A. The mass of one mole of the virus is calculated by multiplying the mass of a single virus by the Avogadro's number which is equal to 6.022 x 10^23. Performing this operation,
mass = (9.0 x 10^-12 mg)(6.022 x 10^23) = 5.42 x 10^12 mg
In correct number of significant figures, the answer would only have to be 5.4 x 10^12 mg.
b. The number of moles of the virus that will have the same mass as the oil tanker is calculated by dividing the mass of the oil tanker by the mass of one mole of the virus. Note that doing division, both would have to have the same units.
n = (3.0 x 10^7 kg) / (5.4 x 10^12 mg)(1 g/1000 mg)(1 kg/1000 g)
Simplifying,
n = 6
In correct number of significant figures, the answer is 6.0.
Answer:
13 years
Explanation:
From;
0.693/t1/2 = 2.303/t log (No/N)
Where;
t1/12 = half-life of Cobalt-60 = 5.3 years
t = time taken = the unknown
No = mass of radioactive material originally present = 5.2 g
N = mass of radioactive material at time t = 0.95 g
Substituting values;
0.693/5.3 = 2.303/t log (5.2/0.95)
0.131 = 2.303/t(0.738)
0.131 = 1.6996/t
t = 1.6996/0.131
t = 12.97 years
t = 13 years(to the nearest whole number)
Answer:
(A) fossil is the most useful.
It serves as an index fossil. The seashell is a great index fossil because it only appears in the sandstone layer in multiple locations in the same layer of time.
Explanation:
Answer:
the electricity dissipated, because of the time.
Explanation:
electricity dissipates.
<span>For equation A + 3B + 2C ---> 2D,
1 mole of A will produce 2 moles of D
3 moles of B will produce 2 moles of D, so 1 mole of B will produce 2/3 moles of D
2 moles of C will produce 2 moles of D, so 1 mole of C will produce 1 mole of D
If only 1 mole of B is present, only 2/3 moles of D can be produced. This is regardless of the number of moles of A and C. B is the limiting reactant and the maximum number of moles of D expected is 2/3.</span>
