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_{A}](https://tex.z-dn.net/?f=N_%7BA%7D)
=![\frac{N}{6.023*10^23}](https://tex.z-dn.net/?f=%5Cfrac%7BN%7D%7B6.023%2A10%5E23%7D)
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>
Energy lost to condense = 803.4 kJ
<h3>Further explanation</h3>
Condensation of steam through 2 stages:
1. phase change(steam to water)
2. cool down(100 to 0 C)
1. phase change(condensation)
Lv==latent heat of vaporization for water=2260 J/g
![\tt Q=300\times 2260=678000~J](https://tex.z-dn.net/?f=%5Ctt%20Q%3D300%5Ctimes%202260%3D678000~J)
2. cool down
c=specific heat for water=4.18 J/g C
![\tt Q=300\times 4.18\times (100-0)=125400](https://tex.z-dn.net/?f=%5Ctt%20Q%3D300%5Ctimes%204.18%5Ctimes%20%28100-0%29%3D125400)
Total heat =
![\tt 678000+125400=803400~J](https://tex.z-dn.net/?f=%5Ctt%20678000%2B125400%3D803400~J)
Calculating for the moles of H+
1.0 L x (1.00 mole / 1 L ) = 1 mole H+
From the given balanced equation, we can use the stoichiometric ratio to solve for the moles of PbCO3:
1 mole H+ x (1 mole PbCO3 / 2 moles H+) = 0.5 moles PbCO3
Converting the moles of PbCO3 to grams using the molecular weight of PbCO3
0.5 moles PbCO3 x (267 g PbCO3 / 1 mole PbCO3) = 84.5 g PbCO3
Answer:
A:force times an object displacement
Explanation: