10d - 6 = 4d - 15 - 3d10d - 6 = d - 159d = -9d=-1
It is an algebraic expression that can be simplified to: 27ce
90 = 9.0 x 10^(1)
Collect like terms: 11c+6d-5c-8d = 6c-2d
To simplify 8d -5 +7d -9d, combine the numbers with the ds and don't forget the -5. 8d + 7d - 9d = 6d Simplified, it is 6d - 5.
3c - 4d = 6d + 4c Subtract 4c from both sides: -c - 4d = 6d Add 4d to both sides: -c = 10d or c = - 10d
They can be simplified to 10d
14d + 20e
Ok, if you have Sapling h.w....the answer is... ions with 6d electrons..Fe2+, Ru2+, Os2+ ions with 10d electrons..Zn2+, Cd2+ the question is just asking...which element has d^6 and d^10 in its electronic configuration?
It is an algebraic expression that can be simplified to: 27ce
p2+10d+7
If -6d = 42 then 6d = -42. then divide by 6, d = -7
5de+9e+7d+10e=5de+7d+19e
9,000,000
220.180
9E-05x
-6d = -42-6(7) = -42d = 7