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3 years ago

A sample of nitrogen gas with a volume of 180 cm at a temperature of

Chemistry

1 answer:

maksim [4K]3 years ago

4 0

Answer:

The final pressure of gas is 82.64 KNm⁻²

Explanation:

Given data:

Initial volume of gas = 180 cm³

Temperature of gas = 27°C

Initial pressure = 101 KNm⁻²

Final volume = 220 cm³

Final pressure = ?

Solution:

The given problem will be solved through the Boly's law,

"The volume of given amount of gas is inversely proportional to its pressure by keeping the temperature and number of moles constant"

Mathematical expression:

P₁V₁ = P₂V₂

P₁ = Initial pressure

V₁ = initial volume

P₂ = final pressure

V₂ = final volume

Now we will put the values in formula,

P₁V₁ = P₂V₂

101 KNm⁻² × 180 cm³ = P₂ × 220 cm³

P₂ = 18180 KNm⁻². cm³/220 cm³

P₂ = 82.64 KNm⁻²

The final pressure of gas is 82.64 KNm⁻².

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Andre45 [30]

Answer:

The answer to your question is:

Explanation:

1. How many grams of B are present in 3.35 grams of boron tribromide ?

________ grams B.

MW BBr₃ = 251 g

251 g of BBr₃ ---------------------- 11 g of B

3.35 g ----------------------- x

x = (3.35 x 11) / 251 = 0.147 g of B

2. How many grams of boron tribromide contain 4.69 grams of Br ?

________grams boron tribromide.

MW BBr₃ = 251g

251g of BBr₃ ----------------- 80 g of Br

x ----------------- 4.69 g

x = (4.69 x 251)/ 80 = 14.71 g of BBr₃

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________grams N.

MW NF₃ = 71 g

71 g of NF₃ ----------------- 14 g of N

4.11 g ---------------- x

x = (4.11 x 14) / 71 = 0.81 g of N

4. How many grams of nitrogen trifluoride contain 3.07 grams of F ?

________grams nitrogen trifluoride.

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71 g of NF₃ --------------------- 19 g of F

x --------------------- 3.07 g

x = (3.07 x 71) / 19 = 11.5 g of NF₃

5.How many grams of Co3+ are present in 1.16 grams of cobalt(III) iodide?

________grams Co3+.

MW CoI₃ = 440 g

440 g of CoI₃ ------------------ 59 g of Co

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________grams cobalt(III) iodide.

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x = (2.28 x 440) / 59

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5 0

3 years ago

22.4 liters of a gas at a constant temperature of zero Celsius and a pressure of 1 atmosphere is compressed to a volume of 6.2 l

zhannawk [14.2K]

<h2>Hello!</h2>

The answer is: The new pressure of the gas is 3.6 atm.

From the statement we know that the gas is kept at a constant temperature of 0°C, so, if the gas keeps a constant temperature we can use the Boyle's Law to solve this problem.

The Boyle's Law states that:

$P_{1}V_{1}=P_{2}V_{2}$

Where,

P is the pressure of the gas.

V is the volume of the gas.

So, the given information is:

$V_{1}=22.4L\\P_{1}=1atm\\V_{2}=6.2L$

Now, substituting it into the Boyle's Law equation to calculate the new pressure, we have:

$P_{1}V_{1}=P_{2}V_{2}\\\\1atm*22.4L=P_{2}*6.2L\\\\P_{2}=\frac{1atm*22.4L}{6.2L}=3.6atm$

So, the new pressure is 3.6 atm.

Have a nice day!

4 0

4 years ago

The reaction of solid aluminum with hydrochloric acid is used to make hydrogen gas in a laboratory experiment. The reaction is 2

Gemiola [76]

Answer:

0.003088 moles of hydrogen gas were formed .

Explanation:

Pressure at which hydrogen gas is collected at 20°C = 768.0 Torr

Vapor pressure of water at 20°C = 17.5 Torr

Total pressure = Vapor pressure of water + Partial pressure of hydrogen gas

Partial pressure of hydrogen gas:

Total pressure - Vapor pressure of water

= 768.0 Torr - 17.5 Torr = 750.5 Torr = 0.987 atm

(1 Torr = 0.001315 atm)

Pressure of hydrogen gas =P = 0.986 atm

Temperature at which gas was collected ,T= 20°C = 293.15 K

Volume of the gas ,V= 75.3 mL = 0.0753 L

Moles of hydrogen gas = n

PV=nRT (An ideal gas equation)

$n=\frac{PV}{RT}=\frac{0.987 atm\times 0.0753 L}{0.0821 atm L/mol K\times 293.15 K}=0.003088 mol$

0.003088 moles of hydrogen gas were formed .

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4 years ago

A living cell with a tonicity (solute concentration) equivalent to 0.9% NaCl is placed in a solution containing 2% NaCl. Assume

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Answer:

This question is incomplete

Explanation:

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tino4ka555 [31]

Answer:

Explanation:

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4 years ago