0mL of 0.150 M nitrous acid and 20.0mL of 0.175M sodium nitrate.
Nitrous acid is a weak acid and nitrate ion is its conjugate base.
What is a buffer?It is a solution used to resist abrupt changes in pH when acids or bases are added.
Step 1: Calculate the moles of each species.We do so by multiplying the molar concentration by the volume in liters.
HNO₂: 0.150 mol/L × 0.0450 L = 6.75 × 10⁻³ mol
NaNO₂: 0.175 mol/L × 0.0200 L = 3.50 × 10⁻³ mol
Step 2: Calculate the total volume of the mixture.The total volume will be the sum of the volumes of each solution.
V = 45.0 mL + 20.0 mL = 65.0 mL = 0.0650 L
Step 3: Calculate the molar concentration of each species in the mixture.HNO₂: 6.75 × 10⁻³ mol/0.0650 L = 0.104 M
NaNO₂: 3.50 × 10⁻³ mol/0.0650 L = 0.0538 M
Step 4: Calculate the pH of the buffer.We can calculate the pH of a buffer system using Henderson-Hasselbach's equation.
pH = pKa + log [NaNO₂]/[HNO₂]
pH = 3.16 + log 0.0538/0.104 = 2.87
The pH of a buffer made by combining 45.0 mL of 0.150M nitrous acid and 20.0mL of 0.175M sodium nitrate is 2.87.
Learn more about buffers here: https://brainly.com/question/24188850
What is the balanced equation for the reaction of aqueous cesium sulfate and aqueous barium perchlorate?
Answer:
The balanced chemical reaction is given as:
[tex]Cs_2SO_4(aq)+Ba(ClO_4)_2(aq)\rightarrow BaSO_4(s)+2CsClO_4(aq)[/tex]
Explanation:
When aqueous cesium sulfate and aqueous barium perchlorate are mixed together it gives white precipitate barium sulfate and aqueous solution od cesium perchlorate.
The balanced chemical reaction is given as:
[tex]Cs_2SO_4(aq)+Ba(ClO_4)_2(aq)\rightarrow BaSO_4(s)+2CsClO_4(aq)[/tex]
According to reaction, 1 mole of cesium sulfate reacts with 1 mole of barium perchlorate to give 1 mole of a white precipitate of barium sulfate and 2 moles of cesium perchlorate.
Emission of which one of the following leaves both atomic number and mass number unchanged?
(a) positron
(b) neutron
(c) alpha particle
(d) gamma radiation
(e) beta particle
Answer: Gamma Radiation
Explanation:
The emission of Gamma rays does not cause a change in both the atomic and mass number. They are electromagnetic radiation.
The radiations that leaves without changing the atomic mass and atomic number of the particle have been gamma radiations. Thus, option D is correct.
Radiations have been the energy that has been evolved by the particles during energy transitions. The nuclear decay results with the release of the energy from the particle resulting in the change in the atomic mass.
The electromagnetic radiations have been capable of emitting the radiation without changing the mass and atomic number of the element. The gamma radiations have been the electromagnetic radiations. Thus, option D is correct.
For more information about the emissions, refer to the link:
https://brainly.com/question/517329
Which of the following happens to a molecule of an object when the object is heated? (1 point)
Answer:
They get more energy, so they vibrate!
Explanation:
Which of the following is a salt that will form from the combination of a strong base with a weak acid?
Select the correct answer below:
A. NaHCO3
B. H2O
C. CH3CO2H
D. NH4Cl
Answer:
A. NaHCO₃
Explanation:
NaHCO₃ ⇒ NaOH + H₂CO₃
NaOH is a strong base and H₂CO₃ is a weak acid. Therefore, NaHCO₃ is a salt of a strong base-weak acid reaction. The salt is basic because carbonic acid (H₂CO₃) is a weak acid so it remains undissociated. So, there is a presence of additional OH⁻ ions that makes the solution basic.
Hope that helps.
alculate the pH of the solution, upon addition of 0.035 mol of NaOH to the original buffer. Express your answer using two decimal places.
A 1.0-L buffer solution contains 0.100 mol HC2H3O2 and 0.100 mol NaC2H3O2. The value of Ka for HC2H3O2 is 1.8×10−5.
Calculate the pH of the solution, upon addition of 0.035 mol of NaOH to the original buffer.
Answer:
The pH of this solution = 5.06
Explanation:
Given that:
number of moles of CH3COOH = 0.100 mol
volume of the buffer solution = 1.0 L
number of moles of NaC2H3O2 = 0.100 mol
The objective is to Calculate the pH of the solution, upon addition of 0.035 mol of NaOH to the original buffer.
we know that concentration in mole = Molarity/volume
Then concentration of [CH3COOH] = [tex]\mathtt{ \dfrac{0.100 \ mol}{ 1.0 \ L }}[/tex] = 0.10 M
The chemical equation for this reaction is :
[tex]\mathtt{CH_3COOH + OH^- \to CH_3COO^- + H_2O}[/tex]
The conjugate base is CH3COO⁻
The concentration of the conjugate base [CH3COO⁻] is = [tex]\mathtt{ \dfrac{0.100 \ mol}{ 1.0 \ L }}[/tex]
= 0.10 M
where the pka (acid dissociation constant)for CH3COOH = 4.74
If 0.035 mol of NaOH is added to the original buffer, the concentration of NaOH added will be = [tex]\mathtt{ \dfrac{0.035 \ mol}{ 1.0 \ L }}[/tex] = 0.035 M
The ICE Table for the above reaction can be constructed as follows:
[tex]\mathtt{CH_3COOH \ \ \ + \ \ \ \ OH^- \ \ \to \ \ CH_3COO^- \ \ \ + \ \ \ H_2O}[/tex]
Initial 0.10 0.035 0.10 -
Change -0.035 -0.035 + 0.035 -
Equilibrium 0.065 0 0.135 -
By using Henderson-Hasselbalch equation:
The pH of this solution = pKa + log [tex]\mathtt{\dfrac{CH_3COO^-}{CH_3COOH}}[/tex]
The pH of this solution = 4.74 + log [tex]\mathtt{\dfrac{0.135}{0.065}}[/tex]
The pH of this solution = 4.74 + log (2.076923077 )
The pH of this solution = 4.74 + 0.3174
The pH of this solution = 5.0574
The pH of this solution = 5.06 to two decimal places