Answer:
<em>Answer Below</em>
Explanation:
Percent composition by element
<u>Element </u> <u>Symbol</u> <u>Mass Percent</u>
Aluminium <u>Al</u> 34.590%
Hydrogen <u>H</u> 3.877%
Oxygen <u>O</u> 61.533%
Answer:
2Al + 3CuSO4 → Al2(SO4)3 + 3Cu
Explanation:
I THINK
First find the oxidation states of the various atoms:
<span>in Cr2O2 2- Cr @ +1; In NH3 N @ +3; in CrO3 Cr @ +3, N2 N @ 0 </span>
<span>Note that N gained electrons, ie, was reduced; Cr was oxidized </span>
<span>Now there is a problem, because B has NH4+ which the problem did not, and is not balanced, showing e- in/out </span>
<span>B.NH4+ → N2 </span>
<span>Which of the following is an oxidation half-reaction? </span>
<span>A.Sn 2+ →Sn 4+ + 2e- </span>
<span>Sn lost electrons so it got oxidized</span>
The best way to accurately determine the pair with the highest electronegativity difference is by using their corresponding electronegativity values. For the each of the choices, the difference is:
A. H-S = 2.5 - 2.1 = 0.4
B. H-Cl = 3 - 2.1 = 0.9
C. N-H = 3 - 2.1 = 0.9
D. O-H = 3.5 - 2.1 = 1.4
E. C-H = 2.5 - 2.1 = 0.4
As show, D. has the highest difference. Without looking at their values, you can also determine the pair with the highest difference by taking note of the trend of electronegativity on the periodic table. Electronegativity increases as you go right a group and up a period. This makes oxygen the most electronegative element among the other elements paired with hydrogen.
I think it’s water evaporation I’m not sure tho
