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Basile [38]
3 years ago
13

What is the molar mass of (NH4)3 PO4? 113g, 121g, 149g, 339g

Chemistry
1 answer:
allochka39001 [22]3 years ago
7 0
(NH4)3PO4 :

N = 14 * 3 =  42
H = 1 * 12 = 12
P = 31 * 1 = 31
O = 16 * 4 = 64
-------------------------
42+12+31+64 =  149 g / mol

Hope this helps!.


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Which of the following gases is not an important component of soil?
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Phosgene, COCl2, gained notoriety as a chemical weapon in World War I. Phosgene is produced by the reaction of carbon monoxide w
jek_recluse [69]

The question is incomplete, here is the complete question:

Phosgene, COCl_2, gained notoriety as a chemical weapon in World War I. Phosgene is produced by the reaction of carbon monoxide with chlorine:  

CO(g)+Cl_2(g)\rightleftharpoons COCl_2(g)

The value of K_c for this reaction is 5.79 at 570 K. What are the equilibrium partial pressures of the three gases if a reaction vessel initially contains a mixture of the reactants in which p_{CO}=p_{Cl_2}=0.265atm and p_{COCl_2}=0.000atm ?

<u>Answer:</u> The equilibrium partial pressure of CO, Cl_2\text{ and }COCl_2 is 0.257 atm, 0.257 atm and 0.008 atm respectively.

<u>Explanation:</u>

The relation of K_c\text{ and }K_p is given by:

K_p=K_c(RT)^{\Delta n_g}

K_p = Equilibrium constant in terms of partial pressure

K_c = Equilibrium constant in terms of concentration = 5.79

\Delta n_g = Difference between gaseous moles on product side and reactant side = n_{g,p}-n_{g,r}=1-2=-1

R = Gas constant = 0.0821\text{ L. atm }mol^{-1}K^{-1}

T = Temperature = 570 K

Putting values in above equation, we get:

K_p=5.79\times (0.0821\times 570)^{-1}\\\\K_p=0.124

We are given:

Initial partial pressure of CO = 0.265 atm

Initial partial pressure of chlorine gas = 0.265 atm

Initial partial pressure of phosgene = 0.00 atm

The given chemical equation follows:

                      CO(g)+Cl_2(g)\rightleftharpoons COCl_2(g)

<u>Initial:</u>            0.265      0.265

<u>At eqllm:</u>        0.265-x    0.265-x        x

The expression of K_p for above equation follows:

K_p=\frac{p_{COCl_2}}{p_{CO}\times p_{Cl_2}}

Putting values in above equation, we get:

0.124=\frac{x}{(0.265-x)\times (0.265-x)}\\\\x=0.0082,8.59

Neglecting the value of x = 8.59 because equilibrium partial pressure cannot be greater than initial pressure

So, the equilibrium partial pressure of CO = (0.265-x)=(0.265-0.008)=0.257atm

The equilibrium partial pressure of Cl_2=(0.265-x)=(0.265-0.008)=0.257atm

The equilibrium partial pressure of COCl_2=x=0.008atm

Hence, the equilibrium partial pressure of CO, Cl_2\text{ and }COCl_2 is 0.257 atm, 0.257 atm and 0.008 atm respectively.

6 0
3 years ago
What is the specific heat capacity of a metal if it requires 177.5 J to change the temperatures of 15.0g of the metal from 25.00
Vikentia [17]

Answer:

The specific heat capacity of a metal is 1.31 J/g°C = C

Explanation:

A classical excersise of calorimetry to apply this formula:

Q = m . C . ΔT

177.5 J = 15 g . C (34°C - 25°C)

177.5 J = 15g . 9°C . C

177.5 J /15g . 9°C = C

1.31 J/g°C = C

3 0
3 years ago
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