Answer:
Explanation:
Hello,
In this case, since the neutralization reaction between HCl and NaOH is:
We notice a 1:1 molar ratio, for that reason, at the equivalence point we find:
Thus in terms in molarities one could compute the concentration of HCl in the old bottle for the used NaOH for the neutralization as:
This value is lower than 37% HCl that in molarity is about 12 M, such difference is due to its high volatility.
Best regards.
Answer is 67.8%
To find the percent yield the equation is actual yield over theoretical yield, and multiply by 100. In this problem it is 4.06996 / 6 = 0.678 x 100 = 67.8%
To solve this question, we will use Graham's law which states that:
(R1 / R2) ^ 2 = M2 / M1 where
R1 and R2 are the rates of effusion and M1 and M2 are the molar masses of the two gases.
From the periodic table, we can calculate the molar mass of O2 as follows:
molar mass of O2 = 2*16 = 32 grams
Therefore we have:
R1 / R2 = Ry / RO2 = 1/2
M1 is My we want to get
M2 is molar mass of O2 = 32 grams
Substitute in the above equation to get the molar mass of y as follows:
(1/2) ^2 = (32/My)
1/4 = 32/My
My = 32*4 = 128
Therefore, molar mass of gas y = 128 grams
To prepare 350 mL of 0.100 M solution from a 1.50 M
solution, we simply have to use the formula:
M1 V1 = M2 V2
So from the formula, we will know how much volume of the
1.50 M we actually need.
1.50 M * V1 = 0.100 M * 350 mL
V1 = 23.33 mL
So we need 23.33 mL of the 1.50 M solution. We dilute it
with water to a volume of 350 mL. So water needed is:
350 mL – 23.33 mL = 326.67 mL water
Steps:
1. Take 23.33 mL of 1.50 M solution
<span>2. Add 326.67 mL of water to make 350 mL of 0.100 M
solution</span>
