Cr2(SO4)3(aq) + 3(NH4)2CO3(aq) → 3(NH4)2SO4(aq) + Cr2(CO3)3(s)
<span>Ionic: 2Cr+3 + 3SO4^-2 + 6NH4+ + 3CO3^-2 ----> 6NH4+ + 3SO4^-2 + Cr2(CO3)3 (spectator ions are NH4+, SO4^-2) </span>
<span>Net Ionic: 2Cr^+3(aq) + 3CO3^-2(aq) -------> Cr2(CO3)3(s) </span>
Answer:
The answer is: D
Explanation:
A. They have the same number of electron energy shells. Is false, all the elements are in different periods so, they have different number of lectron energy shells.
B. They are all Halogens. No, is wrong, halogens are F, Cl ,Br
C. They have the same number of electrons. is wrong, if they had the same number of electrons they must be they same element and they aren't the same.
D. They are all Noble gases. Yes, it's true they are noble gases, they are the first group in the periodic table from the right.
Two precursor alkenes
H₃C CH₃
I I
H₂C=C-CH-CH₃ 2,3-dimethyl-1-butene
H₃C CH₃
I I
H₃C-CH=CH-CH₃ 2,3-dimethyl-2-butene
alkane
H₃C CH₃
I I
H₃C-CH-CH-CH₃ 2,3-dimethylbutane
H₃C CH₃ H₃C CH₃
I I I I
H₂C=C-CH-CH₃ + H₂ → H₃C-CH-CH-CH₃
H₃C CH₃ H₃C CH₃
I I I I
H₂C-C=CH-CH₃ + H₂ → H₃C-CH-CH-CH₃
Answer:
Check the explanation
Explanation:
AT = A0 e(-T/H)
... where A0 is the starting activity, AT is the activity at some time T, and H is the half-life, in units of T.
Substituting what we know, we get...
0.71 = (1) e(-T/5730)
Solve for T...
loge(0.71) = -T/5730
T = -loge(0.71)(5730)
T = 1962 (conservatively rounded, T = 2000)
similarly for all
for aboriginal charcoal
0.28 = (1) e(-T/5730)
Solve for T...
loge(0.28) = -T/5730
T = -loge(0.28)(5730)
T = 7294 (conservatively rounded, T = 7000)
for mayan headdress
0.89 = (1) e(-T/5730)
Solve for T...
loge(0.89) = -T/5730
T = -loge(0.89)(5730)
T = 667 (conservatively rounded, T = 700)
for neanderthal
0.05 = (1) e(-T/5730)
Solve for T...
loge(0.05) = -T/5730
T = -loge(0.05)(5730)
T = 17165 (conservatively rounded, T = 17000)
