Q1. A 0.5 g mixture of Cu2O and CuO contain 0.425 g Cu. What is the mass of CuO in the mixture?
Q2. A chemist wants to prepare 0.5 M HCl. The label of a commercial hydrochloric acid shows that S.G. =1.18 and concentration of HCl was 37%. Calculate the volume of commercial hydrochloric acid dose the chemist requires to make up 1.5 L of the dilute acid.
Q3. Relate the rate of decomposition of NO2 to the rate of formation of O2 for the following reaction.
Thanks
#1. I work these as two equations and solve them simultaneously.
Let X = mass CuO
and Y = mass Cu2O
--------------------
equation 1 is X + Y = 0.5
Equation 2 comes from this. The grams copper from CuO + copper from Cu2O = 0.425g. With MM standing for molar mass and AM for atomic mass:
Grams Cu from CuO is (AM Cu*X/MM CuO)
Grams Cu from Cu2O is (2*AM Cu*Y/MM Cu2O)
Put that together for equation 2 as follows:
(AM Cu*X/MM CuO) + (2*AM Cu*Y/MM Cu2O) = 0.425
Solve equation 1 and equation 2 for X = grams CuO
Post your work if you get stuck.
0.212 g CuO
#2. Determine the molarity (M) of the concentrated acid.
1000 mL x 1.18 g/mL x 0.3 x (1/36.5) = ?M My estimated value for this is 12M but you need to refine the answer.
Then use the dilution formula of
mL1 x M1 = mL2 x M2
mL1 x 12 = 1,500 x 0.5
Solve for mL of the 12 M HCl to use.
To prepare you drop ? of the conc HCl into a 1.5 volumetric flask and make to the mark with distilled water. Mix thoroughly and stopper.
#1=0.216GRAMS
Thank you very much!
for #2
Why the equation need to x 1000 ?
Thank
Q1. To find the mass of CuO in the mixture, we need to first calculate the amount of Cu present in the mixture using the given information. We are told that the mass of Cu in the mixture is 0.425 g. The molar mass of Cu is 63.55 g/mol, so we can calculate the number of moles of Cu using the formula:
moles of Cu = mass of Cu / molar mass of Cu
moles of Cu = 0.425 g / 63.55 g/mol
Next, we need to calculate the number of moles of CuO present in the mixture. Since the total mass of the mixture is 0.5 g, we can find the mass of CuO by subtracting the mass of Cu from the total mass:
mass of CuO = total mass of mixture - mass of Cu
mass of CuO = 0.5 g - 0.425 g
Now, we can calculate the moles of CuO using its molar mass (79.55 g/mol):
moles of CuO = mass of CuO / molar mass of CuO
moles of CuO = (0.5 g - 0.425 g) / 79.55 g/mol
Finally, to find the mass of CuO in the mixture, we multiply the moles of CuO by its molar mass:
mass of CuO = moles of CuO * molar mass of CuO
mass of CuO = [(0.5 g - 0.425 g) / 79.55 g/mol] * 79.55 g/mol
Therefore, the mass of CuO in the mixture is equal to the result obtained from the calculation.
Q2. To calculate the volume of commercial hydrochloric acid required to prepare 1.5 L of a 0.5 M HCl solution, we can use the formula:
M1V1 = M2V2
Where:
M1 is the initial concentration of the commercial hydrochloric acid,
V1 is the volume of the commercial hydrochloric acid needed,
M2 is the final concentration of the diluted HCl solution (0.5 M),
V2 is the final volume of the diluted HCl solution (1.5 L).
We are given that the concentration of the commercial hydrochloric acid is 37%, which means that 100 mL of the acid contains 37 mL of HCl (the rest is water).
First, we need to determine the concentration of the commercial hydrochloric acid in terms of moles per liter (M). To do this, we divide the percentage concentration by 100 and multiply by the density of the solution to convert it to moles per liter. The density is determined using the specific gravity (S.G.) provided.
concentration of HCl in the commercial acid = (37% / 100) * (1.18 g/mL * 1000 mL/L) / (36.4611 g/mol)
Next, we can use the formula M1V1 = M2V2 to calculate the volume of the commercial hydrochloric acid needed:
V1 = (M2V2) / M1
V1 = (0.5 mol/L * 1.5 L) / concentration of HCl in the commercial acid
Therefore, the volume of the commercial hydrochloric acid required is equal to the result obtained from the calculation.
Q3. The rate of decomposition of NO2 can be related to the rate of formation of O2 by examining the balanced chemical equation for the reaction involved. The balanced chemical equation for the decomposition of NO2 is:
2NO2(g) → 2NO(g) + O2(g)
From this equation, we can see that for every 2 moles of NO2 decomposed, 1 mole of O2 is formed. This means that the rate of decomposition of NO2 is directly related to the rate of formation of O2.
In other words, if the rate of decomposition of NO2 is known, we can use the stoichiometry of the reaction to determine the rate of formation of O2. For example, if the rate of decomposition of NO2 is measured to be 2 mol/s, then the rate of formation of O2 would be 1 mol/s.
However, it's important to note that the rate of a reaction is determined experimentally and can be influenced by various factors such as temperature, concentration, and catalysts. Therefore, the specific rate of decomposition of NO2 and the rate of formation of O2 would need to be determined through experimentation.