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2 years ago

Which pollution effects the environment the most?

Chemistry

2 answers:

Anuta_ua [19.1K]2 years ago

7 0

ANSWER :

<h3>A. air pollution </h3>

Extra Information :

❥ Air pollution can damage crops and trees in a variety of ways.

❥ Ground-level ozone can lead to reductions in agricultural crop and commercial forest yields, reduced growth and survivability of tree seedlings, and increased plant susceptibility to disease, pests and other environmental stresses (such as harsh weather).

Hope it helps ya...!!♡

Aleks04 [339]2 years ago

5 0

Answer:

Air pollution effect the most

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2 years ago

A balloon containing helium gas expands from 230

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The answer for the following problem is mentioned below.

Therefore the final moles of the gas is 14.2 × $10^{-4}$ moles.

Explanation:

Given:

Initial volume ( $V_{1}$ ) = 230 ml

Final volume ( $V_{2}$ ) = 860 ml

Initial moles ( $n_{1}$ ) = 3.8 × $10^{-4}$ moles

To find:

Final moles ( $n_{2}$ )

We know;

According to the ideal gas equation;

P × V = n × R × T

where;

P represents the pressure of the gas

V represents the volume of the gas

n represents the no of the moles of the gas

R represents the universal gas constant

T represents the temperature of the gas

So;

V ∝ n

$\frac{V_{1} }{V_{2} }$ = $\frac{n_{1} }{n_{2} }$

where,

( $V_{1}$ ) represents the initial volume of the gas

( $V_{2}$ ) represents the final volume of the gas

( $n_{1}$ ) represents the initial moles of the gas

( $n_{2}$ ) represents the final moles of the gas

Substituting the above values;

$\frac{230}{860}$ = $\frac{3.8 * 10^-4}{n_{2} }$

$n_{2}$ = 14.2 × $10^{-4}$ moles

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2 years ago

5.11 g of MgSO₄ is placed into 100.0 mL of water. The water's temperature increases by 6.70°C. Calculate ∆H, in kJ/mol, for the

cupoosta [38]

Answer: Thus ∆H, in kJ/mol, for the dissolution of MgSO₄ is -66.7 kJ

Explanation:

To calculate the entalpy, we use the equation:

$q=mc\Delta T$

where,

q = heat absorbed by water = ?

m = mass of water = ${\text {volume of water}}\times {\text {density of water}}=100.0ml\times 1.00g/ml=100.0g$

c = heat capacity of water = 4.186 J/g°C

$\Delta T$ = change in temperature = $6.70^0C$

$q=100.0g\times 4.184J/g^0C\times 6.70^0C=2803.3J=2.8033kJ$

Sign convention of heat:

When heat is absorbed, the sign of heat is taken to be positive and when heat is released, the sign of heat is taken to be negative.

The heat absorbed by water will be equal to heat released by $MgSO_4$

To calculate the number of moles, we use the equation:

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

Given mass = 5.11 g

Molar mass = 120 g/mol

Putting values in above equation, we get:

$\text{Moles of }MgSO_4=\frac{5.11g}{120g/mol}=0.042mol$

0.042 moles of $MgSO_4$ releases = 2.8033 kJ

1 mole of $MgSO_4$ releases = $\frac{2.8033 kJ}{0.042}\times 1=66.7kJ$

Thus ∆H, in kJ/mol, for the dissolution of MgSO₄ is -66.7 kJ

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3 years ago