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dimulka [17.4K]
3 years ago
6

In run 1, you mix 7.9 mL of the 43 g/L MO solution (MO molar mass is 327.33 g/mol), 3.13 mL of the 0.040 M SnCl2 in 2.0 M HCl so

lution, 5.49 mL of 2.0 M HCl solution, and 3.43 mL of 2.0M NaCl solution. What is the [H3O+]? Remember that there is a contribution of H3O+ from two solutions.
Chemistry
1 answer:
blsea [12.9K]3 years ago
5 0

Answer:

Concentration of H3O⁺  [H3O⁺] = 0.864 M

Explanation:

Given that:

The mass concentration of MO = 43 g/L

The volume of MO = 7.9 mL = 7.9 × 10⁻³ L

Recall that

The mass number of MO = Mass concentration of MO × Volume of MO

The mass number of MO = (43 g/L) * (7.9 × 10⁻³ L)

The mass number of MO =  0.3397 g

number of  moles of MO = (mass number of MO) / (molar mass of MO)

number of  moles of MO = (0.3397 g) / (327.33 g/mol)

moles of MO = 0.00104 mol

The total volume = 7.9 mL + 3.13 mL + 5.49 mL + 3.43 mL

The total volume = 19.95 mL = 19.95 × 10⁻³ L

Concentration of MO [MO} =(number of moles of MO) / (total volume)

[MO] = 0.00104 mol  /  19.95 × 10⁻³ L

[MO] = 5.2130 × 10⁻⁸ M

the number of moles of H3O⁺ = molarity of HCl in the solution × the volume of HCl in solution

the number of moles of H3O⁺ = [(2.0 M) * (3.13 mL)] + [(2.0 M) * (5.49 mL)]

the number of moles of H3O⁺ = 17.24 mmol

Concentration of H3O⁺  [H3O⁺] = (the number of moles of H3O⁺) / (total volume)

Concentration of H3O⁺  [H3O⁺] = (17.24 mmol) / (19.95 mL)

Concentration of H3O⁺  [H3O⁺] = 0.864 M

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Match the following words.
azamat

Answer:

1. active site

2. substrate.

3. activity

4. cofactor

5. complex

6. coenzyme.

Explanation:

1. A substrate binds to an enzyme at the active site where the reaction occurs.

2. In a catalyzed reaction, a reactant is often called a  substrate.

3  activity is a measure of how fast an enzyme can convert the reagent to the  product.

4. An inorganic substance necessary for the function of some enzymes is called a cofactor.

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3 years ago
What will most likely happen in the absence of a vacuole?
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The cell will not store food, water, nutrients, and waste.

Explanation:

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3 years ago
A 8.00 L tank at 26.9 C is filled with 5.53 g of dinitrogen difluoride gas and 17.3 g of sulfur hexafluoride gas. You can assume
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Answer:

mole fraction of N_2 O = 0.330

mole of fraction SF_4 = 0.669

PRESSURE OF N_2 O = 39127.053 Pa

pressure of SF_4 = 792126.36

Total pressure   = 118253.413 Pa

Explanation:

Given data:

volume of tank 8 L

Weight of dinitrogen difluoride gas 5.53 g

weight of sulphur hexafluoride gas 17.3 g

Amount of N_2 O = \frac{5.53}{14*2 + 16} = 0.1256 mol

amount of SF_4 = \frac{17.3}{32.1 + 19*4} = 0.254 mol

mole fraction of N_2 O = \frac{0.1256}{0.1256 + 0.254} = 0.330

mole of fractionSF_4 = \frac{0.254}{0.1256 + 0.254} = 0.669

PV = nRT

P of N_2 O = \frac{0.1256 *8.31 (273 + 26.9}{0.008} = 39127.053 Pa

mole of SF_4=\frac{0.254 *8.31*(273+26.9)}{.008} = 79126.36 Pa

Total pressure  = 39127.053 + 79126.36 = 118253.413 Pa

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Answer:

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Explanation:

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