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vovangra [49]
3 years ago
14

In the important industrial process for producing ammonia (the Haber Process), the overall reaction is: N2(g) + 3H2(g) → 2NH3(g)

+ 100.4 kJ A yield of NH3 of approximately 98% can be obtained at 200°C and 1,000 atmospheres of pressure. What is the ΔH in kJ of heat released per mole of NH3(g) formed?
A) 50.2 kJ
B) -50.2 kJ
C) 100.4 kJ
D) -100.4 kJ
Chemistry
1 answer:
irakobra [83]3 years ago
3 0
Answer is: <span>B) -50.2 kJ.

Balanced chemical reaction: </span>N₂(g) + 3H₂(g) → 2NH₃(g) ΔH = -<span>100.4 kJ.
This is exothermic reaction, because heat is released and energy is include as product of chemical reaction.
Make proportion, two moles of ammonia released 100.4 kJ of heat, then one mole of ammonia released:
2 mol(NH</span>₃) : (-100.4 kJ) = 1 mol : ΔH.
ΔH = -50.2 kJ; <span>heat released per mole of NH</span>₃.
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For the reaction, Cl2 + 2KBr --&gt; 2KCl + Br2, how many moles of potassium chloride, KCl, are produced from 102g of potassium b
Sedaia [141]

Solution :

From the balanced chemical equation, we can say that 1 moles of KBr will produce 1 moles of KCl .

Moles of KBr in 102 g of potassium bromide.

n = 102/119.002

n = 0.86 mole.

So, number of miles of KCl produced are also 0.86 mole.

Mass of KCl produced :

m = 0.86 \times Molar \  mass \ of \  KCl\\\\m = 0.86 \times 74.5 \ gram \\\\m = 64.07\  gram

Hence, this is the required solution.

5 0
3 years ago
Consider the reaction. 2 Pb ( s ) + O 2 ( g ) ⟶ 2 PbO ( s ) An excess of oxygen reacts with 451.4 g of lead, forming 367.5 g of
Art [367]

Answer : The percent yield of the reaction is, 75.6 %

Solution : Given,

Mass of Pb = 451.4 g

Molar mass of Pb = 207 g/mole

Molar mass of PbO = 223 g/mole

First we have to calculate the moles of Pb.

\text{ Moles of }Pb=\frac{\text{ Mass of }Pb}{\text{ Molar mass of }Pb}=\frac{451.4g}{207g/mole}=2.18moles

Now we have to calculate the moles of PbO

The balanced chemical reaction is,

2Pb(s)+O_2(g)\rightarrow 2PbO(s)

From the reaction, we conclude that

As, 2 mole of Pb react to give 2 mole of PbO

So, 2.18 mole of Pb react to give 2.18 mole of PbO

Now we have to calculate the mass of PbO

\text{ Mass of }PbO=\text{ Moles of }PbO\times \text{ Molar mass of }PbO

\text{ Mass of }PbO=(2.18moles)\times (223g/mole)=486.1g

Theoretical yield of PbO = 486.1 g

Experimental yield of PbO = 367.5 g

Now we have to calculate the percent yield of the reaction.

\% \text{ yield of the reaction}=\frac{\text{ Experimental yield of }PbO}{\text{ Theoretical yield of }PbO}\times 100

\% \text{ yield of the reaction}=\frac{367.5g}{486.1g}\times 100=75.6\%

Therefore, the percent yield of the reaction is, 75.6 %

3 0
3 years ago
What limitation is placed on electrons in the bohr model of the atom?
Olegator [25]
I thinking the limitation is that a shifting electron will always move from a more excited states to a less excited state. Electrons could not circle the nucleus because they would lose energy by emitting electromagnetic radiation and spiral into the nucleus. In addition Bohr was not able to explain electrons orbits of large atom w/many electrons.
4 0
3 years ago
A helium filled ballon had a volume of 8.50 L on the ground at 20.0 C and a pressure of 750.0 Torr. After the ballon was release
Marrrta [24]

Answer:

V_2=12.1L

Explanation:

Hello!

In this case, according to the given data of volume, pressure and temperature, it is possible to infer this problem can be solved via the combined gas law:

\frac{P_1V_1}{T_1} =\frac{P_2V_2}{T_2}

Thus, regarding the question, we evidence we need V2, but first we make sure the temperatures are in Kelvins:

T_1=20+273=293K\\\\T_2=-40+273=233K

Then, we obtain:

V_2=\frac{P_1V_1T_2}{T_1P_2}\\\\V_2=\frac{0.987atm*8.50L*233K}{293K*0.550atm}\\\\V_2=12.1L

Best regards!

5 0
3 years ago
Solve the following mole-mass problem. Refer to the periodic table. How much mass does 0.50 mole of copper contain?
andre [41]
It is 31.75g of copper is contained in .50 mole of copper.
7 0
3 years ago
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