<span>0 °C. 8 °C. 29 °C. 15 °C.</span>
Answer:
0.499atm
Explanation:
The formula is
P1/V1 = P2/V2
so:
1.26atm/7.40L = P2/2.93L
then:
(1.26atm/7.40L)*2.93L = P2
= 0.4988918911atm
the answer must have 3 sig figs
<h2>Natural Abundance for 10B is 19.60%</h2>
Explanation:
- The natural isotopic abundance of 10B is 19.60%.
- The natural isotopic abundance of 11B is 80.40%.
- The isotopic masses of boron are 10.0129 u and 11.009 u respectively.
For calculation of abundance of both the isotopes -
Supposing it was 50/50, the average mass would be 10.5, so to increase the mass we need a more percentage of 11.
Determining it as an equation -
10x + 11y= 10.8
x+y=1 (ratio)
10x + 10y = 10
By taking the denominator away from the numerator
we get;
y = 0.8
x + y = 1
∴ x = 0.2
To get percentages we need to multiply it by 100
So, the calculated abundance is 80% for 11 B and 20% 10 B.
The correct answer is 96.80 grams
Explanation:
The density of a substance shows the total mass the substance contains in 1 mL or 1 cm3 as density is calculated by using the formula D= M (mass) / V (volume). This implies, Iron contains 7.87 grams per milliliter. Moreover, this value and formula can be used to calculate the mass or volume of any other sample. The process to calculate the mass of an iron sample with a volume of 12.3 mL is shown below.
D = M ÷ V
7.87 mL = x ÷ 12.3 mL - x represents the missing value. Now find the value of x by solving the equation
x = 7.87 · 12.3
x = 96.80
This means a sample of 12.3 mL contains a mass of 96.80 grams. Also, you can know this value is correct because if you divide the mass by the value the density is the same (96.80 grams ÷ 12.3 mL = 7.87 g/mL)
1) How old is a bone in which the Carbon-14 in it has undergone 8 half-lives?
Using the graph form the picture you count 8 times the halving of C¹⁴ and you arrive at 45600 years.
2) In the process of radiocarbon dating, the fixed period of radioactive decay used to determine age is called the half-life.
3) A certain byproduct in nuclear reactors, 210Po, decays to become 206Pb. After a time period of about 276 days, only about 25% of an original sample of 210Po remains. The remainder has decayed to 206Pb. Determine the approximate half-life of 210Po.
What the problem is telling you is that at 276 days only 25% original sample remains. If you divide the number of days by two the quantity of original sample will be multiplied by two, and you will have 138 days and 50% of original sample. This is the answer because the the half-life of a isotope is the time in which 50% of original quantity of radioactive atoms will disintegrate.
