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hammer [34]
3 years ago
6

KHP is a monoprotic acid which provides one H+ ion. How would your results be affected if a diprotic acid (such as sulfuric acid

, H2SO4) was used which provides two H+ ions? Calculate the molarity of the NaOH solution if 17.50 mL of N a O H was titrated against 10.50 mL of 0.75 M H2SO4.
Chemistry
1 answer:
kari74 [83]3 years ago
7 0

Explanation:

A.

In a diprotic acid, 2 moles of H+ ions is released. Therefore, number of moles of H+ in a diprotic acid = 2 × number of moles of H+ of monoprotic acid.

B.

Equation of the reaction

2NaOH + H2SO4 --> Na2SO4 + 2H2O

Number of moles of H2SO4 = molar concentration × volume

= 0.75 × 0.0105

= 0.007875 moles.

By stoichiometry, since 1 mole of H2SO4 reacts with 2 moles of NaOH. Therefore, number of moles of NaOH = 2 × 0.007875

= 0.01575 moles.

Molar concentration of NaOH = number of moles ÷ volume

= 0.01575 ÷ 0.0175

= 0.9 M of NaOH.

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What is an acid<br>Define the term reactivity action<br>What is a compound​
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How many grams of potassium (K) contain 5.11 x 10^22 atoms of potassium?
Darina [25.2K]

The atomic mass of K is 39

from Avogadro's law

39g of K contains 6.02x10^23 atoms

therefore if

39=6.02x19^23

X=5.11×10^22

making X the subject of the formula

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2 years ago
Ocean water contains 3.5% NaCl by mass. How many grams of sodium chloride can be obtained from 750 g of seawater
Masteriza [31]

26.25g

Explanation:

Given parameters:

Percentage of NaCl in ocean water = 3.5%

Mass of given seawater = 750g

Unknown:

Mass of NaCl in the seawater = ?

Solution:

  This is a simple percentage problem:

We have been given that ocean water contains 3.5% of NaCl;

 Therefore;

Mass of NaCl in seawater = Percentage of NaCl in ocean water x Mass of seawater

Mass of NaCl in seawater = \frac{3.5}{100} x 750g = 26.25g

Learn more:

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3 0
3 years ago
6 NaOH + 2 Al ???? 2 Na3AlO3 + 3 H2 How much aluminum is required to produce 17.5 grams of hydrogen? How many moles of NaOH are
ANTONII [103]

Answer:

a) 157.5 grams of aluminum.

b) 1 mol

c) 9 g

Explanation:

The reaction is :

6 NaOH + 2Al ---> 2 Na_{3}AlO_{3} + 3H_{2}

As per balanced equation

a) 3 moles of hydrogen will be produced from two moles of aluminium.

The atomic mass of aluminium = 27

therefore

3X2 grams of hydrogen is produced from 2 X 27 grams of Al

1 gram of hydrogen will be produced from \frac{2X27}{3X2}= 9g

therefore 17.5 will be produced from = 9X 17.5 = 157.5 grams of aluminum.

b) as per balanced equation three moles or six gram of hydrogen is produced from 6 moles of NaOH.

Therefore 1 g of hydrogen will be produced from =\frac{6}{6}

or 1 gram will be prepared from = 1 mole

c) from balanced equation three moles are produced from two moles of Al (27X2 = 54 g).

thus from 54  grams gives 6 grams of hydrogen

1 grams will give = \frac{54X1}{6}= 9 g

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3 years ago
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