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3 years ago

Which of the following is not a chemistry topic?

Chemistry

2 answers:

Tju [1.3M]3 years ago

6 0

I’m pretty sure it’s D

stiks02 [169]3 years ago

3 0

Well anything linked to the ocean is biology, its the study of life. but in this situation chemistry is the study of mostly the chemical formulas so i would say D.

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The force that one massive object exerts to attract another object is called .

Ghella [55]

Answer:

Gravity

Explanation:

The force that one massive object exerts to attract another object is called gravity.

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3 years ago

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Which is the best method for organizing information collected in an investigation? Plz help

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Construct a table i think

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4 years ago

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The reaction 2NO(g)+O2(g)−→−2NO2(g) is second order in NO and first order in O2. When [NO]=0.040M, and [O2]=0.035M, the observed

Oksanka [162]

Answer:

(a) The rate of disappearance of $O_{2}$ is: $4.65*10^{-5}$ M/s

(b) The value of rate constant is: 0.83036 $M^{-2}s^{-1}$

(d) The rate will increase by a factor of 3.24

Explanation:

The rate of a reaction can be expressed in terms of the concentrations of the reactants and products in accordance with the balanced equation.

For the given reaction:

$2NO(g)+O_{2}->2NO_{2}$

rate = $-\frac{1}{2} \frac{d}{dt}[NO]$ = $-\frac{d}{dt}[O_{2}]$ = $\frac{1}{2}\frac{d}{dt}[NO_{2}]$ -----(1)

According to the question, the reaction is second order in NO and first order in $O_{2}$ .

Then we can say that, rate = k $[NO]^{2}[O_{2}]$ -----(2)

where k is the rate constant.

The rate of disappearance of NO is given:

$-\frac{d}{dt}[NO]$ = $9.3*10^{-5}$ M/s.

(a) From (1), we can get the rate of disappearance of $O_{2}$ .

Rate of disappearance of $O_{2}$ = $-\frac{d}{dt}[O_{2}]$ = (0.5)*( $9.3*10^{-5}$ ) M/s = $4.65*10^{-5}$ M/s.

(b) The rate of the reaction can be obtained from (1).

rate = $-\frac{1}{2} \frac{d}{dt}[NO]$ = (0.5)*( $9.3*10^{-5}$ )

rate = $4.65*10^{-5}$ M/s

The value of rate constant can be obtained by using (2).

rate constant = k = $\frac{rate}{[NO]^{2}[O_{2}]}$

k = $\frac{4.65*10^{-5}}{(0.040)^{2}(0.035)}$ = 0.83036 $M^{-2}s^{-1}$

k = $\frac{rate}{[NO]^{2}[O_{2}]}$

Substituting the units of rate as M/s and concentrations as M, we get:

$\frac{Ms^{-1} }{M^{3}}$ = $M^{-2}s^{-1}$

(d) The reaction is second order in NO. Rate is proportional to square of the concentration of NO.

$rate\alpha [NO]^{2}$

If the concentration of NO increases by a factor of 1.8, the rate will increase by a factor of $(1.8)^{2}$ = 3.24

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3 years ago

How many “dots” would be around the central atom in the Lewis structure of S?

agasfer [191]

B) 6
one above one below and 2 on the left and right sides

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A cheetah can run 112km/h over a 100-m distance. what is this speed in meters per second

Naya [18.7K]

First convert the 112 km/hr ratio into m/s (meters per second). To do this you multiply 112 km with 1000 m/km (since there's 1000 m in one km). You get 112000 m. Then multiply 1 hr with 60 min/hr (since there's 60 min in one hr. You get 60 min, but you want seconds, so multiply 60 min with 60 s/min to get 3600 s. There you go! Your answer is the speed of 112000m/3600s, but you can simplify that to 31.11m/s (since the answer should be in ? meters per 1 second.

Also, the "100-m-distance" part of the question is just to throw you off, because one particular speed obviously stays constant over any distance. Hope that helps :)

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4 years ago