Answer:

Explanation:
We are asked to find the volume of a solution given the moles of solute and molarity.
Molarity is a measure of concentration in moles per liter. It is calculated using the following formula:

We know there are 0.14 moles of potassium chloride (KCl), which is the solute. The molarity of the solution is 1.8 molar or 1.8 moles of potassium chloride per liter.
- moles of solute = 0.14 mol KCl
- molarity= 1.8 mol KCl/ L
- liters of solution=x
Substitute these values/variables into the formula.

We are solving for x, so we must isolate the variable. First, cross multiply. Multiply the first numerator and second denominator, then the first denominator and second numerator.



Now x is being multiplied by 1.8 moles of potassium chloride per liter. The inverse operation of multiplication is division, so we divide both sides by 1.8 mol KCl/L.


The units of moles of potassium chloride cancel.


The original measurements of moles and molarity have 2 significant figures, so our answer must have the same. For the number we found, that is the thousandth place. The 7 in the ten-thousandth place tells us to round the 7 up to a 8.

There are approximately <u>0.078 liters of solution.</u>
The answer is (2). Heat always flows down the temperature gradient, from high temperature to low temperature. Therefore, since the person is the warmest, heat flows from the person to both the ice and the air. Additionally, since the air is warmer than the ice, heat flows from the air to the ice.
Answer:they have the same number of electrons in the valence shell
Explanation:
Answer:
2nd order.
Explanation:
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Answer:
0.186M
Explanation:
First, we need to obtain the moles of nitric acid that are given for each solution. Then, we need to divide these moles in total volume (120mL + 20mL = 140mL = 0.140L) to obtain molarity:
<em>Moles Nitric acid:</em>
0.0200L * (0.100mol / L) = 0.00200 moles
0.120L * (0.200mol / L)= 0.02400 moles
Total moles: 0.02400moles + 0.00200moles = 0.026 moles of nitric acid
Molarity: 0.026 moles / 0.140L
<h3>0.186M</h3>