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2 years ago

NaI(aq) + Cl2(g) → Identify reaction type, predict products, and balance

Chemistry

1 answer:

mestny [16]2 years ago

7 0

Answer:

1) Redox reaction.

2) NaCl + l₂

3)2NaI + Cl₂ → 2NaCl + l₂

Explanation:

Chemical equation:

NaI + Cl₂ → NaCl + l₂

Balanced chemical equation:

2NaI + Cl₂ → 2NaCl + l₂

The type of reaction is oxidation reduction reaction.

The oxidation reduction reactions are called redox reaction. These reactions are take place by gaining or losing the electrons and oxidation state of elements are changed.

Oxidation:

Oxidation involve the removal of electrons and oxidation state of atom of an element is increased.

Reduction:

Reduction involve the gain of electron and oxidation number is decreased.

Consider the above given reaction:

2NaI⁻¹ + Cl⁰₂ → 2NaCl⁻¹ + l⁰₂

Iodine is oxidized while chlorine is reduced.

Product is:

Sodium chloride and iodine gas is produced.

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Answer:

C. effusion because there is a movement of a gas through a small opening into a larger volume

Explanation:

Effusion makes fluid/gas molecules move to the container with less pressure or larger volume. In diffusion, the movement should work two ways even though one side might receive more. But in effusion, the movement is rather one way.

This case shows how effusion work because its not the concentration that makes the balls moving to the bottom part of the container. No ball moving from bottom container to top either.

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2 years ago

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2. If 2.50g of sodium hydroxide is being reacted with 4.30g of magnesium chloride, how many grams of magnesium hydroxide would b

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Answer:

1.822 g of magnesium hydroxide would be produced.

Explanation:

Balanced reaction: $2NaOH+MgCl_{2}\rightarrow Mg(OH)_{2}+2NaCl$

Compound Molar mass (g/mol)

NaOH 39.997

$MgCl_{2}$ 95.211

$Mg(OH)_{2}$ 58.3197

So, 2.50 g of NaOH = $\frac{2.50}{39.997}$ mol of NaOH = 0.0625 mol of NaOH

4.30 g of $MgCl_{2}$ = $\frac{4.30}{95.211}$ mol of $MgCl_{2}$ = 0.0452 mol of $MgCl_{2}$

According to balanced equation-

2 mol of NaOH produce 1 mol of $Mg(OH)_{2}$

So, 0.0625 mol of NaOH produce $(\frac{0.0625}{2})$ mol of NaOH or 0.03125 mol of NaOH

1 mol of $MgCl_{2}$ produces 1 mol of $Mg(OH)_{2}$

So, 0.0452 mol of $MgCl_{2}$ produce 0.0452 mol of $Mg(OH)_{2}$

As least number of moles of $Mg(OH)_{2}$ are produced from NaOH therefore NaOH is the limiting reagent.

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2 years ago

In the first step of glycolysis, the given two reactions are coupled. reaction 1:reaction 2:glucose+Pi⟶glucose-6-phosphate+H2OAT

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Answer: Reaction 1 is non spontaneous.

Explanation:

According to Gibb's equation:

$\Delta G=\Delta H-T\Delta S$

$\Delta G$ = Gibbs free energy

$\Delta H$ = enthalpy change

$\Delta S$ = entropy change

T = temperature in Kelvin

When $\Delta G$ = +ve, reaction is non spontaneous

$\Delta G$ = -ve, reaction is spontaneous

$\Delta G$ = 0, reaction is in equilibrium

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2 years ago

Four different solutions (I, II, III, and IV) are labeled on the pH scale below. A p H scale is shown, fading from red at the le

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2 years ago

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A 2.0% (w/v) solution of sodium hydrogen citrate, Na2C6H6O7, which also contains 2.5% (w/v) of dextrose, C6H12O6, is used as an

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Answer:

0.0847M is molarity of sodium hydrogen citrate in the solution

Explanation:

The 2.0%(w/v) solution of sodium hydrogen citrate contains 2g of the solute in 100mL of solution. To find the molarity of the solution we need to convert the mass of solute to moles using molar mass and the mL of solution to Liters because molarity is the ratio between moles of sodium hydrogen citrate and liters of solution.

Moles Na2C6H6O7:

Molar Mass:

2Na: 2*22.99g/mol: 45.98g/mol

6C: 6*12.01g/mol: 72.01g/mol

6H: 6*1.008g/mol: 6.048g/mol

7O: 7*16g/mol: 112g/mol

45.98g/mol + 72.01g/mol + 6.048g/mol + 112g/mol = 236.038g/mol

Moles of 2g:

2g * (1mol / 236.038g) = 8.473x10⁻³ moles

Liters solution:

100mL * (1L / 1000mL) = 0.100L

Molarity:

8.473x10⁻³ moles / 0.100L =

<h3>0.0847M is molarity of sodium hydrogen citrate in the solution</h3>

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