What 5 equipment do you needs to mix a chemical solution

Blizzard [7]

Answer:

o temperature increases

5 0

3 years ago

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

Pavlova-9 [17]

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.

4 0

3 years ago

A higher than normal high tide is called

alina1380 [7]

It is called a spring tide

8 0

3 years ago

A cough syrup contains 5.0% ethyl alcohol, c2h5oh, by mass. if the density of the solution is 0.9928 g/ml, determine the molarit

WARRIOR [948]

To answer the question above, let us a basis of the 1000 mL or 1 L.
volume = (0.9928 g/mL)(1000mL) = 992.8 g
Then, determine the mass of the alcohol by multiplying the total mass by the decimal equivalent of 5%.
mass of alcohol = 0.05(992.8 g) = 49.64 g
Then, determine the number of moles of ethyl alcohol by dividing the mass of alcohol by the molar mass (46 g/mol).
n = 49.64 g/ (46 g/mol) = 1.08 mol
Then, divide the number of moles by the volume (our basis is 1 L)
molarity = 1.08 mol/ 1 L = 1.08 M

5 0

3 years ago

There are 100.0 grams of each reactant available determine the limiting reactant in this equation

Romashka [77]

Since you have not included the chemical reaction I will explain you in detail.

1) To determine the limiting agent you need two things:

- the balanced chemical equation

- the amount of every reactant involved as per the chemical equation

2) The work is:

- state the mole ratios of all the reactants: these are the ratios of the coefficientes of the reactans in the balanced chemical equation.

- determine the number of moles of each reactant with this formula:

number of moles = (mass in grams) / (molar mass)

- set the proportion with the two ratios (theoretical moles and actual moles)

- compare which reactant is below than the stated by the theoretical ratio.

3) Example: determine the limiting agent in this reaction if there are 100 grams of each reactant:

i) Chemical equation: H₂ + O₂ → H₂O

ii) Balanced chemical equation: 2H₂ + O₂ → 2H₂O

iii) Theoretical mole ration of the reactants: 2 moles H₂ : 1 mol O₂

iv) Covert 100 g of H₂ into number of moles

n = 100g / 2g/mol = 50 mol of H₂

v) Convert 100 g of O₂ to moles:

n = 100 g / 32 g/mol = 3.125 mol

vi) Actual ratio: 50 mol H₂ / 3.125 mol O₂

vii) Compare the two ratios:

2 mol H₂ / 1 mol O ₂ < 50 mol H₂ / 3.125 mol O₂

Conclusion: the actual ratio of H₂ to O₂ is greater than the theoretical ratio, meaning that the H₂ is in excess respect to the O₂. And that means that O₂ will be consumed completely while some H₂ will remain without react.

Therefore, the O₂ is the limiting reactant in this example.

7 0

3 years ago