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VladimirAG [237]
3 years ago
7

A candy bar has ____ energy. A. Kinetic B. Potential

Chemistry
2 answers:
Elodia [21]3 years ago
6 0

Answer:

Potential energy

Explanation:

A candy bar has carbs, sugars, and calories which are potentially going to be used as energy in a body.  

NARA [144]3 years ago
3 0

Answer:

B. Potential

Explanation:

Because the candy bar is sitting still.

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Calculate the standard potential for a cell that employs the over-all cell reaction: 2Al(s) + 3 I2(s)→2 Al+3 + 6 I- From reducti
Valentin [98]

Answer:

Explanation:

2Al(s) + 3 I₂(s)   →    2 Al⁺³    +     6 I⁻

Aluminium is oxidised and iodine is reduced .

so cell potential = Ereduction - Eoxidation

Al⁺³ + 3e = Al          -  1.66 V

I₂ + 2 e = 2 I⁻             0.54 V

=  .54 - ( - 1.66 )

= 1.66 + .54

= 2.2  V

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PH is a logarithmic scale used to indicate the hydrogen ion concentration, [H+], of a solution: pH=−log[H+] Due to the autoioniz
ruslelena [56]

Answer:

  • A) pH = 2.42
  • B) pH = 12.00

Explanation:

<em>The dissolution of HCl is  HCl → H⁺ + Cl⁻</em>

  • To solve part A) we need to calculate the concentration of H⁺, to do that we need the moles of H⁺ and the volume.

The problem gives us V=2.5 L, and the moles can be calculated using the molecular weight of HCl, 36.46 g/mol:

0.35g_{HCl}*\frac{1mol_{HCl}}{36.46g_{HCl}} *\frac{1molH^{+}}{1mol HCl} = 9.60*10⁻³ mol H⁺

So the concentration of H⁺ is

[H⁺] = 9.60*10⁻³ mol / 2.5 L = 3.84 * 10⁻³ M

pH = -log [H⁺] = -log (3.84 * 10⁻³) = 2.42

  • <em>The dissolution of NaOH is  NaOH → Na⁺ + OH⁻</em>
  • Now we calculate [OH⁻], we already know that V = 2.0 L, and a similar process is used to calculate the moles of OH⁻, keeping in mind the molecular weight of NaOH, 40 g/mol:

0.80g_{NaOH}*\frac{1mol_{NaOH}}{40g_{NaOH}} *\frac{1molOH^{-}}{1mol NaOH}= 0.02 mol OH⁻

[OH⁻] = 0.02 mol / 2.0 L = 0.01

pOH = -log [OH⁻] = -log (0.01) = 2.00

With the pOH, we can calculate the pH:

pH + pOH = 14.00

pH + 2.00 = 14.00

pH = 12.00

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