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2 years ago

What is the volume of 2.0 g of cs2 vapor (molar mass = 76.15 g/mol) at 70°c and 726 mm hg? l?

1 answer:

riadik2000 [5.3K]2 years ago

5 0

To find the volume occupied by 2.0 g of CS₂ gas we can use the ideal gas law equation
PV = nRT
where P - pressure - 726 mmHg x 133 Pa/mmHg = 96 558 Pa
V - volume
n - number of moles = mass / molar mass
number of CS₂ moles = 2.0 g / 76.15 g/mol = 0.026 mol
R - universal gas constant - 8.314 Jmol⁻¹K⁻¹
T - temperature in kelvin - 70 °C + 273 = 343 K
substituting these values in the equation
96 558 Pa x V = 0.026 mol x 8.314 Jmol⁻¹K⁻¹ x 343 K
V = 0.768 L
the volume occupied is 0.768 L

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For the Zn - Cu^2+ voltaic cell Zn(s) + Cu^2+(aq, 1M) + Cu(s) E degree _cell = 1.10 V Given that the standard reduction potentia

Fittoniya [83]

Answer : The value of $E^o_{(Cu^{2+}/Cu)}$ is, 0.34 V

Explanation :

Here, copper will undergo reduction reaction will get reduced. Zinc will undergo oxidation reaction and will get oxidized.

The oxidation-reduction half cell reaction will be,

Oxidation half reaction: $Zn\rightarrow Zn^{2+}+2e^-$

Reduction half reaction: $Cu^{2+}+2e^-\rightarrow Cu$

Oxidation reaction occurs at anode and reduction reaction occurs at cathode. That means, gold shows reduction and occurs at cathode and chromium shows oxidation and occurs at anode.

The overall balanced equation of the cell is,

$Zn+Cu^{2+}\rightarrow Zn^{2+}+Cu$

To calculate the $E^o_{(Cu^{2+}/Cu)}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

$E^o_{cell}=E^o_{(Cu^{2+}/Cu)}-E^o_{(Zn^{2+}/Zn)}$

Putting values in above equation, we get:

$1.10V=E^o_{(Cu^{2+}/Cu)}-(-0.76V)$

$E^o_{(Cu^{2+}/Cu)}=0.34V$

Hence, the value of $E^o_{(Cu^{2+}/Cu)}$ is, 0.34 V

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2 years ago

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2 moles of sodium chloride are dissolved in 8 liters of water. What is its molarity (M) ?

otez555 [7]

Answer:

0.25M

Explanation:

Since molarity is defined as the amount of substance in 1 liter of a solution, and amount in chemistry refers to the number of moles, molarity can be found using the equation below:

Molarity= number of moles ÷ volume in liters

Given: number of moles= 2

Volume= 8L

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2 years ago

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2 years ago

Consider the reaction given below.

Drupady [299]

Answer:

K = 0.167 s⁻¹

Explanation:

1) Rate law, at a given temperature:

Since all the data are obtained at the same temperature, the equilibrium constant is the same.

Since only reactants A and B participate in the reaction, you assume that the form of the rate law is:

r = K [A]ᵃ [B]ᵇ

2) Use the data from the table

Since the first and second set of data have the same concentration of the reactant A, you can use them to find the exponent b:

r₁ = (1.50)ᵃ (1.50)ᵇ = 2.50 × 10⁻¹ M/s

r₂ = (1.50)ᵃ (2.50)ᵇ = 2.50 × 10⁻¹ M/s

Divide r₂ by r₁: [ 2.50 / 1.50] ᵇ = 1 ⇒ b = 0

Use the first and second set of data to find the exponent a:

r₁ = (1.50)ᵃ (1.50)ᵇ = 2.50 × 10⁻¹ M/s

r₃ = (3.00)ᵃ (1.50)ᵇ = 5.00 × 10⁻¹ M/s

Divide r₃ by r₂: [3.00 / 1.50]ᵃ = [5.00 / 2.50]

2ᵃ = 2 ⇒ a = 1

3) Write the rate law

r = K [A]¹ [B]⁰ = K[A]

This means, that the rate is independent of reactant B and is of first order respect reactant A.

4) Use any set of data to find K

With the first set of data

r = K (1.50 M) = 2.50 × 10⁻¹ M/s ⇒ K = 0.250 M/s / 1.50 M = 0.167 s⁻¹

Result: the rate constant is K = 0.167 s⁻¹

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