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olya-2409 [2.1K]
2 years ago
9

A 32.0 mL sample of hydrogen is collected over water at 20.0°C and 750.0 mm Hg. What is the volume of the dry gas at STP?

Chemistry
1 answer:
Luba_88 [7]2 years ago
5 0

bro thanks for the points tho

Explanation:

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Here’s a simplified explanation.

The <em>protons</em> in the nucleus <em>repel each other</em>. The <em>neutrons provide the “glue”</em> that holds the nucleus together and prevents it from flying apart.

The “glue” is the strong nuclear force. It is strong but extremely short range. It falls off extremely rapidly as the p-n distance increases.

A <em>neon atom</em> has 10 protons. There are three stable isotopes, with 10, 11, and 12 neutrons.

With fewer than 10 protons, the glue is not strong enough to hold the nucleus together.

If there are more than 12 neutrons, the average p-n distance is great enough that the glue has again become too weak.

<em>Gold</em> has one stable isotope. It contains 79 protons and 118 neutrons.

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Explanation:

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3 years ago
Find the pH of the equivalence point and the volume (mL) of 0.0372 M NaOH needed to reach the equivalence point in the titration
elixir [45]

Answer:

8.54

Explanation:

At equivalence point :  

42.2 X 0.052 = Vol. NaOH X 0.0372

Vol of NaOH = 2.1944/0.0372 = 58.99 ml

So volume of NaOH recquired to reach equivalence point = 58.99 ml

Number of miliimoles of CH3COOH = molarity X volume in ml = 42.2 X 0.052             = 2.1944 millimoles

Number of millimoles of NaOH = 58.99 X 0.0372 = 2.1944

Now CH₃COOH and NaOH reacts to give CH₃COONa according to the reaction :

CH₃COOH + NaOH ------> CH₃COONa + H₂O

1 mole of CH₃COOH reacts with 1 mole of NaOH to give 1 mole of CH₃COONa  

So 2.1944 millimoles of CH₃COOH will react with 2.1944 millimoles of NaOH to give 2.1944 millimoles of CH₃COONa

So all the acid (CH₃COOH) and base (NaOH) has been converted into salt (CH₃COONa) so there is no acid or base left.

Now molarity of CH₃COONa = number of millimoles of CH₃COONa/total volume in ml = 2.1944/(58.99 + 42.2) = 2.1944/101.19 = 0.02169 M

So using the hydrolysis equation :  

pH = 1/2 [ pKw + pKa + log c ]  

Ka for acetic acid = 1.75 X 10⁻⁵  

so pKa = -log (1.75 X 10⁻⁵) = 4.74  

Kw = 10⁻¹⁴

so pKw = -log 10⁻¹⁴ = 14

c = 0.02169  

so log c = log 0.02169 = -1.66  

putting the values....  

pH = 1/2 [14 + 4.74 - 1.66 ]  

pH = 1/2 [ 17.08] = 8.54

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