An Ionic bond is the result of transfer of electrons between atoms
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)
Consider the isomerization of butane with equilibrium constant is 2.5 .The system is originally at equilibrium with :
[butane]=1.0 M , [isobutane]=2.5 M
If 0.50 mol/L of butane is added to the original equilibrium mixture and the system shifts to a new equilibrium position, what is the equilibrium concentration of each gas?
Answer:
The equilibrium concentration of each gas:
[Butane] = 1.14 M
[isobutane] = 2.86 M
Explanation:
Butane ⇄ Isobutane
At equilibrium
1.0 M 2.5 M
After addition of 0.50 M of butane:
(1.0 + 0.50) M -
After equilibrium reestablishes:
(1.50-x)M (2.5+x)
The equilibrium expression will wriiten as:
![K_c=\frac{[Isobutane]}{[Butane]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BIsobutane%5D%7D%7B%5BButane%5D%7D)
x = 0.36 M
The equilibrium concentration of each gas:
[Butane]= (1.50-x) = 1.50 M - 0.36M = 1.14 M
[isobutane]= (2.5+x) = 2.50 M + 0.36 M = 2.86 M
Answer:
0.99 kg O₂
1.9 kg SO₂
Explanation:
Let's consider the reaction between sulfur and oxygen to form sulfur dioxide.
S + O₂ → SO₂
The mass ratio of S to O₂ is 32.07:32.00. The mass of oxygen required to react with 1 kg of sulfur is:
1 kg S × (32.00 kg O₂/32.07 kg S) = 0.998 kg O₂
The mass ratio of S to SO₂ is 32.07:64.07. The mass of sulfur dioxide formed when 1 kg of sulfur is burned is:
1 kg S × (64.07 kg SO₂/32.07 kg S) = 1.99 kg SO₂
The answer is 1023 particles
