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rjkz [21]
4 years ago
9

At which point was neutralization complete for the acid that is being titrated below?

Chemistry
2 answers:
Hatshy [7]4 years ago
8 0

Answer:

D.

Explanation:

  • From the shown titration curve, It is a titration curve for a diprotic acid (H₂A).
  • The neutralization of the first proton is at point A.
  • The neutralization of the second proton (complete neutralization) is at point D.

<em>So, the right choice is: D.</em>

Rom4ik [11]4 years ago
8 0

D on e2020

extra words to get to 20 letters

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How much energy is required to heat 87.1 g acetone (molar mass=58.08 g/mol) from a solid at -154.0°C to a liquid at -42.0°C? The
WARRIOR [948]

Answer:

The answer to the question above is

The energy required to heat 87.1 g acetone from a solid at -154.0°C to a liquid at -42.0°C = 29.36 kJ

Explanation:

The given variables are

ΔHfus = 7.27 kJ/mol

Cliq = 2.16 J/g°C

Cgas = 1.29 J/g°C

Csol = 1.65 J/g°C

Tmelting = -95.0°C.

Initial temperature = -154.0°C

Final temperature = -42.0°C?

Mass of acetone = 87.1 g

Molar mass of acetone = 58.08 g/mol

Solution

Heat required to raise the temperature of solid acetone from -154 °C to -95 °C or 59 °C is given by

H = mCsolT = 87.1 g* 1.65 J/g°C* 59 °C = 8479.185 J

Heat required to melt the acetone at -95 °C = ΔHfus*number of moles =

But number of moles = mass÷(molar mass) = 87.1÷58.08 = 1.5

Heat required to melt the acetone at -95 °C =1.5 moles*7.27 kJ/mol = 10.905 kJ

The heat required to raise the temperature to -42 degrees is

H = m*Cliq*T = 87.1 g* 2.16 J/g°C * 53 °C = 9971.21 J

Total heat = 9971.21 J + 10.905 kJ + 8479.185 J = 29355.393 J = 29.36 kJ

The energy required to heat 87.1 g acetone from a solid at -154.0°C to a liquid at -42.0°C is 29.36 kJ

4 0
3 years ago
What volume of 12.0 M sulfuric acid is needed to prepare 300.0 mL of 2.50 M solution?
Zepler [3.9K]

Answer:

n=c*V

=2.50*300x10^-3

=.75 moles of hcl

v=n/c

=.75/12

=.0625 L

convert to mL

=62.5mL

4 0
4 years ago
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