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>
Answer:
No
Explanation:
There are no interactive forces (i.e., attraction or repulsion) between the particles of a gas.
The average kinetic energy of gas particles is proportional to the absolute temperature of the gas, and all gases at the same temperature have the same average kinetic energy.
There is no effect. When you observe a substance's physical properties, you're not doing anything that would catalyze a chemical change. For example, you most often observe with your eyes, and/or, if safe, you waft the substance's smell to your nose. Neither of these observations will effect the substance's physical properties.
Answer:
2500 Calories IS REQUIRED TO HEAT 50 g OF WATER FROM 25 C TO 75 C
Explanation:
Specific heat of water = 1 cal/gC
Mass = 50 g
Temperature change = 75 C - 25 C = 50 C
How many calories are required to heat the 50 g of water?
Heat = mass * specific heat * temperature change
Heat = 50 * 1 * 50
Heat = 2500 Calories
In other words, 2500 calories is required to heat 50 g of water from 25 C to 75 C.
