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3 years ago

If a plane travels at a speed of 600 km/h, how long would it take to travel a distance of 300 km

Chemistry

2 answers:

liubo4ka [24]3 years ago

5 0

Answer: 0.5 hour / 30 minutes / 1800s

Explanation: The problem can be expressed through the following equation:

v = d/t

Derive to find t:

t = d/ v

= 300 km /600 km/ h

= 0.5 hr

0.5 hr x 60 mins/ 1 hr = 30 mins

Or in seconds

30 mins x 60 s/ 1 min = 1800 s

iogann1982 [59]3 years ago

4 0

Answer:

30 minutes

Explanation:

If the plane is going 600 km/h

divide the length and time and it would be half of that, so 300 km would take 30 mins.

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3 years ago

Whats the voltage of CuCl2 + Zn -> ZnCl2 + Cu

gtnhenbr [62]

Answer:

Approximately $1.10\; {\rm V}$ under standard conditions.

Explanation:

Equation for the overall reaction:

${\rm CuCl_{2}}\, (aq) + {\rm Zn}\, (s) \to {\rm ZnCl_{2}} \, (aq) + {\rm Cu}\, (s)$ .

Write down the ionic equation for this reaction:

$\begin{aligned}& {\rm Cu^{2+}}\, (aq) + 2\; {\rm Cl^{-}}\, (aq) + {\rm Zn}\, (s)\\ & \to {\rm Zn^{2+}} \, (aq) + 2\; {\rm Cl^{-}}\, (aq) + {\rm Cu}\, (s)\end{aligned}$ .

The net ionic equation for this reaction would be:

${\rm Cu^{2+}}\, (aq) + {\rm Zn}\, (s) \to {\rm Zn^{2+}}\, (aq) + {\rm Cu}\, (s)$ .

In this reaction:

Zinc loses electrons and was oxidized (at the anode): ${\rm Zn}\, (s) \to {\rm Zn^{2+}}\, (aq) + 2\, {\rm e^{-}}$ .
Copper gains electrons and was reduced (at the cathode): ${\rm Cu^{2+}}\, (aq) + 2\, {\rm e^{-}} \to {\rm Cu} \, (s)$ .

Look up the standard potentials for each half-reaction on a table of standard reduction potentials.

Notice that ${\rm Zn}\, (s) \to {\rm Zn^{2+}}\, (aq) + 2\, {\rm e^{-}}$ is oxidation and is likely not on the table of standard reduction potentials. However, the reverse reaction, ${\rm Zn^{2+}}\, (aq) + 2\, {\rm e^{-}} \to {\rm Zn}\, (s)$ , is reduction and is likely on the table.

$E(\text{anode}) = -0.7618\; {\rm V}$ for ${\rm Zn^{2+}}\, (aq) + 2\, {\rm e^{-}} \to {\rm Zn}\, (s)$ , and
$E(\text{cathode}) = 0.3419\; {\rm V}$ for ${\rm Cu^{2+}}\, (aq) + 2\, {\rm e^{-}} \to {\rm Cu} \, (s)$ .

The reduction potential of ${\rm Zn}\, (s) \to {\rm Zn^{2+}}\, (aq) + 2\, {\rm e^{-}}$ would be $-E(\text{anode}) = -(-0.7618\; {\rm V}) = 0.7618\; {\rm V}$ , the opposite of the reverse reaction ${\rm Zn^{2+}}\, (aq) + 2\, {\rm e^{-}} \to {\rm Zn}\, (s)$ .

The standard potential of the overall reaction would be the sum of the standard potentials of the two half-reactions:

$\begin{aligned} E^{\circ} &= E^{\circ}(\text{cathode}) + (-E^{\circ}(\text{anode})) \\ &= 0.3419 - (-0.7618\; {\rm V}) \\ &\approx 1.10\; {\rm V}\end{aligned}$ .