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Nimfa-mama [501]
3 years ago
6

The reaction of sulfuric acid (H2SO4) with potassium hydroxide (KOH) is described by the equation below. Suppose 0.05 L of KOH w

ith unknown concentration is placed in a flask with bromthymol blue indicator. A solution of 0.10 M H2SO4 is dripped into the KOH solution. After exactly 0.033 L of H2SO4 is added, the indicator changes from blue to yellow. What is the concentration of the KOH.
H2SO4+2KOH--->K2SO4+2H2O
Chemistry
2 answers:
Dmitry_Shevchenko [17]3 years ago
7 0

chem eqn is H2SO4+2KOH--->K2SO4+2H2O

1x H2SO4 reacts with 2x KOH

0.033L of 0.10M H2SO4 has 0.033 x 0.10 = 0.0033 moles

2x 0.0033 = 0.0066 moles of KOH

conc of KOH = moles/vol = 0.0066/0.05

=0.132M

strojnjashka [21]3 years ago
3 0

From the chemical equation given:

H2SO4+2KOH--->K2SO4+2H2O

the two reactants, H2SO4 and KOH, are in 1:2 stoichiometric ratio.

No. of moles of KOH = 2* no. of moles of H2SO4

=2*0.1*0.033

The concentration of KOH = no. of moles / volume

=2*0.1*0.033/0.05

=0.132M

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Copper metal (Cu) reacts with silver nitrate (AgNO3) in aqueous solution to form Ag and Cu(NO3)2. An excess of AgNO3 is present.
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31.75 g Cu | 1 mol Cu | 2 mol Ag | 107.9 g Ag         6851.65
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There's also a shorter way to do this: Notice the molar ratio from Cu to Ag, which is 1:2. When you plug in 31.75 into your molar mass for Cu, it equals 1/2 mol. That also means that you have 1 mol Ag because of the ratio, qhich you can then plug into your molar mass, getting 107.9 as well.
3 0
3 years ago
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Can anybody can solve this PLEASE
strojnjashka [21]

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Explanation:

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8 0
3 years ago
If 45 mL of water are added to 250 mL of a 0.75 M K2SO4 solution, what will the molarity of the diluted solution be?
krok68 [10]

Answer:

\large\boxed{\large\boxed{0.64M}}

Explanation:

When you form a <em>diluted solution</em> from a mother (concentrated) solution, the moles of solute are determined by the mother solution.

The main equation is:

Molarity=\dfrac{\text{moles of solute}}{\text{volume of the solution in liters}}

Then, since the moles of solute is the same for both the mother solution and the diluted solution:

          \text{Molarity mother solution }\times\text{ volume mother solution}=\\\\=\text{Molarity diluted solution }\times\text{ volume diluted solution}

Substitute and solve for the molarity of the diluted solution:

           250mL\times 0.75M=(45mL+250mL)\times M\\\\\\M=\dfrac{250mL\times 0.75M}{295mL}=0.64M

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