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Mars2501 [29]
2 years ago
12

In May 2016, William Trubridge broke the world record in free diving (diving underwater without the use of supplemental oxygen)

by diving to a depth of 124 m. Assume that he takes a breath that fills his lungs to 3.6 L at the surface of the water (1.0 atm). Calculate the volume that this same amount of air will occupy in his lungs when he reaches a depth of 124 m (13.3 atm). Enter the answer in units of liters.
Chemistry
1 answer:
Brrunno [24]2 years ago
8 0

Answer:

The volume that this same amount of air will occupy in his lungs when he reaches a depth of 124 m is - 0.27 L.

Explanation:

Using Boyle's law  

{P_1}\times {V_1}={P_2}\times {V_2}

Given ,  

V₁ = 3.6 L  

V₂ = ?

P₁ = 1.0 atm

P₂ = 13.3 atm

Using above equation as:

{P_1}\times {V_1}={P_2}\times {V_2}

{1.0\ atm}\times {3.6\ L}={13.3\ atm}\times {V_2}

{V_2}=\frac{{1.0}\times {3.6}}{13.3}\ L

{V_2}=0.27\ L

<u>The volume that this same amount of air will occupy in his lungs when he reaches a depth of 124 m is - 0.27 L.</u>

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Answer:

75

Explanation:

4 0
2 years ago
Suppose you wanted to make a buffer of exactly ph 7.00 using kh2po4 and na2hpo4. if the final solution was 0.10 m in kh2po4, wha
OleMash [197]

Answer:- 0.138 M

Solution:- The buffer pH is calculated using Handerson equation:

pH=pKa+log(\frac{base}{acid})

KH_2PO_4 acts as a weak acid and Na_2HPO_4 as a base which is pretty conjugate base of the weak acid we have.

The acid hase two protons(hydrogen) where as the base has only one proton. So, we could write the equation as:

H_2PO_4^-\rightleftharpoons H^++HPO_4^-^2

Phosphoric acid gives protons in three steps. So, the above equation is the second step as the acid has only two protons and the base has one proton.

So, we will use the second pKa value. The acid concentration is given as 0.10 M and we are asked to calculate the concentration of the base to make a buffer of exactly pH 7.00.

Let's plug in the values in the equation:

7.00=6.86+log(\frac{base}{0.10})

7.00-6.86=log(\frac{base}{0.10})

0.14=log(\frac{base}{0.10})

Taking antilog:

10^0^.^1^4=\frac{base}{0.10}

1.38=\frac{base}{0.10}

On cross multiply:

[base] = 1.38(0.10)

[base] = 0.138

So, the concentration of the base that is Na_2HPO_4 required to make the buffer is 0.138M.

5 0
2 years ago
When 500.0 g of water is decomposed by electrolysis and the yield of hydrogen is only 75.3%, how much hydrogen chloride can be m
Evgen [1.6K]

The amount of hydrogen chloride that can be made is 1064 g

Why?

The two reactions are:

2H₂O → 2H₂ + O₂ 75.3 % yield

H₂ + Cl₂ → 2HCl 69.8% yield

We have to apply a big conversion factor to go from grams of water (The limiting reactant), to grams of HCl, the final product. We have to be very careful with the coefficients and percentage yields!

500.0gH_2O*\frac{1moleH_2O}{18.01 gH_2O}*\frac{2 moles H_2}{2 moles H_2O}*\frac{2.015g H_2}{1 mole H_2}*\frac{75.3 actual g}{100 theoretical g}=42.12 g H_2

42.12H_2*\frac{1 mole H_2}{2.015gH_2}*\frac{2 moles HCl}{1 mole H_2}*\frac{36.46g}{1 mole HCl}*\frac{69.8 actualg}{100 theoreticalg} =1064gHCl

Have a nice day!

#LearnwithBrainly

7 0
2 years ago
How many moles of lead (ii) chromate are in 51 grams of this substance? answer in units of mol?
maria [59]

Mass of lead (II) chromate is 51 g. The molecular formula is PbCrO_{4} and its molar mass is 323.2 g/mol

Number of moles can be calculated using the following formula:

n=\frac{m}{M}

Here, m is mass and M is molar mass.

Putting the values,

n=\frac{(51 g}{323.1937 g/mol}=0.1578 mol

Therefore, number of moles of lead (II) chromate will be 0.1578 mol.

5 0
1 year ago
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2.Robert Hooke was the first to use the word “cell.”

3.Anton van Leeuwenhoek observed small organisms he called “animalcules.”

4.Matthias Schleiden theorized that plants are made of cells.

5.Theodor Schwann theorized that animals are made of cells.

6.Rudolf Virchow theorized that cells come from other pre-existing cells.

3 0
2 years ago
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