According to <span>Gay-Lussac's Law the temperature and Pressure are directly proportional to each other if the amount and volume of given gas are kept constant.
Mathematically for initial and final states it is expressed as,

P</span>₁ / T₁ = P₂ / T₂ ----- (1)
Data Given;
P₁ = 1.5 atm

T₁ = 35 °C + 273 = 308 K

P₂ = ?

T₂ = 0 °C + 273 = 273 K

Solving Eq. 1 for P₂,

P₂ = P₁ T₂ / T₁

Putting values,
P₂ = (1.5 atm × 273 K) ÷ 308 K

P₂ = 1.32 atm
Result:
As the temperature is decreased so the pressure also decreases from 1.5 atm to 1.32 atm. Therefore the bag will contract.

5 0

4 years ago

If 3.0 atm of pure HN3(g) is decomposed initially, what is the final total pressure in the reaction container? What are the part

lawyer [7]

This is an incomplete question, here is a complete question.

Hydrogen azide, HN₃, decomposes on heating by thefollowing unbalanced reaction:

$HN_3(g)\rightarrow N_2(g)+H_2(g)$

If 3.0 atm of pure HN₃ (g) is decomposed initially,what is the final total pressure in the reaction container? Whatare the partial pressures of nitrogen and hydrogen gas? Assume thatthe volume and temperature of the reaction container are constant.

Answer : The partial pressure of $N_2$ and $H_2$ gases are, 4.5 atm and 1.5 atm respectively.

Explanation :

The given unbalanced chemical reaction is:

$HN_3(g)\rightarrow N_2(g)+H_2(g)$

This reaction is an unbalanced chemical reaction because in this reaction number of hydrogen and nitrogen atoms are not balanced on both side of the reaction.

In order to balance the chemical equation, the coefficient '2' put before the $HN_3$ and the coefficient '3' put before the $N_2$ then we get the balanced chemical equation.

The balanced chemical reaction will be,

$2HN_3(g)\rightarrow 3N_2(g)+H_2(g)$

As we are given:

The pressure of pure $HN_3$ = 3.0 atm

$p_{Total}=2\times p_{HN_3}=2\times 3.0atm=6.0atm$

From the reaction we conclude that:

Number of moles of $N_2$ = 3 mol

Number of moles of $H_2$ = 1 mol

Now we have to calculate the mole fraction of $N_2$ and $H_2$

$\text{Mole fraction of }N_2=\frac{\text{Moles of }N_2}{\text{Moles of }N_2+\text{Moles of }H_2}=\frac{3}{3+1}=0.75$

and,

$\text{Mole fraction of }H_2=\frac{\text{Moles of }H_2}{\text{Moles of }N_2+\text{Moles of }H_2}=\frac{1}{3+1}=0.25$

Now we have to calculate the partial pressure of $N_2$ and $H_2$

According to the Raoult's law,

$p_i=X_i\times p_T$

where,

$p_i$ = partial pressure of gas

$p_T$ = total pressure of gas = 6.0 atm

$X_i$ = mole fraction of gas

$p_{N_2}=X_{N_2}\times p_T$

$p_{N_2}=0.75\times 6.0atm=4.5atm$

and,

$p_{H_2}=X_{H_2}\times p_T$

$p_{H_2}=0.25\times 6.0atm=1.5atm$

Thus, the partial pressure of $N_2$ and $H_2$ gases are, 4.5 atm and 1.5 atm respectively.

8 0

3 years ago

What is the kb of a 0.0200 m ( at equilibrium) solution of methyl amine, ch3nh2, that has a ph of 11.40?

marin [14]

Answer: Kb = 3.15 × 10 ⁻⁴

Explanation:

This is how you calculate Kb for this reaction.

1) Equilibrium equation:

CH₃NH₂ + H₂O ⇄ CH₃NH₃⁺ + OH⁻

2) Kb = [CH₃NH₃⁺] [OH⁻] / [CH₃NH₂] ↔ all the spieces in equilibrium

3) From the stoichiometry [CH₃NH₃⁺] = [OH⁻]

Then, Kb = [OH⁻] [OH⁻] / [CH₃NH₂] = [OH⁻]² / [CH₃NH₂]

4) You get [OH⁻] from the pH in this way:

pOH + pOH = 14 ⇒ pOH = 14 - pH = 14 - 11.40 = 2.60

pOH = - log [OH⁻] = 2.60 ⇒ [OH⁻] = 10^(-2.6) = 0.002512

5) [CH₃NH₂] in equilibrium is given: 0.0200M

6) Now compute:

Kb = (0.002512)² / 0.0200 = 3.15 × 10 ⁻⁴

8 0

3 years ago