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anzhelika [568]
3 years ago
9

Identify the reactant that gets oxidized in the following reaction.

Chemistry
1 answer:
Nat2105 [25]3 years ago
6 0

Answer:

E) This is not an Oxidation/ Reduction type of reaction

Explanation:

Oxidation involves the loss of electrons while reduction involves the addition of electrons. None of the elements change their oxidation number in the reaction, so this is not an oxidation/reduction type of reaction.

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Calculate the ph of the solution resulting by mixing 20.0 ml of 0.15 m hcl with 20.0 ml of 0.10 m koh
Aliun [14]

Answer:

1.60.

Explanation:

  • The no. of millimoles of HCl = MV = (0.15 M)(20.0 mL) = 3.0 mmol.
  • The no. of millimoles of KOH = MV = (0.10 M)(20.0 mL) = 2.0 mmol.

<em>Since the no. of millimoles of HCl is larger than that of KOH. The solution is acidic.</em>

<em></em>

∴ M of remaining HCl [H⁺] remaining = (NV)HCl - (NV)KOH/V total = (3.0 mmol) - (2.0 mmol) / (40.0 mL) = 0.025 M.

∵ pH = - log[H⁺]

<em>∴ pH = - log[H⁺] </em>= - log(0.025) = <em>1.602 ≅ 1.60.</em>

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atroni [7]

<u>Answer: </u>The amount of heat released is 84 calories.

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The equation used to calculate the amount of heat released or absorbed, we use the equation:

Q= m\times c\times \Delta T

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Q = heat gained  or released = ? Cal

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Putting values in above equation, we get:

\Delta T={\text{Change in temperature}}=(10-50)^oC=-40^oC  

Q=10g\times 0.21Cal/g^oC\times-40^oC

Q = -84 Calories

Hence, the amount of heat released is 84 calories.

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