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docker41 [41]
2 years ago
9

Consider the titration of sulfuric acid with sodium hydroxide. What volume (mL) of a 2.658M NaOH solution is required to fully t

itrate 10.00mL of a 0.426M H2SO4 solution
Chemistry
1 answer:
KatRina [158]2 years ago
8 0

The volume of base required to completely neutralize the acid is 3.2 mL of NaOH.

The equation of the reaction is;

2NaOH(aq) + H2SO4(aq) -----> Na2SO4(aq) + 2H2O(l)

From the question;

Concentration of acid CA = 0.426M

Concentration of base  CB = 2.658M

Volume of acid VA = 10.00mL

Volume of base VB = ?

Number of moles of acid NA = 1

Number of moles of base NB = 2

Using the relation;

CAVA/CBVB = NA/NB

CAVANB = CBVBNA

VB = CAVANB/CBNA

VB =  0.426M × 10.00mL × 2/ 2.658M × 1

VB = 3.2 mL

Learn more: brainly.com/question/6111443

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The relative atomic mass of Chlorine is 35.45. Calculate the percentage abundance of the two isotopes of Chlorine, 35Cl and 37Cl
nata0808 [166]

Answer:

35Cl = 75.9 %

37Cl = 24.1 %

Explanation:

Step 1: Data given

The relative atomic mass of Chlorine = 35.45 amu

Mass of the isotopes:

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37Cl = 36.96590258 amu

Step 2: Calculate percentage abundance

35.45 = x*34.96885269 + y*36.96590258

x+y = 1  x = 1-y

35.45 = (1-y)*34.96885269 + y*36.96590258

35.45 = 34.96885269 - 34.96885269y +36.96590258y

0.48114731 = 1,99704989‬y

y = 0.241 = 24.1 %

35Cl = 34.96885269 amu = 75.9 %

37Cl = 36.96590258 amu = 24.1 %

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