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masha68 [24]
3 years ago
7

HOW FAST CAN YOU ANSWER

Chemistry
1 answer:
Alex777 [14]3 years ago
5 0

Answer:

option "B" is correct (substance 2)

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In the following reaction, what is the quantity of heat (in kJ) released when 5.87 moles of CH₄ are burned?
IRISSAK [1]

Taking into account the definition of enthalpy of a chemical reaction, the quantity of heat released when 5.87 moles of CH₄ are burned is 4,707.74 kJ.

The enthalpy of a chemical reaction as the heat absorbed or released in a chemical reaction when it occurs at constant pressure. That is, the heat of reaction is the energy that is released or absorbed when chemicals are transformed into a chemical reaction.

The enthalpy is an extensive property, that is, it depends on the amount of matter present.

In this case, the balanced reaction is:

CH₄ (g) + 2 O₂ (g) → CO₂ (g) + 2 H₂O(g)

and the enthalpy reaction ∆H° has a value of -802 kJ/mol.

This equation indicates that when 1 mole of CH₄ reacts with 2 moles of O2, 802 kJ of heat is released.

When 5.87 moles of CH₄ are burned, then you can apply the following rule of three: if 1 mole of CH₄ releases 802 kJ of heat, 5.87 moles of CH₄ releases how much heat?

heat=\frac{5.87 molesof CH_{4}x802 kJ}{1 mol of CH_{4} }

<u><em>heat= 4,707.74 kJ</em></u>

Finally, the quantity of heat released when 5.87 moles of CH₄ are burned is 4,707.74 kJ.

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5 0
3 years ago
A _________________ is the path of energy transfer from producer to consumers.
Verizon [17]

Answer: Food chain

Explanation: none

3 0
3 years ago
Can someone please help it is due tonight!
kompoz [17]
Reactants left products right
4 0
3 years ago
Read 2 more answers
What is the percentage composition when 10g of magnesium combines with 4g of nitrogen?
Ivanshal [37]
The %  composition   when 10g of magnesium combine   with  4g of   nitrogen  is  71.43%   magnesium   and  28.57 %  nitrogen

               calculation

%
  composition  =  mass  of an element  / total mass  x100
mass  of magnesium = 10 g
mass of nitrogen  =  4g

calculate  the  total  mass  used

=  10g of  Magnesium  + 4 g of  nitrogen = 14 grams

%   composition for  magnesium  is therefore  =  10/14  x100 = 71.43 %

%  
composition  for  nitrogen  is therefore = 4 /14  x100  =   28.57 %
7 0
3 years ago
The first step in industrial nitric acid production is the catalyzed oxidation of ammonia. Without a catalyst, a different react
murzikaleks [220]

<u>Answer:</u> The value of K_{eq} is 4.84\times 10^{-5}

<u>Explanation:</u>

We are given:

Initial moles of ammonia = 0.0280 moles

Initial moles of oxygen gas = 0.0120 moles

Volume of the container = 1.00 L

Concentration of a substance is calculated by:

\text{Concentration}=\frac{\text{Number of moles}}{\text{Volume}}

So, concentration of ammonia = \frac{0.0280}{1.00}=0.0280M

Concentration of oxygen gas = \frac{0.0120}{1.00}=0.0120M

The given chemical equation follows:

                  4NH_3(g)+3O_2(g)\rightleftharpoons 2N_2(g)+6H_2O(g)

<u>Initial:</u>        0.0280        0.0120

<u>At eqllm:</u>    0.0280-4x   0.0120-3x   2x       6x

We are given:

Equilibrium concentration of nitrogen gas = 3.00\times 10^{-3}M=0.003

Evaluating the value of 'x', we get:

\Rightarrow 2x=0.003\\\\\Rightarrow x=0.0015M

Now, equilibrium concentration of ammonia = 0.0280-4x=[0.0280-(4\times 0.0015)]=0.022M

Equilibrium concentration of oxygen gas = 0.0120-3x=[0.0120-(3\times 0.0015)]=0.0075M

Equilibrium concentration of water = 6x=(6\times 0.0015)]=0.009M

The expression of K_{eq} for the above reaction follows:

K_{eq}=\frac{[H_2O]^6\times [N_2]^2}{[NH_3]^4\times [O_2]^3}

Putting values in above expression, we get:

K_{eq}=\frac{(0.009)^6\times (0.003)^2}{(0.022)^4\times (0.0075)^3}\\\\K_{eq}=4.84\times 10^{-5}

Hence, the value of K_{eq} is 4.84\times 10^{-5}

4 0
3 years ago
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