Answer:
The number of moles of salt in one tablespoon is = <u>0.11 mole</u>
<u>Grams </u>cancel each other.
Explanation:
<u>Moles</u> : It is the unit of quantity . It is the mass of the substance present in exactly 12g of C-12.

<u>Moles Calculation:</u>
Given mass = 6.37 gram
Molar mass = 58.5 g/mol

= 0.1088
= 0.11 mole
<u>Units calculation</u>



<u>g ang g cancels each other </u>
moles = moles
<u>Hence unit = gram (g ) cancel each other.</u>
Answer:
Mass percent N₂ = 89%
Mass percent H₂ = 11%
Explanation:
First we <u>use PV=nRT to calculate n</u>, which is the total number of moles of nitrogen and hydrogen:
- 1.03 atm * 7.45 L = n * 0.082 atm·L·mol⁻¹·K⁻¹ * 305 K
So now we know that
- MolH₂ + MolN₂ = 0.307 mol
and
- MolH₂ * 2 g/mol + MolN₂ * 28 g/mol = 3.49 g
So we have a <u>system of two equations and two unknowns</u>. We use algebra to solve it:
Express MolH₂ in terms of MolN₂:
- MolH₂ + MolN₂ = 0.307 mol
Replace that value in the second equation:
- MolH₂ * 2 g/mol + MolN₂ * 28 g/mol = 3.49
- (0.307-MolN₂) * 2 + MolN₂ * 28 = 3.49
- 0.614 - 2MolN₂ + 28molN₂ = 3.49
Now we calculate MolH₂:
- MolH₂ + MolN₂ = 0.307 mol
Finally, we convert each of those mol numbers to mass, to <u>calculate the mass percent</u>:
- N₂ ⇒ 0.111 mol * 28 g/mol = 3.108 g N₂
- H₂ ⇒ 0.196 mol * 2 g/mol = 0.392 g H₂
Mass % N₂ = 3.108/3.49 * 100% = 89.05% ≅ 89%
Mass % H₂ = 0.392/3.49 * 100% = 11.15% ≅ 11%
Answer:
Average atomic mass = 15.86 amu.
Explanation:
Given data:
Number of atoms of Z-16.000 amu = 205
Number of atoms of Z-14.000 amu = 15
Average atomic mass = ?
Solution:
Total number of atoms = 205 + 15 = 220
Percentage of Z-16.000 = 205/220 ×100 = 93.18%
Percentage of Z-14.000 = 15/220 ×100 = 6.82 %
Average atomic mass = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass) / 100
Average atomic mass = (93.18×16.000)+(6.82×14.000) /100
Average atomic mass = 1490.88 + 95.48 / 100
Average atomic mass = 1586.36 / 100
Average atomic mass = 15.86 amu.
Explanation:
To solve this question, we will use the Clayperon Equation:
P.V = n.R.T
where:
P = 101.28 kPa
1 atm = 101,325 Pa
x atm = 101,280 Pa
x = 1 atm
V = 37.058 L
n = we don't know
R = 0.082 atm.L/K.mol
T = -139.88 ºC = -139.88+273.15 = 133.27 K
1*37.058 = n*0.082*133.27
n = 0.29 moles
Answer: 0.29 moles