The solid compound, K2SO4 contains a cation called K+ and an anion called SO42-. In this case, there are 2 atoms of potassium, 1 atom of sulfur and 4 moles of oxygen. The compound also contains ionic bonds because of the composing non-metals and metal.
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)
➡ ANSWER
☑ <em><u>C</u></em><em><u>.</u></em><em><u> </u></em><em><u>3.5 105</u></em><em><u> </u></em><em><u>Hz</u></em>
The area where the periodic table has the highest electronegativity are Groups 1 and Groups 7 because they are close on giving away an electron and the other is close to becoming stable.
Answer:
After 5 second 25% C-15 will remain.
Explanation:
Given data:
Half life of C-15 = 2.5 sec
Original amount = 100%
Sample remain after 5 sec = ?
Solution:
Number of half lives = T elapsed / half life
Number of half lives = 5 sec / 2.5 sec
Number of half lives = 2
At time zero = 100%
At first half life = 100%/2 = 50%
At second half life = 50%/2 = 25%
Thus after 5 second 25% C-15 will remain.
