Explanation:
Carbon monoxide and hydrogen gas reacts together to form methanol:
CO + 2H2 => CH3OH
Since 0.266mol * 2 = 0.532mol > 0.524mol, the limiting reactant here is hydrogen and therefore there will be 0.524mol / 2 = 0.262mol of methanol.
<span>.750 moles X (6.02 x10^23 atoms/1 mol)= 4.52 X 10^23 atoms is the answer </span>
Answer:
As long as it is a blank solution of the reagent, the Absorbance will be 0 regardless of the path length.
Explanation:
Absorbance of light by a reagent of concentration c, is given as
A = εcl
A = Absorbance
ε = molar absorptivity
c = concentration of reagent.
l = length of light path or length of the solution the light passes through.
So, if all.other factors are held constant, If a sample for spectrophotometric analysis is placed in a 10-cm cell, the absorbance will be 10 times greater than the absorbance in a 1-cm cell.
But the reagent blank solution is called a blank solution because it lacks the given reagent. A blank solution does not contain detectable amounts of the reagent under consideration. That is, the concentration of reagent in the blank solution is 0.
Hence, the Absorbance is subsequently 0. And increasing or decreasing the path length of light will not change anything. As long as it is a blank solution of the reagent, the Absorbance will be 0 regardless of the path length.
Hope this Helps!!!
Answer:19.2
Explanation: if see you have to times it and show it with times
Answer:
The percent by mass of copper in the mixture was 32%
Explanation:
The ammount of HNO₃ used is:
mol HNO₃ = volume * concentration
mol HNO₃ = 0.015 l * 15.8 mol/l
mol HNO₃ = 0.237 mol
According to the reaction, 4 mol HNO₃ will react with 1 mol Cu and produce 1 mol Cu²⁺. Since we have 0.237 mol HNO₃, the amount of Cu that could react would be (0.237 mol HNO₃ * 1 mol Cu / 4 mol HNO₃) 0.06 mol. This reaction would produce 0.060 mol Cu²⁺, however, only 0.010 mol Cu²⁺ were obtained, indicating that only 0.010 mol Cu were present in the mixture. This means that the acid was in excess, so we can assume that all copper present in the mixture has reacted.
Since 0.010 mol of Cu²⁺ were produced, the amount of Cu was 0.01 mol.
1 mol of Cu has a mass of 63.5 g, then 0.01 mol has a mass of:
0.01 mol Cu * 63.5 g / 1 mol = 0.635 g.
Since this amount was present in 2.00 g mixture, the amount of copper in 100 g of the mixture will be:
100 g(mixture) * 0.635 g Cu / 2 g(mixture) = 32 g
Then, the percent by mass of Cu (which is the mass of Cu in 100 g mixture) is 32%
