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Svet_ta [14]
3 years ago
15

2.At 35°C, a small sample of methane gas (CH4) has a volume of 1.5 liters. The temperature of the methane gas is slowly cooled t

o 17°C. What is the new volume of the methane sample?​
Chemistry
1 answer:
defon3 years ago
5 0

Answer:

V₂ = 1.41 L

Explanation:

Given data:

Initial temperature = 35°C (35 +273.15 K = 308.15 K)

Initial volume = 1.5 L

Final temperature = 17°C (17+273.15 K = 290.15 K)

Final volume = ?

Solution:

The given problem will be solve through the Charles Law.

According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.

Mathematical expression:

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume  

T₂ = Final temperature

Now we will put the values in formula.

V₁/T₁ = V₂/T₂

V₂ = V₁T₂/T₁  

V₂ = 1.5 L × 290.15 K / 308.15 k

V₂ = 435.23 L.K / 308.15 k

V₂ = 1.41 L

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Decide which element probably forms a compound with chlorine that has a chemical formula most and least similar to the chemical
valentinak56 [21]

Answer:

phosphorus - least similar to barium

beryllium - most similar to barium

Explanation:

The formula of the chloride formed between barium and chlorine is BaCl2.

Barium is a metallic element and has the valency of +2. Beryllium is also a group two metal with a valency of +2. Hence the compound formed between beryllium and chlorine is BeCl2

Phosphorus is a nonmetal and forms a completely different chloride from BaCl2. The chloride of phosphorus is PCl5 and PCl3. These are least similar to BaCl2

6 0
3 years ago
The decomposition of dinitrogen pentoxide, N2O5, to NO2 and O2 is a first-order reaction. At 60°C, the rate constant is 2.8 × 10
Sati [7]

Answer:

a. 113 min

Explanation:

Considering the equilibrium:-

                   2N₂O₅ ⇔ 4NO₂ + O₂

At t = 0        125 kPa

At t = teq     125 - 2x      4x        x

Thus, total pressure = 125 - 2x + 4x + x = 125 - 3x

125 - 3x = 176 kPa

x = 17 kPa

Remaining pressure of N₂O₅ = 125 - 2*17 kPa = 91 kPa

Using integrated rate law for first order kinetics as:

[A_t]=[A_0]e^{-kt}

Where,  

[A_t] is the concentration at time t

[A_0] is the initial concentration

Given that:

The rate constant, k = 2.8\times 10^{-3} min⁻¹

Initial concentration [A_0] = 125 kPa

Final concentration [A_t] = 91 kPa

Time = ?

Applying in the above equation, we get that:-

91=125e^{-2.8\times 10^{-3}\times t}

125e^{-2.8\times \:10^{-3}t}=91

-2.8\times \:10^{-3}t=\ln \left(\frac{91}{125}\right)

t=113\ min

3 0
3 years ago
Which describes the current model of the atom? (2 points)
Harman [31]
A. I hope this helps!
4 0
3 years ago
Read 2 more answers
A gaseous mixture of O2 and N2 contains 37.8% nitrogen by mass. What is the partial pressure of oxygen in the mixture if the tot
kondor19780726 [428]

Answer: The partial pressure of oxygen in the mixture if the total pressure is 525 mmHg is 310 mm Hg

Explanation:

mass of nitrogen = 37.8 g

mass of oxygen = (100-37.8) g = 62.2 g

Using the equation given by Raoult's law, we get:

p_A=\chi_A\times P_T

p_{O_2} = partial pressure of O_2 = ?

\chi_{O_2} = mole fraction of O_2=\frac{\text{Moles of }O_2}{\text{Total moles}}

P_{T} = total pressure of mixture  = 525 mmHg

{\text{Moles of }O_2}=\frac{\text {Given mass}}{\text {Molar mass}}=\frac{62.2g}{32g/mol}=1.94moles

{\text{Moles of }N_2}=\frac{\text {Given mass}}{\text {Molar mass}}=\frac{37.8g}{28g/mol}=1.35moles

Total moles = 1.94 + 1.35 = 3.29 moles

\chi_{O_2}=\frac{1.94}{3.29}=0.59

p_{O_2}=\chi_{O_2}\times P_T=0.59\times 525=310mmHg

Thus the partial pressure of oxygen in the mixture if the total pressure is 525 mmHg is 310 mm Hg

7 0
3 years ago
Aluminum boils at 2467°C. Aluminum’s boiling point in Kelvin is 2194.<br> T or F
dusya [7]
That is false because aluminum melts at 2,470C
7 0
3 years ago
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