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stich3 [128]
3 years ago
9

Nitrogen dioxide decomposes at 300°C via a second-order process to produce nitrogen monoxide and oxygen according to the followi

ng chemical equation.
2 NO2(g) → 2 NO(g) + O2(g).
A sample of NO2(g) is initially placed in a 2.50-L reaction vessel at 300°C. If the half-life and the rate constant at 300°C are 11 seconds and 0.54 M-1 s-1, respectively, how many moles of NO2 were in the original sample?

I just need an idea on how to approach this problem. ...?
Chemistry
1 answer:
qaws [65]3 years ago
4 0
This problem can be simplified by assuming that the reaction took place in a rigid vessel, that is, the volume is held constant. It is also safe to assume that the reaction is isothermal. The first step here is to derive an expression for the dependency of the reactant concentration with time. Since it is stated that the reaction is second order we start with

-dC/dt = k C^2

where

C is the amount of reactant (NO2)
t is time 
k is the reaction rate constant

The negative sign indicates that the concnetration is decreasing with time. Solving the equation, we get

1/Co - 1/C = kt

where Co is the initial amount of NO2

Now we are given the half life which is the time in which the amount of NO2 is halved, that is, C = 0.5Co. Therefore we can solve the initial amount of NO2 (Co) by substituting t = 11 s, k = 0.54 M-1s-1 and C = 0.5Co

Hope this helps
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In aqueous solution, hypobromite ion, BrO-, reacts to produce bromate ion, BrO3 -, and bromide ion, Br-, according to the follow
maw [93]

Answer:

22.73s

Explanation:

The reaction is a second order reaction, we know this by observing the unit of the slope.

rate constant = k = 0.056 M-1s-1

the initial concentration of BrO- [A]o = 0.80 M

time = ?

Final concentration [A]t= one-half of 0.80 M = 0.40M

1 / [A]t = kt + 1 / [A]o

1 / 0.40 = 0.056 * t + 1 / 0.80

t = (2.5 - 1.25) / 0.056

t = 22.73s

5 0
3 years ago
Aqueous sodium phosphate and aqueous iron (III) chloride react to produce aqueous sodium chloride and solid iron (III) phosphate
mojhsa [17]

Answer:

this one is hard

Explanation:

but it's iron because the sodium so yea there u go.

4 0
2 years ago
The question is in the picture below
Rus_ich [418]

Answer:

\Delta\text{H}_1+2\Delta\text{H}_2-\Delta\text{H}_3

Explanation:

Hess's Law of Constant Heat Summation states that if a chemical equation can be written as the sum of several other chemical equations, the enthalpy change of the first chemical equation is equal to the sum of the enthalpy changes of the other chemical equations. Thus, the reaction that involves the conversion of reactant A to B, for example, has the same enthalpy change even if you convert A to C, before converting it to B. Regardless of how many steps it takes for the reactant to be converted to the product, the enthalpy change of the overall reaction is constant.

With Hess's Law in mind, let's see how A can be converted to 2C +E.

\bf{\text{A} \rightarrow 2\text{B}}                  (Δ\text{H}_1)  -----(1)

Since we have 2B, multiply the whole of II. by 2:

\bf{2\text{B} \rightarrow 2\text{C} +2\text{D}}       (2Δ\text{H}_2) -----(2)

This step converts all the B intermediates to 2C +2D. This means that the overall reaction at this stage is \text{A} \rightarrow 2\text{C} +2\text{D}.

Reversing III. gives us a negative enthalpy change as such:

\bf{2\text{D} \rightarrow \text{E}}                  (-Δ\text{H}_3) -----(3)

This step converts all the D intermediates formed from step (2) to E. This results in the overall equation of \text{A} \rightarrow 2\text{C} +\text{E}, which is also the equation of interest.

Adding all three together:

\text{A} \rightarrow 2\text{C}+\text{E}            (\bf{\Delta\text{H}_1+2\Delta\text{H}_2-\Delta\text{H}_3 })

Thus, the first option is the correct answer.

Supplementary:

To learn more about Hess's Law, do check out: brainly.com/question/26491956

4 0
1 year ago
What happens to excess heat produced in the muscle during contraction and relaxation
Afina-wow [57]

Answer:

Calcium is then pumped back into the sarcoplasmic reticulum breaking the link between actin and myosin.

Explanation:

Actin and myosin return to their unbound state causing the muscle to relax. Alternatively relaxation (failure) will also occur when ATP is no longer available.

3 0
2 years ago
The carbon-carbon double bond in ethene is ________ and ________ than the carbon-carbon triple bond in ethyne.
lbvjy [14]

Answer:

weaker and longer

Explanation:

Since there are 3 bonds in ethyne in comparision with the 2 bonds of ethyne between carbon atoms, they are attracted more to each other → the bond gets shorter . And since there are one more bond that supports the union → the bond gets stronger

thus the carbon-carbon double bond in ethene is weaker and longer than the carbon-carbon triple bond in ethyne

3 0
3 years ago
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