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3 years ago

How does convection occur in the troposphere?

Chemistry

2 answers:

densk [106]3 years ago

5 0

Answer:

the warm air rises and cool air sinks

Explanation:

Tamiku [17]3 years ago

4 0

Answer:

the warm air rises and cool air sinks

Explanation:

The surface heated air expands as it warms, becomes less dense than surrounding cooler air and rises as buoyant and turbulent bubbles. This is convection and is the main process by which the troposphere mixes and heats. Although convection stirs and mixes the troposphere, the higher it is the colder it becomes.

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Which statement describes the bonds in iron sulfate, FeSO4?

Harlamova29_29 [7]

Answer: Would be D. Fe and S have an ionic bond, while S and O have covalent.

Hope that helps.

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2 years ago

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Why is excess magnesium oxide used in the formation of magnesium sulfate crystals?<br>

Tom [10]

I really don't know what the answer is

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2 years ago

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For which of the following reactions is S° > 0. Choose all that apply. N2(g) + 3H2(g) 2NH3(g) 2C2H6(g) + 7O2(g) 4CO2(g) + 6H2

Lostsunrise [7]

Answer: $2C_2H_6(g)+7O_2(g)\rightarrow 4CO_2(g)+6H_2O(g)$

$NH_4I(s)\rightarrow NH_3(g)+HI(g)$

$2H_2O(g)+2SO_2(g)\rightarrow 2H_2S(g)+3O_2(g)$

Explanation:

Entropy is the measure of randomness or disorder of a system.

A system has positive value of entropy if the disorder increases and a system has negative value of entropy if the disorder decreases.

1. $N_2(g)+3H_2(g)\rightarrow 2NH_3(g)$

As 4 moles of gaseous reactants are changing to 2 moles of gaseous products, the randomness is decreasing and the entropy is negative

2. $2C_2H_6(g)+7O_2(g)\rightarrow 4CO_2(g)+6H_2O(g)$

As 9 moles of gaseous reactants are changing to 10 moles of gaseous products, the randomness is increasing and the entropy is positive.

3. $NH_4I(s)\rightarrow NH_3(g)+HI(g)$

As 1 mole of solid reactants is changing to 2 moles of gaseous products, the randomness is increasing and the entropy is positive.

4. $2H_2O(g)+2SO_2(g)\rightarrow 2H_2S(g)+3O_2(g)$

As 4 moles of gaseous reactants is changing to 5 moles of gaseous products, the randomness is increasing and the entropy is positive

5. $2NO(g)+2H_2(g)\rightarrow N_2(g)+2H_2O(l)$

As 4 moles of gaseous reactants is changing to 1 moles of gaseous products, the randomness is decreasing and the entropy is negative.

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3 years ago

A student analyzes soil samples from her backyard. She organizes the samples into three categories, based on their pH. The three

Sedaia [141]

I think the best answer is the last option. A scatter plot is the appropriate type of graph for the student to use to show the percent samples per group. This plot is somewhat similar to line graphs. However, they are use for a specific purpose which is to show the relationship between two parameters. In this case, the correlation between pH and the percent of samples.

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3 years ago

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The ideal gas heat capacity of nitrogen varies with temperature. It is given by:

hammer [34]

Answer:

A) 1059 J/mol

B) 17,920 J/mol

Explanation:

Given that:

Cp = 29.42 - (2.170*10^-3 ) T + (0.0582*10^-5 ) T2 + (1.305*10^-8 ) T3 – (0.823*10^-11) T4

R (constant) = 8.314

We know that:

$C_p=C_v+R$

We can determine $C_v$ from above if we make $C_v$ the subject of the formula as:

$C_v$ $=C_p-R$

$C_V = 29.42-(2.7*10^{-3})T+(5.82*10^{-7})T2-(1.305*10^{-8})T3-(8.23*10^{-12})T4-8.314$

$C_V = 21.106-(2.7*10^{-3})T+(5.82*10^{-7})T2-(1.305*10^{-8})T3-(8.23*10^{-12})T4$

A).

The formula for calculating change in internal energy is given as:

$dU=C_vdT$

If we integrate above data into the equation; it implies that:

$U2-U1=\int\limits^{500}_{450}(21.106-(2.7*10^{-3})T+(5.82*10^{-7})T2-(1.305*10^{-8})T3-(8.23*10^{-12})T4\,) du$

$U2-U1=\int\limits^{500}_{450}(21.106-(2.7*10^{-3})T/1+(5.82*10^{-7})T2/2-(1.305*10^{-8})T3/3-(8.23*10^{-12})T4/4\,)$

$U2-U1= 1059J/mol$

Hence, the internal energy that must be added to nitrogen in order to increase its temperature from 450 to 500 K = 1059 J/mol.

B).

If we repeat part A for an initial temperature of 273 K and final temperature of 1073 K.

then T = 273 K & T2 = 1073 K

∴

$U2-U1=\int\limits^{500}_{450}(21.106-(2.7*10^{-3})T/1+(5.82*10^{-7})T2/2-(1.305*10^{-8})T3/3-(8.23*10^{-12})T4/4\,)$

$U2-U1=\int\limits^{500}_{450}(21.106-(2.7*10^{-3})273/1+(5.82*10^{-7})1073/2-(1.305*10^{-8})T3/3-(8.23*10^{-12})T4/4\,)$

$U2-U1= 17,920 J/mol$

3 0

3 years ago