Percent (%) Composition of CuO
Cu = 1 x 50g - Multiply by one as there is one Cu
O = 1 x 12.5g - Multiply by one as there is one O
CuO = 62.5g
% for Cu = 50g over 62.5 multiplied by 100 = 80%
% for O = 12.5g over 62.5 multiplied by 100 = 20%
Final Answer :
<em>Percent (%) Composition of CuO = </em>80% (Cu) & 20% (O)
Answer: The given statement is true.
Explanation:
When we increase the amount of solvent which is water in this case then it means there will occur an increase in the molecules. Hence, there will be more number of collisions to take place with increase in number of molecules.
Therefore, more is the amount of interaction taking place between the molecules of a solution more will be its rate of hydrolysis.
Thus, we can conclude that the statement increasing the amount of water in which the sugar is dissolved will increase the frequency of collisions between the sucrose molecules and the water molecules resulting in an increase in the rate of hydrolysis, is true.
<span>n a metallic bond, atoms of the metal are surrounded by a constantly moving "sea of electrons." This moving sea of electrons enables the metal to conduct electricity.
Answer: Metallic</span>
<span> C.The results of the Michelson-Morley experiment did not fit the theory of the luminiferous ether, so the theory had to be rejected. </span>
A polar molecule<span> has a net dipole as a result of the opposing charges (i.e. having partial positive and partial negative charges) from </span>polar<span> bonds arranged asymmetrically. Water (H</span>2<span>O) is an example of a </span>polar molecule<span> since it has a slight positive charge on one side and a slight negative charge on the other.</span>