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chubhunter [2.5K]

1 year ago

9

A horse gallops along a racecourse at a speed of 40 km/h. It takes the horse 3 minutes to run the length of the track.

1 answer:

Natali [406]1 year ago

7 0

To convert minutes to hours we divide the minutes by 60. So if we divide 3 by 60 we get 0.05 hours.

<h3>How to convert minutes into hour?</h3>

We know that in hour, there are 60 minutes so if we go from minutes to hours then we have to divide the number by 60 and when we go from hours to minutes we multiply with the same 60 number.

So we can conclude that to convert minutes to hours we divide the minutes by 60. So if we divide 3 by 60 we get 0.05 hours.

Learn more about hour here: brainly.com/question/291457

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Helpp please I would really appreciate it!!!

KIM [24]

Answer:

I got 3/8, hope this helps.

Explanation:

4 0

1 year ago

Does adding 1 mol of NaCl to 1 kg of water lower the vapor pressure of water to the same extent, a lesser extent, or a greater e

igor_vitrenko [27]

Answer:

Adding 1 mol of NaCl to 1 kg of water lower the vapor pressure of water <em><u>to the same extent</u></em> by adding 1 mol of $C_06H_{12}O_6$ to 1 kg of water.

Explanation:

1) Moles of NaCl , $n_1=1 mol$

Mass of water = m= 1 kg = 1000 g

Moles of water = $n_2=\frac{1000 g}{18 g/mol}=55.55 mol$

Vapor pressure of the solution = $p$

Vapor pressure of the pure solvent that is water = $p_o=17.5 Torr$

Mole fraction of solute(NaCl)= $\chi_1=\frac{n_1}{n_1+n_2}$

$\frac{p_o-p}{p_o}=\frac{n_1}{n_1+n_2}$

$\frac{17.5 Torr-p}{17.5 Torr}=\frac{1 mol}{1 mol+55.55 mol}$

$p=17.19 Torr$

The vapor pressure for the NaCl solution at 17.19 Torr.

2) Moles of sucrose , $n_1=1mol$

Mass of water = m = 1 kg = 1000 g

Moles of water = $n_2=\frac{1000 g}{18 g/mol}=55.55 mol$

Vapor pressure of the solution = $p'$

Vapor pressure of the pure solvent that is water = $p_o=17.5 Torr$

Mole fraction of solute ( glucose)= $\chi_1=\frac{n_1}{n_1+n_2}$

$\frac{p_o-p}{p_o}=\frac{n_1}{n_1+n_2}$

$\frac{17.5 Torr-p}{17.5 Torr}=\frac{1 mol}{1 mol+55.55 mol}$

$p'=17.19 Torr$

The vapor pressure for the glucose solution at 17.19 Torr.

p = p' = 17.19 Torr

Adding 1 mol of NaCl to 1 kg of water lower the vapor pressure of water to the same extent by adding 1 mol of $C_06H_{12}O_6$ to 1 kg of water.

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(4) total number of valence electrons, because they exist in the same group.

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dsp73

Answer:

cellular respiration

Explanation:

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