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Blababa [14]
2 years ago
9

Is kp or kc larger (both at 300 k) for the reaction 2 nh3(g) ⇀↽ n2(g) + 3 h2(g) ? 1. they are both the same. 2. kc is larger. 3.

kp is larger?
Chemistry
1 answer:
fgiga [73]2 years ago
5 0
The equation is 2 NH3 (g) ⇀↽ N2 (g) + 3 H2 (g) 
Difference in the number of moles delta n = ((3 + 1) - 2) = 4 - 2 = 2 
We have an equation Kp= Kc (R x T) ^ (delta n); R is constant and T = 300 K 
Kp / Kc = (R x T) ^2 Based on the temperature value (300 K), we can conclude that Kp is Larger.
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Sliva [168]

Answer:

\large \boxed{\text{1763 psi}}

Explanation:

We can use Dalton's Law of Partial Pressures:

Each gas in a mixture of gases equals its pressure independently of the other gases

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4 0
3 years ago
a radioisotope has a half life of three hours how much of a 120g sample remains after 9 hours. what will the g amount be after 3
aksik [14]

\huge\boxed{♔︎Answer♔︎}

60g after 3 hours, 30g after 6 hours and 15g after 9 hours

Explanation:

Weight of the radioactive sample = 120g

half life time period = 3 hours

(a) The weight of sample after 3 hours

\textsf{ No. of half lives} =  \sf  \cancel\frac{3}{3}  = 1

The fraction of sample left

\sf  { \frac{1}{2} }^{1}  =  \frac{1}{2}

Mass of the sample left

\sf \frac{1}{2}  \times 120 =   \cancel\frac{120}{2}  = 60g

<u>6</u><u>0</u><u>g</u><u> </u><u>of</u><u> </u><u>sample</u><u> </u><u>is</u><u> </u><u>left</u><u> </u><u>after</u><u> </u><u>3</u><u> </u><u>hours</u>

(b) The weight of sample after 6 hours

\textsf{ No. of half lives} =  \sf   \cancel\frac{6}{3}  = 2

The fraction of the sample left

\sf   { \frac{1}{2} }^{2}  =  \frac{1}{2}  \times  \frac{1}{2}  =  \frac{1}{4}

Mass of the sample left

\sf \frac{1}{4}  \times 120 =   \cancel\frac{120}{4}  = 30g

<u>3</u><u>0</u><u>g</u><u> </u><u>of</u><u> </u><u>sample</u><u> </u><u>is</u><u> </u><u>left</u><u> </u><u>after</u><u> </u><u>6</u><u> </u><u>hours</u>

(c) The weight of sample after 9 hours

\textsf {No. of half lives} =  \sf   \cancel\frac{9}{3}  = 3

The fraction of sample left

\sf { \frac{1}{2} }^{3}  =  \frac{1}{2}  \times  \frac{1}{2}  \times  \frac{1}{2} =  \frac{1}{8}

Mass of sample left

\sf  \frac{1}{8}  \times 120 =  \cancel \frac{120}{8}  = 15g

<u>1</u><u>5</u><u>g</u><u> </u><u>of</u><u> </u><u>sample</u><u> </u><u>is</u><u> </u><u>left</u><u> </u><u>after</u><u> </u><u>9</u><u> </u><u>hours</u><u>.</u>

3 0
2 years ago
Aspirin is one of the Aesthetic Values of biodiversity<br><br> true<br> false
Trava [24]

Answer:

true

Explanation:

6 0
2 years ago
Un recipiente cerrado, de 4,25 L, con tapa móvil, contiene H2S(g) a 740 Torr y 50,0°C. Se introduce en ese recipiente N2(g) a te
yulyashka [42]

Answer:

n_{N_2}=6.41mol

Explanation:

¡Hola!

En este caso, teniendo en cuenta la información dada por el problema, inferimos que primero se debe usar la ecuación del gas ideal con el fin de calcular las moles de gas que se encuentran al inicio del experimento:

PV=nRT\\\\n=\frac{RT}{PV} \\\\n=\frac{0.08206\frac{atm*L}{mol*K}*(50.0+273.15)K}{740/760atm*4.25L}\\\\n=6.41mol

Seguidamente, usamos la ley de Avogadro para calcular las moles finales, teniendo el cuenta que el volumen final es el doble del inicial (8.50 L):

n_2=\frac{6.41mol*8.50L}{4.25L}\\\\n_2=12.82mol

Quiere decir que las moles de N2(g) que se agregaron son:

n_{N_2}=12.81mol-6.41mol\\\\n_{N_2}=6.41mol

¡Saludos!

8 0
3 years ago
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