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ludmilkaskok [199]
3 years ago
6

According to the following reaction, how many moles of hydrobromic acid are necessary to form 0.274 moles bromine?

Chemistry
1 answer:
Naily [24]3 years ago
8 0

Answer:

0.548 moles of HBr are required

Explanation:

Given data:

Number of moles of  hydrobromic acid = ?

Moles of bromine formed = 0.274 mol

Solution:

Chemical equation:

2HBr     →   H₂  + Br₂

Now we will compare the moles of HBr with Br₂.

               Br₂           :           HBr

               1               :            2

               0.274       :           2×0.274=0.548

Thus, 0.548 moles of HBr are required.

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3 years ago
A 55.0-mg sample of al(oh)3 is reacted with 0.200m hcl. how many milliters of the acid are needed to neutralize the al(oh)3?
chubhunter [2.5K]
Al(OH)₃ + 3HCl = AlCl₃ + 3H₂O

C=0.200 mmol/mL
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m{Al(OH)₃}=55.0 mg

n{Al(OH)₃}=m{Al(OH)₃}/M{Al(OH)₃}

3n{Al(OH)₃}=n(HCl)=CV

3m{Al(OH)₃}/M{Al(OH)₃}=CV

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V=3*55.0/[0.200*78.0]=10.6 mL
8 0
4 years ago
Boron has an atomic mass of 10.8 amu. It is known that naturally occurring boron is composed of two isotopes, boron-10 and boron
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B-10 = 19.9%
B-11 = 80.1%

Abundance of B10 = x
Abundance of B11= y
You know x+y = 1 because there are only the 2 isotopes.
Y= 1-x

10.01294x + 11.00931 (1-x) = 10.811
10.01294x + 11.00931 - 11.00931x = 10.811 - 0.99016x = -0.198
X = 0.200

Check:
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20% B10. 80% B11
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