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Ivanshal [37]
3 years ago
14

A sample of gas has a volume of 1.16L at a temperature of 23C. At what temperature will the gas have a volume of 1.25L?

Chemistry
1 answer:
Masja [62]3 years ago
7 0

Answer:

T₂ = 319.13 K

Explanation:

Given data:

Initial volume of gas = 1.16 L

Initial temperature = 23°C

Final temperature = ?

Final volume = 1.25 L

Solution:

Initial temperature = 23°C (23+273.15 = 296.15 K)

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

T₂  = V₂ T₁ / V₁

T₂ = 1.25 L × 296.15 K / 1.16 L

T₂ = 370.19 L.K /  1.16 L

T₂ = 319.13 K

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How many atoms are there in 3.559*10^-6 mol of krypton?
algol [13]

Answer:

The answer is

<h2>2.143 \times  {10}^{18}  \:  \: atoms</h2>

Explanation:

To find the number of atoms given the number of moles we use the formula

N = n × L

where

N is the number of entities

n is the number of moles

L is the Avogadro's constant which is

6.02 × 10²³ entities

From the question

n = 3.559 \times  {10}^{ - 6}  \: mol

Substitute the values into the above formula and solve

That's

<h3>N  = 3.559 \times  {10}^{ - 6}  \times 6.02 \times  {10}^{23}  \\  = 2.1425 \times  {10}^{18}</h3>

We have the final answer as

<h3>N  = 2.143 \times  {10}^{18}  \:  \: atoms</h3>

Hope this helps you

3 0
3 years ago
What is the percent yield of CuS for the following reaction given that you start with 15.5 g of Na2S and 12.1 g CuSO4? The actua
notka56 [123]

Answer:

(B) 42.1%

Explanation:

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Given: For Na_2S

Given mass = 15.5 g

Molar mass of Na_2S = 78.0452 g/mol

<u>Moles of Na_2S = 15.5 g / 78.0452 g/mol = 0.1986 moles</u>

Given: For CuSO_4

Given mass = 12.1 g

Molar mass of CuSO_4 = 159.609 g/mol

<u>Moles of CuSO_4 = 12.1 g / 159.609 g/mol = 0.0758 moles</u>

According to the given reaction:

Na_2S+CuSO_4\rightarrow Na_2SO_4+CuS

1 mole of Na_2S react with 1 mole of CuSO_4

0.1986 mole of Na_2S react with 0.1986 mole of CuSO_4

Available moles of CuSO_4 = 0.0758 moles

Limiting reagent is the one which is present in small amount. Thus, CuSO_4 is limiting reagent. (0.0758 < 0.1986)

The formation of the product is governed by the limiting reagent. So,

1 mole of CuSO_4 gives 1 mole of CuS

0.0758 mole of CuSO_4 gives 0.0758 mole of CuS

Molar mass of CuS = 95.611 g/mol

Mass of CuS = Moles × Molar mass = 0.0758 × 95.611 g = 7.2473 g

Theoretical yield = 7.2473 g

Given experimental yield = 3.05 g

% yield = (Experimental yield / Theoretical yield) × 100 = (3.05/7.2473) × 100 = 42.1 %

<u>Option B is correct.</u>

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