The balanced equation for the above reaction is as follows;
Mg + 2HCl ---> MgCl₂ + H₂
stoichiometry of HCl to MgCl₂ is 2:1
we have been told that Mg is in excess therefore HCl is the limiting reactant
number of HCl moles reacted - 0.100 mol/L x 0.0256 L = 0.00256 mol
according to molar ratio, number of MgCl₂ moles formed - 0.00256/2
Therefore number of MgCl₂ moles formed - 0.00128 mol
mass of MgCl formed - 0.00128 mol x 95.20 g/mol = 0.122 g
Answer:

Explanation:
The molar mass of uranium-235 is 235 g/mol. So one mole of uranium-235 has a mass of 235 g. Put differently 6.022×10^23 atoms of uranium-235 have a mass of 235 g. Knowing that, how can we use that to find the mass of one atom?
mass of one atom = 
Answer:
percent yield = 40.6 %
Explanation:
The question asks to determine the percent yield, which can be defined as:
percent yield = 
where the actual yield is how much product was obtained, in this case 12.5 g of CCl₂F₂, and the theoretical yield is how much product could be obtained with the given reactants theoretically.
So we know already the actual yield, we need to <em>calculate the theoretical yield.</em>
First we need to <em>write the reaction chemical equation</em>:
CCl₄ + HF → CCl₂F₂ + HCl
and <em>balance the equation</em>:
CCl₄ + 2 HF → CCl₂F₂ + 2 HCl
In the equation we can see that <em>for every mol of CCl₄ we should get 1 mol of CCl₂F₂</em> (molar ratio 1:1). So if we <u>calculate the moles of CCl₄</u> in the given 39.2 g of CCl₄ we could know how many moles of CCl₂F₂ (assuming HF is in excess).
- Moles of CCl₄ = mass CCl₄ / molar mass CCl₄
- Molar Mass CCl₄ = 12.011 + 4 * 35.45 = 153.811 g/mol
- Moles of CCl₄ = 39.2 g / 153.811 g/mol = 0.2549 moles
From the molar ratio we know:
Moles of CCl₂F₂ = moles of CCl₄ = 0.2549 moles
Now we need to <u>convert these moles into grams</u> to get the theoretical yield of CCl₂F₂ in grams:
- mass CCl₂F₂ = moles CCl₂F₂ * molar mass CCl₂F₂
- Molar Mass CCl₂F₂ = 12.011 + 2 * 35.45 + 2 * 18.998 = 120.907 g/mol
- Mass CCl₂F₂ = 0.2549 moles * 120.907 g/mol = 30.81 g
- Theoretical yield CCl₂F₂ = 30.81 g
Percent yield = (12.5 g / 30.81 g) * 100 = 40.6 %
Answer:
I’m pretty sure the answer is blind spot one or the actual distance of traffic I’ll asume actually distance traffic
Explanation: