Answer:
Explanation:
Toe change the retention factor of a TLC analysis, you can change your solvent for a more or less polar one, depending on your analyte. You can use a mix of solvents too.
You can also change the your method to visualize the spots, you can use fluorescent compounds that can only be seen in black light, you can use Iodine, Bromine and so on.
Answer:
4 g after 58.2 years
0.0156 After 291 years
Explanation:
Given data:
Half-life of strontium-90 = 29.1 years
Initially present: 16g
mass present after 58.2 years =?
Mass present after 291 years =?
Solution:
Formula:
how much mass remains =1/ 2n (original mass) ……… (1)
Where “n” is the number of half lives
to find n
For 58.2 years
n = 58.2 years /29.1 years
n= 2
or 291 years
n = 291 years /29.1 years
n= 10
Put values in equation (1)
Mass after 58.2 years
mass remains =1/ 22 (16g)
mass remains =1/ 4 (16g)
mass remains = 4g
Mass after 58.2 years
mass remains =1/ 210 (16g)
mass remains =1/ 1024 (16g)
mass remains = 0.0156g
2Ca + 2H2O —-> Ca(OH)2 + H2
Answer:
220.42098 amu
Explanation:
(220 .9 X .7422) + (220 X .0.1278) + (218.1 X 0.13) = 220.42098 amu
These are weighted averages.
So, we will take mass of one and multiply by abundance percentage that is provided and add them together.
In order to calculate the average atomic mass, we have to convert the percentages of abundance to decimals. So, you get
(220 .9 X .7422) + (220 X .0.1278) + (218.1 X 0.13) = 220.42098 amu
The balanced chemical reaction for this would be written as:
2Mg + O2 = 2MgO
We use this reaction and the amount of the reactant given to calculate for the amount of magnesium oxide that is produced. We do as follows:
1.5 g Mg (1 mol / 24.31 g) ( 2 mol MgO / 2 mol Mg ) (40.30 g /1 mol ) = 2.49 g MgO produced
