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rusak2 [61]
3 years ago
15

Write a scientific argument that addresses the question: Which suspect is most likely to have made the hydrofluoric acid? First,

state your claim. Then, use evidence to support your claim. For each piece of evidence you use, explain how the evidence supports your claim. As you write, refer back to your completed Modeling Tool sheets and Science Seminar Evidence Cards.
We need 2 reasons why its either Pat or Tracy. I need a paragraph on it.
ILL GIVE BRANLIEST TO THE BEST ANSWER THATS A PARAGRAPH AND GIVES 2 REASONS WHY ITS EITHER PAT OR TRACY!!

Chemistry
1 answer:
aniked [119]3 years ago
7 0

Answer:

Pat is most likely to have made the hydrofluoric acid.

Explanation:

How is hydrofluoric acid made?

It is manufactured by heating purified fluorspar (calcium fluoride) with concentrated sulfuric acid to produce the gas, which is then condensed by cooling or dissolving in water. ... The acid hydrolysis of fluorite-containing minerals generates an impure gas stream consisting of sulfur dioxide, water and HF.

that's all I got sorry

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CH4 with pressure 1 atm and volume 10 liter at 27°C is passed into a reactor with 20% excess oxygen, how many moles of oxygen is
BaLLatris [955]

Answer : The moles of O_2 left in the products are 0.16 moles.

Explanation :

First we have to calculate the moles of CH_4.

Using ideal gas equation:

PV=nRT

where,

P = pressure of gas = 1 atm

V = volume of gas = 10 L

T = temperature of gas = 27^oC=273+27=300K

n = number of moles of gas = ?

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

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

(1atm)\times (10L)=n\times (0.0821L.atm/mol.K)\times (300K)

n=0.406mole

Now we have to calculate the moles of O_2.

The balanced chemical reaction will be:

CH_4+2O_2\rightarrow CO_2+2H_2O

From the balanced reaction we conclude that,

As, 1 mole of CH_4 react with 2 moles of O_2

So, 0.406 mole of CH_4 react with 2\times 0.406=0.812 moles of O_2

Now we have to calculate the excess moles of O_2.

O_2 is 20 % excess. That means,

Excess moles of O_2 = \frac{(100 + 20)}{100} × Required moles of O_2

Excess moles of O_2 = 1.2 × Required moles of O_2

Excess moles of O_2 = 1.2 × 0.812 = 0.97 mole

Now we have to calculate the moles of O_2 left in the products.

Moles of O_2 left in the products = Excess moles of O_2 - Required moles of O_2

Moles of O_2 left in the products = 0.97 - 0.812 = 0.16 mole

Therefore, the moles of O_2 left in the products are 0.16 moles.

7 0
3 years ago
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