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TEA [102]
4 years ago
5

Define a function ComputeGasVolume that returns the volume of a gas given parameters pressure, temperature, and moles. Use the g

as equation PV = nRT, where P is pressure in Pascals, V is volume in cubic meters, n is number of moles, R is the gas constant 8.3144621 ( J / (mol*K)), and T is temperature in Kelvin.
Chemistry
1 answer:
andre [41]4 years ago
7 0

Answer: A function can be defined as a relation in which one thing is dependent on another for its value.

Explanation: Given R = 8.314J/mol*k

PV = nRT

V = nRT/ P

V = 8.314RT / P (cm^3)

Volume of gas =(( 8.314 * R* T) / P ) cm3

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4 years ago
Hand lighters contain butane gas. If a hand lighter contains butane gas at 2.50 atm of
Lesechka [4]

Answer:

3.35 atm

Since P₂ > 3.00 atm, the lighter would explode.

Explanation:

Step 1: Given data

  • Initial pressure of butane gas (P₁): 2.50 atm
  • Initial temperature of butane gas (T₁): 293 K
  • Final pressure of butane gas (P₂): ?
  • Final temperature of butane gas (T₂): 393 K

Step 2: Calculate the final pressure of butane gas

If we assume ideal behavior, we can calculate the final pressure of butane gas using Gay Lussac's law.

P₁/T₁ = P₂/T₂

P₂ = P₁ × T₂/T₁

P₂ = 2.50 atm × 393 K/293 K = 3.35 atm

Since P₂ > 3.00 atm, the lighter would explode.

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3 years ago
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3 years ago
3. Calculate the molecular or formula mass of each of the following: (a) Pa (c) Ca(NO3)2 (d) CH3CO2H (acetic acid)​
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Answer:

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3 0
3 years ago
Balance each of the following in BASE: Show work
kupik [55]

Answer : The balanced chemical equation in a basic solution are,

(A) 2CrO_4^{2-}+8H_2O+3Cu\rightarrow 2Cr(OH)_3+4OH^-+3Cu(OH)_2

(B) 2NO_2+2OH^-\rightarrow NO_3^-+H_2O+NO_2^-

(C) 4Zn+7OH^-+NO_3^-+6H_2O\rightarrow 4[Zn(OH)_4]^{2-}+NH_3

(D) Br_2+12OH^-+5Br_2\rightarrow 2BrO_3^-+6H_2O+10Br^-

Explanation :

<u>(A) The given chemical reaction is :</u>

CrO_4^{2-}(aq)+Cu(s)\rightarrow Cr(OH)_3(s)+Cu(OH)_2(s)

The oxidation-reduction half reaction will be :

Oxidation : Cu\rightarrow Cu^{2+}+2e^-

Reduction : CrO_4^{2-}+4H_2O+3e^-\rightarrow Cr^{3+}+8OH^-

In order to balance the electrons, we multiply the oxidation reaction by 3 and reduction reaction by 2 then added both equation, we get the balanced redox reaction.

The balanced chemical equation in a basic solution will be,

2CrO_4^{2-}+8H_2O+3Cu\rightarrow 2Cr^{3+}+16OH^-+3Cu^{2+}

or,

2CrO_4^{2-}+8H_2O+3Cu\rightarrow 2Cr(OH)_3+4OH^-+3Cu(OH)_2

<u>(B) The given chemical reaction is :</u>

NO_2(g)\rightarrow NO_3^-(aq)+NO_2^-(aq)

The oxidation-reduction half reaction will be :

Oxidation : NO_2+2OH^-\rightarrow NO_3^-+H_2O+1e^-

Reduction : NO_2+1e^-\rightarrow NO_2^-

The electrons in both the reactions are balanced. Now added both equation, we get the balanced redox reaction.

The balanced chemical equation in a basic solution will be,

2NO_2+2OH^-\rightarrow NO_3^-+H_2O+NO_2^-

<u>(C) The given chemical reaction is :</u>

Zn(s)+NO_3^-(aq)\rightarrow Zn(OH)_4^{2-}(aq)+NH_3(g)

The oxidation-reduction half reaction will be :

Oxidation : Zn+4OH^-\rightarrow [Zn(OH)_4]^{2-}+2e^-

Reduction : NO_3^-+6H_2O+8e^-\rightarrow NH_3+9OH^-

In order to balance the electrons, we multiply the oxidation reaction by 4 and then added both equation, we get the balanced redox reaction.

The balanced chemical equation in a basic solution will be,

4Zn+7OH^-+NO_3^-+6H_2O\rightarrow 4[Zn(OH)_4]^{2-}+NH_3

<u>(D) The given chemical reaction is :</u>

Br_2(l)\rightarrow BrO_3^-(aq)+Br^-(aq)

The oxidation-reduction half reaction will be :

Oxidation : Br_2+12OH^-\rightarrow 2BrO_3^-+6H_2O+10e^-

Reduction : Br_2+2e^-\rightarrow 2Br^-

In order to balance the electrons, we multiply the reduction reaction by 5 and then added both equation, we get the balanced redox reaction.

The balanced chemical equation in a basic solution will be,

Br_2+12OH^-+5Br_2\rightarrow 2BrO_3^-+6H_2O+10Br^-

7 0
3 years ago
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