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Gnoma [55]
3 years ago
12

A worker is told her chances of being killed by a particular process are 1 in every 300 years. Should the worker be satisfied or

alarmed? What is the FAR (assuming normal working hours) and the deaths per person per year? What should her chances be, assuming an average chemical plant?
Chemistry
1 answer:
Pavlova-9 [17]3 years ago
4 0

Answer:

(a) Yes, he should be worried. The Fatal accident rate (FAR) is too high according to standars of the industry. This chemical plant has a FAR of 167, where in average chemical plants the FAR is about 4.

(b) FAR=167 and Death poer person per year = 0.0033 deaths/year.

(c) The expected number of fatalities on a average chemical plant are one in 12500 years.

Explanation:

Asumming 50 weeks of work, with 40 hours/week, we have 2000 work hours a year.

In 300 years we have 600,000 hours.

With these estimations, we have (1/600,000)=1.67*10^(-6) deaths/hour.

If we have 2000 work hours a year, it is expected 0.0033 deaths/year.

1.67*10^{-6} \frac{deaths}{hour}*2000 \frac{hours}{year}=0.0033 deaths/year

The Fatal accident rate (FAR) can be expressed as the expected number of fatalities in 100 millions hours (10^(8) hours).

In these case we have calculated 1.67*10^(-6) deaths/hour, so we can estimate FAR as:

FAR=1.67*10^{-6} \frac{deaths}{hour}*10^{8}  hours=1.67*10^{2} =167

A FAR of 167 is very high compared to the typical chemical plants (FAR=4), so the worker has reasons to be worried.

If we assume FAR=4, as in an average chemical plant, we expect

4\frac{deaths}{10^{8} hour} *2000\frac{hours}{year}=8*10^{-5} \frac{deaths}{year}

This is equivalent to say

\frac{1}{8*10^{-5} } \frac{years}{death}=1.25*10^{4} \frac{years}{death} =12500 \, \frac{years}{death}

The expected number of fatalities on a average chemical plant are one in 12500 years.

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Calculate the mass percent (m/m) of a solution prepared by dissolving 51.56 g of NaCl in 164.2 g of H2O. Express your answer to
denis-greek [22]

Answer:

"23.896%" is the right answer.

Explanation:

The given values are:

Mass of NaCl,

= 51.56 g

Mass of H₂O,

= 165.6 g

As we know,

⇒  Mass of solution = Mass \ of \ (NaCl+H_2O)

                                 = 51.56+164.2

                                 = 215.76 \ g

hence,

⇒ Mass \ percent =\frac{Mass \ of \ NaCl}{Mass \ of \ solution}\times 100

                           =\frac{51.56}{215.76}\times 100

                           =23.896 \ percent

4 0
2 years ago
For the diprotic weak acid h2a, ka1 = 3.2 × 10-6 and ka2 = 6.1 × 10-9. what is the ph of a 0.0650 m solution of h2a? what are th
Stolb23 [73]
Given:

Diprotic weak acid H2A:
 
Ka1 = 3.2 x 10^-6
Ka2 = 6.1 x 10^-9. 
Concentration = 0.0650 m 

Balanced chemical equation:

H2A ===> 2H+  + A2- 
0.0650       0        0
-x                2x       x
------------------------------
0.065 - x     2x      x

ka1 = 3.2 x 10^-6 = [2x]^2 * [x] / (0.065 - x)

solve for x and determine the concentration at equilibrium. 


5 0
3 years ago
Calculate the enthalpy of formation of butane, C4H10, using the balanced chemical
melisa1 [442]

Answer:

-125.4

Explanation:

Target equation is 4C(s) + 5H2(g) = C4H10

These are the data equations for enthalpy of combustion

  1. C(s) + O2(g) =O2(g) -393.5 kJ/mol * 4
  2. H2(g) + ½O2(g) =H20(l) = 285.8 kJ/mol * 5
  3. 2CO2(g) + 3H2O(l) = 13/2O2 (g) + C4H10 - 2877.1 reverse

To get target equation multiply data equation 1 by 4; multiply equation 2 by 5; and reverse equation 3, so...

Calculate 4(-393.5) + 5(-285.8) + 2877.6 and you should get the answer.

7 0
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Determine how many ml of water you need to remove, by evaporation, if you have a 500 ml of 10.20 M HNO3 dilute solution and you
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The total volume of water that would be removed will be 75 mL

<h3>Dilution equation</h3>

Using the dilution equation:

M1V1 = M2V2

In this case, M1 = 500 mL, V1 = 10.20 M, M2 = 12 M

Substitute:

V2 = 500 x 10.20/12

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The final volume in order to arrive at 12 M HNO3 would be 425 mL from the initial 500 mL. Thus, the total amount of water that will be removed by evaporation can be calculated as:

500 - 425 = 75 mL

More on dilution can be found here: brainly.com/question/7208939

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