Answer:
pH change is -0.07
Explanation:
Using H-H equation for acetic acid:
pH = pKa + log [Acetate salt] / [Acetic acid]
Replacing:
pH = 4.74 + log[1.188M] / [1.188M]
pH = 4.74
The HCl reacts with sodium acetate producing acetic acid, thus:
HCl + CH₃COONa → CH₃COOH + NaCl
That means the final moles of sodium acetate are initial moles - moles of HCl and moles of acetic acid are initial moles + moles of HCl.
As the volume of the buffer is 1.0L, initial moles of both substances are 1.188moles. After reaction, the moles are:
sodium acetate: 1.188mol - 0.1mol = 1.088mol
Acetic acid: 1.188mol + 0.1mol = 1.288mol
Using again H-H equation:
pH = 4.74 + log[1.088M] / [1.288M]
pH = 4.67
pH change is: 4.67 - 4.74 = -0.07
Answer:
Hydrogen and Oxygen
Explanation:
also it can be written as H2O
The correct answer would be <span>5.43 x 10^-4</span>
Answer:
C2H4Cl2
Explanation:
Firstly, we know that the compound contains only three elements. These are carbon, hydrogen and oxygen. We have the percentage compositions of carbon and hydrogen, thus we need the one for chlorine. To get the one for chlorine, we simply subtract that of carbon and hydrogen from a total of 100%.
Hence percentage composition of chlorine = 100 - 24.27 - 4.07 = 71.66%
Now, we divide the percentage compositions by the atomic masses. The atomic masses of carbon, hydrogen and chlorine are 12, 35.5 and 1 respectively. We go on to the divisions as follows.
C = 24.27/12 = 2.0225
H = 4.07/1 = 4.07
Cl = 71.66/35.5 = 2.02
We then go on to divide each by the smallest which is 2.02
C = 2.0225/2.02 = 1
H = 4.07/2.02 = 2
Cl = 2.02/2.02 = 1
Hence the empirical formula is CH2Cl
Now, since the molecular mass is 98.95, we need to calculate the molecular formula
Hence, [CH2Cl]n = 98.95
[12 + 2(1) + 35.5]n = 98.95
[12 + 2 + 35.5]n = 98.95
49.5n = 98.95
n = 98.95/49.5 = 2
The molecular formula is thus [CH2Cl]2 = C2H4Cl2
<span>To find: Sample error in percent
Solution:
Formula:
((Experimental value-theoretical value)/theoretical value)*100
where, theoretical value = 100°c
and, experimental value = 102°c (sample 1) 99.2°c (sample 2)
Sample error (in percentage) when boiling level was 102°c = ((102°c-100°c)/100°c)*100 = 2%
Sample error (in percentage) when boiling level was 99.2°c = ((99.2°c-100°c)/100°c)*100 = -0.8%</span>
