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3 years ago

Dust particles enter the atmosphere when wind picks them up off the ground and carries the upward.

Chemistry

1 answer:

goldfiish [28.3K]3 years ago

6 0

The statement seems to be true, as per my knowledge.

Hope I helped!! xx

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Calculate the amount of heat required to completely sublime 55.0 g of solid dry ice CO2 at its sublimation temperature. The heat

aliina [53]

Answer:

40.4 kJ

Explanation:

Step 1: Given data

Mass of CO₂ (m): 55.0 g

Heat of sublimation of CO₂ (ΔH°sub): 32.3 kJ/mol

Step 2: Calculate the moles corresponding to 55.0 g of CO₂

The molar mass of CO₂ is 44.01 g/mol.

n = 55.0 g × 1 mol/44.01 g = 1.25 mol

Step 3: Calculate the heat (Q) required to sublimate 1.25 moles of CO₂

We will use the following expression.

Q = n × ΔH°sub

Q = 1.25 mol × 32.3 kJ/mol = 40.4 kJ

7 0

3 years ago

This is for chemistry:/

Andreyy89

Answer:

Its the bottom one :) ......

5 0

3 years ago

An object has a mass of 3.50 grams and a density of 5.61 g/mL. What is the volume of this object? *

Elina [12.6K]

Answer:

Explanation:

The volume of a substance when given the density and mass can be found by using the formula

$volume = \frac{mass}{density} \\$

From the question we have

$volume = \frac{3.50}{5.61} \\ = 0.6238859...$

We have the final answer as

Hope this helps you

5 0

3 years ago

A representative particle of carbon dioxide is

KonstantinChe [14]

Carbon atom with iron and helium

6 0

3 years ago

Read 2 more answers

Consider the balanced equation for the following reaction:

Bad White [126]

<u>Answer:</u> The theoretical yield of the lithium chlorate is 1054.67 grams

<u>Explanation:</u>

To calculate the mass for given number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Actual moles of lithium chlorate = 9.45 moles

Molar mass of lithium chlorate = 90.4 g/mol

Putting values in above equation, we get:

$9.45mol=\frac{\text{Actual yield of lithium chlorate}}{90.4g/mol}\\\\\text{Actual yield of lithium chlorate}=(9.45mol\times 90.4g/mol)=854.28g$

To calculate the theoretical yield of lithium chlorate, we use the equation:

$\%\text{ yield}=\frac{\text{Actual yield}}{\text{Theoretical yield}}\times 100$

Actual yield of lithium chlorate = 854.28 g

Percentage yield of lithium chlorate = 81.0 %

Putting values in above equation, we get:

$81=\frac{854.28g}{\text{Theoretical yield of lithium chlorate}}\times 100\\\\\text{Theoretical yield of lithium chlorate}=\frac{854.28\times 100}{81}=1054.67g$

Hence, the theoretical yield of the lithium chlorate is 1054.67 grams

7 0

3 years ago