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Brrunno [24]
2 years ago
14

I need answers to question 1,2,3

Chemistry
1 answer:
sashaice [31]2 years ago
6 0

Answer:

1. 0.125 mole

2. 42.5 g

3. 0.61 mole

Explanation:

1. Determination of the number of mole of NaOH.

Mass of NaOH = 5 g

Molar mass of NaOH = 23 + 16 + 1

= 40 g/mol

Mole of NaOH =?

Mole = mass /molar mass

Mole of NaOH = 5/40

Mole NaOH = 0.125 mole

2. Determination of the mass of NH₃.

Mole of NH₃ = 2.5 moles

Molar mass of NH₃ = 14 + (3×1)

= 14 + 3

= 17 g/mol

Mass of NH₃ =?

Mass = mole × molar mass

Mass of NH₃ = 2.5 × 17

Mass of NH₃ = 42.5 g

3. Determination of the number of mole of Ca(NO₃)₂.

Mass of Ca(NO₃)₂ = 100 g

Molar mass of Ca(NO₃)₂ = 40 + 2[14 + (3×16)]

= 40 + 2[14 + 48]

= 40 + 2[62]

= 40 + 124

= 164 g/mol

Mole of Ca(NO₃)₂ =?

Mole = mass /molar mass

Mole of Ca(NO₃)₂ = 100 / 164

Mole of Ca(NO₃)₂ = 0.61 mole

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Alexus [3.1K]

Answer:

By a factor of 12

Explanation:

For the reaction;

A + 2B → products

The rate law is;

rate = k[A]²[B]

As you can see, the rate is proportional to the square of the concentration of  A  and the of the concentration of  B .

Let's say initially, [A] = x, [B] = y

The rate law in this case is equal to;

rate1 = k. x².y

Now you double the concentration of A and triple the concentration of B.

[A] = 2x, [B] = 3y

The new rate law is given as;

rate2 = k . (2x)². (3y)

rate2 = k . 4x² . 3y

rate2 = 12 k . x² . y

Comparing rate 2 and rate 1, the ratio is given as; rate 2/ rate 1 = 12

Therefore the rate has increased by a factor of 12.

5 0
2 years ago
How do I write 0.03438 m in scientific notation using 3 digits?
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2 years ago
If 15.0 mL of phosphoric acid completely neutralizes 38.5 mL of 0.150 mol/L calcium hydroxide, what is the concentration of the
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Answer:

Let me give it a try.

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Balancing this reaction

2H3PO4 + 3Ca(OH)2 == Ca3(PO4)2 + 6H2O.

Moles= Molarity x Volume

Volume = 38.5ml = 0.0385L

Moles of Ca hydroxide = 0.150m/L x 0.0385L

(Notice the units canceling out...leaving moles).

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From balanced reaction...

3moles of Ca(OH)2 completely reacts with 2moles of H3PO4

0.005775moles of Ca(OH)2 would completely react with....

= 0.005775 x 2/(3)

=0.00385moles of H3PO4.

Now we're looking for its Concentration in Mol/L

Molarity=Moles of solute/Volume of solution(in L)

Volume of solution assuming no other additions to the reaction = 15ml + 38.5ml =53.5ml =0.0535L

Molarity = 0.00385/0.0535

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If this is wrong

then Simply Try The formula for Mixing of solutions

C1V1 = C2V2

0.15 x 38.5 = C2 x (15+38.5)

C2 = 0.11M/L.

7 0
2 years ago
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