Atomic mass Ar => 39.948 a.m.u
39.948 g --------------- 6.02x10²³ atoms
?? g -------------------- 3.8x10²⁴ atoms
(3.8x10²⁴) x 39.948 / 6.02x10²³ => 250 g
hope this helps!
Answer:
0.733 mol.
Explanation:
- From the balanced equation:
<em>2Fe₂O₃ + C → Fe + 3CO₂,</em>
It is clear that 1.0 moles of Fe₂O₃ react with 1.0 mole of C to produce 1.0 mole of Fe and 3.0 moles of CO₂.
- Since Fe₂O₃ is in excess, C will be the limiting reactant.
<u><em>Using cross multiplication:</em></u>
1.0 mole of C produces → 3.0 moles of CO₂, from the stichiometry.
??? mole of C produces → 2.2 moles of CO₂.
∴ The no. of moles of C needed to produce 2.2 moles of CO₂ = (1.0 mole of C) (2.2 mole of CO₂) / (3.0 mole of CO₂) = 0.733 mol.
Answer:
The mass percent of potassium is 39%
Option C is correct
Explanation:
Step 1: Data given
Atomic mass of K = 39.10 g/mol
Atomic mass of H = 1.01 g/mol
Atomic mass of C = 12.01 g/mol
Atomic mass of O = 16.0 g/mol
Step 2: Calculate molar mass of KHCO3
Molar mass KHCO3 = 39.10 + 12.01 + 1.01 + 3*16.0
Molar mass KHCO3 = 100.12 g/mol
Step 3: Calculate mass percent of potassium (K)
%K = (atomic mass of K / molar mass of KHCO3) * 100%
%K = (39.10 / 100.12) * 100%
%K = 39.05 %
The mass percent of potassium is 39%
Option C is correct
Answer: 1.4x10-3 g N2O4
Explanation: First convert molecules of N2O4 to moles using Avogadro's Number. Then convert moles to mass using the molar mass of N2O4.
9.2x10^18 molecules N2O4 x 1 mole N2O4 / 6.022x10²³ molecules N2O4
= 1.53x10-5 moles N2O4
1.53x10-5 moles N2O4 x 92 g N2O4/ 1 mole N2O4
= 1.4x10-3 g N2O4
Answer:
Given that W=mg:
The weight of the box would be 50 N taking the value of 'g' as 10ms-2. Taking the value of 'g' as 9.8ms-2, the weight of the box would be 49 N.
(N = Newtons)
