Answer:
The correct answer is 0.10.
Explanation:
Based on the given question, in a buffer solution of 1 liter, the molarity of acetic acid is 1.420 M and the molarity of sodium acetate is 0.67. The pKa value of acetic acid given is 4.74, now the pH of buffer is,
pH of buffer = pKa + log ([CH3COONa]/[CH3COOH])
= 4.74 + log (0.67/1.420)
= 4.74 + (-0.326)
= 4.41
Now 0.10100 mol of HCl is added, the HCl reacts with sodium acetate to give,
CH3COONa + HCl = CH3COOH + NaCl
Now the concentration of CH3COONa becomes = 0.67-0.101 = 0.57 M, and the new concentration of CH3COOH becomes = 1.420 + 0.101 = 1.52 M
Now the new pH will be,
= pKa + log (0.57/1.52)
= 4.74 + (-0.426)
= 4.31
The pH change is 4.41-4.31 = 0.10
Answer:
78.14 % B; 21.86 % H
Step-by-step explanation:
The formula of diborane is B₂H₆.
The masses of each element are:
2B = 2 × 10.81 = 21.62 u
6H = 6 × 1.008 = <u> 6.048 u
</u>
TOTAL = 27.67 u
% B = 21.62/27.67 × 100
= 78.14 %
% H = 6.048/ 27.67
= 21.86 %
To determine the empirical formula of the compound given, we need to determine the ratio of each element in the compound. To do that, we need to know the amount of carbon, hydrogen and oxygen atoms in the compound. We determine these from the given amounts of the products after burning.
Given:
2.5 g vanillin
5.79 g CO2
1.18 g H2O
Solution:
mol C = 5.79 g CO2 ( 1 mol CO2 / 44.01 g CO2) ( 1 mol C / 1 mol CO2) = 0.132 mol C
mol H = 1.18 g H2O ( 1 mol g H2O / 18.02 g H2O) ( 2 mol H / 1 mol H2O) = 0.131 mol H
We convert them into grams,
0.132 mol C (12.01 g / 1 mol) = 1.59 g
0.131 mol H (1.01 g / 1 mol) = 0.132 g
mass of C and H in vanillin = 1.722 g
mass of O = 2.5 - 1.722 = 0.778 g O
mol O = 0.778 g O ( 1 mol O / 16 g ) = 0.049 mol O
Dividing the number of moles of each element with the smallest value of number of mole, we will have the empirical formula:
moles ratio
C 0.132 / 0.049 2.69 x 3 = 8
H 0.131 / 0.049 2.67 x 3 = 8
O 0.049 / 0.049 1 x 3 = 3
The empirical formula would be C8H8O3.
Answer:
most of its mass is in the nucleus
Explanation:
the nucleus consists of protons and nutrons