In balanced equation there are same number of atoms in each element on both sides of the equation. unbalanced equation is when there are different number of atoms in each element on the both sides
<span>the balanced equation for the reaction is as follows
Na</span>₂<span>SO</span>₄<span> + BaCl</span>₂<span> ----> 2NaCl + BaSO</span>₄
<span>stoichiometry of Na</span>₂<span>SO</span>₄<span> to BaCl</span>₂<span> is 1:1
first we need to find out which the limiting reactant is
limiting reactant is fully used up in the reaction.
number of Na2So4 moles - 0.5 mol number of BaCl2 moles - 60 g / 208 g/mol = 0.288 mol
since molar ratio is 1:1 equal number of moles of both reactants should react with each other
therefore BaCl2 is the limiting reactant and Na2SO4 is in excess. amount of product formed depends on number of limiting reactant present.
stoichiometry of BaCl</span>₂<span> to BaSO</span>₄<span> is 1:1.
therefore number of BaSO4 moles formed - 0.288 mol</span>
Answer:
ΔTb = 0.66 C
Explanation:
Given
Mass of KBr = 185 g
Mass of water = 1.2 kg
Kb = 0.51 C/m
Explanation:
The change in boiling point (ΔTb) is given by the product of molality (m) of the solution and the boiling point constant (Kb)


[tex]\Delta T_{b}= 0.51 C/m * 1.296 m = 0.66 C[\tex]
It was empty , dark , and cold
Hydrogen bonds together the atoms in a water molecule. Hope I helped!