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amid [387]
2 years ago
9

A sample of ammonia gas at 75°c and 445 mm hg has a volume of 16.0 l. what volume will it occupy if the pressure rises to 1225 m

m hg at constant temperature? 1 atm = 760 mm hg.
Chemistry
1 answer:
ioda2 years ago
7 0
In this kind of exercises, you should  use the "ideal gas" rules: PV = nRT
P should be in Pascal: 
445mmHg = 59328Pa
1225mmHg = 163319Pa

V should be in cubic meter:
16L = 0.016 m3

R = \frac{PV}{nT} = constant
\frac{P1 V1}{n T} = \frac{P2 V2}{n T}
==> P1 * V1 = P2 * V2
V2 = \frac{P1 V1}{P2} = \frac{445 0.016}{1225}
V2 = 0.00581 m3 = 5.81 L


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The reaction below has an equilibrium constant kp=2.2×106 at 298 k. 2cof2(g)⇌co2(g)+cf4(g) calculate kp for the reaction below.
AnnZ [28]

<span>I believe the correct 2nd reaction is:</span>

cof2(g)⇌1/2 co2(g)+1/2 cf4(g)

where we can see that it is exactly one-half of the original

Therefore the new Kp is:

new Kp = (old Kp)^(1/2)

new Kp = (2.2 x 10^6)^(1/2)

<span>new Kp = 1,483.24 </span>

8 0
2 years ago
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A solution of SO2 in water contains 0.00023 g of SO2 per liter of solution. What is the concentration of SO2 in ppm? in ppb?
Mnenie [13.5K]

Answer:

= 230 ppb

Explanation:

Considering that;

1ppm = 1mg/L  

Then;

0.00023g = 0.23mg  

Therefore;

0.00023 g/L = 0.23 mg/L

0.23 mg/L = 0.23 ppm

1 ppm = 100 ppb

Therefore;

0.23 ppm = 0.23 ×1000

                = 230 ppb

5 0
2 years ago
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Write a balanced half-reaction describing the oxidation of gaseous dihydrogen to aqueous hydrogen cations.
xz_007 [3.2K]
The oxidation state of hydrogen gas is 0 and oxidation state of hydrogen cation is +1.
There’s an increase in oxidation number therefore it’s an oxidation reaction.
Oxidation reactions give out electrons. The masses and charges on both sides should be balanced
Half reaction is
H2 —> 2H+ +2e
8 0
2 years ago
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Given the following reactions Fe2O3 (s) + 3CO (s) → 2Fe (s) + 3CO2 (g) ΔH = -28.0 kJ 3Fe (s) + 4CO2(s) → 4CO (g) + Fe3O4(s) ΔH =
Taya2010 [7]

Answer: The enthalpy of the reaction is -109 kJ

Explanation:

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

Fe_2O_3(s)+3CO(s)\rightarrow 2Fe(s)+3CO_2(g)   \Delta H=-28.0kJ  \times 3    

3Fe_2O_3(s)+9CO(s)\rightarrow 6Fe(s)+9CO_2(g)   \Delta H=-84.0kJ     (1)

3Fe(s)+4CO_2(s)\rightarrow 4CO(g)+Fe_3O_4(s) \Delta H=+12.5kJ  \times 2  

6Fe(s)+8CO_2(s)\rightarrow 8CO(g)+2Fe_3O_4(s) \Delta H=+25.0kJ     (2)

The final reaction is:

Subtracting (2) from (1):

3Fe_2O_3(s)+CO(g)\rightarrow CO_2(g)+2Fe_3O_4(s) \Delta H=-84.0-(+25.0)=-109kJ

Thus the enthalpy of the reaction is -109 kJ

7 0
2 years ago
Before the fire, the forest consists of large trees. After the fire, there is only ash. Explain what the law of conservation of
Tasya [4]

Answer: explained below

Explanation:

Matter can change form through physical and chemical changes, but through any of these changes, matter is conserved. The same amount of matter exists before and after the change—none is created or destroyed.

5 0
2 years ago
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