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tester [92]
3 years ago
13

17. What is the molarity of ZnCl2 that forms when 15 grams of Zn reacts completely with CuCl, if a

Chemistry
1 answer:
GalinKa [24]3 years ago
7 0

Answer:

1.31M

Explanation:

Based on the reaction, 1 mole of Zn produce 1 mole of ZnCl₂. As the reaction occurs completely, the moles of Zn added = Moles of ZnCl₂ produced. To find molarity we need the moles of ZnCl₂ and the volume of the solution in liters:

<em>Moles Zn = Moles ZnCl₂ -Molar mass Zn: 65.38g/mol-:</em>

15g * (1mol / 65.38g) = 0.23 moles of ZnCl₂

<em>Volume in Liters:</em>

175mL * (1L / 1000mL) = 0.175L

The molarity is:

0.23moles / 0.175L

<h3>1.31M</h3>
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Answer:

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Explanation:

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11) The bombardment of U-235 with neutrons leads to the reaction;

U\frac{235}{92}  + n\frac{1}{0}---> I\frac{138}{53}  + Y\frac{95}{39} +3n \frac{1}{0}

Hence

a = 92, b= 95, c= 53

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3 years ago
What is the ph of a 0.01 m solution of the strong acid hno3 in water?
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A 25.0-mL sample of 0.150 M hydrocyanic acid is titrated with a 0.150 M NaOH solution. The Ka of hydrocyanic acid is 4.9 × 10-10
lara [203]

Answer:

The pOH = 1.83

Explanation:

Step 1: Data given

volume of the sample = 25.0 mL

Molarity of hydrocyanic acid = 0.150 M

Molarity of NaOH = 0.150 M

Ka of hydrocyanic acid = 4.9 * 10^-10

Step 2: The balanced equation

HCN + NaOH → NaCN + H2O

Step 3: Calculate the number of moles hydrocyanic acid (HCN)

Moles HCN = molarity * volume

Moles HCN = 0.150 M * 0.0250 L

Moles HCN = 0.00375 moles

Step 3: Calculate moles NaOH

Moles NaOH = 0.150 M * 0.0305 L

Moles NaOH = 0.004575 moles

Step 4: Calculate the limiting reactant

0.00375 moles HCN will react with 0.004575 moles NaOH

HCN is the limiting reactant. It will completely be reacted. There will react 0.00375 moles NaOH. There will remain 0.004575 - 0.00375 = 0.000825 moles NaOH

Step 5: Calculate molarity of NaOH

Molarity NaOH = moles NaOH / volume

Molarity NaOH = 0.000825 moles / 0.0555 L

Molarity NaOH = 0.0149 M

Step 6: Calculate pOH

pOH = -log [OH-]

pOH = -log (0.0149)

pOH = 1.83

The pOH = 1.83

6 0
3 years ago
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