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musickatia [10]
3 years ago
15

How many millimeters of 1.25 M Copper(2) sulfate, CuSO4, will contain 55.75g CuSo4

Chemistry
1 answer:
Shkiper50 [21]3 years ago
8 0

Answer:

280 mL

Explanation:

Given data:

Mass of CuSO₄ = 55.75 g

Molarity of CuSO₄ = 1.25 M

Volume of CuSO₄ = ?

Solution:

First of all we will calculate the number of moles.

Number of moles:

Number of moles = mass / molar mass

Number of moles = 55.75 g/ 159.6 g/mol

Number of moles = 0.35 mol

Volume of CuSO₄:

Molarity = number of moles / volume in litter

1.25 M = 0.35 mol / volume in litter

Volume in litter = 0.35 mol / 1.25 M

Volume in litter = 0.28 L

Volume in milliliter:

0.28 L × 1000 mL/ 1 L = 280 mL

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Explanation:

compound: Substance that is made up of more than one type of atom bonded together

mixture: a combination of two or more elements of compound which are not chemically bonded

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Select all statements that correctly describe hemoglobin and myoglobin structure. a. Molecular oxygen binds irreversibly to the
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Answer:

c. By itself, heme is not a good oxygen carrier. It must be part of a larger protein to prevent oxidation of the iron.

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Explanation:

The differences between hemoglobin and myoglobin are most important at the level of quaternary structure. Hemoglobin is a tetramer composed of two each of two types of closely related subunits, alpha and beta. Myoglobin is a monomer (so it doesn't have a quaternary structure at all). Myoglobin binds oxygen more tightly than does hemoglobin. This difference in binding energy reflects the movement of oxygen from the bloodstream to the cells, from hemoglobin to myoglobin.

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3 years ago
What is the formula for the compound by calcium ions and chloride ions?
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Acid & Base Calculations Calculate the hydronium ion concentration for each. Tell whether it is an acid or a base. 1. pH = 5
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<u>Explanation:</u>

pH is the negative logarithm of hydronium ion concentration present in a solution.

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  • If the solution has low hydrogen ion concentration, then the pH will be high and the solution will be basic. The pH range of basic solution is 7.1 to 14
  • The solution having pH equal to 7 is termed as neutral solution.

To calculate the pH of the solution, we use equation:

pH=-\log[H_3O^+]     ......(1)

To calculate the pOH of the solution, we use the equation:

pH + pOH = 14                ........(2)

  • <u>For 1:</u>

We are given:

pH = 5.54

Putting values in equation 1, we get:

5.54=-\log[H_3O^+]

[H_3O^+]=2.88\times 10^{-6}M

Now, putting values in equation 2, we get:

14 = 5.54 + pOH

pOH = 8.46

The solution is acidic in nature.

  • <u>For 2:</u>

We are given:

pOH = 9.7

Putting values in equation 2, we get:

14 = 9.7 + pH

pH = 4.3

Now, putting values in equation 1, we get:

4.3=-\log[H_3O^+]

[H_3O^+]=5.012\times 10^{-5}M

The solution is acidic in nature.

  • <u>For 3:</u>

We are given:

pH = 7.0

Putting values in equation 1, we get:

7.0=-\log[H_3O^+]

[H_3O^+]=1.00\times 10^{-7}M

Now, putting values in equation 2, we get:

14 = 7.0 + pOH

pOH = 7.0

The solution is neither acidic nor basic in nature.

  • <u>For 4:</u>

We are given:

pH = 12.9

Putting values in equation 1, we get:

12.9=-\log[H_3O^+]

[H_3O^+]=1.26\times 10^{-13}M

Now, putting values in equation 2, we get:

14 = 12.9 + pOH

pOH = 1.1

The solution is basic in nature.

  • <u>For 5:</u>

We are given:

pOH = 1.2

Putting values in equation 2, we get:

14 = 1.2 + pH

pH = 12.8

Now, putting values in equation 1, we get:

12.8=-\log[H_3O^+]

[H_3O^+]=1.58\times 10^{-13}M

The solution is basic in nature.

  • <u>For 6:</u>

We are given:

[H_3O^+]=1\times 10^{-5}M

Putting values in equation 1, we get:

pH=-\log(1\times 10^{-5})

pH=5

Now, putting values in equation 2, we get:

14 = 5 + pOH

pOH = 9

The solution is acidic in nature.

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3 years ago
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