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MArishka [77]
3 years ago
14

How many grams of water are made from the reaction of 4.0 grams of hydrogen gas with the given reaction 2H2+O2--->2H2O

Chemistry
1 answer:
Sergeeva-Olga [200]3 years ago
4 0

Answer:

The mass of water is 36 g.

Explanation:

Mass of hydrogen = 4 g

Mass of water = ?

Solution:

First of all we will write the balance chemical equation:

2H₂ + O₂  →  2H₂O

Number of moles of hydrogen = mass / molar mass

Number of moles of hydrogen = 4 g/ 2 g/mol

Number of moles of hydrogen = 2 mol

Now we compare the moles of water with hydrogen from balance chemical equation.

                               H₂     :     H₂O

                                2      :       2

Mass of water = moles × molar mass

Mass of water =  2 mol × 18 g/mol

Mass of water =  36 g

If the water oxygen is in excess than mass of water would be 36 g.

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The boiling point of HF is higher than the boiling point of H_2, and it is higher than the boiling point of F_2.

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Only London Forces are formed - Therefore more energy is required to break the intermolecular forces in HF than in the other hydrogen halides and so HF has a higher boiling point.

H_2 and F_2 will only have intra-molecular attractions and there will be no hydrogen bonds present in them. As a result, their boiling point will be lower.

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1 year ago
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How many atoms in the pictured molecule can form hydrogen bonds with water molecules anatomy?
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0.10 M potassium chromate is slowly added to a solution containing 0.20 M AgNO3 and 0.20 M Ba(NO3)2. What is the Ag+ concentrati
erastova [34]

Answer:

[Ag^{+}]=4.2\times 10^{-2}M

Explanation:

Given:

[AgNO3] = 0.20 M

Ba(NO3)2 = 0.20 M

[K2CrO4] = 0.10 M

Ksp of Ag2CrO4 = 1.1 x 10^-12

Ksp of BaCrO4 = 1.1 x 10^-10

BaCrO_4 (s)\leftrightharpoons  Ba^{2+}(aq)\;+\;CrO_{4}^{2-}(aq)

Ksp=[Ba^{2+}][CrO_{4}^{2-}]

1.2\times 10^{-10}=(0.20)[CrO_{4}^{2-}]

[CrO_{4}^{2-}]=\frac{1.2\times 10^{-10}}{(0.20)}= 6.0\times 10^{-10}

Now,

Ag_{2}CrO_4(s) \leftrightharpoons  2Ag^{+}(aq)\;+\;CrO_{4}^{2-}(aq)

Ksp=[Ag^{+}]^{2}[CrO_{4}^{2-}]

1.1\times 10^{-12}=[Ag^{+}]^{2}](6.0\times 10^{-10})

[Ag^{+}]^{2}]=\frac{1.1\times 10^{-12}}{(6.0\times 10^{-10})}= 1.8\times 10^{-3}

[Ag^{+}]=\sqrt{1.8\times 10^{-3}}=4.2\times 10^{-2}M

So, BaCrO4 will start precipitating when [Ag+] is 4.2 x 1.2^-2 M

                       

7 0
3 years ago
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