Answer:
I believe
Explanation:
I believe that the answer is B, but I could be wrong. I think it's How much work can be done in a given time, because, In physics, power is the amount of energy transferred or converted per unit time.
Answer:
D. when the number of moles of acid is exactly equal to the number of moles of base.
Explanation:
<em>Regarding options A. and E</em>., pKa and pKb would only be taken into consideration if the titrations were of <em>weak</em> acids and bases. However it is possible to have a titration of monoprotic acids and bases with strong acids and bases.
Another way of looking at the answer is identifying <em>which one best describes the equivalence point</em>.
Answer: Option (3) is the correct answer.
Explanation:
A reaction due to which there occurs change in chemical composition of reactants is known as a chemical reaction.
For example, evolution of heat when two substances combine together shows a chemical reaction has taken place between the substances.
Whereas a reaction which does not bring any changes in chemical composition of the reactants is known as a physical reaction.
For example, boiling point, melting point, mass, volume etc are all physical properties.
Thus, we can conclude that out of the given options a substance melts when sitting in the Sun, is not an indication of a chemical reaction.
A) Limiting reactant
You need the molar ratios (from the balanced chemical equation) and the molar masses of each compound (from the atomic masses)
a) Molar ratios:
6 mol HF : 1 mol SiO2 : 1 mol H2SiF6
2) Molar masses:
Atomic masses:
H: 1 g/mol
F: 19 g/mol
Si: 28 g/mol
O: 16g/mol
=>
HF:1g/mol + 19 g/mol = 20 g/mol
SiO2: 28g/mol + 2*16g/mol = 60 g/mol
H2SiF6: 2*1g/mol + 28g/mol + 6*19g/mol = 144g/mol
3) convert data in grams to moles
21.0 g SiO2 / 60 g/mol = 0.35 mol SiO2
70.5 g HF / 20 g/mol = 3.525 mol HF
4) Use the theorical ratios to deduce which is in excess and which is the limiting reactant.
6 mol HF / 1mol SiO2 < 3.525 mol HF / 0.35 mol SiO2 ≈ 10
=> There is more HF than the needed to react with 0.35mol of SiO2 =>
SiO2 is the limiting reactant (HF is in excess)
b) Mass of excess reactant.
1) Calculate how many grams reacted, which requires to calculate first the number of moles that reacted
0.35 mol SiO2 * 6 mol HF / 1 mol SiO2 = 2.1 mol of HF
2.1 mol HF * 20 g/mol = 42 gram of HF
2) Subtract the quantity that reacted from the original quantity:
70.5 g - 42 g = 28.5 g of HF in excess
c) Theoretical yield of H2SiF6
1 mol of SiO2 ; 1 mol of H2SiF6 => 0.35 mol SiO2 : 0.35 mol H2SiF6
Convert those moles to grams: 0.35 mol * 144 g/mol = 50.4 grams
d) % yield
% yield = actual yield / theoretical yield * 100 = 45.8 / 50.4 * 100 = 90.87%
