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amid [387]
2 years ago
13

Cuando se quema 1 mol de metano –o sea, 16 g–, se desprenden 802

Chemistry
1 answer:
Anvisha [2.4K]2 years ago
3 0

Answer:

1 gramo de metano aporta 50.125 kilojoules.

1 gramo de metano aporta 48.246 kilojoules.

Explanation:

La cantidad de energía liberada por la combustión de una unidad de masa del hidrocarburo (Q), en kilojoules por mol, es igual a la cantidad de energía liberada por mol de compuesto (\bar {Q}), en kilojoules por mol, dividido por su masa molar (M), en gramos por mol:

Q = \frac{\bar Q}{M} (1)

A continuación, analizamos cada caso:

Metano

Q = \frac{802\,\frac{kJ}{mol} }{16\,\frac{g}{mol} }

Q = 50.125\,\frac{kJ}{g}

1 gramo de metano aporta 50.125 kilojoules.

Octano

Q = \frac{5500\,\frac{kJ}{mol} }{114\,\frac{g}{mol} }

Q = 48.246\,\frac{kJ}{mol}

1 gramo de metano aporta 48.246 kilojoules.

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If 507 g FeCl2 were used up in the reaction FeCl2 + 2NaOH Fe(OH)2(s) + 2NaCl, how many grams of NaCl would be made?
Ket [755]
Number of moles of FeCl2 used = mass/ molar mass 
Number of moles = 507/126.751 = 4.
If one mole of Fe reacts with two moles of sodium 
Then 4 moles of Fe produces 8 moles of sodium.
Number of moles of sodium = mass/molar mass
Molar mass of sodium chloride = 23 +35.5 = 58.5 g/mol
Hence mass = 8 * 58.5 = 468 g. Hence Option A.

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3 years ago
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Answer:

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Explanation:

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3 years ago
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can anyone explain to me how to balance a chemical equation? im having trouble in understanding how i keep getting it wrong.
Virty [35]
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People take antacids, such as milk of magnesia, to reduce the discomfort of acid stomach or heartburn. The recommended dose of m
Llana [10]

Answer:

V_{HCl}=0.208L=208mL

Explanation:

Hello,

In this case, since the chemical reaction is:

2HCl+Mg(OH)_2\rightarrow MgCl_2+2H_2O

We can see that hydrochloric acid and magnesium hydroxide are in a 2:1 mole ratio, which means that the neutralization point, we can write:

n_{HCl}=2*n_{Mg(OH)_2}

In such a way, the moles of magnesium hydroxide (molar mass 58.3 g/mol) in 500 mg are:

n_{Mg(OH)_2}=500mg*\frac{1g}{1000mg}*\frac{1mol}{58.3g}  =0.00858mol

Next, since the pH of hydrochloric acid is 1.25, the concentration of H⁺ as well as the acid (strong acid) is:

[H^+]=[HCl]=10^{-pH}=10^{-1.25}=0.0562M

Then, since the concentration and the volume define the moles, we can write:

[HCl]*V_{HCl}=2*n_{Mg(OH)_2}

Therefore, the neutralized volume turns out:

V_{HCl}=\frac{2*0.00858mol}{0.0562\frac{mol}{L} }\\ \\V_{HCl}=0.208L=208mL

Best regards.

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Freezing point

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