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The answer for the following problem is mentioned below.
- <u><em>Therefore number of molecules(N) present in the calcium phosphate sample are 19.3 × 10^23 molecules.</em></u>
Explanation:
Given:
mass of calcium phosphate (
) = 125.3 grams
We know;
molar mass of calcium phosphate (
) = (40×3) + 3 (31 +(4×16))
molar mass of calcium phosphate (
) = 120 + 3(95)
molar mass of calcium phosphate (
) = 120 +285 = 405 grams
<em>We also know;</em>
No of molecules at STP conditions(
) = 6.023 × 10^23 molecules
To solve:
no of molecules present in the sample(N)
We know;
N÷
=
N =(405×6.023 × 10^23) ÷ 125.3
N = 19.3 × 10^23 molecules
<u><em>Therefore number of molecules(N) present in the calcium phosphate sample are 19.3 × 10^23 molecules</em></u>
1) Carbon-13:
Proton-6 Neutron-7 Electron-6
2)Atomic mass of element X:
(55*10+56*20+57*70)/100=56.6
We need to keep in mind that the compound is neutral.
H2SO3
2(+1)+S+3(-2)=0 (since its neutral)
2+S-6=0
S-4=0
S=4
Therefore the oxidation number for sulfur is +4.