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kari74 [83]
3 years ago
15

In general what two classes of materials dissolve in water

Chemistry
1 answer:
Ivenika [448]3 years ago
3 0
<span>Polar covalent & Ionic are the two materials

</span>
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What is the oxidizing agent in the reaction 2MnO4-(aq) 10I-(aq) 16H (aq) ---&gt; 2Mn2 (aq) 5I2(aq) 8H2O(l)
faltersainse [42]

Correct Question: what is the oxidizing agent in the reaction.

2MnO4–(aq) +10Cl–(aq) + 16H+(aq) --------> 5Cl2(g) + 2Mn2+(aq) +8H2O(l)

Answer: MnO4-is the oxidizing agent

Explanation:

In the reaction 2MnO4–(aq) +10Cl–(aq) + 16H+(aq) --------> 5Cl2(g) + 2Mn2+(aq) +8H2O(l)

Oxidizing agent oxidizes other molecules while the themselves get reduced.

oxidizing agents give away Oxygen to other compounds.

MnO4-is the oxidizing agent because

On the reactants side

Oxidation number of Mn in 2MnO4- is +7

Oxidation number of Cl- is -1

On the products side

Oxidation number of Mn is +2

While oxidation number of Cl is zero

Therefore the oxidizing agent is 2MnO4 because is oxidizes Chlorine from -1 to 0 while itself got reduced from oxidation state of +7 to +2

4 0
3 years ago
A method used by the U.S. Environmental Protection Agency (EPA) for determining the concentration of ozone in air is to pass the
baherus [9]

Answer: 1. 9.08\times 10^{-6} moles

2. 90 mg

Explanation:

O_3(g)+2NaI(aq)+H_2O(l) \rightarrow O_2(g)+I_2(s)+2NaOH(aq)

According to stoichiometry:

1 mole of ozone is removed by 2 moles of sodium iodide.

Thus 4.54 \times 10^{-6} moles of ozone is removed by =\frac{2}{1}\times 4.54 \times 10^{-6}=9.08\times 10^{-6} moles of sodium iodide.

Thus 9.08\times 10^{-6} moles of sodium iodide are needed to remove 4.54\times 10^{-6} moles of O_3

2. \text{Number of moles of ozone}=\frac{0.01331g}{48g/mol}=0.0003moles

According to stoichiometry:

1 mole of ozone is removed by 2 moles of sodium iodide.

Thus 0.0003 moles of ozone is removed by =\frac{2}{1}\times 0.0003=0.0006 moles of sodium iodide.

Mass of sodium iodide= moles\times {\text {molar mass}}=0.0006\times 150g/mol=0.09g=90mg    (1g=1000mg)

Thus 90 mg of sodium iodide are needed to remove 13.31 mg of O_3.

3 0
3 years ago
There are two opposing processes that occur in a solution in contact with undissolved solute. These are dissolving and ________.
Alex777 [14]

Answer:

c. crystallization

Explanation:

The opposing process that occur in a solution in contact with undissolved solute are dissolution and crystallization.

In the dissolution process the solid substance coverts into liquid state and mixes with solution. Whereas in Crystallization the the chemical is converted from the liquid solution to solid crystal state.

3 0
3 years ago
what is the total displacement of a dog that runs north 80 meters to chase a ball, returns to its master, and runs south 20 mete
vladimir1956 [14]
The total displacement will be 20 meters.
7 0
3 years ago
The pH of a 0.25 M aqueous solution of hydrofluoric acid, HF, at 25.0 °C is 2.03. What is the value of Ka for HF?
Mazyrski [523]
Last option that is none of above is right answer.
8 0
3 years ago
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