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9966 [12]
3 years ago
6

Determine the potential deaths resulting from the following exposure to chlorine:

Chemistry
1 answer:
jekas [21]3 years ago
6 0

Answer:

For each scenario as following:

A. 3 Potential deaths by chlorine exposure

B. 1 Potential deaths by chlorine exposure

C. 3 Potential deaths by chlorine exposure

Explanation:

According to Freitag, 1941 Chlorine exposure can be lethal at the concentration of 34-51 ppm in a time of 1h-1.5h. The answers are based on his reference.

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liberstina [14]
The universe comes into existence is first
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The universe begins expanding is third
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Atomic nuclei form is last

I'm almost certain that is correct. Do not take my word for this.
6 0
3 years ago
Read 2 more answers
What volume of water vapor would be produced from the combustion of 815.74 grams of propane (C3H8) with 1,006.29 grams of oxygen
d1i1m1o1n [39]

3940.2 is the volume of water vapour that would be produced from the combustion of 815.74 grams of propane (C_3H_8) with 1,006.29 grams of oxygen gas, under a pressure of 1.05 atm and a temperature of 350. degrees C.

<h3>What is an ideal gas equation?</h3>

The ideal gas law (PV = nRT) relates the macroscopic properties of ideal gases. An ideal gas is a gas in which the particles (a) do not attract or repel one another and (b) take up no space (have no volume).

Stoichiometric calculations:

C_3H_8(g) + 5 O_2(g)→ 3 CO_2(g) + 4 H_2O(g)

From the equation of the reaction, the mole ratio of propane to oxygen is 1:5.

Mole of 815.74 grams of propane = \frac{ 815.74}{44.1 }

Mole of 815.74 grams of propane = 18.49750567 moles

Mole of  1,006.29 grams of oxygen =\frac{ 1,006.29}{32 }

Mole of  1,006.29 grams of oxygen = 31.4465625 moles

Going by the mole ratio, it appears propane is limiting while oxygen is in excess.

From the equation, 1 mole of propane produces 4 moles of water vapour. Thus, the equivalent mole of water vapour will be:

18.49750567 moles x 4 = 73.99 moles.

Using the ideal gas equation:

PV = nRT

v = (73.99  x 0.08206 x 623) ÷ 0.96

v =  3940.2

Hence, 3940.2 is the volume of water vapour that would be produced from the combustion of 815.74 grams of propane (C_3H_8) with 1,006.29 grams of oxygen gas, under a pressure of 1.05 atm and a temperature of 350. degrees C.

Learn more about the ideal gas here:

brainly.com/question/27691721

#SPJ1

7 0
2 years ago
In a chemical equation the sum of the masses
monitta
The sum of the masses of the reactants must equal the sum of the masses of the products; as required by 
the Principle of Conservation of Mass. 
5 0
3 years ago
A sample of water vapor has a volume of 3.15 L, a pressure of 2.40 atm, and a temperature of 325 K. What is the new temperature,
lara31 [8.8K]

Answer:

The answer to your question is:   T2 = 235.44 °K

Explanation:

Data

V1 = 3.15 L                    V2 = 2.78 L

P1 = 2.40 atm               P2 = 1.97 atm

T1 = 325°K                    T2 = ?

Formula

\frac{P1V1}{T1} = \frac{P2V2}{T2}

Process

            T2 = (P2V2T1) / (P1V1)

            T2 = (1.97x 2.78x 325) / (2.40 x 3.15)

            T2 = 1779.895 / 7.56

            T2 = 235.44 °K

4 0
2 years ago
When electrons are added to the outer most shell of a carbon atom what does it form
bagirrra123 [75]
When electrons are added to the outermost shell of a carbon atom, it forms an anion.
3 0
2 years ago
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