Ask question

Ask question

All categories

3 years ago

8

Consider the reaction below. How much heat is absorbed if 5.00 moles of nitrogen react

1 answer:

bekas [8.4K]3 years ago

3 0

Explanation:

The given chemical reaction is:

$2 N_2 (g) + O_2(g) -> 2 N_20 (g) delta Hrxn= +163.2 kJ$

When two moles of nitrogen reacts with oxygen, it requires 163.2kJ of energy.

When 5.00 mol of nitrogen requires how much energy?

$5.00 mol x \frac{163.2 kJ }{2 mol} \\=408 kJ$

Hence, the answer is 408 kJ of heat energy is required.

You might be interested in

Water vapor is condensed steam. <br><br> True <br> False

adelina 88 [10]

False, because water vapor, water vapour or aqueous vapor, is the gaseous phase of water. It is one state of water within the hydrosphere. Water vapor can be produced from the evaporation or boiling of liquid water or from the sublimation of ice. Unlike other forms of water, water vapor is invisible.

8 0

3 years ago

Which of the following amino acids may be considered hydrophobic at physiological ph?

soldier1979 [14.2K]

Answer:

Aspartic acid may be considered hydrophobic at physiological.

3 0

3 years ago

What is the volume of 2.1 moles of chlorine gas (Cl2) at standard temperature and pressure (STP)?

lukranit [14]

1 mol --- 22,4 dm³
2,1 mol --- X
X = 2,1×22,4
X = 47,04 dm³ = 47,04 L

C)

3 0

3 years ago

Read 2 more answers

Calculate the equilibrium constant for the following reaction: Co2+ (aq) + Zn(s> CO (s) + Zn2+ (aq)

Simora [160]

<u>Answer:</u> The $K_{eq}$ of the reaction is $1.73\times 10^{16}$

<u>Explanation:</u>

For the given half reactions:

Oxidation half reaction: $Zn(s)\rightarrow Zn^{2+}+2e^-;E^o_{Zn^{2+}/Zn}=-0.76V$

Reduction half reaction: $Co^{2+}+2e^-\rightarrow Co(s);E^o_{Co^{2+}/Co}=-0.28V$

Net reaction: $Zn(s)+Co^{2+}\rightarrow Zn^{2+}+Co(s)$

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

Putting values in above equation, we get:

$E^o_{cell}=-0.28-(-0.76)=0.48V$

To calculate equilibrium constant, we use the relation between Gibbs free energy, which is:

$\Delta G^o=-nfE^o_{cell}$

and,

$\Delta G^o=-RT\ln K_{eq}$

Equating these two equations, we get:

$nfE^o_{cell}=RT\ln K_{eq}$

where,

n = number of electrons transferred = 2

F = Faraday's constant = 96500 C

$E^o_{cell}$ = standard electrode potential of the cell = 0.48 V

R = Gas constant = 8.314 J/K.mol

T = temperature of the reaction = $25^oC=[273+25]=298K$

$K_{eq}$ = equilibrium constant of the reaction = ?

Putting values in above equation, we get:

$2\times 96500\times 0.48=8.314\times 298\times \ln K_{eq}\\\\K_{eq}=1.73\times 10^{16}$

Hence, the $K_{eq}$ of the reaction is $1.73\times 10^{16}$

8 0

3 years ago

In this model of a molecule of ammonia, NH3, how many covalent bonds are represented?

Verdich [7]

I think the answer is three but i can't be sure.

4 0

3 years ago