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stira [4]
3 years ago
5

8. How long does it take a 19 kW steam engine to do 6.8 x 107 J of work?

Chemistry
1 answer:
kykrilka [37]3 years ago
8 0

Answer:

3.58 × 10³ seconds or 59.65 minutes

Explanation:

We are given;

  • Power of a steam engine as 19 kW
  • Work done as 6.8 × 10^7 Joules

We are required to calculate time;

We need to know that;

  • Power = Work done ÷ time
  • Power is the rate of doing work.

19 kW = 19,000 Watt

Rearranging the formula;

time = Work done ÷ power

       =  6.8 × 10^7 Joules ÷ 19,000 watt

       = 3578.95 seconds

        = 3.58 × 10³ seconds

But, 60 seconds = 1 minute

Therefore, t = 59.65 minutes

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3 years ago
Be sure to answer all parts. The percent by mass of bicarbonate (HCO3−) in a certain Alka-Seltzer product is 32.5 percent. Calcu
pochemuha

Answer : The volume of CO_2 will be, 514.11 ml

Explanation :

The balanced chemical reaction will be,

HCO_3^-+HCl\rightarrow Cl^-+H_2O+CO_2

First we have to calculate the  mass of HCO_3^- in tablet.

\text{Mass of }HCO_3^-\text{ in tablet}=32.5\% \times 3.79g=\frac{32.5}{100}\times 3.79g=1.23175g

Now we have to calculate the moles of HCO_3^-.

Molar mass of HCO_3^- = 1 + 12 + 3(16) = 61 g/mole

\text{Moles of }HCO_3^-=\frac{\text{Mass of }HCO_3^-}{\text{Molar mass of }HCO_3^-}=\frac{1.23175g}{61g/mole}=0.0202moles

Now we have to calculate the moles of CO_2.

From the balanced chemical reaction, we conclude that

As, 1 mole of HCO_3^- react to give 1 mole of CO_2

So, 0.0202 mole of HCO_3^- react to give 0.0202 mole of CO_2

The moles of CO_2 = 0.0202 mole

Now we have to calculate the volume of CO_2 by using ideal gas equation.

PV=nRT

where,

P = pressure of gas = 1.00 atm

V = volume of gas = ?

T = temperature of gas = 37^oC=273+37=310K

n = number of moles of gas = 0.0202 mole

R = gas constant = 0.0821 L.atm/mole.K

Now put all the given values in the ideal gas equation, we get :

(1.00atm)\times V=0.0202 mole\times (0.0821L.atm/mole.K)\times (310K)

V=0.51411L=514.11ml

Therefore, the volume of CO_2 will be, 514.11 ml

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