Answer:
87.9%
Explanation:
Balanced Chemical Equation:
HCl + NaOH = NaCl + H2O
We are Given:
Mass of H2O = 9.17 g
Mass of HCl = 21.1 g
Mass of NaOH = 43.6 g
First, calculate the moles of both HCl and NaOH:
Moles of HCl: 21.1 g of HCl x 1 mole of HCl/36.46 g of HCl = 0.579 moles
Moles of NaOH: 43.6 g of NaOH x 1 mole of NaOH/40.00 g of NaOH = 1.09 moles
Here you calculate the mole of H2O from the moles of both HCl and NaOH using the balanced chemical equation:
Moles of H2O from the moles of HCl: 0.579 moles of HCl x 1 mole of H2O/1 mole of HCl = 0.579 moles
Moles of H2O from the moles of NaOH: 1.09 moles of HCl x 1 mole of H2O/1 mole of NaOH = 1.09 moles
From the calculations above, we can see that the limiting reagent is HCl because it produced the lower amount of moles of H2O. Therefore, we use 0.579 moles and NOT 1.09 moles to calculate the mass of H2O:
Mass of H2O: 0.579 moles of H2O x 18.02 g of H2O/1 mole of H2O = 10.43 g
% yield of H2O = actual yield/theoretical yield x 100= 9.17 g/10.43 g x 100 = 87.9%
Answer:
3
Explanation:
Nonmetals are only on the right side (other than hydrogen of course).
Malleable means to be able to be hammered out of shape with no cracking and luster is shiny (metallic properties)
Most likely a transition metal
Answer: 64.71 L. The balloon will have expanded.
Explanation: 1st, sea level is equal to 1 atm. You will then use Boyle’s Law: P1V1=P2V2.
P1= 1 atm
V1= 55 L
P2= 0.85 atm
V2= ?
Now insert the numbers for the letters:
(1)(55)=(0.85)(V2)
55=0.85V2
Divide by 0.85 on both sides:
V2=64.71L
Now, according to Boyle’s Law, when pressure decreases, volume increases. That’s why the balloon inflates rather than deflates.