The balanced equation for the above neutralisation reaction is as follows;

2KOH + H₂SO₄ --> K₂SO₄ + 2H₂O

stoichiometry of KOH to H₂SO₄ is 2:1

neutralisation is the reaction between H⁺ ions and OH⁻ ions to form water which is neutral

number of KOH moles - 1.56 mol

2 mol of KOH require 1 mol of H₂SO₄ for neutralisation

therefore 1.56 mol of KOH require - 1/2 x 1.56 mol = 0.78 mol

0.78 mol of H₂SO₄ are required for neutralisation

**Answer:**

**Explanation:**

The balanced equation is

I₂(g) + Br₂(g) ⇌ 2IBr(g)

**Data:
**

Kc = 8.50 × 10⁻³

n(IBr) = 0.0600 mol

V = 1.0 L

**1. Calculate [IBr]
**

**2. Set up an ICE table.
**

**3. Calculate [I₂]
**

**4. Convert the temperature to kelvins
**

T = (150 + 273.15) K = 423.15 K

**5. Calculate p(I₂)
**

Answer:

energy required is 0.247kJ

Explanation:

The formula to use is Energy = nRdT;

Where n is number of mole

R is the molar gas constant

dT is the change in temperature

n = reacting mass of mercury / molar mass of mercury = 27.4/200.59 = 0.137

dT = final temperature - initial temperature = 376.20 - 158.30 = 217.90K

R = 8.314Jper mol per Kelvin

Energy = 0.137 x 8.314 x 217.90 = 247.12J

Energy in kJ= 247.12/1000= 0.247kJ

**Answer:**

**Explanation:**

The last number in a measured number is always an estimated digit.