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3 years ago

Ultimately the carbon molecules in pyruvate end up as what molecules

Chemistry

1 answer:

jeyben [28]3 years ago

3 0

CO2 Thats your answer

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Nitrogen (N2) and hydrogen (H2) react to make NH3 according to the following

Schach [20]

The number of moles of NH3 that could be made would be 0.5 moles

<h3>Stoichiometric reactions</h3>

From the balanced equation of the reaction:

N2 (g) + 3 H2(g) ----> 2NH3 (g)

The mole ratio of N2 to H2 is 1:3

Thus, for 0.50 moles of N2, 1.5 moles of H2 should be present. But 0.75 moles of H2 was allowed to react. Meaning that H2 is limiting in this case.

Mole ratio of H2 and NH3 = 3:2

Thus for 0.75 moles H2, the mole of NH3 that would be produced will be:

2 x 0.75/3 = 0.5 moles

More on stoichiometric calculations can be found here: brainly.com/question/8062886

3 0

2 years ago

The pen on seismograph swings freely.

Sever21 [200]

This statment is false, the pen on a seismograph does not swing freely.

7 0

3 years ago

Day 2 code word

bagirrra123 [75]

Answer:

chloroplast

Explanation:

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3 years ago

The geocentric theory was once accepted as an explanation of how objects were positioned in our solar system. Which statement be

dusya [7]

Answer: c

Explanation:

4 0

3 years ago

Given the two reactions H2S(aq)⇌HS−(aq)+H+(aq), K1 = 9.57×10−8, and HS−(aq)⇌S2−(aq)+H+(aq), K2 = 1.46×10−19, what is the equilib

Dvinal [7]

<u>Answer:</u> The value of $K_c$ for the final reaction is $7.16\times 10^{25}$

<u>Explanation:</u>

The given chemical equations follows:

<u>Equation 1:</u> $H_2S(aq.)\rightleftharpoons HS^-(aq.)+H^(aq.);K_1$

<u>Equation 2:</u> $HS^-(aq.)\rightleftharpoons S^{2-}(aq.)+H^(aq.);K_2$

The net equation follows:

$S^{2-}(aq.)+2H^+(aq.)\rightleftharpoons H_2S(aq.);K_c$

As, the net reaction is the result of the addition of reverse of first equation and the reverse of second equation. So, the equilibrium constant for the net reaction will be the multiplication of inverse of first equilibrium constant and the inverse of second equilibrium constant.

The value of equilibrium constant for net reaction is:

$K_c=\frac{1}{K_1}\times \frac{1}{K_2}$

We are given:

$K_1=9.57\times 10^{-8}$

$K_2=1.46\times 10^{-19}$

Putting values in above equation, we get:

$K_c=\frac{1}{(9.57\times 10^{-8})}\times \frac{1}{(1.46\times 10^{-19})}=7.16\times 10^{25}$

Hence, the value of $K_c$ for the final reaction is $7.16\times 10^{25}$

5 0

3 years ago