The five main branches of chemistry are organic chemistry, inorganic chemistry, analytical chemistry, physical chemistry and biochemistry. Chemistry can be further divided into many sub-branches that may fall under more than one of the main branches.
Answer:
I guess A, I am not sure...
Answer:
C: The actual yield depends on the reaction conditions, but the theoretical yield varies only with reactant amounts
Explanation:
Looking at the options, the correct one is Option C because the actual yield usually depends on the conditions of the reaction, while the theoretical yield usually varies with only the amount of reactant.
<u>Analysing the Question:</u>
We are given a 250 mL solution of 0.5M K₂Cr₂O₇
Which means that we have:
0.5 Mole in 1L of the solution
0.125 moles in 250 mL of the solution <em>[dividing both the numbers by 4]</em>
<em />
<u>Mass of K₂Cr₂O₇ in the given solution:</u>
Molar mass of K₂Cr₂O₇(Potassium Dichromate) = 194 g/mol
<em>we know that we have 0.125 moles in the 250 mL solution provided</em>
Mass = Number of moles * Molar mass
Mass = 0.125 * 194
Mass = 36.75 grams
Answer:
%N = 25.94%
%O = 74.06%
Explanation:
Step 1: Calculate the mass of nitrogen in 1 mole of N₂O₅
We will multiply the molar mass of N by the number of N atoms in the formula of N₂O₅.
m(N): 2 × 14.01 g = 28.02 g
Step 2: Calculate the mass of oxygen in 1 mole of N₂O₅
We will multiply the molar mass of O by the number of O atoms in the formula of N₂O₅.
m(O): 5 × 16.00 g = 80.00 g
Step 3: Calculate the mass of 1 mole of N₂O₅
We will sum the masses of N and O.
m(N₂O₅) = m(N) + m(O) = 28.02 g + 80.00 g = 108.02 g
Step 4: Calculate the percent composition of N₂O₅
We will use the following expression.
%Element = m(Element)/m(Compound) × 100%
%N = m(N)/m(N₂O₅) × 100% = 28.02 g/108.02 g × 100% = 25.94%
%O = m(O)/m(N₂O₅) × 100% = 80.00 g/108.02 g × 100% = 74.06%
