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horsena [70]
2 years ago
9

1. A sample of a gas is occupying a 500 mL container at a pressure of 1.4 atm and a temperature of 25 °C.

Chemistry
1 answer:
Alex787 [66]2 years ago
4 0

The new volume of a sample of gas initially occupying a 500 mL container at a pressure of 1.4 atm and a temperature of 25 °C is 281.01mL.

<h3>How to calculate volume?</h3>

The volume of a gas can be calculated using the following formula:

P1V1//T1 = P2V2/T2

Where;

  • P1 = initial pressure
  • P2 = final pressure
  • V1 = initial volume
  • V2 = final volume
  • T1 = initial temperature
  • T2 = final temperature

1.4 × 500/298 = 2.7 × V2/323

2.348 × 323 = 2.7V2

758.72 = 2.7V2

V2 = 281.009mL

Therefore, the new volume of a sample of gas initially occupying a 500 mL container at a pressure of 1.4 atm and a temperature of 25 °C is 281.01mL.

Learn more about volume at: brainly.com/question/1578538

#SPJ1

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Calculate the number of grams of solute in 500.0 mL of 0.189 M KOH.
KIM [24]

Answer : The number of grams of solute in 500.0 mL of 0.189 M KOH is, 5.292 grams

Solution : Given,

Volume of solution = 500 ml

Molarity of KOH solution = 0.189 M

Molar mass of KOH = 56 g/mole

Formula used :

Molarity=\frac{\text{Mass of KOH}\times 1000}{\text{Molar mass of KOH}\times \text{Volume of solution in ml}}

Now put all the given values in this formula, we get the mass of solute KOH.

0.189M=\frac{\text{Mass of KOH}\times 1000}{(56g/mole)\times (500ml)}

\text{Mass of KOH}=5.292g

Therefore, the number of grams of solute in 500.0 mL of 0.189 M KOH is, 5.292 grams

7 0
3 years ago
Read 2 more answers
For the reaction, Cl2 + 2KBr --&gt; 2KCl + Br2, how many moles of potassium chloride, KCl, are produced from 102g of potassium b
Sedaia [141]

Solution :

From the balanced chemical equation, we can say that 1 moles of KBr will produce 1 moles of KCl .

Moles of KBr in 102 g of potassium bromide.

n = 102/119.002

n = 0.86 mole.

So, number of miles of KCl produced are also 0.86 mole.

Mass of KCl produced :

m = 0.86 \times Molar \  mass \ of \  KCl\\\\m = 0.86 \times 74.5 \ gram \\\\m = 64.07\  gram

Hence, this is the required solution.

5 0
3 years ago
VA 19.75-g sample was heated by 12.35 calories. The specific heat of the sample is 0.125 cal/g°C. What was the initial temperatu
SOVA2 [1]

Answer:

31.9 °C  

Explanation:

The formula for the heat q absorbed by an object is

q = mCΔT where ΔT = (T₂ - T₁)

Data:

q = 12.35 cal

m = 19.75 g

C = 0.125 cal°C⁻¹g⁻¹

T₂ = 37.0 °C

Calculations

(a) Calculate ΔT

q = mCΔT

12.35 cal = 19.25 g × 0.125 cal°C⁻¹g⁻¹ × ΔT

12.35 = 2.406ΔT °C⁻¹  

ΔT  = 12.35/(2.406 °C⁻¹) = 5.13 °C

(b) Calculate T₂

ΔT = T₂ - T₁

T₁ = T₂ - ΔT = 37.0 °C - 5.13 °C = 31.9 °C

The original temperature was 31.9 °C.

 

6 0
3 years ago
From these four cycles which are water cycle, carbon cycle, nitrogen cycle, and phosphorus cycle which cycle has more nutrients
agasfer [191]

Answer:

Nitrogen cycle

Explanation:

7 0
3 years ago
What is the volume of an object with the mass of 7.9 grams in the density of 2.28g/ml.
Schach [20]

Answer:

3.7mL is the volume of the object

Explanation:

To convert the mass of any object to volume we must use density that is defined as the ratio between mass of the object and the space that is occupying. For an object that weighs 7.9g and the density is 2.28g/mL, the volume is:

7.9g * (1mL / 2.28mL) =

<h3>3.7mL is the volume of the object</h3>
7 0
3 years ago
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