Answer:
There are often not more than one or two independent variables tested in an experiment.
Answer:
Antibacterial specifically refers to an inhibition of growth of bacteria, whereas antibiotics refer to a general inhibition of growth of microorganisms, which could include bacteria and viruses.
Answer:
The concentration of methyl isonitrile will become 15% of the initial value after 10.31 hrs.
Explanation:
As the data the rate constant is not given in this description, However from observing the complete question the rate constant is given as a rate constant of 5.11x10-5s-1 at 472k .
Now the ratio of two concentrations is given as
Here C/C_0 is the ratio of concentration which is given as 15% or 0.15.
k is the rate constant which is given as 
So time t is given as
So the concentration will become 15% of the initial value after 10.31 hrs.
The empirical formula for a compound is KClO3
Explanation
find the moles of each element
moles = % composition/molar mass
molar mass of of potassium =39g/mol ,chlorine = 35.5 g/mol, oxygen =16 g/mol
moles of potassium = 31.9 / 39 = 0.818 moles
moles of chlorine = 28.9/35.5 = 0.814 moles
moles of oxygen = 39.2/ 16 = 2.45 moles
find the mole ratio by dividing with the smallest mole = 0.814 moles
potassium = 0.818/0.814 =1
chlorine = 0.814/0.814 = 1
oxygen = 2.45 /0.814 =3
the empirical formula is therefore = KClO3
Answer:
The percent isotopic abundance of C- 12 is 98.93 %
The percent isotopic abundance of C- 13 is 1.07 %
Explanation:
we know there are two naturally occurring isotopes of carbon, C-12 (12u) and C-13 (13.003355)
First of all we will set the fraction for both isotopes
X for the isotopes having mass 13.003355
1-x for isotopes having mass 12
The average atomic mass of carbon is 12.0107
we will use the following equation,
13.003355x + 12 (1-x) = 12.0107
13.003355x + 12 - 12x = 12.0107
13.003355x- 12x = 12.0107 -12
1.003355x = 0.0107
x= 0.0107/1.003355
x= 0.0107
0.0107 × 100 = 1.07 %
1.07 % is abundance of C-13 because we solve the fraction x.
now we will calculate the abundance of C-12.
(1-x)
1-0.0107 =0.9893
0.9893 × 100= 98.93 %
98.93 % for C-12.
