<span>A) This solution was not basic when it was heated in part 3. ( in part 3 i convertedCu(OH)2 to CuO).
Incorrectly low, because not all copper compounds will precipitate out
B) A slightly blue solution was decanted from Cu in part V. (in part 5 i reduced Cu(H20)6 ions with zink)
Incorrectly low, because some copper were thrown away
C) In part 5 the water in the beaker boiled away, exposing the evaporating dish to excess heat (same as above).
incorrectly high, because other compounds might be present as well </span>
The answer to your question is A.
Answer:
0.85 mole
Explanation:
Step 1:
The balanced equation for the reaction of CaCl2 to produce CaCO3. This is illustrated below:
When CaCl2 react with Na2CO3, CaCO3 is produced according to the balanced equation:
CaCl2 + Na2CO3 -> CaCO3 + 2NaCl
Step 2:
Conversion of 85g of CaCO3 to mole. This is illustrated below:
Molar Mass of CaCO3 = 40 + 12 + (16x3) = 40 + 12 + 48 = 100g/mol
Mass of CaCO3 = 85g
Moles of CaCO3 =?
Number of mole = Mass /Molar Mass
Mole of CaCO3 = 85/100
Mole of caco= 0.85 mole
Step 3:
Determination of the number of mole of CaCl2 needed to produce 85g (i.e 0. 85 mole) of CaCO3.
This is illustrated below :
From the balanced equation above,
1 mole of CaCl2 reacted to produced 1 mole of CaCO3.
Therefore, 0.85 mole of CaCl2 will also react to produce 0.85 mole of CaCO3.
From the calculations made above, 0.85 mole of CaCl2 is needed to produce 85g of CaCO3
Hey there,
Answer:
4 valence electrons.
Hope this helps :D
<em>~Top☺</em>
I literally hate chem but I think it’s ionic, I’m not not completely sure but it kinda sounds about right . Not the best anwser haha hope it kinda helps lol
