Answer:
i took the test but failed and just got that question right so b
Explanation:
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>
Elastic collisions meant that there will be no change in the total momentum of the system and surrounding after collision.
<u>Explanation:</u>
Collision is the process where two or more bodies hit each other. So when they hit each other, transfer of energy will be occurring between them. In other words, the momentum can be changed after the collision. There are two kinds of collision.
- elastic collision
- inelastic collision
The elastic collision are those which will not undergo any change in the total momentum after collision. This means that the sum of momentum of the objects before collision will be equal to the sum of momentum of the objects after collision. Thus, there will be no change in the total momentum of the object after collision in case of elastic collision.
He Rydberg formula can be extended for use with any hydrogen-like chemical elements.
<span>1/ λ = R*Z^2 [ 1/n1^2 - 1/n2^2] </span>
<span>where </span>
<span>λ is the wavelength of the light emitted in vacuum; </span>
<span>R is the Rydberg constant for this element; R 1.09737x 10^7 m-1 </span>
<span>Z is the atomic number, for He, Z =2; </span>
<span>n1 and n2 are integers such that n1 < n2 </span>
<span>The energy of a He+ 1s orbital is the opposite to the energy needed to ionize the electron that is </span>
<span>taking it from n = 1 (1/n1^2 =1) to n2 = ∞ (1/n2^2 = 0) </span>
<span>.: 1/ λ = R*Z^2 = 1.09737x 10^7*(2)^2 </span>
<span>λ = 2.278*10^-8 m </span>
<span>E = h*c/λ </span>
<span>Planck constant h = 6.626x10^-34 J s </span>
<span>c = speed of light = 2.998 x 10^8 m s-1 </span>
<span>E = (6.626x10^-34*2.998 x 10^8)/(2.278*10^-8) = 8.72*10^-18 J ion-1 </span>
<span>Can convert this value to kJ mol-1: </span>
<span>(8.72*10^-18*6.022 x 10^23)/1*10^3 = 5251 kJ mol-1 </span>
<span>Lit value: RP’s secret book: 5240.4 kJ mol-1 (difference is due to a small change in R going from H to He+) </span>
<span>So energy of the 1s e- in He+ = -5251 kJ mol-1</span>
