Answer:
never true is the answer
explanation:
compounds are formed by two or more elements that are chemically combined with one another. They can be broken down only be chemical or electro chemical means not by physical means.
hence, the answer i have given above is correct
please mark as brainliest
Answer:- As per the question is asked, 35.0 moles of acetylene gives 70 moles of carbon dioxide but if we solve the problem using the limiting reactant which is oxygen then 67.2 moles of carbon dioxide will form.
Solution:- The balanced equation for the combustion of acetylene is:

From the balanced equation, two moles of acetylene gives four moles of carbon dioxide. Using dimensional analysis we could show the calculations for the formation of carbon dioxide by the combustion of 35.0 moles of acetylene.

= 
The next part is, how we choose 35.0 moles of acetylene and not 84.0 moles of oxygen.
From balanced equation, there is 2:5 mol ratio between acetylene and oxygen. Let's calculate the moles of oxygen required to react completely with 35.0 moles of acetylene.

= 
Calculations shows that 87.5 moles of oxygen are required to react completely with 35.0 moles of acetylene. Since only 84.0 moles of oxygen are available, the limiting reactant is oxygen, so 35.0 moles of acetylene will not react completely as it is excess reactant.
So, the theoretical yield should be calculated using 84.0 moles of oxygen as:

= 
Answer:
The electrolyte carries positively charged lithium ions from the anode to the cathode and vice versa through the separator. The movement of the lithium ions creates free electrons in the anode which creates a charge at the positive current collectors
Answer:
(a) 
(b)


(c) 
Explanation:
Hello,
(a) In this case, with the given formula we easily compute the mass of gold contained in the sovereign as shown below:

(b) Now, by knowing the density of gold and copper, 19.32 and 8.94 g/cm³ respectively, we compute each volume, by also knowing that the rest of the coin contains copper:


(c) Finally, the volume is computed by dividing the total mass over the total volume containing both gold and copper:

Best regards.
64: C
65: D because weathering is defined as rock being eroded/changed
hope i could help