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kompoz [17]
3 years ago
5

Question 10

Chemistry
1 answer:
fomenos3 years ago
7 0

Answer:0.00061L He

Explanation:Avogadro constant, N = 6.02214076*10^23 per mole

at STP 1 mole = 22.71L

1.64*10^19 He = ( 1.64*10^19/N)*22.71 = 1.64*10^19*[3.77*10^-23] = 6.18*10^-4L

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In the citric acid cycle, atp molecules are produced by _____.
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Can a non-polar molecule contain polar bonds? Explain your answer with an example.
Rom4ik [11]

Answer:

Yes

Explanation:

It is possible that a non polar molecule contain polar bonds.

Consider the example of CO₂:

In carbon dioxide the atoms are arranged in angle of 180 degree and molecule is linear. So in this way the net dipole movement becomes zero and molecule is non polar.

But when we see the electronegativity of both bonded atoms the oxygen is more electronegative than carbon. The oxygen atom have partial negative charge and carbon have partial positive charge because shared pair of electron is more attracted towards the oxygen. Thus bond become polar because one end is partial positive and other is partial negative. Thus a non polar carbon dioxide molecule have polar bond.

8 0
3 years ago
A buffer solution contains 0.240 M ammonium chloride and 0.499 M ammonia. If 0.0565 moles of perchloric acid are added to 250 mL
snow_lady [41]

Answer:

The pH of the resulting solution is 9.02.

Explanation:

The initial pH of the buffer solution can be found using the Henderson-Hasselbalch equation:

pOH = pKb + log(\frac{[NH_{4}Cl]}{[NH_{3}]})  

pOH = -log(1.78\cdot 10^{-5}) + log(\frac{0.240}{0.499}) = 4.43        

pH = 14 - pOH = 14 - 4.43 = 9.57

Now, the perchloric acid added will react with ammonia:

n_{NH_{3}} = 0.499moles/L*0.250 L - 0.0565 moles = 0.0683 moles

Also, the moles of ammonium chloride will increase in the same quantity according to the following reaction:

NH₃ + H₃O⁺ ⇄ NH₄⁺ + H₂O    

n_{NH_{4}Cl} = 0.240 moles/L*0.250L + 0.0565 moles = 0.1165 moles

Finally, we can calculate the pH of the resulting solution:

pOH = pKb + log(\frac{[NH_{4}Cl]}{[NH_{3}]})  

pOH = -log(1.78\cdot 10^{-5}) + log(\frac{0.1165 moles/0.250 L}{0.0683 moles/0.250 L}) = 4.98  

pH = 14 - 4.98 = 9.02

Therefore, the pH of the resulting solution is 9.02.

I hope it helps you!

5 0
3 years ago
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