Which information should be added to the chart in order find out who ran a greater distance?

Chemistry

1 answer:

Arisa [49]3 years ago

6 0

Answer:

This question is incomplete, the complete question is:

Nancy and Hiyang are training for a race. They entered some of their training notes in a chart. Which information should be added to the chart in order find out who ran a greater distance?

The answer is C). the units used to measure distance each day

Explanation:

According to the question, Nancy and Hiyang are training for a race that involves them recording the distance they ran in a chart. Distance, as a quantity, is measured using different S.I units like metres, kilometers, miles, centimeters, etc.

However, in order to accurately discover whether Nancy or Hiyang ran a greater distance as recorded in their chart, the units used to measure distance each day must be included. This is because the unit of a quantity determines how big or small it is in comparison to another. For example, 20metres is not the same as 20centimetres.

If the unit they used in measuring their distance is not included, it will be impossible to tell what is being measured, talkless of who ran a greater distance

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if i add 25 ml of water to 135 ml of a 0.25 M NaOH solution what will the molarity of the diluted solution be

DENIUS [597]

Answer:

0.21 M. (2 sig. fig.)

Explanation:

The molarity of a solution is the number of moles of the solute in each liter of the solution. The unit for molarity is M. One M equals to one mole per liter.

How many moles of NaOH in the original solution?

$n = c \cdot V$ ,

where

$n$ is the number of moles of the solute in the solution.
$c$ is the concentration of the solution. $c = 0.25 \;\text{M} = 0.25\;\text{mol}\cdot\textbf{L}^{-1}$ for the initial solution.
$V$ is the volume of the solution. For the initial solution, $V = 135\;\textbf{mL} = 0.135\;\textbf{L}$ for the initial solution.

$n = c\cdot V = 0.25\;\text{mol}\cdot\textbf{L}^{-1} \times 0.135\;\textbf{L} = 0.03375\;\text{mol}$ .

What's the concentration of the diluted solution?

$\displaystyle c = \frac{n}{V}$ .

$n$ is the number of solute in the solution. Diluting the solution does not influence the value of $n$ . $n = 0.03375\;\text{mol}$ for the diluted solution.
Volume of the diluted solution: $25\;\text{mL} + 135\;\text{mL} = 160\;\textbf{mL} = 0.160\;\textbf{L}$ .

Concentration of the diluted solution:

$\displaystyle c = \frac{n}{V} = \frac{0.03375\;\text{mol}}{0.160\;\textbf{L}} = 0.021\;\text{mol}\cdot\textbf{L}^{-1} = 0.021\;\text{M}$ .

The least significant number in the question comes with 2 sig. fig. Keep more sig. fig. than that in calculations but round the final result to 2 sig. fig. Hence the result: 0.021 M.

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lana [24]

I'm guessing D or C, remember that the noble gases cannot combine

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