This question is dealing with the half-life of carbon-14 which can be used to determine the age of a substance according to the following equation:
t = [ln(N/No)/(-ln2)] · t1/2
N = # of carbon-14 atoms presently = 250 atoms
No = # of carbon-14 atoms initially = 1000 atoms
t1/2 = half-life = 5730 years for carbon-14
We can now input all of the information into the formula to find the age of the fossil:
t = [ln (250/1000)/-ln2] x 5730 years
t = 11460 years
The fossil should be found to be roughly 11,460 years old.
The speed of light in a vacuum stands at “exactly 299,792,458 metres per second“. The reason today we can put an exact figure on it is because the speed of light in a vacuum is a universal constant that has been measured with lasers; and when an experiment involves lasers, it's hard to argue with the results.
Answer:
a: 1
b: 4.5x10⁻⁴
c: 1.125x10⁻⁶
[H₃O⁺] = 0.000859M
Explanation:
As HNO₂ is a weak acid, its equilibrium in water is:
HNO₂(aq) + H₂O(l) ⇄ H₃O⁺(aq) + NO₂⁻(aq)
Equilibrium constant, ka, is defined as:
ka = 4.5x10⁻⁴ = [H₃O⁺] [NO₂⁻] / [HNO₂] <em>(1)</em>
Equilibrium concentration of each specie are:
[HNO₂] = 0.00250M - x
[H₃O⁺] = x
[NO₂⁻] = x
Replacing in (1):
4.5x10⁻⁴ = x × x / 0.00250M - x
1.125x10⁻⁶ - 4.5x10⁻⁴x = x²
0 = x² + 4.5x10⁻⁴x - 1.125x10⁻⁶
As the quadratic equation is ax² + bx + c = 0
Coefficients are:
a: 1
b: 4.5x10⁻⁴
c: 1.125x10⁻⁶
Now, solving quadratic equation:
x = -0.0013 → False answer, there is no negative concentrations.
<em>x = 0.000859</em>
As [H₃O⁺] = x; <em>[H₃O⁺] = 0.000859M</em>
I hope it helps!
C and D
surface water is evaporated by the sun and concentrated the ocean salt and oceans are diluted when rivers flow into them
Density= mass/volume
volume=mass/density
volume= 40.0g/1.114g per mL
volume= 35.90664273 mL
volume = 35.9 mL
