Answer:
Ethyl ethanoate is formed
Answer:
Evaporation is a technique used to separate out homogeneous mixtures where there is one or more dissolved salts.
Explanation:
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Their positive charge is located in a small region that is called the nucleus.
Explanation:
Ernest Rutherford in his gold foil experiment was able to demonstrate that the nucleus is made up of positive charges which occupies a small and tiny nucleus.
Rutherford bombarded a thin gold foil with alpha particles from a radioactive source.
- He observed that all the particles passed through but a small portion was deflected back.
- This led to his proposition of the nuclear model the atom.
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Ernest Rutherford brainly.com/question/1859083
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Answer:
96.5 grams is the mass of the silver block.
Explanation:
Heat lost by iron will be equal to heat gained by the water
Mass of iron = 
Specific heat capacity of silver= 
Initial temperature of the silver= 
Final temperature = 
Mass of water= 
Specific heat capacity of water= 
Initial temperature of the water = 
Final temperature of water =
On substituting all values:
we get, 
96.5 grams is the mass of the silver block.
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>
