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Andreas93 [3]
2 years ago
10

A gas has a volume of 350 mL at 45 oK. If the volume changes to 400 mL, what is the new temperature?

Chemistry
1 answer:
elena-14-01-66 [18.8K]2 years ago
4 0

Answer:

T₂ = 39.4 °K

Explanation:

Because you are only dealing with volume and temperature, you can use Charles' Law to find the missing value. The formula looks like this:

V₁ / T₁ = V₂ / T₂

In this formula, "V₁" and T₁" represent the initial volume and temperature. "V₂" and "T₂" represent the final volume and temperature. You have been given values for all of the variables except for "T₂". Therefore, by plugging these values into the formula, you can simplify to find the answer.

V₁ = 350 mL                     T₁ = 45 °K

V₂ = 400 mL                     T₂ = ?

V₁ / T₁ = V₂ / T₂                                        <----- Charles' Law formula

(350 mL)(45 °K) = (400 mL)T₂                <----- Insert values into variables

15750 = (400 mL)T₂                                <----- Multiply left side

39.4 = T₂                                                 <----- Divide both sides by 400

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3 0
4 years ago
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adelina 88 [10]

Answer:

A.

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Explanation:

Ductility

6 0
3 years ago
Okay so this is half math half chemistry...
AleksandrR [38]
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3 0
3 years ago
5.1169 mol of Ne is held at 0.9148 atm and 911 K. What is the volume of its container in liters?
sveticcg [70]

By applying the Boyle's equation and substituting our given data the volume of the container was found to be 418.14 Litres

<h3>Boyle's  Law</h3>

Given Data

  • number of moles of Ne = 5.1169 mol
  • Pressure = 0.9148 atm
  • Temperature = 911 K

We know that the relationship between pressure and temperature is given as

PV = nRT

R = 0.08206

Making the volume subject of formula we have

V= nRT/P

Substituting our given data to find the volume we have

V = 5.1169*0.08206*911/0.9148

V = 382.522353554/0.9148

V = 418.14 L

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brainly.com/question/469270

5 0
2 years ago
56.5 g of sodium nitrate is dissolved in water to make 627 g of solution . what is the percent sodium nitrate in the solution
Tamiku [17]

Answer:

9.01%

Explanation:

The following data were obtained from the question:

Mass of sodium nitrate, NaNO3 = 56.5g

Mass of solution = 627g

The percentage composition of sodium nitrate, NaNO3 in the solution can be obtained as follow:

Percentage composition of NaNO3 = Mass of NaNO3/mass of solution x 100

Percentage composition of NaNO3 = 56.5/627 x 100 = 9.01%

Therefore, the percentage composition of sodium nitrate, NaNO3 in the solution is 9.01%

8 0
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