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valentina_108 [34]

3 years ago

7

How many mol of water (H2O) are produced from 74.4 grams of ammonium sulfate, (NH)2SO,?

1 answer:

otez555 [7]3 years ago

8 0

Answer:

yeah

Explanation:

How many mol of water (H2O) are produced from 74.4 grams of ammonium sulfate, (NH)2SO,?

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El concepto de orbital es equivalente al concepto de órbita propuesto por bohr

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<em>réponse</em>: <em>verdadero</em>

<em>Explication</em>:

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4 years ago

If the diet is deficient in one or more of the _____amino acids, the body must break down existing proteins to provide the amino

AysviL [449]

Essential amino acids
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3 years ago

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A student heats a LaTeX: 2.5 g sample of calcium carbonate in a test tube. What mass of calcium oxide remains in the test tube w

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Answer:

2.5

Explanation:

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3 years ago

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H2(g) + F2(g)2HF(g) Using standard thermodynamic data at 298K, calculate the entropy change for the surroundings when 2.20 moles

abruzzese [7]

<u>Answer:</u> The value of $\Delta S^o$ for the surrounding when given amount of hydrogen gas is reacted is -31.02 J/K

<u>Explanation:</u>

Entropy change is defined as the difference in entropy of all the product and the reactants each multiplied with their respective number of moles.

The equation used to calculate entropy change is of a reaction is:

$\Delta S^o_{rxn}=\sum [n\times \Delta S^o_{(product)}]-\sum [n\times \Delta S^o_{(reactant)}]$

For the given chemical reaction:

$H_2(g)+F_2(g)\rightarrow 2HF(g)$

The equation for the entropy change of the above reaction is:

$\Delta S^o_{rxn}=[(2\times \Delta S^o_{(HF(g))})]-[(1\times \Delta S^o_{(H_2(g))})+(1\times \Delta S^o_{(F_2(g))})]$

We are given:

$\Delta S^o_{(HF(g))}=173.78J/K.mol\\\Delta S^o_{(H_2)}=130.68J/K.mol\\\Delta S^o_{(F_2)}=202.78J/K.mol$

Putting values in above equation, we get:

$\Delta S^o_{rxn}=[(2\times (173.78))]-[(1\times (130.68))+(1\times (202.78))]\\\\\Delta S^o_{rxn}=14.1J/K$

Entropy change of the surrounding = - (Entropy change of the system) = -(14.1) J/K = -14.1 J/K

We are given:

Moles of hydrogen gas reacted = 2.20 moles

By Stoichiometry of the reaction:

When 1 mole of hydrogen gas is reacted, the entropy change of the surrounding will be -14.1 J/K

So, when 2.20 moles of hydrogen gas is reacted, the entropy change of the surrounding will be = $\frac{-14.1}{1}\times 2.20=-31.02J/K$

Hence, the value of $\Delta S^o$ for the surrounding when given amount of hydrogen gas is reacted is -31.02 J/K

7 0

3 years ago

Explain how chemists overcome the problem of low yield In industry

mrs_skeptik [129]

High temperature and pressure produce the highest rate of reaction. However, this must be balanced with the high cost of the energy needed to maintain these conditions. Catalysts increase the rate of reaction without affecting the yield. This can help create processes which work well even at lower temperatures.

I hope this helps you.

8 0

4 years ago