I will say this is True….?
I believe the answer is Canada!
Can not be D or B and A is not enough so C is right.
Answer:

Explanation:
Ba(OH)₂ + 2HCl ⟶ BaCl₂ + H₂O
V/mL: 249
c/mol·L⁻¹: 0.0443 0.285
1. Calculate the moles of Ba(OH)₂

2. Calculate the moles of HCl
The molar ratio is 2 mol HCl:1 mol Ba(OH)₂

3. Calculate the volume of HCl

Answer : The balanced chemical reaction will be:

Explanation :
Single replacement reaction : A chemical reaction in which the more reactive element replace the less reactive element.
It is represented as,

In this reaction, A is more reactive element and B is less reactive element.
As per question, when gold (IV) iodide react with bromine to give gold (IV) bromide and iodine.
The balanced chemical reaction will be:
