Assuming the M is molarity, molarity is equal to mols/volume in liters.
So it would be 50M NaCl = X mols/ 1.0L
50M * 1.0L = X mols
X = 50
50 Mols of NaCl
Now that you have mols, you can then convert mols to grams by doing 50mols NaCl * (58.44 g NaCl/ 1 mol NaCl)
= 2922 grams of NaCl
Answer:
The Kelvin temperature scale reflects the relationship between temperature and average kinetic energy.
Explanation:
The Kelvin temperature of a substance is directly equal to the average kinetic energy of the particles of a substance.
Answer: The average atomic mass of oxygen is 15.999 amu
Explanation:
Mass of isotope O-16 = 15.995 amu
% abundance of isotope O-16= 99.759 % = 
Mass of isotope O-17 = 16.995 amu
% abundance of isotope O-17 = 0.037% = 
Mass of isotope O-18 = 17.999 amu
% abundance of isotope O-18 = 0.204% = 
Formula used for average atomic mass of an element :
![A=\sum[(15.995\times 0.99759)+(16.995\times 0.00037)+(17.999 \times 0.00204)]](https://tex.z-dn.net/?f=A%3D%5Csum%5B%2815.995%5Ctimes%200.99759%29%2B%2816.995%5Ctimes%200.00037%29%2B%2817.999%20%5Ctimes%200.00204%29%5D)
Thus the average atomic mass of oxygen is 15.999 amu
Answer:
<h2>The first thing to do here is to use the molarity and the volume of the initial solution to figure out how many grams of copper(II) chloride it contains.</h2><h2 /><h2>133</h2><h2>mL solution</h2><h2>⋅</h2><h2>1</h2><h2>L</h2><h2>10</h2><h2>3</h2><h2>mL</h2><h2>⋅</h2><h2>7.90 moles CuCl</h2><h2>2</h2><h2>1</h2><h2>L solution</h2><h2>=</h2><h2>1.051 moles CuCl</h2><h2>2</h2><h2 /><h2>To convert this to grams, use the compound's molar mass</h2><h2 /><h2>1.051</h2><h2>moles CuCl</h2><h2>2</h2><h2>⋅</h2><h2>134.45 g</h2><h2>1</h2><h2>mole CuCl</h2><h2>2</h2><h2>=</h2><h2>141.31 g CuCl</h2><h2>2</h2><h2 /><h2>Now, you know that the diluted solution must contain </h2><h2>4.49 g</h2><h2> of copper(II) chloride. As you know, when you dilute a solution, you increase the amount of solvent while keeping the amount of solute constant.</h2><h2 /><h2>This means that you must figure out what volume of the initial solution will contain </h2><h2>4.49 g</h2><h2> of copper(II) chloride, the solute.</h2><h2 /><h2>4.49</h2><h2>g</h2><h2>⋅</h2><h2>133 mL solution</h2><h2>141.32</h2><h2>g</h2><h2>=</h2><h2>4.23 mL solution</h2><h2>−−−−−−−−−−−−−− </h2><h2 /><h2>The answer is rounded to three sig figs.</h2><h2 /><h2>You can thus say that when you dilute </h2><h2>4.23 mL</h2><h2> of </h2><h2>7.90 M</h2><h2> copper(II) chloride solution to a total volume of </h2><h2>51.5 mL</h2><h2> , you will have a solution that contains </h2><h2>4.49 g</h2><h2> of copper(II) chloride.</h2>
