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4 years ago

How would adding the catalyst nitrogen monoxide (NO) affect this reaction?

Chemistry

2 answers:

marin [14]4 years ago

5 0

<u>Answer:</u> The correct answer is Option A.

<u>Explanation:</u>

A catalyst is defined as the substance which increases the rate of the reaction without actually participating in the reaction.

This substance decreases the activation energy of the reactants, so as to proceed the reaction at a faster rate by increasing their effective collisions.

It does not affect the concentration of the reactants.

For the given reaction:

$2SO_2(g)+O_2(g)\rightarrow 2SO_3(g)$

If we add NO, which is acting as a catalyst will increase the rate of the reaction to produce the products which is $SO_3$

Hence, the correct answer is Option A.

Margarita [4]4 years ago

5 0

Answer:

Explanation:

Because NO would increase the rate

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If your weight is 120 pounds

denis23 [38]

Explanation:

Weight is 120 pounds and mass is 54 kg

The mass of an object is same everywhere. If we were on the moon, its mass will remain to be 54 kg.

Weight of an object is given by :

$W=m\times g$

g is acceleration due to gravity on that surface

For moon, it is 1/6 the gravitational force on Earth, g = 1.63 m/s²

1 pound = 4.45 N

It means my weight is 533.78 N

On the surface of Moon is will be :

$W=\dfrac{533.78}{1.63}\\W=327.47\ N$

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3 years ago

Calculate the change in the standard entropy of the system, delta s degree for the synthesis of ammonia from N_2(g) and H_2(g) a

Svet_ta [14]

Answer:

$\Delta S^{0}$ for the reaction is -198.762 J/K

Explanation:

$N_{2}(g)+3H_{2}(g)\rightarrow 2NH_{3}(g)$

Standard change in entropy for the system ( $\Delta S^{0}$ ) is given by-

$\Delta S^{0}=[2moles\times S^{0}(NH_{3})_{g}]-[1mole\times S^{0}(N_{2})_{g}]-[3\times S^{0}(H_{2})_{g}]$

where $S^{0}$ represents standard entropy.

Here $S^{0}(NH_{3})_{g}=192.45J/(K.mol)$ , $S^{0}(N_{2})_{g}=191.61J/(K.mol)$ and $S^{0}(H_{2})_{g}=130.684J/(K.mol)$

So, $\Delta S^{0}=[2\times 192.45]-[1\times 191.61]-[3\times 130.684]J/K=-198.762J/K$

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3 years ago

What is the empirical formula for propene C3H6

USPshnik [31]

The empirical formula is the simplest formula attainable while maintaining the ratio so it will be CH2.

Explanation:

The empirical formula of a chemical compound is the simplistic positive integer ratio of atoms being in a compound. A simple example of this thought is that the empirical formula of sulfur monoxide, or SO, would simply be SO, as is the empirical formula of disulfur dioxide, S2O2.

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4 years ago

How much energy do individual photons of 470 nm light have

IceJOKER [234]

<h3>Answer:</h3>

4.227 × 10^-19 Joules

<h3>Explanation:</h3>

Energy of a photon of light is calculated by the formula;

E = hf, where h is the plank's constant, 6.626 × 10^-34 J-s and f is the frequency.

But, f = c/λ

Where, c is the speed of light (2.998 × 10⁸ m/s), and λ is the wavelength.

Given the wavelength is 470 nm or 4.7 × 10^-7 m

Therefore;

E = hc/λ

= (6.626 × 10^-34 J-s × 2.998 × 10^8 m/s) ÷ 4.7 × 10^-7 m

= 4.227 × 10^-19 Joules

Therefore, the energy of a photon with 470 nm is 4.227 × 10^-19 Joules

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4 years ago

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Kryger [21]

Answer:

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Definition : -

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3 years ago