So let's convert this amount of mL to grams:
Then we need to convert to moles using the molar weight found on the periodic table for mercury (Hg):
Then we need to convert moles to atoms using Avogadro's number:
![\frac{6.022*10^{23}atoms}{1mole} *[8.135*10^{-2}mol]=4.90*10^{22}atoms](https://tex.z-dn.net/?f=%20%5Cfrac%7B6.022%2A10%5E%7B23%7Datoms%7D%7B1mole%7D%20%2A%5B8.135%2A10%5E%7B-2%7Dmol%5D%3D4.90%2A10%5E%7B22%7Datoms%20)
So now we know that in 1.2 mL of liquid mercury, there are 
present.
The correct answer is (3)
I-131 and P-32
The explanation:
according to attached table:
- we can see that the half life of p 32 is 14.28d (more than one hour)
- and the half life of I-131 is 8.021 d
(more than one hour)
and They both have β- decay mode and with half-lives greater than hour.
I believe your answer is 23.
Credit: answers.yahoo.com
Hope this helps!
Answer:
a) The concentration of drug in the bottle is 9.8 mg/ml
b) 0.15 ml drug solution + 1.85 ml saline.
c) 4.9 × 10⁻⁵ mol/l
Explanation:
Hi there!
a) The concentration of the drug in the bottle is 294 mg/ 30.0 ml = 9.8 mg/ml
b) The drug has to be administrated at a dose of 0.0210 mg/ kg body mass. Then, the total mass of drug that there should be in the injection for a person of 70 kg will be:
0.0210 mg/kg-body mass * 70 kg = 1.47 mg drug.
The volume of solution that contains that mass of drug can be calculated using the value of the concentration calculated in a)
If 9.8 mg of the drug is contained in 1 ml of solution, then 1.47 mg drug will be present in (1.47 mg * 1 ml/ 9.8 mg) 0.15 ml.
To prepare the injection, you should take 0.15 ml of the concentrated drug solution and (2.0 ml - 0.15 ml) 1.85 ml saline
c) In the injection there is a concentration of (1.47 mg / 2.0 ml) 0.735 mg/ml.
Let´s convert it to molarity:
0.735 mg/ml * 1000 ml/l * 0.001 g/mg* 1 mol/ 15000 g = 4.9 × 10⁻⁵ mol/l
