Answer:
N2I6 = 789 g
N2I6 = 8.2x1022 molecules N2I6 x 1 mole/6.02x1023 molecules = 1.36x10-1 moles = 0.136 moles
N2I6=0.136molesx789g/mole=107g=110g
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- Cutiepatutie ☺❀❤
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>
Answer:
Explanation:
Bronsted Base is an H+ acceptor
No good answer Bronstead base does not accept hydroxide or electrons
There are 4, 1 Carbon 4 Fluorine