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Novosadov [1.4K]
3 years ago
9

A chemical engineer must report the average volume of a certain pollutant produced by the plants under her supervision. Here are

the data she has been given by each plant:
Plant Volume of pollutant
Cross Creek 10.88 L
Oglala 0.92 L
Platte 15.82 L

What average volume should the chemical engineer report? Be sure your answer has the correct number of significant digits.
Chemistry
1 answer:
Mkey [24]3 years ago
4 0

Answer:

9.2 L is the average volume should the chemical engineer report.

Explanation:

Volume of pollutant from Cross creek plants  10.88 L

Volume of pollutant from Oglala plants = 0.92 L

Volume of pollutant from Platte plants = 15.82 L

Average volume pollutants will be given by :

=\frac{10.88 L+0.92 L+15.82 L}{3}=9.2067 L\approx 9.2 L

9.2 L is the average volume should the chemical engineer report.

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3 years ago
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A engineer measures the peak power output is 0.3227MW what is the peak power output in kilowatts ?
Lana71 [14]
<h3>Answer:</h3>

322.7 kW

<h3>Explanation:</h3>
  • Power refers to the rate at which work is done.
  • Therefore; Power = Work done ÷ time
  • It is measured in joules per seconds or Watts

In this case, we are required to convert 0.3227 MW to kilowatts

We need to know that;

  • 10^6 watts = 1 Megawatts(MW)
  • 10^3 Watts = 1 kilowatts (kW)

Therefore;

10^3 kW = 1 MW

Therefore, the suitable conversion factor is 10^3kW/MW

Hence;

0.3227 MW is equivalent to;

   = 0.3227 MW × 10^3kW/MW

   = 322.7 kW

Thus, the peak power output is 322.7 kW

3 0
3 years ago
The equilibrium constant, Kc, for the following reaction is 0.967 at 650 K. 2NH3(g) N2(g) 3H2(g) When a sufficiently large sampl
AlekseyPX

Answer: Concentration of NH_3 in the equilibrium mixture is 0.31 M

Explanation:

Equilibrium concentration of H_2 = 0.729 M

The given balanced equilibrium reaction is,

                 2NH_3(g)\rightleftharpoons N_2(g)+3H_2(g)

Initial conc.            x                0           0

At eqm. conc.     (x-2y) M     (y) M   (3y) M

The expression for equilibrium constant for this reaction will be:

3y = 0.729 M

y = 0.243 M

K_c=\frac{[y]\times [3y]^3}{[x-2y]^2}

Now put all the given values in this expression, we get :

K_c=\frac{0.243\times (0.729)^3}{(x-2\times 0.243)^2}

0.967=\frac{0.243\times (0.729)^3}{(x-2\times 0.243)^2}

x=0.80

concentration of NH_3 in the equilibrium mixture = 0.80-2\times 0.243=0.31

Thus concentration of NH_3 in the equilibrium mixture is 0.31 M

3 0
3 years ago
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