Molarity is a unit of concentration defined as the number of moles of solute (the substance being dissolved) per volume of solution (the solvent in which all the solute is dissolved). Mathematically, molarity is expressed as
.
In this question, we are given the molarity and the volume of a solution of NaCl. We can use this information to calculate the number of moles of NaCl present in the solution. Rearranging the equation to solve for moles of solute, we multiply the molarity by the volume of the solution (the units for volume must be in liters, so 125 mL is expressed as the equivalent 0.125 L):

The question asks for the number of moles of NaCl in this solution, so number 3 would be correct.
The answer is
C. condensation
Sugar dissolving in warm water because when the sugar dissolves you can get the sugar back by letting the water evaporate.,
Because the catalyst is not really part of the reaction. it is something that speed up a reaction by lowering the energy need for the reaction to take place. however, in the end the catalyst is brought back to its initial state. that's why it is long lasting
Answer: 27.09 ppm and 0.003 %.
First, <u>for air pollutants, ppm refers to parts of steam or gas per million parts of contaminated air, which can be expressed as cm³ / m³. </u>Therefore, we must find the volume of CO that represents 35 mg of this gas at a temperature of -30 ° C and a pressure of 0.92 atm.
Note: we consider 35 mg since this is the acceptable hourly average concentration of CO per cubic meter m³ of contaminated air established in the "National Ambient Air Quality Objectives". The volume of these 35 mg of gas will change according to the atmospheric conditions in which they are.
So, according to the <em>law of ideal gases,</em>
PV = nRT
where P, V, n and T are the pressure, volume, moles and temperature of the gas in question while R is the constant gas (0.082057 atm L / mol K)
The moles of CO will be,
n = 35 mg x
x
→ n = 0.00125 mol
We clear V from the equation and substitute P = 0.92 atm and
T = -30 ° C + 273.15 K = 243.15 K
V = 
→ V = 0.0271 L
As 1000 cm³ = 1 L then,
V = 0.0271 L x
= 27.09 cm³
<u>Then the acceptable concentration </u><u>c</u><u> of CO in ppm is,</u>
c = 27 cm³ / m³ = 27 ppm
<u>To express this concentration in percent by volume </u>we must consider that 1 000 000 cm³ = 1 m³ to convert 27.09 cm³ in m³ and multiply the result by 100%:
c = 27.09
x
x 100%
c = 0.003 %
So, <u>the acceptable concentration of CO if the temperature is -30 °C and pressure is 0.92 atm in ppm and as a percent by volume is </u>27.09 ppm and 0.003 %.