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3 years ago

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What can you tell about the location of the Sun from this illustration of the Milky Way? A) It is outside of the Milky Way. B) I

t is near an outer arm of the Milky Way. C) It is on the right side of the Milky Way. D) It is within the Central Bulge of the Milky Way. Hint

1 answer:

MakcuM [25]3 years ago

7 0

It is within the central bulge

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Balance the following equation; then determine how many grams of CO2 will be produced when 40.0 mol of CO reacts with Fe2O3.

sashaice [31]

Fe2O3 + 3CO --------> 2Fe + CO2

1 : 3 : 2 : 3

13.3 <--- 40 ------> 26.6 ---> 40 ( mol)

n = m/M

m CO2 = n.M = 13.3 . 40 = 532 ( g)

p/s : i hope that this will help ( cause i'm not really good at english :}}} )

6 0

3 years ago

Explain why it is not a good idea to throw an arsenal canon to fire. Which gas law applies?

Feliz [49]

Answer:

is a class of heavy military ranged weapons built to launch munitions far beyond the range and power of infantry firearms. This development continues today; modern self-propelled artillery vehicles are highly mobile weapons of great versatility generally providing the largest share of an army's total firepower.

Explanation:

it could explode

6 0

2 years ago

Temperature is a measure of how much heat you have in, or around you. is it true or false

allochka39001 [22]

Your answer is

true

hope it helps you

7 0

2 years ago

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 $M^{-2}s^{-1}$

(c) The units of rate constant is: $M^{-2}s^{-1}$

(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 $M^{-2}s^{-1}$

(c) The units of the rate constant can be obtained from (2).

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

5 0

3 years ago

The balanced chemical equation for the reaction of copper (Cu) and silver nitrate (AgNO3) is shown below.

lesantik [10]

Remark
The balance numbers in front of Ag and AgNO3 are both 2. That number is in moles.

Rule: if the moles are the same in the equation, then whatever you are given for one, will be the same for the other. So you have 0.854 moles of Ag. You will also have 0.854 moles of AgNO3

Answer: 0.854 <<<<<

8 0

3 years ago

Read 2 more answers