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lorasvet [3.4K]
3 years ago
13

ELETE

Chemistry
1 answer:
yKpoI14uk [10]3 years ago
6 0

Answer:

5 L

Explanation:

Use Charles law and rearrange formula

Change C to K

- Hope that helped! Please let me know if you need further explanation.

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Identify the limiting reactant when 9.65-g H2SO4 reacts with 6.10-g of NaOH.
Masteriza [31]

Answer:

3.56

Explanation:

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2 years ago
Functions of the skin other than thermoregulation​
german

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Protection and sensation

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Help me with this question please !!<br> A<br> B <br> C<br> D
RUDIKE [14]

Answer:

c becuase I'm not to good at this but I'm pretty sure

Explanation:

I think

7 0
2 years ago
what is the difference between the number of electrons in an atom of selenium and the number of electrons in an atom of aluminum
igomit [66]

Answer:

The atomic number of Selenium is 34. This means that Selenium possesses 34 electrons.

The atomic number of Aluminium is 13. This means that Aluminium has 13 electrons.

Hence, there is a difference of 21 between the number of electrons in an atom of selenium and the number of electrons in an atom of aluminium.

Selenium has 6 electrons in it's outer most shell whereas aluminium has 3 electrons in its outer most shell. As a result, aluminium will have a greater tendency to lose one of its outer most electrons to become stable.

7 0
3 years ago
What is the pH of a 3.40 mM of acetic acid, 500 mL in which 50 mL of 1.00 M NaOAc has been added? Ka HOAc is 1.8 x 10-5? What pe
Deffense [45]

Answer:

a) pH = 4.213

b) % dis = 2 %

Explanation:

Ch3COONa → CH3COO- + Na+

CH3COOH ↔ CH3COO- + H3O+

∴ Ka = 1.8 E-5 = ([ CH3COO- ] * [ H3O+ ]) / [ CH3COOH ]

mass balance:

⇒ <em>C</em> CH3COOH + <em>C</em> CH3COONa = [ CH3COOH ] + [ CH3COO- ]

<em>∴ C </em>CH3COOH = 3.40 mM = 3.4 mmol/mL * ( mol/1000mmol)*(1000mL/L)

∴ <em>C</em> CH3COONa = 1.00 M = 1.00 mol/L = 1.00 mmol/mL

⇒ [ CH3COOH ] = 4.4 - [ CH3COO- ]

charge balance:

⇒ [ H3O+ ] + [ Na+ ] = [ CH3COO- ] + [ OH- ]....is negligible [ OH-], comes from water

⇒ [ CH3COO- ] = [ H3O+ ] + 1.00

⇒ Ka = (( [ H3O+ ] + 1 )* [ H3O+ ]) / ( 3.4 - [ H3O+])) = 1.8 E-5

⇒ [ H3O+ ]² + [ H3O+ ] = 6.12 E-5 - 1.8 E-5 [ H3O+ ]

⇒ [ H3O+ ]² + [ H3O+ ] - 6.12 E-5 = 0

⇒  [ H3O+ ] = 6.12 E-5 M

⇒ pH = - Log [ H3O+ ] = 4.213

b) (% dis)* mol acid = <em>C</em> CH3COOH = 3.4

∴ mol CH3COOH = 500*3.4 = 1700 mmol = 1.7 mol

⇒ % dis = 3.4 / 1.7 = 2 %

7 0
3 years ago
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