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2 years ago

.A gas occupies 25,3 mL at a pressure of 152 kPa. Find the volume if the pressure is

Chemistry

1 answer:

Strike441 [17]2 years ago

5 0

Answer:

47.36mL

Explanation:

Using Boyles law equation, which states that:

P1V1 = P2V2

Where;

V1 = initial volume (mL)

V2 = final volume (mL)

P1 = initial pressure (atm)

P2 = final pressure (atm)

Based on the provided information, V1 = 25.3mL, P1 = 152 kPa, V2 = ?, P2 = 0.804atm

First, we need to convert 152kPa to atm by dividing by 101

1kPa = 0.0099atm

152kPa = 1.505atm

P1V1 = P2V2

1.505 × 25.3 = 0.804 × V2

38.08 = 0.804V2

V2 = 38.08/0.804

V2 = 47.36mL

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2c+02=2CO2. The moles of co2 produced when 0.25 moles of O2 react is?

Sophie [7]

<h3>Answer:</h3>

$\displaystyle 0.5 \ mol \ CO_2$

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
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Addition
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<h3>Explanation:</h3>

<u>Step 1: Define</u>

[RxN - Balanced] 2C + O₂ → 2CO₂

[Given] 0.25 moles O₂

[Solve] moles CO₂

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[RxN] 1 mol O₂ → 2 mol CO₂

<u>Step 3: Stoichiometry</u>

[DA] Set up: $\displaystyle 0.25 \ moles \ O_2(\frac{2 \ mol \ CO_2}{1 \ mol \ O_2})$
[DA] Multiply/Divide [Cancel out units]: $\displaystyle 0.5 \ mol \ CO_2$

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2 years ago

In a single displacement reaction between Sodium Phosphate and Barium, how much of each product (in grams) will be formed from 1

mixer [17]

Answer:

14.6 g of barium phosphate

3.35 g of sodium metal

Explanation:

2Na3PO4(aq) + 3Ba(s) -------> Ba3(PO4)2(aq) + 6Na(s)

The first step in any such reaction is to but down the balanced reaction equation according to the stoichiometry of the reaction.

The two products formed are barium phosphate and sodium metal.

Number of moles of barium corresponding to 10.0g of barium = mass of barium/ molar mass of barium

Molar mass of barium = 137.327 g

Number of moles of barium = 10/137.327

Number of moles of barium = 0.0728 moles

For barium phosphate;

3 moles of barium yields 1 mole of barium phosphate

0.0728 moles yields 0.0728 moles × 1/3 = 0.0243 moles of barium phosphate

Molar mass of barium phosphate = 601.93 g/mol

Therefore mass of barium phosphate = 0.0243 moles × 601.93 g/mol = 14.6 g of barium phosphate

For sodium metal

3 moles of barium yields 6 moles of sodium metal

0.0728 moles of barium yields 0.0728 × 6 / 3 = 0.1456 moles of sodium

Molar mass of sodium metal= 23 gmol-1

Mass of sodium metal= 0.1456g × 23 gmol-1 = 3.35 g of sodium metal

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madam [21]

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3 years ago