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1 year ago

What is the coefficient for oh−(aq) when s2−(aq) no3−(aq) → s(s) no(g) is balanced in basic aqueous solution?

Chemistry

1 answer:

omeli [17]1 year ago

5 0

From the balanced redox equation of the reaction, the coefficient of OH⁻ is 8.

<h3>What is the balanced redox equation of the reaction?</h3>

A redox equation is the equation of a redox reaction in which oxidation and reduction occurs simultaneously.

The given redox reaction takes place in a basic solution

The balanced redox equation of the reaction is given below:

4 H₂O(l) + 3 S²⁻(aq) + 2 NO₃⁻(aq) ---> 3 S(s) + 2 NO + 8 OH⁻(aq)

In conclusion, a redox equation is balanced when oxidation and reduction occur to the same extent.

Learn more about redox equations at: brainly.com/question/26750732

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Consider the reaction Mg(s) + I2 (s) → MgI2 (s) Identify the limiting reagent in each of the reaction mixtures below:

Lapatulllka [165]

Answer:

a) Nor Mg, neither I2 is the limiting reactant.

b) I2 is the limiting reactant

c) Mg is the limiting reactant

d) Mg is the limiting reactant

e) Nor Mg, neither I2 is the limiting reactant.

f) I2 is the limiting reactant

g) Nor Mg, neither I2 is the limiting reactant.

h) I2 is the limiting reactant

i) Mg is the limiting reactant

Explanation:

Step 1: The balanced equation:

Mg(s) + I2(s) → MgI2(s)

For 1 mol of Mg we need 1 mol of I2 to produce 1 mol of MgI2

a. 100 atoms of Mg and 100 molecules of I2

We'll have the following equation:

100 Mg(s) + 100 I2(s) → 100MgI2(s)

This is a stoichiometric mixture. Nor Mg, neither I2 is the limiting reactant.

b. 150 atoms of Mg and 100 molecules of I2

We'll have the following equation:

150 Mg(s) + 100 I2(s) → 100 MgI2(s)

I2 is the limiting reactant, and will be completely consumed. There will be consumed 100 Mg atoms. There will remain 50 Mg atoms.

There will be produced 100 MgI2 molecules.

c. 200 atoms of Mg and 300 molecules of I2

We'll have the following equation:

200 Mg(s) + 300 I2(s) →200 MgI2(s)

Mg is the limiting reactant, and will be completely consumed. There will be consumed 200 I2 molecules. There will remain 100 I2 molecules.

There will be produced 200 MgI2 molecules.

d. 0.16 mol Mg and 0.25 mol I2

We'll have the following equation:

Mg(s) + I2(s) → MgI2(s)

Mg is the limiting reactant, and will be completely consumed. There will be consumed 0.16 mol of I2. There will remain 0.09 mol of I2.

There will be produced 0.16 mol of MgI2.

e. 0.14 mol Mg and 0.14 mol I2

We'll have the following equation:

Mg(s) + I2(s) → MgI2(s)

This is a stoichiometric mixture. Nor Mg, neither I2 is the limiting reactant.

There will be consumed 0.14 mol of Mg and 0.14 mol of I2. there will be produced 0.14 mol of MgI2

f. 0.12 mol Mg and 0.08 mol I2

We'll have the following equation:

Mg(s) + I2(s) → MgI2(s)

I2 is the limiting reactant, and will be completely consumed. There will be consumed 0.08 moles of Mg. There will remain 0.04 moles of Mg.

There will be produced 0.08 moles of MgI2.

g. 6.078 g Mg and 63.455 g I2

We'll have the following equation:

Mg(s) + I2(s) → MgI2(s)

Number of moles of Mg = 6.078 grams / 24.31 g/mol = 0.250 moles

Number of moles I2 = 63.455 grams/ 253.8 g/mol = 0.250 moles

This is a stoichiometric mixture. Nor Mg, neither I2 is the limiting reactant.

There will be consumed 0.250 mol of Mg and 0.250 mol of I2. there will be produced 0.250 mol of MgI2

h. 1.00 g Mg and 2.00 g I2

We'll have the following equation:

Mg(s) + I2(s) → MgI2(s)

Number of moles of Mg = 1.00 grams / 24.31 g/mol = 0.0411 moles

Number of moles I2 = 2.00 grams/ 253.8 g/mol = 0.00788 moles

I2 is the limiting reactant, and will be completely consumed. There will be consumed 0.00788 moles of Mg. There will remain 0.03322 moles of Mg.

There will be produced 0.00788 moles of MgI2.

i. 1.00 g Mg and 2.00 g I2

We'll have the following equation:

Mg(s) + I2(s) → MgI2(s)

Number of moles of Mg = 1.00 grams / 24.31 g/mol = 0.0411 moles

Number of moles I2 = 20.00 grams/ 253.8 g/mol = 0.0788 moles

Mg is the limiting reactant, and will be completely consumed. There will be consumed 0.0411 moles of Mg. There will remain 0.0377 moles of I2.

There will be produced 0.0411 moles of MgI2.

4 0

3 years ago

a mixture containing 21.4 g of ice. (0.00 c) and 75.3 g of water (55.3 c) is placed in an insulatated container

Leto [7]

58874889879879797fc8755555555555555555555555555555555555555555555511111111111111111111111111111111113333333333333333333222222222222220000000000000000000000000000000000000000000000..................

7 0

3 years ago

For the following reaction, KcKc = 255 at 1000 KK.

bonufazy [111]

Answer :

The equilibrium concentration of CO is, 0.016 M

The equilibrium concentration of Cl₂ is, 0.034 M

The equilibrium concentration of COCl₂ is, 0.139 M

Explanation :

The given chemical reaction is:

$CO(g)+Cl_2(g)\rightleftharpoons COCl_2(g)$

Initial conc. 0.1550 0.173 0

At eqm. (0.1550-x) (0.173-x) x

As we are given:

$K_c=255$

The expression for equilibrium constant is:

$K_c=\frac{[COCl_2]}{[CO][Cl_2]}$

Now put all the given values in this expression, we get:

$255=\frac{(x)}{(0.1550-x)\times (0.173-x)}$

x = 0.139 and x = 0.193

We are neglecting value of x = 0.193 because equilibrium concentration can not be more than initial concentration.

Thus, we are taking value of x = 0.139

The equilibrium concentration of CO = (0.1550-x) = (0.1550-0.139) = 0.016 M

The equilibrium concentration of Cl₂ = (0.173-x) = (0.173-0.139) = 0.034 M

The equilibrium concentration of COCl₂ = x = 0.139 M

5 0

3 years ago

What is the ability to transfer thermal energy to something else

Fofino [41]

Answer:

.Heat energy can be transferred from one object to another. The transfer or flow due to the difference in temperature between the two objects is called heat.

Explanation:

The transfer of thermal energy is called heat. In this process, thermal energy moves through a substance. For example, an ice cube has heat energy and so does a glass of lemonade.

6 0

3 years ago

An object has a total mechanical energy of 150 J. At point A, the object has a kinetic energy

nika2105 [10]

Answer:

60 J

Explanation:

The law of conservation of energy states that energy is neither created nor destroyed, just converted into different forms. This means the total mechanical energy of the object at point A will be the same as the total mechanical energy at point B, and the question tells us the total of that mechanical energy is 150 J. Note we are assuming no energy is lost from the system as heat.

At point B, if the potential energy is 90 J, the remainder of the 150 J total must be kinetic energy. KE = 150 J - 90 J = 60 J.

8 0

3 years ago