The effusion rate is 1.125 cm/sec for ammonia.
How to find effusion rate ?
Effusion rate (r1) HCl = 43.2 cm/min
Molar mass (m2) NH3 =17.04g/mole
Molar mass (m1) HCl =36.46g/mole
- Substitute the molar masses of the gases into Graham's law and solve for the ratio.
firstly convert 43.2 cm/min into cm/sec i.e., 0.72 cm/sec
Then,
0.72/r2 =√17.04/36.46
r2= 1.125 cm/sec
Hence, the rate of diffusion of ammonia is 1.125 times faster than the rate of diffusion of hydrogen chloride.
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Answer:
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Answer:
<h2>The first thing to do here is to use the molarity and the volume of the initial solution to figure out how many grams of copper(II) chloride it contains.</h2><h2 /><h2>133</h2><h2>mL solution</h2><h2>⋅</h2><h2>1</h2><h2>L</h2><h2>10</h2><h2>3</h2><h2>mL</h2><h2>⋅</h2><h2>7.90 moles CuCl</h2><h2>2</h2><h2>1</h2><h2>L solution</h2><h2>=</h2><h2>1.051 moles CuCl</h2><h2>2</h2><h2 /><h2>To convert this to grams, use the compound's molar mass</h2><h2 /><h2>1.051</h2><h2>moles CuCl</h2><h2>2</h2><h2>⋅</h2><h2>134.45 g</h2><h2>1</h2><h2>mole CuCl</h2><h2>2</h2><h2>=</h2><h2>141.31 g CuCl</h2><h2>2</h2><h2 /><h2>Now, you know that the diluted solution must contain </h2><h2>4.49 g</h2><h2> of copper(II) chloride. As you know, when you dilute a solution, you increase the amount of solvent while keeping the amount of solute constant.</h2><h2 /><h2>This means that you must figure out what volume of the initial solution will contain </h2><h2>4.49 g</h2><h2> of copper(II) chloride, the solute.</h2><h2 /><h2>4.49</h2><h2>g</h2><h2>⋅</h2><h2>133 mL solution</h2><h2>141.32</h2><h2>g</h2><h2>=</h2><h2>4.23 mL solution</h2><h2>−−−−−−−−−−−−−− </h2><h2 /><h2>The answer is rounded to three sig figs.</h2><h2 /><h2>You can thus say that when you dilute </h2><h2>4.23 mL</h2><h2> of </h2><h2>7.90 M</h2><h2> copper(II) chloride solution to a total volume of </h2><h2>51.5 mL</h2><h2> , you will have a solution that contains </h2><h2>4.49 g</h2><h2> of copper(II) chloride.</h2>
Answer:
Explanation:
H₂SO₄ is a strong acid, which means that most of it ionizes in aqueous solution.
Since it is a diprotic acid (two hydrogen ions) its ionization occurs in two steps:
- H₂SO₄ (aq) → H⁺(aq) + HSO₄⁻(aq)
- HSO₄⁻ (aq) → H⁺(aq) + SO₄²⁻(aq)
Thus, almost all H₂SO₄ has ionized and its final concentration is almost nothing.
After the first ionization, the conentrations of H⁺(aq) and HSO₄⁻ are equal but by the second ionization more H⁺ ions are produced along with SO₄⁻.
You can show it as one step dissociation, assuming 100% dissociation (given this is a strong acid):
By the stequiometry you can build this table:
H₂SO₄ (aq) → 2H⁺(aq) + SO₄²⁻(aq)
Initial A 0 0
Change - x +2x +x
Equilibrium A - x 2x x
As explained, A - x is very low, and 2x is twice x. Thus,
The rank of the concentrations from highest to lowest is:
