Zn = 28.15%
Cl = 30.53%
O = 41.32%
<h3>Further explanation</h3>
Given
Zn(CIO3)2 compound
Required
The % composition
Solution
Ar Zn = 65.38
Ar Cl = 35,453
Ar O = 15,999
MW Zn(CIO3)2 = 232.3
Zn = 65,38/232.3 x 100% = 28.15%
Cl = (2 x 35.453) / 232.3 x 100% = 30.53%
O = (6 x 15.999) / 232.3 x 100% = 41.32%
I believe the answer you're looking for is B. It's Polluted. I hope this helps.
Answer:
10.85 g of water
Explanation:
First we write the balanced chemical equation
Then we calculate the number of moles of nitric acid produced
n(HNO3) = 
According to the balanced equation, water needed in moles is always half the number of moles of HNO3 produced. So since we will produce 1.2044 mol of HNO3, we will need 0.6022 mol of water. Now to calculate what mass that is:
mass(water)=number of moles*molar mass=0.6022mol*18.02g/mol=10.85g
Answer:
M of HI = 5.4 M.
Explanation:
- We have the rule: at neutralization, the no. of millimoles of acid is equal to the no. of millimoles of the base.
<em>(XMV) acid = (XMV) base.</em>
where, X is the no. of (H) or (OH) reproducible in acid or base, respectively.
M is the molarity of the acid or base.
V is the volume of the acid or base.
<em>(XMV) HI = (XMV) Ca(OH)₂.</em>
For HI; X = 1, M = ??? M, V = 25.0 mL.
For Ca(OH)₂, X = 2, M = 1.5 M, V = 45.0 mL.
<em>∴ M of HI = (XMV) Ca(OH)₂ / (XV) HI</em> = (2)(1.5 M)(45.0 mL) / (1)(25.0 mL) = <em>5.4 M.</em>
