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4vir4ik [10]
3 years ago
13

Which of these is an example of chemical separation of a substance?

Chemistry
1 answer:
OleMash [197]3 years ago
6 0

Answer:

Bond broking

Explanation:

Bond broking.

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A chemist dissolves of pure hydroiodic acid in enough water to make up of solution. Calculate the pH of the solution. Be sure yo
Gre4nikov [31]

Answer:

1.76

Explanation:

There is some info missing. I think this is the original question.

<em>A chemist dissolves 660.mg of pure hydroiodic acid in enough water to make up 300.mL of solution. Calculate the pH of the solution. Be sure your answer has the correct number of significant digits.</em>

<em />

Step 1: Calculate the molarity of HI(aq)

M = mass of solute / molar mass of solute × liters of solution

M = 0.660 g / 127.91 g/mol × 0.300 L

M = 0.0172 M

Step 2: Write the acid dissociation reaction

HI(aq) ⇄ H⁺(aq) + I⁻(aq)

HI is a strong acid, so [H⁺] = 0.0172 M

Step 3: Calculate the pH

pH = -log [H⁺]

pH = -log 0.0172

pH = 1.76

5 0
3 years ago
Is this statement true or false? Growth is caused by cells in your body dividing and making new cells.
goblinko [34]

Answer:

true

Explanation:

4 0
3 years ago
In a chemical equation
djverab [1.8K]

Answer:

Answer in explanation

Explanation:

A. This is wrong. The reactants are on the left side of the yields arrow

B. This is wrong. They can only be at one side at a time. They cannot be on both sides at the same time.

C. The products are on the right side of the yields arrow and not at the left side

D. This is correct. The reactants are on the left side of the yields arrow.

E. This is correct. We can have varying numbers of the number of atoms on both side of the yields arrow. The numbers may differ until we decide to balance the equation

4 0
3 years ago
How many grams of MnCl2 need to be added to 5 liters of water to make a 0.25 M solution?
damaskus [11]

Answer:

I think it is 860 g MG Cl2

Explanation:

your welcome

4 0
3 years ago
Calculate the standard enthalpy change for the reaction at 25 ∘ C. Standard enthalpy of formation values can be found in this li
meriva

Answer:

The standard enthalpy change for the reaction at 25^{0}\textrm{C} is -2043.999kJ

Explanation:

Standard enthalpy change (\Delta H_{rxn}^{0}) for the given reaction is expressed as:

\Delta H_{rxn}^{0}=[3mol\times \Delta H_{f}^{0}(CO_{2})_{g}]+[4mol\times \Delta H_{f}^{0}(H_{2}O)_{g}]-[1mol\times \Delta H_{f}^{0}(C_{3}H_{8})_{g}]-[5mol\times \Delta H_{f}^{0}(O_{2})_{g}]

Where \Delta H_{f}^{0} refers standard enthalpy of formation

Plug in all the given values from literature in the above equation:

\Delta H_{rxn}^{0}=[3mol\times (-393.509kJ/mol)]+[4mol\times (-241.818kJ/mol)]-[1mol\times (-103.8kJ/mol)]-[5mol\times (0kJ/mol)]=-2043.999kJ

4 0
3 years ago
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