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3 years ago

How many times higher is the concentration of H+ in the Hubbard Brook sample than in unpolluted rainwater?

Chemistry

1 answer:

Anna [14]3 years ago

8 0

Answer:

1. 7 (a neutral solution)

Answer: 10-7= 0.0000001 moles per liter

2. 5.6 (unpolluted rainwater)

Answer: 10-5.6 = 0.0000025 moles per liter

3. 3.7 (first acid rain sample in North America)

Answer: 10-3.7 = 0.00020 moles per liter

The concentration of H+ in the Hubbard Brook sample is 0.00020/0.0000025, which is 80 times higher than the H+ concentration in unpolluted rainwater.

Explanation:

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4 years ago

How many moles of aluminum are needed to react completely with 1.2 mol of feo

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Answer:

0.8 mol.

Explanation:

The balanced equation for the reaction between Al and FeO is represented as:

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4 years ago

A gas mixture with a total pressure of 745 mmHg contains each of the following gases at the indicated partial pressures: CO2, 24

Setler79 [48]

Answer:

For Part A: The partial pressure of Helium is 218 mmHg.

For Part B: The mass of helium gas is 0.504 g.

Explanation:

For Part A:

We are given:

$p_{CO_2}=245mmHg\\p_Ar}=119mmHg\\p_{O_2}=163mmHg\\P=745mmHg$

To calculate the partial pressure of helium, we use the formula:

$P=p_{CO_2}+p_{Ar}+p_{O_2}+p_{He}$

Putting values in above equation, we get:

$745=245+119+163+p_{He}\\p_{He}=218mmHg$

Hence, the partial pressure of Helium is 218 mmHg.

For Part B:

To calculate the mass of helium gas, we use the equation given by ideal gas:

PV = nRT

or,

$PV=\frac{m}{M}RT$

where,

P = Pressure of helium gas = 218 mmHg

V = Volume of the helium gas = 10.2 L

m = Mass of helium gas = ? g

M = Molar mass of helium gas = 4 g/mol

R = Gas constant = $62.3637\text{ L.mmHg }mol^{-1}K^{-1}$

T = Temperature of helium gas = 283 K

Putting values in above equation, we get:

$218mmHg\times 10.2L=\frac{m}{4g/mol}\times 62.3637\text{ L.mmHg }mol^{-1}K^{-1}\times 283K\\\\m=0.504g$

Hence, the mass of helium gas is 0.504 g.

6 0

3 years ago