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2 years ago

Pls help solving question (d)!!

Chemistry

1 answer:

VARVARA [1.3K]2 years ago

7 0

Answer: 1 C4H6O3 + 1 H2O -> 2 C2H4O2

Explanation:

LEFT HAND SIDE:

C= 4

H=8

O= 4

RIGHT HAND SIDE:

C= 2

H= 4

O= 2

You want the left side to be equal to the right side.

We can multiply each atomic element by 2, to get C4H8O4.

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Question 8 Which one of the following salts does not contain water of crystallisation?

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A gas at 74°C is heated to 120°C so there is pressure reaches 1.79 ATM. What is its initial pressure?

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A collection of 6. 25×1018 electrons has the charge of.

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Explanation:

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1 year ago

Tarnish on copper is the compound CuO. A tarnished copper plate is placed in an aluminum pan of boiling water. When enough salt

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Answer: 2.00 V

Explanation:

The balanced redox reaction is:

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2 years ago

Calculate the pH for each of the following cases in the titration of 25.0 mL of 0.150 M sodium hydroxide with 0.150 M HBr(aq). (

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Answer:

a) pH = 13.176

b) pH = 13

c) pH = 12.574

d) pH = 7.0

e) pH = 1.46

f) pH = 1.21

Explanation:

HBr + NaOH ↔ NaBr + H2O

∴ equivalent point:

⇒ mol acid = mol base

⇒ (Va)*(0.150mol/L) = (0.025L)*(0.150mol/L)

⇒ Va = 0.025 L

a) before addition acid:

NaOH → Na+ + OH-

⇒ C NaOH = 0.150 M

⇒ [ OH- ] = 0.150 M

⇒ pOH = - Log ( 0.150 )

⇒ pOH = 0.824

⇒ pH = 14 - pOH

⇒ pH = 13.176

b) after addition 5mL HBr:

⇒ C NaOH = (( 0.025)*(0.150) - (0.005)*(0.150)) / (0.025 + 0.005) = 0.1 M

⇒ C HBr = (0.005)*(0.150) / ( 0.03 ) = 0.025 M

⇒ [ OH- ] = 0.1 M

⇒ pOH = 1

⇒ pH = 13

c) after addition 15mL HBr:

⇒ C NaOH = ((0.025)*(0.150) - (0.015)*(0.150 ))/(0.04) = 0.0375 M

⇒ C HBr = ((0.015)*(0.150))/(0.04) = 0.0563 M

⇒ [ OH- ] = 0.0375 M

⇒ pOH = 1.426

⇒ pH = 12.574

d) after addition 25mL HBr:

equivalent point:

⇒ [ OH- ] = [ H3O+ ]

⇒ Kw = 1 E-14 = [ H3O+ ] * [ OH- ] = [ H3O+ ]²

⇒ [ H3O+ ] = 1 E-7

⇒ pH = 7.0

d) after addition 40mL HBr:

⇒ C HBr = ((0.04)*(0.150) - (0.025)*(0.150)) / (0.04 + 0.025) = 0.035 M

⇒ [ H3O+ ] = 0.035 M

⇒ pH = 1.46

d) after addition 60mL HBr:

⇒ C HBr = ((0.06)*(0.150) - (0.025)*(0.150)) / (0.06+0.025) = 0.062 M

⇒ [ H3O+ ] = 0.062 M

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