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3 years ago

Why would you rather have hot cocoa than lemonade on a cold day? (The lesson is called heat transfer)

Chemistry

1 answer:

storchak [24]3 years ago

6 0

Answer:

you would more than likely have hot coca.

Explanation:

Because when its cold out you don't want something cold, its common sense lol.

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Which would be the biggest long-term benefit of building a city recycling center? A. building the new recycling building

Shkiper50 [21]

Answer:

Explanation:

garbage being reduced is a far more long term other than the other stupid answer choices

7 0

3 years ago

What was the initial temperature of a gas contained in a 2.0-L bottle at a pressure of 1.00 atm if it was placed under vacuum an

lesya692 [45]

V1/T1 = V2/T2

Substitute the value use ratio and proportion. Use calculator.

V1 = (V2 x T1) / T2

1 is initial, 2 is final

4 0

3 years ago

which of the following requires the most energy to break a bond? a. breaking cl-br bond. b. breaking a n-p bond. c. breaking a o

irina1246 [14]

A. Breaking a Cl-Br bond

6 0

3 years ago

What volume of 24% trichloroacetic acid (tca) is needed to prepare eight 3 ounce bottles of 10% tca solution?

Sever21 [200]

Answer:

295.7 mL of 24% trichloroacetic acid (tca) is needed .

Explanation:

Let the volume of 24% trichloroacetic acid solution be x

Volume of required 10% trichloroacetic acid solution =8 bottles of 3 ounces

= 24 ounces = 709.68 mL

(1 ounces = 29.57 mL)

Amount of trichloroacetic acid in 24% solution of x volume of solution will be equal to amount of trichloroacetic acid in 10% solution of volume 709.68 mL.

$x\times \frac{24}{100}=709.68 mL\times \frac{10}{100}$

x = 295.7 mL

295.7 mL of 24% trichloroacetic acid (tca) is needed .

6 0

3 years ago

Calculate E ° for the half‑reaction, AgCl ( s ) + e − − ⇀ ↽ − Ag ( s ) + Cl − ( aq ) given that the solubility product constant

antoniya [11.8K]

Answer: The value of $E^{o}$ for the half-cell reaction is 0.222 V.

Explanation:

Equation for solubility equilibrium is as follows.

$AgCl(s) \rightleftharpoons Ag^{+}(aq) + Cl^{-}(aq)$

Its solubility product will be as follows.

$K_{sp} = [Ag^{+}][Cl^{-}]$

Cell reaction for this equation is as follows.

$Ag(s)| AgCl(s)|Cl^{-}(0.1 M)|| Ag^{+}(1.0 M)| Ag(s)$

Reduction half-reaction: $Ag^{+} + 1e^{-} \rightarrow Ag(s)$ , $E^{o}_{Ag^{+}/Ag} = 0.799 V$

Oxidation half-reaction: $Ag(s) + Cl^{-}(aq) \rightarrow AgCl(s) + 1e^{-}$ , $E^{o}_{AgCl/Ag}$ = ?

Cell reaction: $Ag^{+}(aq) + Cl^{-}(aq) \rightarrow AgCl(s)$

So, for this cell reaction the number of moles of electrons transferred are n = 1.

Solubility product, $K_{sp} = [Ag^{+}][Cl^{-}]$

= $1.77 \times 10^{-10}$

Therefore, according to the Nernst equation

$E_{cell} = E^{o}_{cell} - \frac{0.0592 V}{n} log \frac{[AgCl]}{[Ag^{+}][Cl^{-}]}$

At equilibrium, $E_{cell}$ = 0.00 V

Putting the given values into the above formula as follows.

$E_{cell} = E^{o}_{cell} - \frac{0.0592 V}{n} log \frac{[AgCl]}{[Ag^{+}][Cl^{-}]}$

$0.00 = E^{o}_{cell} - \frac{0.0592 V}{1} log \frac{1}{[Ag^{+}][Cl^{-}]}$

$E^{o}_{cell} = \frac{0.0592}{1} log \frac{1}{K_{sp}}$

= $0.0591 V \times log \frac{1}{1.77 \times 10^{-10}}$

= 0.577 V

Hence, we will calculate the standard cell potential as follows.

$E^{o}_{cell} = E^{o}_{cathode} - E^{o}_{anode}$

$0.577 V = E^{o}_{Ag^{+}/Ag} - E^{o}_{AgCl/Ag}$

$0.577 V = 0.799 V - E^{o}_{AgCl/Ag}$

$E^{o}_{AgCl/Ag}$ = 0.222 V

Thus, we can conclude that value of $E^{o}$ for the half-cell reaction is 0.222 V.

3 0

3 years ago