Answer:
0.1313 g.
Explanation:
- It is known that at STP, 1.0 mole of ideal gas occupies 22.4 L.
- Suppose that hydrogen behaves ideally and at STP conditions.
<u><em>Using cross multiplication:</em></u>
1.0 mol of hydrogen occupies → 22.4 L.
??? mol of hydrogen occupies → 1.47 L.
∴ The no. of moles of hydrogen that occupies 1.47 L = (1.0 mol)(1.47 L)/(22.4 L) = 6.563 x 10⁻² mol.
- Now, we can get the no. of grams of hydrogen in 6.563 x 10⁻² mol:
<em>The no. of grams of hydrogen = no. of hydrogen moles x molar mass of hydrogen</em> = (6.563 x 10⁻² mol)(2.0 g/mol) = <em>0.1313 g.</em>
To answer this question, you need to know <span>Graham's Law of Effusion/Diffusion formula. In this formula, the rate of diffusion/effusion would be influenced by the mass. As the molecule has bigger mass, the rate should be slower because it will be harder to pass the membrane. The calculation should be:</span>
<span>Rate 1 / Rate 2 = √[M2/M1]
</span>4.11/1= √[M2/2]
M2=33.78 g/mol
Answer:
7
Explanation:
The mass number is neutrons plus protons
The oxidation half-reaction occurs at one electrode (the anode), and the reduction half-reaction occurs at the other (the cathode). When the circuit is closed, electrons flow from the anodeto the cathode.
