<h3>
Answer:</h3>
42960 years
<h3>
Explanation:</h3>
<u>We are given;</u>
- Remaining mass of C-14 in a bone is 0.3125 g
- Original mass of C-14 on the bone is 80.0 g
- Half life of C-14 is 5370 years
We are required to determine the age of the bone;
- Remaining mass = Original mass × 0.5^n , where n is the number of half lives.
Therefore;
0.3125 g = 80.0 g × 0.5^n
3.90625 × 10^-3 = 0.5^n
- Introducing logarithm on both sides;
log 3.90625 × 10^-3 = n log 0.5
Solving for n
n = log 3.90625 × 10^-3 ÷ log 0.5
= 8
- Therefore, the number of half lives is 8
- But, 1 half life is 5370 years
- Therefore;
Age of the rock = 5370 years × 8
= 42960 years
Thus, the bone is 42960 years old
Answer:
The correct answer is "D".
Explanation:
The synthesis consists of extracting the most important data from a text or any source of information. To be able to synthesize it is necessary to read, understand and interpret the message or arguments presented. Once these factors have been extracted, they are brought together to form a coherent text that in a few words explains the original text.
Have a nice day!
Answer:
The energies of combustion (per gram) for hydrogen and methane are as follows: Methane = 82.5 kJ/g; Hydrogen = 162 kJ/g
<em>Note: The question is incomplete. The complete question is given below:</em>
To compare the energies of combustion of these fuels, the following experiment was carried out using a bomb calorimeter with a heat capacity of 11.3 kJ/℃. When a 1.00-g sample of methane gas burned with
<em>excess oxygen in the calorimeter, the temperature increased by 7.3℃. When a 1.00 g sample of hydrogen gas was burned with excess oxygen, the temperature increase was 14.3°C. Compare the energies of combustion (per gram) for hydrogen and methane.</em>
Explanation:
From the equation of the first law of thermodynamics, ΔU = Q + W
Since there is no expansion work in the bomb calorimeter, ΔU = Q
But Q = CΔT
where C is heat capacity of the bomb calorimeter = 11.3
kJ/ºC; ΔT = temperature change
For combustion of methane gas:
Q per gram = (
11.3
kJ/ºC * 7.3°C)/1.0g
Q = 83 kJ/g
For combustion of hydrogen gas:
Q per gram = (
11.3
kJ/ºC * 14.3°C)/1.0g
Q = 162 kJ/g
Answer:
The dissociation equations for NaBr gives Na+ and Br-
The dissociation equations for ZnCl2 gives Zn2+ and 2 Cl-
Explanation:
The following pictures shows that the dissociation of one particle of NaBr produces one particle of Na+ (sodium cation) and one particle of Br- (bromine anion).
The dissociation of one particle of ZnCl2 produces one particle of Zn+2 (Zinc cation) and two particles of Cl- (chlorine anion).
