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IceJOKER [234]
3 years ago
8

What is the ph of a solution made by mixing 25.00 ml of 0.100 m hcl with 40.00 ml of 0.100 m koh? assume that the volumes of the

solutions are additive. what is the ph of a solution made by mixing 25.00 ml of 0.100 m hcl with 40.00 ml of 0.100 m koh? assume that the volumes of the solutions are additive. 1.64 0.64 13.36 12.36?
Chemistry
1 answer:
Digiron [165]3 years ago
3 0
<span>Answer: The HCl and KOH will react until one or the other is gone. As you have a larger volume of an equal concentration of HCl, the KOH will go first. moles HCl = 0.04000 L * 0.100 M = 0.00400 moles moles KOH = 0.02500 L * 0.100 M = 0.00250 moles moles HCl left = 0.00400 - 0.00250 = 0.00150 moles Your total volume is now 65.00 mL, so the [HCl] = 0.00150 moles / 0.06500 L = 0.0231 M = [H+] pH = -log [H+] = -log (0.0231) = 1.64</span>
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To keep the evinermental machinces working

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How many neutrons and protons are there in the nuclei of the following atoms. a. Li-7 b. O-16 c. Th-232 d. Pu-239
ANEK [815]

Answer:

a.

P = 3 , N = 4

b.

P = 8 , N = 8

c.

P = 90 , N = 142

d.

P = 94 , N = 145

Explanation:

Mass number  = Number of protons + Number of neutrons

Also,

Atomic number = Number of protons

<u>Li - 7</u>

Given, Mass number of Lithium = 7

For lithium, atomic number = 3

So,

Number of protons = 3

Number of neutrons = Mass number - Number of protons = 7 - 3 = 4

<u>O - 16</u>

Given, Mass number of oxygen = 16

For oxygen, atomic number = 8

So,

Number of protons = 8

Number of neutrons = Mass number - Number of protons = 16 - 8 = 8

<u>Th - 232</u>

Given, Mass number of Thorium = 232

For Thorium, atomic number = 90

So,

Number of protons = 90

Number of neutrons = Mass number - Number of protons = 232 - 90 = 142

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Given, Mass number of Plutonium = 239

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3 0
3 years ago
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B - What we change

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Dependent Variable - What we measure

Control Variable - what stays the same

Conclusion - what we conclude

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5 0
3 years ago
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pav-90 [236]

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Explanation:

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New pressure of air (P2) = ?

Since pressure and volume are given while temperature is held constant, apply the formula for Boyle's law

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Thus, the new pressure of air when the volume is decreased to 111 L is 0.0014 atm

7 0
3 years ago
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