**Answer:**

121 K

**Explanation:**

**Step 1: Given data**

- Initial volume (V₁): 79.5 mL

- Initial temperature (T₁): -1.4°C

- Final volume (V₂): 35.3 mL

**Step 2: Convert "-1.4°C" to Kelvin**

We will use the following expression.

K = °C + 273.15 = -1.4°C + 273.15 = 271.8 K

**Step 3: Calculate the final temperature of the gas (T₂)**

Assuming ideal behavior and constant pressure, we can calculate the final temperature of the gas using Charles' law.

V₁/T₁ = V₂/T₂

T₂ = V₂ × T₁/V₁

T₂ = 35.3 mL × 271.8 K/79.5 mL = 121 K

Molarity of solution is mathematically expressed as,

M =

We know that volume = mass/density

Given: mass of solution = 100 g, Density = 1.34 g/ml

∴ volume = 100/1.34 = 88.49 ml = 0.08849 l

Also, we know that molecular weight of sucrose = 342.3 g/mol

∴M =

= 6.979 M

Thus, molarity of solution is 6.979 M

**Answer:**

pH = 1.32

**Explanation:**

H₂M + KOH ------------------------ HM⁻ + H₂O + K⁺

This problem involves a weak diprotic acid which we can solve by realizing they amount to buffer solutions. In the first deprotonation if all the acid is not consumed we will have an equilibrium of a wak acid and its weak conjugate base. Lets see:

So first calculate the moles reacted and produced:

n H₂M = 0.864 g/mol x 1 mol/ 116.072 g = 0.074 mol H₂M

54 mL x 1L / 1000 mL x 0. 0.276 moles/L = 0.015 mol KOH

it is clear that the maleic acid will not be completely consumed, hence treat it as an equilibrium problem of a buffer solution.

moles H₂M left = 0.074 - 0.015 = 0.059

moles HM⁻ produced = 0.015

Using the Henderson - Hasselbach equation to solve for pH:

ph = pKₐ + log ( HM⁻/ HA) = 1.92 + log ( 0.015 / 0.059) = 1.325

Notes: In the HH equation we used the moles of the species since the volume is the same and they will cancel out in the quotient.

For polyprotic acids the second or third deprotonation contribution to the pH when there is still unreacted acid ( Maleic in this case) unreacted.

**Answer:**

The law of conservation of mass states that in a closed system, mass is neither created nor destroyed during a chemical or physical reaction. The law of conservation of mass is applied whenever you balance a chemical equation.

**Explanation:**

According to the law of conservation of mass, the mass of the products in a chemical reaction must equal the mass of the reactants.

The law of conservation of mass is useful for a number of calculations and can be used to solve for unknown masses, such the amount of gas consumed or produced during a reaction.

It is applicable in a chemical when the the mass of the products in a chemical reaction is equal to the mass of the reactants.

But it is not applicable in a nuclear fusion as some of the mass is generated as energy.