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4 years ago

I need help on chemistry conversions

Chemistry

1 answer:

zysi [14]4 years ago

6 0

Answer:

The answer to your questions are below

Explanation:

7.- 10 mm ----------------------- 1 cm

x ------------------------ 456.7 cm

x = (456.7 x 10) / 1

x = 4567 mm

8.- 1 m ---------------------- 1000 mm

x ---------------------- 6537.6 mm

x = (6537.6 x 1) / 1000

x = 6.54 m

9.- 1 cm ----------------------- 10 mm

276.55 cm ---------------------- x

x = (276.55 x 10) / 1

x = 2765.5 mm

10.- 1 ml ------------------------ 1 cm³

0.555ml -------------------- x

x = 0.555 cm³

11.- 1 km ------------------------ 1000000 mm

57995 km ----------------------- x

x = (57995 x 10000000) / 1

x = 57995000000 mm

12.- 1 l ----------------------- 1000 cm³

3329 l ------------------- x

x = (3329 x 1000) / 1

x = 3329000 cm³

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4 years ago

Saturated hydrocarbons are generally unreactive except in the presence of _____.

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3 years ago

For the reaction 2Co3+(aq)+2Cl−(aq)→2Co2+(aq)+Cl2(g). E∘=0.483 V what is the cell potential at 25 ∘C if the concentrations are [

sveta [45]

Explanation:

The given data is as follows.

$E^{o}$ = 0.483, $[Co^{3+}]$ = 0.173 M,

$[Co^{2+}]$ = 0.433 M, $[Cl^{-}]$ = 0.306 M,

$P_{Cl_{2}}$ = 9.0 atm

According to the ideal gas equation, PV = nRT

or, P = $\frac{n}{V}RT$

Also, we know that

Density = $\frac{mass}{volume}$

So, P = MRT

and, M = $\frac{P}{RT}$

= $\frac{9.0 atm}{0.0820 L atm/mol K \times 298 K}$

= $\frac{9.0}{24.436}$

= 0.368 mol/L

Now, we will calculate the cell potential as follows.

E = $E^{o} - \frac{0.0591}{n} log \frac{[Co^{2+}]^{2}[Cl_{2}]}{[Co^{3+}][Cl^{-}]^{2}}$

= $0.483 - \frac{0.0591}{2} log \frac{(0.433)^{2}(0.368)}{(0.173)(0.306)^{2}}$

= $0.483 - 0.02955 log \frac{0.0689}{0.0162}$

= $0.483 - 0.02955 \times 0.628$

= 0.483 - 0.0185

= 0.4645 V

Thus, we can conclude that the cell potential of given cell at $25^{o}C$ is 0.4645 V.

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4 years ago

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