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3 years ago

A 365 L balloon is blown up inside at 283 K. It is then taken outside in the hot sun and Heated to 300K. What is the new volume?

Chemistry

1 answer:

asambeis [7]3 years ago

6 0

V₁ = initial Volume of the balloon after it is blown up = 365 L

V₂ = new Volume of the balloon after it is taken outside = ?

T₁ = initial temperature of the balloon = 283 K

T₂ = new temperature of the balloon = 300 K

using the equation

V₁/V₂ = T₁/T₂

365/V₂ = 283/300

V₂ = 387 L

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Ksju [112]

109/8.56=12.7
50+12.7
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Mass= Volume x Density so i divided the mass and density to get the volume. and afterwards i would just add it to the mass to get my final answer

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3 years ago

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Type the correct answer in the box. Express your answer to three significant figures. Calcium nitrate reacts with sodium phospha

RoseWind [281]

6.07 grams is the theoretical yield of calcium phosphate (Ca₃(PO₄)₂).

<h3>How we calculate mass from moles?</h3>

Mass of any substance can be calculated by using moles as:

n = W/M, where

W = required mass

M= molar mass

Given chemical reaction is:

3Ca(NO₃)₂ + 2Na₃PO₄ → 6NaNO₃ + Ca₃(PO₄)₂

From the stoichiometry it is clear that:

3 moles of Ca(NO₃)₂ = produce 1 mole of Ca₃(PO₄)₂

Given mass of Ca(NO₃)₂ = 96.1g

Mole of Ca(NO₃)₂ = 96.1g/164g/mol = 0.5859moles

So, 0.5859 moles of Ca(NO₃)₂ = produce 0.5859×1/3 = 0.0196 moles of Ca₃(PO₄)₂

Required mass of Ca₃(PO₄)₂ will be calculated by using moles as:

W = 0.0196mole × 310g/mole = 6.07 grams

Hence, 6.07 grams is the theoretical yield of calcium phosphate.

To know more about moles, visit the below link:

brainly.com/question/15373263

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2 years ago

What is the percentage yield of O2 if 12.3 g of KClO3 (molar mass 123 g) is decomposed to produce 3.2 g of O2 (molar mass 32 g)

My name is Ann [436]

Answer:

The percentage yield of O2 is 66.7%

Explanation:

Reaction for decomposition of potassium chlorate is:

2KClO₃ → 2KCl + 3O₂

The products are potassium chloride and oxygen.

Let's find out the moles of chlorate.

Mass / Molar mass = Moles

12.3 g / 123 g/mol = 0.1 mol

So ratio is 2:3, 2 moles of chlorate produce 3 mol of oxygen.

Then, 0.1 mol of chlorate may produce (0.1 .3)/ 2 = 0.15 moles

Let's convert the moles of produced oxygen, as to find out the theoretical yield.

0.15 mol . 32 g/ 1mol = 4.8 g

To calculate the percentage yield, the formula is

(Produced Yield / Theoretical yield) . 100 =

(3.2g / 4.8g) . 100 = 66.7 %

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Katarina [22]

Answer:

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elena-s [515]

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