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timofeeve [1]
3 years ago
10

5.22*10^-3 = ?*10^-2

Chemistry
1 answer:
Andrew [12]3 years ago
6 0

x=0.522 is the answer

x times :10^-2=5.22 times 10^{-3}\

simplify 10^-3: 5.22/1000= x times 10^-2 = 5.22/1000

simplify 10^-2 1/100 = x 1/100=5.22/1000

Multiply both sides by 100 100x 1/100 = 5.22x100/1000

simplify 0.522 sorry I took so long I did the best i could to explain it

Hope this helps!

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Luba_88 [7]

Answer:

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Explanation:

6 0
3 years ago
What causes this nuclear reaction to occur
IrinaVladis [17]

Answer:

bombarding it with an energetic particle

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8 0
2 years ago
What mass of precipitate forms when 185.5 ml of 0.533 m naoh is added to 627 ml of a solution that contains 15.8 g of aluminum s
Law Incorporation [45]

Answer:

2,57 g of precipitate.

Explanation:

For the reaction:

6 NaOH + Al₂(SO₄)₃ → 2 Al(OH)₃ + 3 Na₂SO₄

The precipitate is Al(OH)₃.

185,5mL of 0,533M NaOH are:

0,1855L × 0,533M = <em>0,0989 moles NaOH</em>

Moles of Al₂(SO₄)₃ are:

15,8g × \frac{1mol}{342,15g} = <em>0,0462 moles Al₂(SO₄)₃</em>

For the total reaction of 0,0989 moles NaOH with Al₂(SO₄)₃ you need:

0,0989moles NaOH × \frac{1molAl_{2}(SO_{4})_{3}}{6 moles NaOH} = <em>0,0165 moles Al₂(SO₄)₃</em>

As you have <em>0,0462 moles Al₂(SO₄)₃ </em>the limiting reactant is NaOH.

0,0989 moles of NaOH produce:

0,0989moles NaOH × \frac{2molAl(OH)_{3}}{6 moles NaOH} = <em>0,0330 moles of Al(OH)₃</em>

These moles are:

0,0330 moles of Al(OH)₃ × (78 g/mol) = <em>2,57 g of Al(OH)₃ ≡ mass of precipitate</em>

<em></em>

I hope it helps!

<em> </em>

3 0
3 years ago
Help meeeeehahahshshs
natka813 [3]
What grade are you in I’m in 9th ;-;
7 0
3 years ago
How many grams of water are needed to dissolve 27.8 g of ammonium nitrate?
andreyandreev [35.5K]
The moles of ammonium nitrate needed to dissolve 0.35 moles
The moles of water that will react is 0.35 moles as due to ratio
so mass of water will be 0.35 x 18=6.3g 
                                 MASS OF WATER WILL BE 6.3 g
7 0
3 years ago
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