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Lesechka [4]
3 years ago
15

Rank the solutions in order of decreasing [H3O ]. Rank solutions from largest to smallest hydronium ion concentration. To rank i

tems as equivalent, overlap them. a. 0.10 M HNO3.
b. 0.10 M HCN.
c. 0.10 M HNO2.
d. 0.10 M HClO.e. smallest concentration.f. largest concentration.
Chemistry
1 answer:
Usimov [2.4K]3 years ago
5 0

Answer:

0.10M HCN  <  0.10 M HClO  <  0.10 M HNO₂  < 0.10 M HNO₃

Explanation:

We are comparing acids with the same concentration. So what we have to do first is to determine if we have any strong acid and for the rest ( weak acids ) compare them by their Ka´s ( look for them in reference tables ) since we know the larger the Ka, the more Hydronium concentration will be in these solutions at the same concentration.

HNO₃ is a strong acid and will have the largest hydronium concentration.

HCN  Ka = 6.2 x 10⁻¹⁰

HNO₂ Ka = 4.0 x 10⁻⁴

HClO  Ka = 3.0 x 10⁻⁸

The ranking from smallest to largest hydronium concentration will then be:

0.10M HCN  <  0.10 M HClO  <  0.10 M HNO₂  < 0.10 M HNO₃

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Answer:

Physical change: A,C ,D, F, H, I

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3 years ago
How many H2O molecules are in 183.2 grams of H20 gas?
jek_recluse [69]

Answer: There are 61.24 \times 10^{23} molecules present in 183.2 grams of H_{2}O gas.

Explanation:

Given: Mass = 183.2 g

Number of moles is the mass of substance divided by its molar mass.

As molar mass of water is 18 g/mol. Therefore, moles of H_{2}O are calculated as follows.

Moles = \frac{mass}{molar mass}\\= \frac{183.2 g}{18 g/mol}\\= 10.17 mol

According to the mole concept, there are 6.022 \times 10^{22} molecules present in one mole of a substance.

Hence, molecules present in 10.17 moles are calculated as follows.

10.17 mol \times 6.022 \times 10^{23}\\= 61.24 \times 10^{23}

Thus, we can conclude that there are 61.24 \times 10^{23} molecules present in 183.2 grams of H_{2}O gas.

6 0
3 years ago
The molar solubility of silver bromide, AgBr in pure water is 0.0007350 mol/L. What is the
gayaneshka [121]

Answer:

0.000000540

Explanation:

Step 1: Make an ICE chart for the solution of AgBr

"S" represents the molar solubility of AgBr

        AgBr(s) ⇄ Ag⁺(aq) + Br⁻(aq)

I                           0             0

C                          +S          +S

E                           S             S

Step 2: Write the expression for the solubility product constant (Ksp)

Ksp = [Ag⁺] [Br⁻] = S × S

Ksp = S² = (0.0007350)² = 0.000000540

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What is the name of the element?<br> bromine<br> iodine<br> tellurium<br> xenon
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Answer:

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<h2>..................</h2>
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