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3 years ago

How do you balance the equation NO2 + H2 →-> NH3 +H2O using the oxidation number method?

Chemistry

2 answers:

Ksivusya [100]3 years ago

6 0

$NO_2 + H_2 \to NH_3 + H_2 O$

 $2NO_2 + 7H_2 \to 2NH_3 + 4H_2 O$

taurus [48]3 years ago

4 0

NO2 + H2 ----> NH3 + H2O

N(+IV) + 7e -----> N(-III) |2
H2(0) - 2e ----> 2H(+I) |7

after: 2NO2 + 7H2 ----> 2NH3 + 4H2O

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At what temperature would 0.500 moles of gas particles stored in a 100.0 mL container reach a pressure of 15.0 atm?

MArishka [77]

Hey there!

In this case, the equation of Clapeyron is used :

R = 0.082

Volume in liters :

100.0 mL / 1000 => 0.1 L

P * V = n * R * T

15.0 * 0.1 = 0.500 * 0.082 * T

1.5 = 0.041 * T

T = 1.5 / 0.041

T = 36.5 K

Answer C

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3 years ago

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Explain why there might be a change in the density of a forged product as compared to that of the cast blank.

kkurt [141]

Answer:

Forged parts are often tougher than cast parts. This can be determined by performing tensile tests on various areas on the parts. Additionally, the microstructures of forged and cast parts can be used to determine if a part was forged or cast. The microstructure of a cast part will have a more uniform grain structure.

Explanation:

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2 years ago

what happens to the water level of water when you heat it and then set it aside for it to cool? what will happen THIS IS URGENT

IceJOKER [234]

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3 years ago

Calculate the heat absorbed by a sample of water that has a mass of 45.00 g when the temperature increases from 21.0oC to 38.5 o

Elenna [48]

Answer:

The heat absorbed by the sample of water is 3,294.9 J

Explanation:

Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.

The sensible heat of a body is the amount of heat received or transferred by a body when it undergoes a temperature variation (Δt) without there being a change of physical state (solid, liquid or gaseous). Its mathematical expression is:

Q = c * m * ΔT

Where Q is the heat exchanged by a body of mass m, made up of a specific heat substance c and where ΔT is the temperature variation.

In this case:

Q=?
m= 45 g
c= 4.184 $\frac{J}{g*C}$
ΔT= Tfinal - Tinitial= 38.5 C - 21 C= 17.5 C

Replacing:

Q= 4.184 $\frac{J}{g*C}$ * 45 g* 17.5 C

Solving:

Q=3,294.9 J

The heat absorbed by the sample of water is 3,294.9 J

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ohaa [14]

Answer:

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