A scientist who wants to date materials older than 40,000 years would most likely use
Potassium – argon dating or K-AR dating. Potassium-40 is a radioactive isotope
of potassium that decays into argon-40. The half life of potassium-40 is 1.3
billion years far longer than carbon-14.
Answer:
Increase the concentrations,
Decrease the surface area of NaOH (in case it's solid);
Use a catalyst;
Increase the temperature
Explanation:
Reaction rate depends on:
- the concentration of the reactants: the greater the concentration, the greater the rate, since more particles per unit volume will increase the probability of a successful collision leading to a reaction. We may then increase the concentration of NaOH or vinegar solution;
- the surface area of the solid reactant: in case we have NaOH in solid form, we may decrease the size of the crystals to increase the reaction rate. This is because the greater the area, the more particles are likely to collide with it;
- add a catalyst: catalysts increase reaction rate, in this problem, we don't need any catalyst, as this is an acid-base reaction that is relatively quick, a catalyst would decrease the activation energy barrier;
- increase the temperature: higher energy of the particles and a higher velocity of each reactant would increase the probability of a successful collision.
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.
For this problem, the solution is exhibiting some colligative properties since the solute in the solution interferes with some of the properties of the solvent. We use equation for the boiling point elevation for this problem. We do as follows:
<span>
ΔT(boiling point) = (Kb)mi
</span>ΔT(boiling point) = (0.512)(1.3/2.0)(2)
ΔT(boiling point) = 0.67 degrees Celsius
<span>
T(boiling point) = 100 + 0.67 = 100.67 degrees Celsius</span>
Answer:
3.60 ml
Explanation:
First of all we must put down the equation of the reaction. This will serve as a guide to our solution;
KOH(aq) + HBr(aq) -----> KBr(aq) + H2O(l)
The following were given in the question;
Concentration of acid CA= 2M
Volume of acid VA= 18ml
Concentration of base CB= 0.01 M
Volume of base VB= ????
Number of moles of acid NA= 1
Number of moles of base NB= 1
From;
CAVA/CBVB = NA/NB
CAVANB= CBVBNA
Therefore;
VB= CAVANB/CBNA
Substituting values;
VB= 2 × 18 ×1 / 0.01×1
VB= 3.60 ml
Therefore; 3.60 ml of base was used.