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3 years ago

Which rule is being disobeyed by this orbital diagram?

Chemistry

1 answer:

klio [65]3 years ago

5 0

Answer:

Im pretty sure its d. Diagonal rule

Explanation:

That might be wrong

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A gas occupies 3.00 L at standard temperature. The volume at 350 C is 6.84 L. True False

kupik [55]

Answer:

TRUE

Explanation:

Standard temperature is 273K and 350°C is 623K.

V/V' = T/T'

3/6.84 = 273/T'

T' = (273×6.84)/3

=> T' = 623K

So it is true.

7 0

3 years ago

In the first step of the Ostwald process for the synthesis of nitric acid, ammonia is converted to nitric oxide by the high-temp

Paul [167]

Solution: The given balanced equation is:

$4NH_3+5O_2\rightarrow 4NO+6H_2O$

If we have a hypothetical equation:

$A+2B\rightarrow 3C+5D$

Then the rate could be written as:

$rate=-\frac{\Delta [A]}{\Delta t}=-\frac{1}{2}\frac{\Delta [B]}{\Delta t}=\frac{1}{3}\frac{\Delta [C]}{\Delta t}=\frac{1}{5}\frac{\Delta [D]}{\Delta t}$

From above expression one thing could easily be noticed that the coefficients of all are inverted. Also, there is negative sign in front of reactants and positive sign in front of the products. Negative sign stands for rate of consumption where as positive sign stands for rate of formation.

Like the above example, we can write the rate for the given equation and it would be looking as:

$rate=-\frac{1}{4}\frac{\Delta [NH_3]}{\Delta t}=-\frac{1}{5}\frac{\Delta [O_2]}{\Delta t}=\frac{1}{4}\frac{\Delta [NO]}{\Delta t}=\frac{1}{6}\frac{\Delta [H_2O]}{\Delta t}$

Now we can easily answer all the parts of the question.

(a) From above expression, the rate of consumption of $O_2$ related to rate of consumption of $NH_3$ as:

$-\frac{1}{5}\frac{\Delta [O_2]}{\Delta t}=-\frac{1}{4}\frac{\Delta [NH_3]}{\Delta t}$

multiply both sides by -5

$\frac{\Delta [O_2]}{\Delta t}=\frac{5}{4}\frac{\Delta [NH_3]}{\Delta t}$

So, rate of consumption of oxygen is $\frac{5}{4}$ the rate of consumption of ammonia.

(b) The relationship between rate of formation of NO to the rate of consumption of ammonia will be written as:

$\frac{1}{4}\frac{\Delta [NO]}{\Delta t}=-\frac{1}{4}\frac{\Delta [NH_3]}{\Delta t}$

Multiply both sides by 4

$\frac{\Delta [NO]}{\Delta t}=-\frac{\Delta [NH_3]}{\Delta t}$

So, rate of formation of NO equals to the rate of consumption of ammonia.

Now, the rate of formation of $H_2O$ to the rate of consumption of ammonia would be:

$\frac{1}{6}\frac{\Delta [H_2O]}{\Delta t}=-\frac{1}{4}\frac{\Delta [NH_3]}{\Delta t}$

Multiply both sides by 6

$\frac{\Delta [H_2O]}{\Delta t}=-\frac{6}{4}\frac{\Delta [NH_3]}{\Delta t}$

So, the rate of formation of $H_2O$ is $\frac{6}{4}$ times that is 1.5 times to the rate of consumption of ammonia.

5 0

4 years ago

Is this true or false? false true

yulyashka [42]

Answer:

True

Explanation:

6 0

3 years ago

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Is fertilizer a mixture compound or element

egoroff_w [7]

I always thought it was a mixture but can also be a compound

5 0

3 years ago

Balancing equations I need help asap for this please

Nookie1986 [14]

No 8) one mole at all side.
no 9)one mole at N2 and three mole at H2 to get two mole of NH3.
no 10) four mole of Al and three mole of O2 to get two mole of Al2O3.
no 12) two mole of Na and two mole of H2O to get two mole of NaOH and one mole of H2.
no 13) one mole of H2SO4 and two mole of NaOH to get two mole of H2O.

7 0

4 years ago

Read 2 more answers