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snow_tiger [21]
2 years ago
15

The K w for water at 0 ∘ C is 0.12 × 10 − 14 M 2 . Calculate the pH of a neutral aqueous solution at 0 ∘ C . p H = Is a pH = 7.2

5 solution acidic, basic, or neutral at 0 ∘ C ?
Chemistry
1 answer:
labwork [276]2 years ago
5 0

Answer:

pH → 7.46

Explanation:

We begin with the autoionization of water. This equilibrium reaction is:

2H₂O  ⇄   H₃O⁺  +  OH⁻            Kw = 1×10⁻¹⁴         at 25°C

Kw = [H₃O⁺] . [OH⁻]

We do not consider [H₂O] in the expression for the constant.

[H₃O⁺] = [OH⁻] = √1×10⁻¹⁴   →  1×10⁻⁷ M

Kw depends on the temperature

0.12×10⁻¹⁴ = [H₃O⁺] . [OH⁻]  → [H₃O⁺] = [OH⁻]         at 0°C

√0.12×10⁻¹⁴ = [H₃O⁺] → 3.46×10⁻⁸ M

- log [H₃O⁺] = pH

pH = - log 3.46×10⁻⁸ → 7.46

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kati45 [8]

Answer:

a. The original temperature of the gas is 2743K.

b. 20atm.

Explanation:

a. As a result of the gas laws, you can know that the temperature is inversely proportional to moles of a gas when pressure and volume remains constant. The equation could be:

T₁n₁ = T₂n₂

<em>Where T is absolute temperature and n amount of gas at 1, initial state and 2, final states.</em>

<em />

<em>Replacing with values of the problem:</em>

T₁n₁ = T₂n₂

X*7.1g = (X+300)*6.4g

7.1X = 6.4X + 1920

0.7X = 1920

X = 2743K

<h3>The original temperature of the gas is 2743K</h3><h3 />

b. Using general gas law:

PV = nRT

<em>Where P is pressure (Our unknown)</em>

<em>V is volume = 2.24L</em>

<em>n are moles of gas (7.1g / 35.45g/mol = 0.20 moles)</em>

R is gas constant = 0.082atmL/molK

And T is absolute temperature (2743K)

P*2.24L = 0.20mol*0.082atmL/molK*2743K

<h3>P = 20atm</h3>

<em />

7 0
1 year ago
A 0.1153-gram sample of a pure hydrocarbon was burned in a C-H combustion train to produce 0.3986 gram of CO2and 0.0578 gram of
Mars2501 [29]

<u>Answer:</u> The mass of carbon and hydrogen in the sample is 0.1087 g and 0.0066 g respectively and the percentage composition of carbon and hydrogen in the sample is 94.27 % and 5.72 % respectively.

<u>Explanation:</u>

The chemical equation for the combustion of hydrocarbon having carbon and hydrogen follows:

C_xH_y+O_2\rightarrow CO_2+H_2O

where, 'x' and 'y' are the subscripts of carbon and hydrogen respectively.

We are given:

Mass of CO_2=0.3986g

Mass of H_2O=0.0578g

Mass of sample = 0.1153 g

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

  • <u>For calculating the mass of carbon:</u>

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 0.3986 g of carbon dioxide, \frac{12}{44}\times 0.3986=0.1087g of carbon will be contained.

  • <u>For calculating the mass of hydrogen:</u>

In 18g of water, 2 g of hydrogen is contained.

So, in 0.0578 g of water, \frac{2}{18}\times 0.0578=0.0066g of hydrogen will be contained.

To calculate the percentage composition of a substance in sample, we use the equation:

\%\text{ composition of substance}=\frac{\text{Mass of substance}}{\text{Mass of sample}}\times 100      ......(1)

  • <u>For Carbon:</u>

Mass of sample = 0.1153 g

Mass of carbon = 0.1087 g

Putting values in equation 1, we get:

\%\text{ composition of carbon}=\frac{0.1087g}{0.1153g}\times 100=94.27\%

  • <u>For Hydrogen:</u>

Mass of sample = 0.1153 g

Mass of hydrogen = 0.0066 g

Putting values in equation 1, we get:

\%\text{ composition of hydrogen}=\frac{0.0066g}{0.1153g}\times 100=5.72\%

Hence, the mass of carbon and hydrogen in the sample is 0.1087 g and 0.0066 g respectively and the percentage composition of carbon and hydrogen in the sample is 94.27 % and 5.72 % respectively.

8 0
1 year ago
Calculate the mass of 3.75 x 10^23 molecules of CaSO4 Please HelpIn a hurry Ill give brainliest
SOVA2 [1]

Answer:

Mass = 84.82 g

Explanation:

Given data:

Number of molecules of CaSO₄ = 3.75× 10²³

Mass in gram = ?

Solution:

Avogadro number:

The given problem will solve by using Avogadro number.

It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.

The number 6.022 × 10²³ is called Avogadro number.

1 mole = 6.022 × 10²³ molecules

3.75× 10²³ molecule × 1 mol / 6.022 × 10²³ molecules

0.623 mol

Mass in gram:

Mass = number of moles × molar mass

Mass = 0.623 mol × 136.14 g/mol

Mass = 84.82 g

7 0
1 year ago
How many moles of N2 are essential for generating 0.08 moles of Li3N in the given reaction?6Li N2 2Li3N
jeka57 [31]
<span>08 moles Li3N * 1mole N2/2moles Li3N = 0.04 </span>
3 0
2 years ago
Read 2 more answers
Determine the number of neutrons in an atom of Rg-272.
IRISSAK [1]
In order to find the number of neutrons in the atom,

you need to calculate the difference between the top and bottom numbers

which means 272 - 111 = 161

Hope this helps
6 0
2 years ago
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