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tia_tia [17]
3 years ago
5

Bilangan oksidasi vanadium paling tinggi terdapat dalam senyawa..

Chemistry
1 answer:
emmasim [6.3K]3 years ago
8 0
The answer is (b). As, vanadium is attached to five fluoride atoms, each flouride containing -1 oxidation state, hence five fluoride contains -5, to neutralize, vanadium should have +5 oxidation state.

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Ethers is the correct answer

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Can you Convert 75°F to °C
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Answer: 23.8889

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(75°F − 32) × 5/9 = 23.889°C

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How do scientists use radioactive decay to date fossils
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Santiago's car has an overall average speed of 25 cm/s. If he could increase his car's overall average speed by 10%, what would
Evgen [1.6K]

Given:

The overall average speed = 25 cm/s.

The overall average speed increased by 10%.

To find:

The new overall average speed.

Solution:

We have,

Overall average speed = 25 cm/s.

The overall average speed increased by 10%. So, the new average speed is

New overall average speed = Initial average speed + 10% of initial  overall average speed

\text{New overall average speed}=25+\dfrac{10}{100}\times 25

\text{New overall average speed}=25+2.5

\text{New overall average speed}=27.5

Therefore, the new overall average speed is 27.5 cm/s.

3 0
3 years ago
Hydrogen is manufactured on an industrial scale by this sequence of reactions: Write an equation that gives the overall equilibr
RideAnS [48]

The question is incomplete. The complete question is :

Hydrogen is manufactured on an industrial scale by this sequence of reactions:

$CH_3(g) + H_2O(g) \rightleftharpoons CO(g) + 3H_2(g    ) \ \ \ \ \ \ \ \ \ \ K_1$

$CO(g) + H_2O(g) \rightleftharpoons CO_2(g) + H_2(g) \ \ \ \ \ \ \ \ \ \ \ \  K_2$

The net reaction is  :

$CH_4(g) + 2H_2O(g) \rightleftharpoons CO_2(g) + 4H_2(g) \ \ \ \ \ \ \ \ \ K$

Write an equation that gives the overall equilibrium constant K in terms of the equilibrium constants K_1 and K_2. If you need to include any physical constants, be sure you use their standard symbols, which you'll find in the ALEKS Calculator.

Solution :

$CH_3(g) + H_2O(g) \rightleftharpoons CO(g) + 3H_2(g    ) \ \ \ \ \ \ \ \ \ \ K_1$

$K_1 = \frac{[CO][H_2]^3}{[CH_4][H_2O]}$     ...............(1)

$CO(g) + H_2O(g) \rightleftharpoons CO_2(g) + H_2(g) \ \ \ \ \ \ \ \ \ \ \ \  K_2$

$K_2 = \frac{[CO_2][H_2]}{[CO][H_2O]}$  ...................(2)

$CH_4(g) + 2H_2O(g) \rightleftharpoons CO_2(g) + 4H_2(g) \ \ \ \ \ \ \ \ \ K$

$K=\frac{[CO_2][H_2]^4}{[CH_4][H_2O]^2}$

On multiplication of equation (1) and (2), we get

$K_1 \times K_2=\frac{[CO][H_2]^3}{[CH_4][H_2O]} \times \frac{[CO_2][H_2]}{[CO][H_2O]}$

$K_1K_2=\frac{[CO_2][H_2]^4}{[CH_4][H_2O]^2}$  .................(4)

Comparing equation (3) and equation (4), we get

$K=K_1K_2$

4 0
2 years ago
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