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3 years ago

Which of the following pH values indicates the highest concentration of hydronium ions in a solution?

1 answer:

4 0

Hydronium ions are H30+, and pH = -log10[H+]

<span>As H+ ions in solution exist as H3O+ ions, the lower the pH, the higher the concentration of H30+</span>

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What cause a mass defect?

Gelneren [198K]

Answer:

The answer is C.

Explanation:

Mass is converted to the energy binding a nucleus together.

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2 years ago

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Why does the sun appear very large compared to the other stars

Ivanshal [37]

Answer:

the other stars are much farther away from Earth than our sun

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3 years ago

How many liters of a 6.0M solution Fe(SCN)2 are needed to prepare 2.0L of a 0.75M solution?

lubasha [3.4K]

Answer:

0.25L

Explanation:

Using the dilution formula

C1V1=C2V2

C1=6M

V1?

C2=0.75M

V2=2.0L

V1= C2V2/C1

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3 years ago

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What does the chemical formula CaCl, show about the compound it represents? It is made up of one element. It is made up of two e

zavuch27 [327]

Answer:

It is made up of two elements.

Explanation:

To answer the question given above,

We shall determine the number of elements present in CaCl₂. This can be obtained as follow:

CaCl₂ contains calcium (Ca) and chlorine gas (Cl₂).

This implies that CaCl₂ contains two different elements.

Now, considering the options given in the question above, CaCl₂ is made up of two elements.

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2 years ago

If 4.0 g of helium gas occupies a volume of 22.4 L at 0 o C and a pressure of 1.0 atm, what volume does 3.0 g of He occupy under

WINSTONCH [101]

Answer:

the volume occupied by 3.0 g of the gas is 16.8 L.

Explanation:

Given;

initial reacting mass of the helium gas, m₁ = 4.0 g

volume occupied by the helium gas, V = 22.4 L

pressure of the gas, P = 1 .0 atm

temperature of the gas, T = 0⁰C = 273 K

atomic mass of helium gas, M = 4.0 g/mol

initial number of moles of the gas is calculated as follows;

$n_1 = \frac{m_1}{M} \\\\n_1 = \frac{4}{4} = 1$

The number of moles of the gas when the reacting mass is 3.0 g;

m₂ = 3.0 g

$n_2 = \frac{m_2}{M} \\\\n_2 = \frac{3}{4} \\\\n_2 = 0.75 \ mol$

The volume of the gas at 0.75 mol is determined using ideal gas law;

PV = nRT

$PV = nRT\\\\\frac{V}{n} = \frac{RT}{P} \\\\since, \ \frac{RT}{P} \ is \ constant,\ then;\\\frac{V_1}{n_1} = \frac{V_2}{n_2} \\\\V_2 = \frac{V_1n_2}{n_1} \\\\V_2 = \frac{22.4 \times 0.75}{1} \\\\V_2 = 16.8 \ L$

Therefore, the volume occupied by 3.0 g of the gas is 16.8 L.

4 0

3 years ago