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expeople1 [14]
3 years ago
10

To test the effectiveness of a gunpowder mixture, 1 gram was exploded under controlled STP conditions, and the reaction chamber

was found to expand to 310 cm3. What volume would these gases occupy at the temperature produced by an uncontrolled explosion (2,200°C) and 2.1 atmospheres of pressure?
Chemistry
1 answer:
Alekssandra [29.7K]3 years ago
7 0

Answer:

1.3 × 10³ cm³

Explanation:

The gas occupies a volume of V₁ = 310 cm³ under standard temperature and pressure (STP), that is, T₁ = 273.15 K and P₁ = 1.0 atm. In order to find the volume V₂ under different conditions we can use the combined gas law formula.

\frac{P_{1}.V_{1}}{T_{1}} =\frac{P_{2}.V_{2}}{T_{2}} \\V_{2}=\frac{P_{1}.V_{1}.T_{2}}{T_{1}.P_{2}}=\frac{1.0atm\times 310cm^{3} \times 2473K }{273.15K \times 2.1atm} =1.3 \times 10^{3} cm^{3}

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2) Show the calculation of Kc for the following reaction if an initial reaction mixture of 0.800 mole of CO and 2.40 mole of H2
nadezda [96]

Answer:

Kc = 3.90

Explanation:

CO reacts with H_2 to form CH_4 and H_2O. balanced reaction is:

CO(g) + 3H_2 (g) \leftrightharpoons CH_4(g)  +  H_2O(g)

No. of moles of CO = 0.800 mol

No. of moles of H_2 = 2.40 mol

Volume = 8.00 L

Concentration = \frac{Moles}{Volume\ in\ L}

Concentration of CO = \frac{0.800}{8.00} = 0.100\ mol/L

Concentration of H_2 = \frac{2.40}{8.00} = 0.300\ mol/L

                 CO(g) + 3H_2 (g) \leftrightharpoons CH_4(g)  +  H_2O(g)

Initial            0.100      0.300             0   0

equi.            0.100 -x    0.300 - 3x     x    x

It is given that,

at equilibrium H_2O (x) = 0.309/8.00 = 0.0386 M

So, at equilibrium CO = 0.100 - 0.0386 = 0.0614 M

At equilibrium H_2 = 0.300 - 0.0386 × 3 = 0.184 M

At equilibrium CH_4 = 0.0386 M

Kc=\frac{[H_2O][CH_4]}{[CO][H_2]^3}

Kc=\frac{0.0386 \times 0.0386}{(0.184)^3 \times 0.0614} =3.90

8 0
3 years ago
Which amendment repealed prohibition
nexus9112 [7]
It is the 21st amendment
8 0
3 years ago
Find the mass of 4.5 moles of H3PO4
Contact [7]

Hey there!

H₃PO₄

Find molar mass.

H: 3 x 1.008 = 3.024

P: 1 x 30.97 = 30.97

O: 4 x 16 = 64

---------------------------------

                  97.994 grams

The mass of 1 mole of H₃PO₄ is 97.994 grams.

We have 4.5 moles.

97.994 x 4.5 = 440

The mass of 4.5 moles of H₃PO₄ is 440 grams.

Hope this helps!

6 0
3 years ago
Chlorophyll
Leno4ka [110]
Chlorophyll helps the plant by absorbing light energy and turning it into sugars to help the plant grow (photosynthesis)

The correct answer is A.
4 0
3 years ago
How would you prepare 100.0 ml of.400 m CaCl2 from a stock solution of 2.00 M CaCl2?
docker41 [41]
C₀=2 mol/l
c₁=0.400 mol/l
v₁=100.0 ml = 0.1 l

c₁v₁=c₀v₀

v₀=c₁v₁/c₀
v(H₂O)=v₁-v₀

v₀=0.1*0.400/2=0.02 l = 20 ml
v(H₂O)=100 - 20 = 80 ml

It is necessary to mix 20 ml of the feed solution and 80 ml of water.

4 0
3 years ago
Read 2 more answers
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