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nata0808 [166]
3 years ago
6

Determine the [H+] or [OH−] for each of a solutions at 25°C.

Chemistry
1 answer:
levacccp [35]3 years ago
6 0

Answer:

Solution B: [OH-] = 1×10^–6 M

Solution C: [OH-] = 1×10^–6 M

Explanation:

Solution B:

Hydronium ion concentration, [H3O+] = 9.99×10^–9 M

Hydroxide ion concentration, [OH-] = ?

The Hydroxide ion concentration, [OH-] can be obtained as follow:

[H3O+] × [OH-] = 1×10^–14

9.99×10^–9 × [OH-] = 1×10^–14

Divide both side by 9.99×10^–9

[OH-] = 1×10^–14 / 9.99×10^–9

[OH-] = 1×10^–6 M

Therefore, the Hydroxide ion concentration, [OH-] is 1×10^–6 M

Solution C:

Hydronium ion concentration, [H3O+] = 0.000777 M

Hydroxide ion concentration, [OH-] = ?

The Hydroxide ion concentration, [OH-] can be obtained as follow:

[H3O+] × [OH-] = 1×10^–14

0.000777 × [OH-] = 1×10^–14

Divide both side by 0.000777

[OH-] = 1×10^–14 / 0.000777

[OH-] = 1.29×10^–11 M

Therefore, the Hydroxide ion concentration, [OH-] is 1.29×10^–11 M

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A gas has a volume of 31.2 L at a temperature of 28 degrees Celsius and a pressure of 82.6 kPa. How many moles are in the sample
zhenek [66]

1.02 moles of the gas is present in the sample.

The conditions in which the gases deviate from ideal behavior is high pressure and low temperatures.

Correct option is 2.

Explanation:

Data given:

volume of the gas = 31.2 litres

temperature of the gas = 28 degrees or 301.15 K

pressure of the gas = 82.6 kPa or 0.815 atm

R (Gas constant) = 0.0820 Latm/moles K

number of moles =?

From the ideal gas law, we have

PV = nRT

rearranging the equation:

n = \frac{PV}{RT}

n = \frac{0.815 X 31.2}{301.15 X 0.0820}

n = 1.02 moles

At high pressure and low temperature an ideal gas deviates from ideal behaviour. Under high pressure the gas molecules get closer to each other and intermolecular force acts on them as molecules attract each other while in ideal case gas has no attractions in its molecules.

5 0
3 years ago
Suppose a group of volunteers is planning to build a park near a local lake. The lake is known to contain low levels of arsenic
ale4655 [162]

A) 10.75 is the concentration of arsenic in the sample in parts per billion .

B) 7,633.66 kg the total mass of arsenic in the lake that the company have to remove.

C)It will take 1.37 years to remove all of the arsenic from the lake.

Explanation:

A) Mass of arsenic in lake water sample = 164.5 ng

The ppb is the amount of solute (in micrograms) present in kilogram of a solvent. It is also known as parts-per million.

To calculate the ppm of oxygen in sea water, we use the equation:

\text{ppb}=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 10^9

Both the masses are in grams.

We are given:

Mass of arsenic = 164.5 ng = 164.5\times 10^{-9} g

1 ng=10^{-9} g

Volume of the sample = V = 15.3 cm^3

Density of the lake water sample ,d= 1.00 g/cm^3

Mass of sample =  M = d\times V=1.0 g/cm^3\times 15.3 cm^3=15.3 g

ppb=\frac{164.5\times 10^{-9} g}{15.3 g}\times 10^9=10.75

10.75 is the concentration of arsenic in the sample in parts per billion.

B)

Mass of arsenic in 1 cm^3  of lake water = \frac{164.5\times 10^{-9} g}{15.3}=1.075\times 10^{-8} g

Mass of arsenic in 0.710 km^3 lake water be m.

1 km^3=10^{15} cm^3

Mass of arsenic in 0.710\times 10^{15} cm^3 lake water :

m=0.710\times 10^{15}\times 1.075\times 10^{-8} g=7,633,660.130 g

1 g = 0.001 kg

7,633,660.130 g = 7,633,660.130 × 0.001 kg=7,633.660130 kg ≈ 7,633.66 kg

7,633.66 kg the total mass of arsenic in the lake that the company have to remove.

C)

Company claims that it takes 2.74 days to remove 41.90 kilogram of arsenic from lake water.

Days required to remove 1 kilogram of arsenic from the lake water :

\frac{2.74}{41.90} days

Then days required to remove 7,633.66 kg of arsenic from the lake water :

=7,633.66\times \frac{2.74}{41.90} days=499.19 days

1 year = 365 days

499.19 days = \frac{499.19}{365} years = 1.367 years\approx 1.37 years

C)

Company claims that it takes 2.74 days to remove 41.90 kilogram of arsenic from lake water.

Days required to remove 1 kilogram of arsenic from the lake water :

\frac{2.74}{41.90} days

Then days required to remove 7,633.66 kg of arsenic from the lake water :

=7,633.66\times \frac{2.74}{41.90} days=499.19 days

1 year = 365 days

499.19 days = \frac{499.19}{365} years = 1.367 years\approx 1.37 years

It will take 1.37 years to remove all of the arsenic from the lake.

6 0
2 years ago
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gizmo_the_mogwai [7]
4, loses an electron. a plus sign indicates an atom is losing an electron while a minus sign indicates an atom is gaining an electron
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3 years ago
Read 2 more answers
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kogti [31]

Answer:

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b.

\bold{Fe_{2}(SO_{4})_{3}+3Ba(OH)_{2}\rightarrow 3BaSO_{4}+2Fe(OH)_{3}}

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7 0
3 years ago
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What are the parts of the water system? Label the water wheel to show the matter and forms energy that flow through the system.
Irina-Kira [14]

Answer:

Source, processing and distribution are the components of water system.

Explanation:

There are three parts of water system i. e. the source, the processing and distribution. Water is extracted from a source such as underground water, lake or river etc. After extraction this water is transported to the processing unit where it can be purified and after purification it is distributed to all places where it is needed. Potential energy is a form of energy that flows through this water system because the water is extracted from a depth and we know that depth and height refers to potential energy.

8 0
3 years ago
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