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:
8 moles of C
Explanation:
From the question given above, the following equation was obtained:
3A + 2B —> 6C
From the equation above,
3 moles of A reacted to produce 6 moles of C.
Thus, the number of mole of C produced by reacting 4 moles of A can be obtained as follow:
From the equation above,
3 moles of A reacted to produce 6 moles of C.
Therefore, 4 moles of C will react to produce = (4 × 6)/3 = 8 moles of C
Thus, 8 moles of C can be obtained from the reaction of 4 moles of A with excess B
Answer:
8.5155g NH3
Explanation:
the molar mass of NH3 is 17.031 g/mol
0.5 mol NH3 x 17.031 gNH3/1 mol NH3 = 8.5155g NH3
Answer:
lens are the answers, I hope it is right
Answer:
4,38%
small molecular volumes
Decrease
Explanation:
The percent difference between the ideal and real gas is:
(47,8atm - 45,7 atm) / 47,8 atm × 100 = 4,39% ≈ <em>4,38%</em>
This difference is considered significant, and is best explained because argon atoms have relatively <em>small molecular volumes. </em>That produce an increasing in intermolecular forces deviating the system of ideal gas behavior.
Therefore, an increasing in volume will produce an ideal gas behavior. Thus:
If the volume of the container were increased to 2.00 L, you would expect the percent difference between the ideal and real gas to <em>decrease</em>
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I hope it helps!
