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Vlad1618 [11]
2 years ago
14

Find the mass of 4.5 moles of H3PO4

Chemistry
1 answer:
Contact [7]2 years ago
6 0

Hey there!

H₃PO₄

Find molar mass.

H: 3 x 1.008 = 3.024

P: 1 x 30.97 = 30.97

O: 4 x 16 = 64

---------------------------------

                  97.994 grams

The mass of 1 mole of H₃PO₄ is 97.994 grams.

We have 4.5 moles.

97.994 x 4.5 = 440

The mass of 4.5 moles of H₃PO₄ is 440 grams.

Hope this helps!

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Explanation:

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5 0
2 years ago
What can you infer about the volume of an object imagined with a greater mass?
pashok25 [27]
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3 0
3 years ago
Nitrogen and hydrogen combine at a high temperature, in the presence of a catalyst, to produce ammonia. N2(g)+3H2(g)⟶2NH3(g) Ass
yaroslaw [1]

Answer:

moles of ammonia produced = 0.28 moles

Explanation:

The reaction is

N_{2}(g)+3H_{2}(g) --> 2NH_{3}(g)

As per equation, one mole of nitrogen will react with three moles of hydrogen to give two moles of ammonia

So 0.140 moles of nitrogen will react with = 3 X 0.140 moles of Hydrogen

             = 0.42 moles of hydrogen molecule.

this will give 2 X 0.140 moles of ammonia = 0.28 moles of ammonia

the moles of ammonia produced = 0.28 moles

Here the nitrogen is limiting reagent.

4 0
3 years ago
Calculate the entropy change for the surroundings of the reaction below at 350K: N2(g) + 3H2(g) -&gt; 2NH3(g) Entropy data: NH3
krek1111 [17]

Answer : The entropy change for the surroundings of the reaction is, -198.3 J/K

Explanation :

We have to calculate the entropy change of reaction (\Delta S^o).

\Delta S^o=S_{product}-S_{reactant}

\Delta S^o=[n_{NH_3}\times \Delta S^0_{(NH_3)}]-[n_{N_2}\times \Delta S^0_{(N_2)}+n_{H_2}\times \Delta S^0_{(H_2)}]

where,

\Delta S^o = entropy of reaction = ?

n = number of moles

\Delta S^0{(NH_3)} = standard entropy of NH_3

\Delta S^0{(H_2)} = standard entropy of H_2

\Delta S^0{(N_2)} = standard entropy of N_2

Now put all the given values in this expression, we get:

\Delta S^o=[2mole\times (192.5J/K.mole)]-[1mole\times (191.5J/K.mole)+3mole\times (130.6J/K.mole)]

\Delta S^o=-198.3J/K

Therefore, the entropy change for the surroundings of the reaction is, -198.3 J/K

4 0
2 years ago
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Is there choices to this question? cant answer it without choices

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