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3 years ago

5

If it takes 3 hours to drive a distance of 192km on a freeway, what would be your average speed in km/h? You can use a calculato

r.

1 answer:

Bingel [31]3 years ago

4 0

Answer:

The average speed in km/h would be 64km per hour.

Speed = Distance /Time

192/3 = 64

Explanation:

I hope this helps

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Using the periodic table and your knowledge of atomic structure. Compare the number of electrons in a carbon-12 and carbon-14

Brut [27]

Both carbon 12 and carbon 14 have the same no. of Electrons but the difference is carbon 14 has more neutrons

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3 years ago

Read 2 more answers

A Cu2+ solution is prepared by dissolving a 0.4749 g piece of copper wire in acid. The solution is then passed through a Walden

Luda [366]

Answer:

Concentration of $Cr_2O_7^{2-}$ = 0.03101 M

Concentration of $MnO_4^-$ = 0.03721 M

Explanation:

A)

The reduction for $Cr_2O_7^{2-}$ is;

$Cr_2O_7^{2-} + 14 H ^+ _{(aq)} + 6 e^- -----> 2 Cr^{3+} _{(aq)}+7H_2O _{(l)}$

$Cu^+_{(aq)} -----> Cu^{2+} _{(aq)} + 1 e^-$

6 moles of $Cu ^+$ = 1 mole of $Cr_2O_7^{2-}$

number of moles of Cu reacted = $\frac{mass \ of \ Cu \ wire }{ molecular weigh tof \ Cu wire }$

number of moles of Cu reacted = $\frac{0.4749}{63.55}$

number of moles of Cu reacted = 0.00747 mole

number of moles of $Cr_2O_7^{2-}$ reacted = $\frac{0.00747}{6}$

number of moles of $Cr_2O_7^{2-}$ reacted = 0.001245 mole

Concentration of $Cr_2O_7^{2-}$ = $\frac{number \ of moles }{Volume}$

Given that the volume = 40.15 mL = $40.15 *10^{-3}$ ; we have:

Concentration of $Cr_2O_7^{2-}$ = $\frac{0.001245}{40.15*10^{-3}}$

Concentration of $Cr_2O_7^{2-}$ = 0.03101 M

B)

The reduction for $MnO_4^-$ is;

$MnO_4^- + 8H^+ + 5 e^- -----> Mn^{2+} + 4H_2O$

$Cu^+_{(aq)} -----> Cu^{2+} _{(aq)} + 1 e^-$

5 moles of $Cu ^+$ = 1 mole of $Cr_2O_7^{2-}$

number of moles of Cu reacted = $\frac{mass \ of \ Cu \ wire }{ molecular weigh tof \ Cu wire }$

number of moles of Cu reacted = $\frac{0.4749}{63.55}$

number of moles of Cu reacted = 0.00747 mole

number of moles of $MnO_4^-$ reacted = $\frac{0.00747}{5}$

number of moles of $MnO_4^-$ reacted = 0.001494 mole

Concentration of $MnO_4^-$ = $\frac{number \ of moles }{Volume}$

Given that the volume = 40.15 mL = $40.15 *10^{-3}$ ; we have:

Concentration of $MnO_4^-$ = $\frac{0.001494 }{40.15*10^{-3}}$

Concentration of $MnO_4^-$ = 0.03721 M

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4 years ago

A typical volume leaving the large intestine might be 140 mL. If the volume exiting the large intestine were 364 mL, then it wou

RUDIKE [14]

If the volume exiting the large intestine were 364 mL, then it would most likely indicate a case of diarrhea. This condition may lead to dehydration.

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Diarrhea is an unhealthy condition that occurs when the body excretes an excessive amount of water.

Diarrhea must be treated because this condition can lead to dehydration (it may even lead to death in severe cases).

Both diarrhea and vomiting may lead to the loss of body fluids.

Learn more about diarrhea dehydration here:

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3 years ago

Calculate the pressure using the formula P=1.03+ mass on syringe/area of top of syringe

vazorg [7]

Mass divided by volume

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4 years ago

For the reaction 2NH3(g)↽−−⇀3H2(g)+N2(g) 2 NH 3 ( g ) ↽ − − ⇀ 3 H 2 ( g ) + N 2 ( g ) the equilibrium concentrations were found

forsale [732]

Answer:

0.324

Explanation:

The following data were obtained from the question:

Concentration of NH3, [NH3] = 0.25 M

Concentration of H2, [H2] = 0.3 M

Concentration of N2, [N2] = 0.75 M

Equilibrium constant (Kc) =.?

The balanced equation for the reaction is given below:

2NH3 <==> 3H2 + N2

The equilibrium constant, Kc for a given reaction is the ratio of the concentration of the products raised to their coefficient to the concentration of the reactants raised to their coefficient. Thus, the equilibrium constant for the above reaction can be obtained as illustrated below:

Kc = [H2]³ [N2] / [NH3]²

Concentration of NH3, [NH3] = 0.25 M

Concentration of H2, [H2] = 0.3 M

Concentration of N2, [N2] = 0.75 M

Equilibrium constant (Kc) =.?

Kc = [H2]³ [N2] / [NH3]²

Kc = [0.3]³ × [0.75] / [0.25]²

Kc = (0.027 × 0.75) / 0.0625

Kc = 0.02025 / 0.0625

Kc = 0.324

Therefore, the equilibrium constant for the reaction is 0.324

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3 years ago