Answer:
So ²³⁵UF₆ effuses 1.0043 times faster than ²³⁸UF₆
Explanation:
The rate of effusion of two gases A and B can be expressed by <em>Graham's law</em>:
Where M is the molar mass, and in this case A is ²³⁵UF₆ while B is ²³⁸UF₆.
So now we <u>calculate the molar mass of each mass</u>:
²³⁵UF₆ ⇒235 + 6*19 = 349 g/mol
²³⁸UF₆ ⇒238 + 6*19 = 352 g/mol
Putting the data in Graham's law:
= 1.0043
So ²³⁵UF₆ effuses 1.0043 times faster than ²³⁸UF₆.
First, let us write the reaction. We know that the reactants are Aluminum and Sulfuric acid, and one of the products is hydrogen. That means the reaction is a single replacement reaction as shown below:
2 Al + 3 H₂SO₄ = 3 H₂ + Al₂(SO₄)₃
Next, let us determine which of the reactants is limiting:
1.80 g Al (1 mol/26.98 g)(3 mol H₂SO₄/2 mol Al)(98 g H₂SO₄/mol) = 9.807 g
It would need 9.807 g. But the only available amount is 6 g. That means that H₂SO₄ is the limiting reactant. We use 6 g as the basis to know the theoretical yield:
6 g*(1 mol H₂SO₄/98 g)*(3 mol H₂/ 3 mol H₂SO₄)*(2 g /mol H₂) = 0.122 g
Therefore, the percent yield is equal to
Percent yield = (0.112 g/0.122 g)*100
Percent yield = 91.8%
Answer:
The reaction is shifted to the left.
Explanation:
- Le Châtelier's principle states that when there is an dynamic equilibrium, and this equilibrium is disturbed by an external factor, the equilibrium will be shifted in the direction that can cancel the effect of the external factor to reattain the equilibrium.
- When there is a decrease in pressure, the equilibrium will shift towards the side with more no. of moles of gas molecules of the reaction.
- The reactants side (left) has 4.0 moles of gases and the products side (right) has 2.0 moles of gases.
- So, decreasing the pressure will shift the reaction to the side with more moles of gas (left side).
- <em>so, the reaction is shifted to the left.</em>
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gas = methane
burn with O₂ (oxygen)
CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l)
Answer:
How do you find the density of a liquid experiment?
To measure the density of a liquid you do the same thing you would for a solid. Mass the fluid, find its volume, and divide mass by volume. To mass the fluid, weigh it in a container, pour it out, weigh the empty container, and subtract the mass of the empty container from the full container.
