Answer:
Explanation:
% yields = <u>experimental yields</u> / theoretical yields )x 100%
=( 302 ÷ 320 )x 100% = 94.4 %
% error = 100% - 94.37% 5.64%
or u can use this law :
% error =(theoretical value - experimental value )/ theoretical value) x 100%
( 320 - 302 )/ 320 ) x 100% = 5.64%
Answer:
D) 65.7%
Explanation:
Based on the reaction:
2H2(g)+O2(g)⟶2H2O(l)
<em>2 moles of hydrogen produce 2 moles of water assuming an excess of oxygen.</em>
<em />
To find percent yield of the reaction we need to find theoretical yield (The yield assuming all hydrogen reacts producing water). With theoretical yield and actual yield (32.8g H₂O) we can determine percent yield as 100 times the ratio between actual yield and theoretical yield.
<em>Theoretical yield:</em>
Moles of 5.58g H₂:
5.58g H₂ ₓ (1 mol / 2.016g) = 2.768 moles H₂
As 2 moles of H₂ produce 2 moles of H₂O, if all hydrogen reacts will produce 2.768 moles H₂O. In grams:
2.768 moles H₂O ₓ (18.015g / mol) =
49.86g H₂O is theoretical yield
<em>Percent yield:</em>
Percent yield = Actual yield / Theoretical yield ₓ 100
32.8g H₂O / 49.86g ₓ 100 =
65.7% is percent yield of the reaction
<h3>D) 65.7%
</h3>
Answer:
-1
Explanation:
According to this question, the oxidation state/number of H and O in C2H4O is +1 and -2 respectively.
The oxidation state of carbon in the compound can be calculated thus:
Where;
x represents the oxidation number of C
C2H4O = 0 (net charge)
x(2) + 1(4) - 2 = 0
2x + 4 - 2 = 0
2x + 2 = 0
2x = -2
Divide both sides by 2
x = -1
The oxidation number of C in C2H4O is -1.
Answer:
ionization energy and electronegativity
Explanation:
both increase as you go right and up
