Answer:
mass CaI2 = 23.424 Kg
Explanation:
From the periodic table we obtain for CaI2:
⇒ molecular mass CaI2: 40.078 + ((2)(126.90)) = 293.878 g/mol
∴ mol CaI2 = (4.80 E25 units )×(mol/6.022 E23 units) = 79.708 mol CaI2
⇒ mass CaI2 = (79.708 mol CaI2)×(293.878 g/mol) = 23424.43 g
⇒ mass CaI2 = 23.424 Kg
Answer: 
is a reactant; it is present before the reaction occurs.
Explanation:
In a chemical reaction the chemical formulas written before the arrow are described as reactants as they react together to form products which are written after the arrow.
Thus and HCl are reactants here whereas 
, 
and 
are products.
Explanation:
Charles' law gives the relationship between the volume and the temperature of the gas. Mathematically,
Volume ∝ Temperature
i.e. 
We have, V₁ = 1.6 L, T₁ = 278 K, T₂ = 253, V₂=?
So, the new volume is 1.45 L.
Na2S2O3(aq) + 4Cl2( g) + 5H2O = 2NaHSO4(aq) + 8HCl(aq)
1)How many moles of HCl can form from 0.21 mol of Cl2?
0.21 mol Cl2 ( 8 mol HCl / 4 mol Cl2 ) = 0.42 mol HCl
2)How many moles of H2O are required for the reaction of 0.18 mol of Cl2?
0.18 mol Cl2 ( 5 mol H2O / 4 mol Cl2 ) = 0.225 mol H2O
3)How many moles of H2O react if 0.50 mol HCl is formed?
0.50 mol HCl ( 5 mol H2O / 8 mol HCl ) = 0.3125 mol H2O
Moles of Bromine produced = 9 moles
<h3>Further explanation</h3>
Given
9 moles of Chlorine gas
Word equation
Required
Moles of Chlorine produced
Solution
We change the word equation into a chemical equation (with a formula)
Aluminum bromide reacts with chlorine gas to produce Aluminum chloride and bromide gas
2AlBr₃+3Cl₂⇒2AlCl₃+3Br₂
moles Cl₂ = 9
Maybe you mean, <em>how many moles of Bromine can we produce?</em>
From equation, mol ratio Cl₂ : Br₂ = 3 : 3, so mol Br₂=mol Cl₂=9 moles
