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Savatey [412]
3 years ago
8

N the reaction Mg + 2HCl -> MgCl2 + H2 which element’s oxidation number does not change?

Chemistry
1 answer:
BartSMP [9]3 years ago
4 0

Answer:- Oxidation number of Cl does not change as it is -1 on both sides.

Explanations:- oxidation number of Mg on reactant side is 0 as it is in its elemental form(not combined with another element).

Oxidation number  of hydrogen in its compounds is +1, so if H is +1 in HCl the oxidation number of Cl is -1 as the sum has to be zero.

On product side, Mg oxidation number is +2 as the oxidation number of alkaline earth metals in their compounds is +2.

Two Cl are present in magnesium chloride, so if Mg is +2 then Cl is -1.

Oxidation number of H on product side is 0 as it is present in its elemental for, H_2 ,

So, it is only chlorine(Cl) whose oxidation number does not change for the given equation.


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In order to prepare very dilute solutions, a lab technician chooses to perform a series of dilutions instead of measuring a very
SVETLANKA909090 [29]

<u>Answer:</u> The final concentration of potassium nitrate is 5.70\times 10^{-6}M

<u>Explanation:</u>

To calculate the molecular mass of solute, we use the equation used to calculate the molarity of solution:

\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}

We are given:

Mass of potassium nitrate (solute) = 0.360 g

Molar mass of potassium nitrate = 101.1 g/mol

Volume of solution = 500.0 mL

Putting values in above equation, we get:

\text{Molarity of }KNO_3=\frac{0.360\times 1000}{101.1\times 500.0}\\\\\text{Molarity of }KNO_3=7.12\times 10^{-3}M

To calculate the molarity of the diluted solution, we use the equation:

M_1V_1=M_2V_2          .......(1)

  • <u>Calculating for first dilution:</u>

M_1\text{ and }V_1 are the molarity and volume of the concentrated KNO_3 solution

M_2\text{ and }V_2 are the molarity and volume of diluted KNO_3 solution

We are given:

M_1=7.12\times 10^{-3}M\\V_1=10mL\\M_2=?M\\V_2=500.0mL

Putting values in equation 1, we get:

7.12\times 10^{-3}\times 10=M_2\times 500\\\\M_2=\frac{7.12\times 10^{-3}\times 10}{500}=1.424\times 10^{-4}M

  • <u>Calculating for second dilution:</u>

M_2\text{ and }V_2 are the molarity and volume of the concentrated KNO_3 solution

M_3\text{ and }V_3 are the molarity and volume of diluted KNO_3 solution

We are given:

M_2=1.424\times 10^{-4}M\\V_2=10mL\\M_3=?M\\V_3=250.0mL

Putting values in equation 1, we get:

1.424\times 10^{-4}\times 10=M_3\times 250\\\\M_3=\frac{1.424\times 10^{-4}\times 10}{250}=5.70\times 10^{-6}M

Hence, the final concentration of potassium nitrate is 5.70\times 10^{-6}M

8 0
2 years ago
The value of ΔH° for the reaction below is -6535 kJ. ________ kJ of heat are released in the combustion of 16.0 g of C6H6 (l)?
Umnica [9.8K]

Answer:

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Explanation:

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2 years ago
For the reaction between aqueous silver nitrate and aqueous sodium chloride, write each of the following. The products of the re
vova2212 [387]

Answer:

A balanced ionic equation shows the reacting ions in a chemical reaction. These equations can be used to represent what happens in precipitation reactions or displacement reactions.

Precipitation reactions

In a typical precipitation reaction, two soluble reactants form an insoluble product and a soluble product.

For example, silver nitrate solution reacts with sodium chloride solution. Insoluble solid silver chloride and sodium nitrate solution form:

AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq)

The Na+ ions and NO3- ions remain separate in the sodium nitrate solution and do not form a precipitate. Ions that remain essentially unchanged during a reaction are called spectator ions.This means these can be ignored when writing the ionic equation. Only how the solid silver chloride forms is needed to be shown:

Ag+(aq) + Cl-(aq) → AgCl(s)

In a balanced ionic equation:

the number of positive and negative charges is the same

the numbers of atoms of each element on the left and right are the same

Displacement reactions

Displacement reactions take place when a reactive element displaces a less reactive element from one of its compounds.

A common type of displacement reaction takes place when a reactive metal reacts with the salt of a less reactive metal. For example, copper reacts with silver nitrate solution to produce silver and copper(II) nitrate solution:

2AgNO3(aq) + Cu(s) → 2Ag(s) + Cu(NO3)2(aq)

In this reaction, the NO3- ions remain in the solution and do not react - they are the spectator ions in this reaction. So, they can be removed from the ionic equation:

2Ag+(aq) + Cu(s) → 2Ag(s) + Cu2+(aq)

Question

Explain why this ionic equation is balanced:

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Hide answer

There are the same numbers of atoms of each element on both sides of the equation. The total charge on both sides is also the same (zero).

Question

Balance this ionic equation, which represents the formation of a silver carbonate precipitate:

Ag+(aq) + CO32-(aq) → Ag2CO3(s)

Hide answer

2Ag+(aq) + CO32-(aq) → Ag2CO3(s)

Question

Balance this ionic equation, which represents the displacement of iodine from iodide ions by chlorine:

Cl2(aq) + I-(aq) → I2(aq) + Cl-(aq)

Hide answer

Cl2(aq) + 2I-(aq) → I2(aq) + 2Cl-(aq

Explanation:

this will help, I used this for my work x

4 0
2 years ago
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