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snow_tiger [21]

3 years ago

10

If you used the method of initial rates to obtain the order for no2, predict what reaction rates you would measure in the beginn

ing of the reaction for initial concentrations of 0.200 m, 0.100 m, & 0.050 m no2.

1 answer:

luda_lava [24]3 years ago

8 0

Answer:

Rate law is: Rate = k[NO2]^2 = 10 x [NO2]^2

When [NO2] = 0.200 M:

Rate = 10 x 0.200^2 = 0.400 M/s

When [NO2] = 0.100 M:

Rate = 10 x 0.100^2 = 0.100 M/s

When [NO2] = 0.050 M:

Rate = 10 x 0.050^2 = 0.0250 M/s

Explanation:

If you used the method of initial rates to obtain the order for no2, predict what reaction rates you would measure in the beginning of the reaction for initial concentrations of 0.200 m, 0.100 m, & 0.050 m no2.

Rate law is: Rate = k[NO2]^2 = 10 x [NO2]^2

When [NO2] = 0.200 M:

Rate = 10 x 0.200^2 = 0.400 M/s

When [NO2] = 0.100 M:

Rate = 10 x 0.100^2 = 0.100 M/s

When [NO2] = 0.050 M:

Rate = 10 x 0.050^2 = 0.0250 M/s

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Potassium forms ions with a charge of +1 (K+). Iodine forms ions with a charge of -1. Selenium forms ions with a charge of -2 (S

Taya2010 [7]

It is given that the charge on potassium is +1 i.e. Potassium is a cation, K⁺

The charge on selenium is given as -2, i.e. selenium is an anion, Se²⁻

The two ions are shown to combine to form a molecule: KXSeY where X and Y correspond to the proportions in which the two atoms combine.

Since the molecule is neutral, the positive and negative charges need to be balanced. This is possible only if 2 units of K⁺ combines with 1 unit of Se²⁻.

Therefore, the formula would be K₂Se

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3 0

4 years ago

Air containing 0.04% carbon dioxide is pumped into a room whose volume is 6000 ft3. The air is pumped in at a rate of 2000 ft3/m

koban [17]

Here is the full question:

Air containing 0.04% carbon dioxide is pumped into a room whose volume is 6000 ft3. The air is pumped in at a rate of 2000 ft3/min, and the circulated air is then pumped out at the same rate. If there is an initial concentration of 0.2% carbon dioxide, determine the subsequent amount in the room at any time.

What is the concentration at 10 minutes? (Round your answer to three decimal places.

Answer:

0.046 %

Explanation:

The rate-in;

$R_{in}$ $= \frac{0.04}{100}*2000$

$R_{in}$ = 0.8

The rate-out

$R_{out}$ = $\frac{A}{6000}*2000$

$R_{out}$ = $\frac{A}{3}$

We can say that:

$\frac{dA}{dt}=$ $0.8-\frac{A}{3}$

where;

A(0)= 0.2% × 6000

A(0)= 0.002 × 6000

A(0)= 12

$\frac{dA}{dt} +\frac{A}{3} =0.8$

Integration of the above linear equation =

$e^{\int\limits \frac {1}{3}dt } =$ $e^{\frac{1}{3}t$

so we have:

$e^{\frac{1}{3}t}\frac{dA}{dt}} +\frac{1}{3}e^{\frac{1}{3}t}A$ $= 0.8e^{\frac{1}{3}t$

$\frac{d}{dt}[e^{\frac{1}{3}t}A]$ $= 0.8e^{\frac{1}{3}t$

$Ae^{\frac{1}{3}t} =2.4e\frac{1}{3}t +C$

∴ $A(t) = 2.4 +Ce^{-\frac{1}{3}t$

Since A(0) = 12

Then;

$12 =2.4 + Ce^{-\frac{1}{3}}(0)$

$C= 12-2.4$

$C =9.6$

Hence;

$A(t) = 2.4 +9.6e^{-\frac{t}{3}}$

$A(0) = 2.4 +9.6e^{-\frac{10}{3}}$

$A(t) = 2.74$

∴ the concentration at 10 minutes is ;

= $\frac{2.74}{6000}*100$ %

= 0.0456667 %

= 0.046% to three decimal places

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