Answer:
option C is correct (250 g)
Explanation:
Given data:
Half life of carbon-14 = 5700 years
Total amount of sample = 1000 g
Sample left after 11,400 years = ?
Solution:
First of all we will calculate the number of half lives passes during 11,400 years.
Number of half lives = time elapsed/ half life
Number of half lives = 11,400 years/5700 years
Number of half lives = 2
Now we will calculate the amount left.
At time zero = 1000 g
At first half life = 1000 g/2 = 500 g
At second half life = 500 g/2 = 250 g
Thus, option C is correct.
<span>Answer:
For this problem, you would need to know the specific heat of water, that is, the amount of energy required to raise the temperature of 1 g of water by 1 degree C. The formula is q = c X m X delta T, where q is the specific heat of water, m is the mass and delta T is the change in temperature. If we look up the specific heat of water, we find it is 4.184 J/(g X degree C). The temperature of the water went up 20 degrees.
4.184 x 713 x 20.0 = 59700 J to 3 significant digits, or 59.7 kJ.
Now, that is the energy to form B2O3 from 1 gram of boron. If we want kJ/mole, we need to do a little more work.
To find the number of moles of Boron contained in 1 gram, we need to know the gram atomic mass of Boron, which is 10.811. Dividing 1 gram of boron by 10.811 gives us .0925 moles of boron. Since it takes 2 moles of boron to make 1 mole B2O3, we would divide the number of moles of boron by two to get the number of moles of B2O3.
.0925/2 = .0462 moles...so you would divide the energy in KJ by the number of moles to get KJ/mole. 59.7/.0462 = 1290 KJ/mole.</span>
Answer:
2.1 x 10^-2
Explanation:
Divide 8.4 x 10^-3 x 8.4 x 10^-3 by 5.82 x 10^-2 and you end up with 2.1 x 10^-2
