Answer:
0.0826 mol (corrected to 3 sig. fig.)
Explanation:
First, balance the equation:
2NH3 →3H2 + N2
Take the atomic no. of N=14.0, and H=1.0,
no. of moles = mass / molar mass
So, no. of moles of NH3 dissociated = 2.81 / (14.0+1.0x3)
= 0.165294117mol
From the equation, the mole ratio of NH3:N2 = 2:1, meaning for every 2 moles of NH3 dissociated, one mole of N2 is formed.
So, using this ratio, the no. of moles of N2 formed will be 0.165294117 / 2
=0.0826 mol (corrected to 3 sig. fig.)
g Calculate the concentration of sulfate ion when BaSO4 just begins to precipitate from a solution that is 0.0758 M in Ba2+.
Answer:
1.42 × 10⁻⁹ M
Explanation:
Step 1: Given data
Concentration of Ba²⁺ ([Ba²⁺]): 0.0758 MSolubility product constant of BaSO₄ at 25°C: 1.08 × 10⁻¹⁰Step 2: Write the reaction for the solution of BaSO₄
BaSO₄(s) ⇄ Ba²⁺(aq) + SO₄²⁻(aq)
Step 3: Calculate the concentration of SO₄²⁻
We will use the following expression.
Ksp = 1.08 × 10⁻¹⁰ = [Ba²⁺] × [SO₄²⁻]
[SO₄²⁻] = 1.08 × 10⁻¹⁰ / [Ba²⁺] = 1.08 × 10⁻¹⁰ / 0.0758 = 1.42 × 10⁻⁹ M
Calculate the amount of energy absorbed by 45.0 g sample of water to raise its temperature from 18.0C to 48.0 C. The specific heat of water is 4.18 J/g C. 1000 J= 1kj
Answer:
5.643 kJ
Explanation:
The quantity of heat released or absorbed by a substance (Q) is given by the equation:
Q = mcΔT
Where m is the mass of the substance, c is the specific heat of substance and ΔT is the difference between the final temperature and the initial temperature.
Given that:
m = 45 g, Final temperature = 48°C, Initial temperature = 18°C, c = specific heat of water = 4.18 J/g°C
ΔT = Final temperature - Initial temperature = 48°C - 18°C = 30°C
The quantity of heat is:
Q = mcΔT = 45 g × 4.18 J/g°C × 30°C = 5643 J
Q = 5.643 kJ
Calculate Delta G for each reaction using Delta Gf values: answer kJ ...thank you
a) H2(g)+I2(s)--->2HI(g)
b) MnO2(s)+2CO(g)--->Mn(s)+2CO2(g)
c) NH4Cl(s)--->NH3(g)+HCl(g)
Answer:
a) [tex]\Delta G=2.6kJ[/tex]
b) [tex]\Delta G=-979.57kJ[/tex]
c) [tex]\Delta G=264.21kJ[/tex]
Explanation:
Hello,
In this case, in each reaction we must subtract the Gibbs free energy of formation the reactants to the Gibbs free energy of formation of the products considering each species stoichiometric coefficients. In such a way, the Gibbs free energy of formations are:
[tex]\Delta _fG_{H_2}=\Delta _fG_{I_2}=0kJ/mol\\\Delta _fG_{HI}=1.3kJ/mol\\\Delta _fG_{CO_2}=-394.4kJ/mol\\\Delta _fG_{CO}=-137.3 kJ/mol\\\Delta _fG_{NH_3}=16.7 kJ/mol\\\Delta _fG_{HCl}=-95.3kJ/mol\\\Delta _fG_{MnO_2}=465.37kJ/mol\\\Delta _fG_{Mn}=0kJ/mol\\\Delta _fG_{NH_4Cl}=-342.81kJ/mol[/tex]
So we proceed as follows:
a)
[tex]\Delta G=2\Delta _fG_{HI}-\Delta _fG_{H_2}-\Delta _fG_{I_2}\\\\\Delta G=2*1.3\\\\\Delta G=2.6kJ[/tex]
b)
[tex]\Delta G=\Delta _fG_{Mn}+2*\Delta _fG_{CO_2}-\Delta _fG_{MnO_2}-2*\Delta _fG_{CO}\\\\\Delta G=0+2*-394.4-465.37-2*-137.3\\\\\Delta G=-979.57kJ[/tex]
c)
[tex]\Delta G=\Delta _fG_{NH_3}+\Delta _fG_{HCl}-\Delta _fG_{NH_4Cl}\\\\\Delta G=16.7-95.3-(-342.81)\\\\\Delta G=264.21kJ[/tex]
Regards.
Which element has the largest atomic radius? As N P Sb
Answer:
Sb
Explanation:
The periodic trend for atomic radius is that it decreases from left to right and increases from top to bottom, therefore the elements with the larger atomic radius will be the ones which are closest to the bottom left corner of the periodic table. Since all of these elements are in the same group, the one with the largest atomic radius will be the one at the "bottom", and that is Sb.
A mixture of water and graphite is heated to 600 K in a 1 L container. When the system comes to equilibrium it contains 0.17 mol of H2, 0.17 mol of CO, 0.74 mol of H2O, and some graphite. Some O2 is added to the system and a spark is applied so that the H2 reacts completely with the O2.
Find the amount of CO in the flask when the system returns to equilibrium.
Express your answer to two significant figures and include the appropriate units.
Answer:
0.44 moles
Explanation:
Given that :
A mixture of water and graphite is heated to 600 K in a 1 L container. When the system comes to equilibrium it contains 0.17 mol of H2, 0.17 mol of CO, 0.74 mol of H2O, and some graphite.
The equilibrium constant [tex]K_c= \dfrac{[CO][H_2]}{[H_2O]}[/tex]
The equilibrium constant [tex]K_c= \dfrac{(0.17 )(0.17)}{0.74}[/tex]
The equilibrium constant [tex]K_c= 0.03905[/tex]
Some O2 is added to the system and a spark is applied so that the H2 reacts completely with the O2.
The equation for the reaction is :
[tex]H_2 + \dfrac{1}{2}O_2 \to H_2O \\ \\ 0.17 \ \ \ \ \ \ \ \ \ \to0.17[/tex]
Total mole of water now = 0.74+0.17
Total mole of water now = 0.91 moles
Again:
[tex]K_c= \dfrac{[CO][H_2]}{[H_2O]}[/tex]
[tex]0.03905 = \dfrac{[0.17+x][x]}{[0.91 -x]}[/tex]
0.03905(0.91 -x) = (0.17 +x)(x)
0.0355355 - 0.03905x = 0.17x + x²
0.0355355 +0.13095 x -x²
x² - 0.13095 x - 0.0355355 = 0
By using quadratic formula
x = 0.265 or x = -0.134
Going by the value with the positive integer; x = 0.265 moles
Total moles of CO in the flask when the system returns to equilibrium is :
= 0.17 + x
= 0.17 + 0.265
= 0.435 moles
=0.44 moles (to two significant figures)
Water (2190 g ) is heated until it just begins to boil. If the water absorbs 5.83×105 J of heat in the process, what was the initial temperature of the water?
Answer:
The initial temperature was [tex]36.4^\circ \:C[/tex]
Explanation:
[tex]\Delta t=\frac{q}{m\cdot C_s}=\frac{5.83\times10^5}{2190\times 4.184}\\\\=63.6^\circ\:C[/tex]
The temperature difference [tex]=100-63.6=36.4^\circ\:C[/tex]
Best Regards!
M(8,7) is the midpoint of rs. The coordinates of s are (9,5) what is the coordinates of r
Answer:
Coordinate or r = (7,9).
Explanation:
Data obtained from the question include the following:
Mid point = (8,7)
Coordinate of S = (9,5)
Coordinate of r =...?
We shall determine the coordinate of r as follow:
Let the coordinate of r be (x2, y2)
Mid point = (x1 + x2)/2 , (y1 + y2)/2
Mid point = (8,7)
Coordinate of S = (9,5)
x1 = 9
y1 = 5
x2 =?
y2 =?
The value of x2 can be obtained as follow:
8 = (x1 + x2)/2
8 = (9 + x2)/2
Cross multiply
9 + x2 = 2 × 8
9 + x2 = 16
Collect like terms
x2 = 16 – 9
x2 = 7
The value of y2 can be obtained as follow:
5 = (y1 + y2)/2
7 = (5 + y2)/2
Cross multiply
5 + y2 = 2 × 7
5 + y2 = 14
Collect like terms
y2 = 14 – 5
y2 = 9
Coordinate of r = (x2, y2)
Coordinate or r = (7,9)
A sample of neon gas at a pressure of 0.609 atm and a temperature of 25.0 °C, occupies a volume of 19.9 liters. If the gas is compressed at constant temperature to a
volume of 12.7 liters, the pressure of the gas sample will be
atm.
Answer:
The pressure of the gas sample will be 0.954 atm.
Explanation:
Boyle's law states that the pressure of a gas in a closed container is inversely proportional to the volume of the container, when the temperature is constant. That is, if the pressure increases, the volume decreases; conversely if the pressure decreases, the volume increases.
Boyle's law is expressed mathematically as:
Pressure * Volume = constant
o P * V = k
To determine the change in pressure or volume during a transformation at constant temperature, the following is true:
P1 · V1 = P2 · V2
That is, the product between the initial pressure and the initial volume is equal to the product of the final pressure times the final volume.
In this case:
P1= 0.609 atmV1= 19.9 LP2=?V2= 12.7 LReplacing:
0.609 atm* 19.9 L= P2* 12.7 L
Solving:
[tex]P2=\frac{0.609 atm* 19.9 L}{12.7 L}[/tex]
P2= 0.954 atm
The pressure of the gas sample will be 0.954 atm.
Calculate the equilibrium constant at 298 K for the reaction of formaldehyde (CH2O) with hydrogen gas using the following information. CH2O(g) + 2H2(g) LaTeX: \longleftrightarrow⟷ CH4(g) + H2O(g) LaTeX: \DeltaΔH°= –94.9 kJ; LaTeX: \DeltaΔS°= –224.2 J/K A. 1.92 B. 9.17 x 10-6 C. 2.07 x 1028 D. 1.10 x 105 E. 8.08 x 104 F. 3.98 x 1011 Group of answer choices
Answer:
E. 8.08 x 10⁴.
Explanation:
Hello,
In this case, for the reaction:
[tex]CH_2O(g) + 2H_2(g) \rightleftharpoons CH_4(g) + H_2O(g)[/tex]
We can compute the Gibbs free energy of reaction via:
[tex]\Delta G\°=\Delta H\°-T\Delta S\°[/tex]
Since both the entropy and enthalpy of reaction are given at 298 K (standard temperature), therefore:
[tex]\Delta G\°=-94.9kJ-(298K)(-224.2\frac{J}{K}*\frac{1kJ}{1000kJ} )\\\\\Delta G\°=-28.1kJ[/tex]
Then, as the equilibrium constant is computed as:
[tex]K=exp(-\frac{\Delta G\°}{RT} )[/tex]
We obtain:
[tex]K=exp(-\frac{-28.1kJ/mol}{8.314x10^{-3}\frac{kJ}{mol* K}}*298K )\\\\K=8.08 x10^4[/tex]
For which the answer is E. 8.08 x 10⁴.
Best regards,
Sulfur dioxide and oxygen react to form sulfur trioxide during one of the key steps in sulfuric acid synthesis. An industrial chemist studying this reaction fills a
75.0 L tank with 3.8 mol of sulfur dioxide gas and 7.0 mol of oxygen gas, and when the mixture has come to equilibrium measures the amount of sulfur trioxide
gas to be 1.5 mol
Calculate the concentration equilibrium constant for the reaction of sulfur dioxide and oxygen at the final temperature of the mixture. Round your answer to 2
significant digits.
Answer:
[tex]\large \boxed{5.1}[/tex]
Explanation:
1. Initial concentrations of reactants
[SO₂] = (3.8 mol)/(75 L) = 0.0507 mol·L⁻¹
[O₂] = (7.0 mol)/(75 L) = 0.0933 mol·L⁻¹
2. Equilibrium concentration of SO₃
[SO₃] = (1.5 mol)/(75 L) = 0.0200 mol·L⁻¹
3. Set up an ICE table
2SO₂ + O₂ ⇌ 2SO₃
I/mol·L⁻¹: 0.0507 0.0933 0
C/mol·L⁻¹: -2x -x +2x
E/mol·L⁻¹: 0.0507-2x 0.0933-x 2x
4. Calculate x
We know the final concentration of SO₃ is 0.0200 mol·L⁻¹, so
2x = 0.0200
x = 0.0100
5. Find the final concentrations of the reactants
Insert the numbers into the ICE table.
2SO₂ + O₂ ⇌ 2SO₃
I/mol·L⁻¹: 0.0507 0.0933 0
C/mol·L⁻¹: -0.0200 -0.0100 +0.0200
E/mol·L⁻¹: 0.0307 0.0833 0.0200
6. Calculate K
[tex]K_{\text{eq}} = \dfrac{\text{[SO$_{3}$]}^{2}}{\text{[SO}_{2}]^{2}\text{[O$_{2}$]}} = \dfrac{0.0200^{2}}{0.0307^{2}\times0.0833} =\mathbf{5.1}\\\\\text{The value of the equilibrium constant is $\large \boxed{\mathbf{5.1}}$}[/tex]
Liquids A, B, and C are insoluble in one another (i.e., they are immiscible). A, B, and C have densities of 0.780 g/cm3, 1.102 g/cm3 , and 1.040 g/cm3, respectively. Which drawing represents the result of placing all three liquids into the same graduated cylinder?
Answer:
The drawing that represents the result of placing all three liquids into the same graduated cylinder will have the liquid arranged one on top of the other from top to bottom in the order of A, C, B.
Explanation:
The image with the options is not provided in this question, but I can answer this fairly so that you can pick from the question, the correct drawing.
We know that two or more immiscible liquids contained together in a container will always separate in the order of their density from top to bottom, with the densest at the bottom, and the least densest at the top. In this case, liquid A is the least densest, and liquid B is the densest. Liquid A will stay on top, and liquid B will be at the bottom. Liquid C will be in between liquid A and liquid B.
If 35.2 g of magnesium react with excess hydrochloric acid, how many grams of magnesium chloride are produced? Mg(s) + 2 HCl(aq) → MgCl2(aq) + H2(aq)
Answer:
139.33 g of magnesium chloride, MgCl2.
Explanation:
We'll begin by writing the balanced equation for the reaction. This is illustrated below:
Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(aq)
Next, we shall determine the mass of Mg that reacted and the mass of MgCl2 from the balanced equation.
This is illustrated below:
Molar mass of Mg = 24 g/mol
Mass of Mg from the balanced equation = 1 x 24 = 24 g
Molar mass of MgCl2 = 24 + (2x35.5) = 95 g/mol
Mass of MgCl2 from the balanced equation = 1 x 95 = 95 g
From the balanced equation above,
24 g of Mg reacted to produce 95 g of MgCl2.
Finally, we shall determine the mass of MgCl2 produced by reacting 35.2 g of Mg.
This can be obtained as follow:
From the balanced equation above,
24 g of Mg reacted to produce 95 g of MgCl2.
Therefore, 35.2 g of Mg will react to produce = (35.2 x 95)/24 = 139.33 g of MgCl2.
From the calculations made above, 139.33 g of magnesium chloride, MgCl2 were produced.
Answer please and thank you
Answer:
Option B. 30 KJ.
Explanation:
The following data were obtained from the question:
Temperature (T) = 5000 K
Enthalpy change (ΔH) = – 220 kJ/mol
Change in entropy (ΔS) = – 0.05 KJ/mol•K
Gibbs free energy (ΔG) =...?
The Gibbs free energy, ΔG can be obtained by using the following equation as illustrated below:
ΔG = ΔH – TΔS
ΔG = – 220 – (5000 x – 0.05)
ΔG = – 220 – (– 250)
ΔG = – 220 + 250
ΔG = 30 KJ
Therefore, the Gibbs free energy, ΔG is 30 KJ.
If you weighed out 203 mg of the green chloro complex and dissolved it in 24.14 mL of acidic solvent, the molarity of your stock solution would be 0.0295 M. Using your precise value of mass and volume that you entered above, please enter your calculated value for the concentration of the original green chloro complex stock solution in moles per liter.
Mgreen stock =
Answer:
0.00295M
Explanation:
Mass Concentration = mass/vol
= 0.203 g/ 0.02414 L = 8.409 g/L
But molarity = Mass conc / molar mass
∴ Molar mass(mol/L) = mass conc / molarity
= .84909 / 0.0295
= 285.06 g/mol
If 1 mol of green stock - 285.06g
? mol - 0.203 g
= 0.00071213 g
= 0.00071213 g / .2414L = 0.0095 mol/L.
13C NMR is a technique in which the total number of signals represents the number of unique carbon atoms in a molecule. Propose a structure that is consistent with the following data.
a. The IR includes peaks at 1603 and 1495 cm^-1
b. The 13c NMR has a total of 7 signals
c. The compound has one acidic proton.
Answer:
D. Poop Butt.
Explanation:
Based on the given data, we can propose a possible structure that fits the criteria: a. carbonyl group (C=O) and an aromatic ring b. there are seven unique carbon environments. c. Presence of a functional group like a carboxylic acid or phenol .
a. The IR peaks at 1603 [tex]cm^{-1}[/tex] and 1495[tex]cm^{-1}[/tex]suggest the presence of both a carbonyl group (C=O) and an aromatic ring.
b. The 13C NMR having a total of 7 signals indicates that there are seven unique carbon environments in the molecule.
c. Considering the presence of an acidic proton, it suggests the presence of a functional group like a carboxylic acid (COOH) or phenol ([tex]C_6H_5OH[/tex]).
Putting all this information together, a possible structure that fits the data could be benzoic acid ([tex]C_6H_5COOH[/tex]). It contains a benzene ring (giving 6 unique carbon environments), a carbonyl group (giving 1 unique carbon environment), and an acidic proton in the carboxylic acid group. This structure satisfies all the given data.
To know more about carboxylic acid, here
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For each of the processes, determine whether the entropy of the system is increasing or decreasing. The system is underlined.
1. a snowman melts on a spring day
2. a document goes through a paper shredder
3. a water bottle cools down in a refrigerator
4. silver tarnishes
5. dissolved sigar precipitates out of water to form rock candy
A. Entropy is increasing
B. Entropy is decreasing
Entropy is INCREASING when a snowman melts, a document goes through paper shredder, silver tarnishes, while it is DECREASING when dissolved sugar precipitates, water vapor forms droplets and water cools down.
Entropy can be defined as the degree of randomness or disorder of a particular system.
Entropy is equal to zero (0) for a perfectly ordered system.
Heat increases the entropy of the system because more energy excites the molecules and it increases the amount of random activity.
Moreover, the cooling decreases the entropy of the system because molecules are more ordered and it decreases the amount of random activity.
In conclusion, entropy is INCREASING when a snowman melts, a document goes through paper shredder, silver tarnishes, while it is DECREASING when dissolved sugar precipitates, water vapor forms water droplets and the water cools down.
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give an example for photodecomposition reaction
A decomposition reaction occurs when one reactant breaks down into two or more products. This can be represented by the general equation: AB → A + B. Examples of decomposition reactions include the breakdown of hydrogen peroxide to water and oxygen, and the breakdown of water to hydrogen and oxygen.
Mass of flask acid= 98.788
Mass of flask = 98.318
Mass of weak acid???
What is the mass of weak acid?
Answer:
0.460 g
Explanation:
Mass of flask + acid = 98.778 g
Mass of flask = 98.318 g
Mass of acid = 0.460 g
Where possible, classify these systems as reactant-favored or product-favored at 298 K. If the direction cannot be determined from the information given, classify the reaction as "Insufficient information."
A. Reactant-favored
B. Product-favored
C. Insufficient information
1. A(s) + B(g) 2C(g) delta H degree = -109 kJ
2. A(s) + 2B(g) C(g) delta H degree=+271 kJ
3. 2A(g) + B(g) 4C(g) delta H degree=+322 kJ
4. A(g) + 2B(g) 2C(g) delta H degree=-89 kJ
Answer:
There is insufficient information to know direction of these systems
Explanation:
Delta H of a reaction is defined as the amount of energy involved when it occurs. The ΔH < 0 represents the reaction will release energy and ΔH > 0 the reaction will absorb energy.
As you can see, ΔH doesn't give information about the direction of a reaction (Spontaneity). In fact, to know spontaneity of a reaction you must know ΔG involved in this reaction.
As the reactions have ΔH but not ΔG,
There is insufficient information to know direction of these systemsHydrazine, N2H4 , reacts with oxygen to form nitrogen gas and water. N2H4(aq)+O2(g)⟶N2(g)+2H2O(l) If 2.45 g of N2H4 reacts with excess oxygen and produces 0.450 L of N2 , at 295 K and 1.00 atm, what is the percent yield of the reaction?
Answer:
24.15%
Explanation:
According to the given situation the computation of the percent yield of the reaction is shown below:-
PV = NRT = N = [tex]\frac{PV}{RT}[/tex]
Mole of [tex]N_2[/tex] = [tex]\frac{PV}{RT}[/tex]
= [tex]\frac{1\times 0.450}{0.0821\times 295}[/tex]
= [tex]\frac{0.450}{24.2195}[/tex]
= 0.0186
Mole of [tex]N_2H_4 = \frac{2.45}{32}[/tex]
= 0.077
Now, the percentage of yield is
= [tex]\frac{Practical\ yield}{Theoretical\ yield}\times 100[/tex]
= [tex]\frac{0.0186}{0.077}\times 100[/tex]
= 24.15%
Therefore for computing the percentage of yield we simply divide the practical yield by theoretical yield and multiply with 100 so that we can get the result into the percentage form.
When the following molecular equation is balanced using the smallest possible integer coefficients, the values of these coefficients are:
P2O5 (s) + H2O (l) =H3PO4 (aq)
The balanced chemical equation for the reaction between hydrogen sulfide and oxygen is:
2H2S(g) + 3O2(g) =2H2O(l) + 2SO2(g)
We can interpret this to mean:
3moles of oxygen and_______moles of hydrogen sulfide react to produce______moles of water and_______ moles of sulfur dioxide.
Answer:
1. The coefficients are: 1, 3, 2
2. From the balanced equation, we obtained the following:
3 moles oxygen, O2 reacted.
2 moles of Hydrogen sulfide, H2S reacted.
2 moles of water were produced.
2 moles of sulphur dioxide, SO2 were produced.
Explanation:
1. Determination of the coefficients of the equation.
This is illustrated below:
P2O5(s) + H2O(l) <==> H3PO4(aq)
There are 2 atoms of P on the left side and 1 atom on the right side. It can be balance by putting 2 in front of H3PO4 as shown below:
P2O5(s) + H2O(l) <==> 2H3PO4(aq)
There are 2 atoms of H on the left side and 6 atoms on the right side. It can be balance by putting 3 in front of H2O as shown below:
P2O5(s) + 3H2O(l) <==> 2H3PO4(aq)
Now the equation is balanced.
The coefficients are: 1, 3, 2.
2. We'll begin by writing the balanced equation for the reaction. This is given below:
2H2S(g) + 3O2(g) => 2H2O(l) + 2SO2(g)
From the balanced equation above,
3 moles of oxygen, O2 reacted with 2 moles of Hydrogen sulfide, H2S to produce 2 moles of water, H2O and 2 moles of sulphur dioxide, SO2.
Which of the following is an important intermediate in the mechanism of the reaction
in the box?
Answer:
Explanation:
.
11. In TLC analysis of ferrocene and acetylferrocene (on silica TLC plate) which prediction is correct: A) ferrocene is more polar and moves higher up the plate (higher Rf value) B) Acetylferrocene is more polar and moves higher up the plate (higher Rf value) C) ferrocene is less polar and moves higher up the plate (higher Rf value) D) Acetylferrocene is less polar and moves higher up the plate (higher Rf value)
Answer:
Alternative C would be the correct choice.
Explanation:
The dual compounds were evaluated on something like a TLC plate through three separate additives in conducting a TLC study of ferrocene versus acetylferrocene.The polar as well as nonpolar ferrocene where nonpolar is about 0.63 with the maximum [tex]R_f[/tex] value, and indeed the polar is somewhere around 0.19 with [tex]R_f[/tex].TLC plate (30:1 toluene/ethanol) established with.The other three choices are not related to the given circumstances. So that option C would be the appropriate choice.
A soda manufacturing company is experimenting with changing the taste of its product as the concentration of carbon dioxide changes. To track their results, they must determine how concentration changes with pressure. The concentration of CO2 under a partial pressure of 0.719 atm is 429.7 ppm. At what pressure (in atm) would the CO2 need to be so that the concentration of CO2 is 235.3 ppm at the same temperature
Answer:
0.394 atm
Explanation:
Mathematically, when we increase the pressure of a gas, we are increasing its concentration and when we decrease the pressure, we are decreasing its concentration.l at same temperature
What this means is that pressure and concentration are directly proportional.
Representing concentration by c and pressure by p, we have;
P1/C1 = P2/C2
From the question;
P1 = 0.719 atm
P2 = ?
C1 = 429.7 ppm
C2 = 235.3 ppm
Now, we can rewrite the equation to be;
P1C2/C1 = P2
Substituting the values we have;
0.719 * 235.3/429.7 = 0.394 atm
The pH of a solution prepared by mixing 40.00 mL of 0.10 M NH3 with 50.00 mL of 0.10 M NH4Cl and 30mL of 0.05 M H2SO4 is 5.17. Assume that the volume of the solutions are additive . What would be the Ka for NH4
Answer:
Following are the answer to this question:
Explanation:
The value of pH solution is =5.17 So, the p^{OH}:
[tex]p^{OH}[/tex]=14-56.17
=8.823
The volume of the [tex]NH_{3}[/tex] = 40.00 ml
convert into the liter= 0.040L
The value of the concentrated [tex]NH_{3}[/tex] =0.10 M
The volume of the [tex]NH_{4}Cl[/tex]= 50.00 ml
convert into the liter= 0.050L
The value of concentrated [tex]NH_{4}Cl[/tex]= 0.10 M
The volume of the [tex]H_{2}So_{4}[/tex]= 30 ml
convert into the liter= 0.030L
The value of concentrated [tex]H_2So_4[/tex]=0.05 M
Calculating total volume=(0.40+0.050+0.030)
=0.120 L
calculating the new concentrated value of [tex]NH_3[/tex] = [tex]\frac{0.10\times 0.040}{0.120}= 0.33 \ M[/tex]
calculating the new concentrated value of [tex]NH_4Cl[/tex]= [tex]\frac{0.050\times 0.10}{0.120}= 0.04166 \ M[/tex]calculating the new concentrated value of [tex]H_2So_4= \frac{0.030\times 0.05}{0.120}= 0.0125 \ M[/tex] when 1 mol [tex]H_2So_4[/tex] produced 2 mols [tex]H^{+}[/tex] so, 0.0125 in [tex]H_2So_4[/tex]produced:
[tex]=4 \times (2 \times 0.0125) \ mol H^{+}\\\\= 0.025 mol H^{+}[/tex]
create the ICE table:
[tex]NH_3 \ \ \ \ \ \ \ \ + H^{+} \ \ \ \ \ \ \longrightarrow NH_4^{+}[/tex]
I (m) 0.033(m) 0.025 0.04166
C -0.025 -0.025 + 0.025
E 8.3\times 10^{-3} 0 0.0667
now calculating pH:
when ph= 8.83:
[tex]P^{H}= p^{kb}|+ \log\frac{[NH_4^{+}]}{[NH_3]}\\\\8.83=p^{kb}+\log\frac{0.0667}{8.3 \times 10^{-3}}\\\\p^{kb}=8.83-0.9069\\\\ \ \ \ =7.7231 \\\\\ The P^{kb} \ for \ NH_3 \ is =7.7231\\\\\ The P^{kb} \ for N^{+}H_4=14-7.7231\\\\\ \ \ \ \ \ =6.2769[/tex]
A buffer with a pH of 3.98 contains 0.23 M of sodium benzoate and 0.38 M of benzoic acid. What is the concentration of [H3O+] in the solution after the addition of 0.058 mol HCl to a final volume of 1.3 L? Assume that any contribution of HCl to the volume is negligible g
Answer:
New pH = 3.84
Explanation:
First of all we may think that if the buffer has pH 3.98 and we're adding H⁺, pH's buffer will be lower, as the [H⁺] is been increased.
Let's determine the moles of each compound:
0.23 M . 1.3L = 0.299 moles of NaBz
0.38 M . 1.3L = 0.494 moles of HBz
We add 0.058 of HCl, which is the same as 0.058 moles of H⁻
HCl → H⁺ + Cl⁻
As we add the moles of protons, these are going to react to the Bz⁻
In the buffer system we have these dissociations:
NaBz → Na⁺ + Bz⁻
HBz → H⁺ + Bz⁻
So, as we add protons, we have a new equilibrium:
Bz⁻ + H⁺ ⇄ HBz
In 0.299 0.058 0.494
Eq 0.241 - 0.552
Protons are substracted to benzoate, so the [HBz] is now higher than before. We calculate the new pH, with the Henderson Hasselbach equation
pH = pKa + log (Bz⁻/HBz)
pH = 4.20 + log (0.241 / 0.552) → 3.84
A chemist adds of a M barium chlorate solution to a reaction flask. Calculate the mass in grams of barium chlorate the chemist has added to the flask. Round your answer to significant digits.
The given question is incomplete, the complete question is:
A chemist adds 200.0 ml of a 0.52M barium chlorate (Ba(CIO3)2) solution to a reaction flask. Calculate the mass in grams of barium chlorate the chemist has added to the flask. Round your answer to significant digits.
Answer:
The correct answer is 32 grams.
Explanation:
Based on the given solution, the molarity of barium chlorate solution given is 0.52 M, this shows that the solution will comprise 0.52 moles in 1 L or 1000 ml of the solution.
Therefore, in 200 ml, it will comprise 0.52/1000 × 200 moles of Ba(ClO₃)₂,
= 0.52/1000 × 200 = 0.104 moles
The molecular mass of Ba(ClO₃)₂ is 304.23 gram per mole
So, the mass of Ba(ClO₃)₂ in 0.104 moles will be,
= 304.23 g/mol × 0.104
= 31.639 grams or 32 grams.
Convert cm/S^2 to km/h^
2
Answer:
The answer to this question is 0.072km/h
bleaching powder reaction, mechanism, use
Answer:
Bleaching Powder's chemical formula is CaOCl2 and is called Calcium Oxychloride. It is prepared on dry slaked lime by chlorine gas. 2. ... It gives calcium chloride, chlorine and water when bleaching powder reacts with hydrochloric acid.
Explanation:
Draw a Lewis structure for one important resonance form of HBrO4 (HOBrO3). Include all lone pair electrons in your structure. Do not include formal charges in your structure.
Answer:
The Lewis structure is attached with the answer -
Explanation:
Lewis structure or Lewis dot diagram are diagrams or representation of showing the bonding between different or same atoms of a molecule in any and also shows lone pairs of electrons that may exist in the molecule as dots.
HBrO₄ is bromine oxoacid which is also known as perbromic acid. It is a unstable inorganic compound.
The Lewis structure is attached in form of image with representation of lone pairs of electrons.