Answer: 0.25 mol
Explanation:
Use the formula n=N/NA
n= number of mols
N = number of particles
Nᵃ = Avogadros constant = 6.02 x
So, n=
The 10 to the power of 23 cancels out and you are left with 1.505/6.02, which is approximately 1/4. This is the same as 0.25 mol.
Hope this helped :)
An oxide is a molecule that has 2 Oxygen atoms in its empirical formula. Stoichiometry would be used as the ratio of Oxygen to Dioxides is 3:2. So 52.29 would be multiplied by 3/2. The answer is 78.435 Mol of CO2 are formed.
The term sensitivity in Analytical Chemistry is "the slope of the calibration curve or a function of analyte concentration or amount".
<u>Answer:</u> Option B
<u>Explanation:</u>
In a sample, the little amounts of substances can be accurately evaluated by a method is termed as "Analytical sensitivity". This detect a target analyte like an antibody or antigen, process is considered as potential of a test to and generally demonstrated as the analyte's minimum detectable concentration.
The acceptable diagnostic sensitivity is not guaranteed by high analytical sensitivity. The percentage of individuals who have a given disarray who are identified by the method as positive for the disarray is known as "Diagnostic sensitivity".
Answer:
1.428 moles
Explanation:
If 0.0714 moles of N2 gas occupies 1.25 L space,
how many moles of N2 have a volume of 25.0 L?
Assume temperature and pressure stayed constant.
we experience it 0.0714 moles: 1.25L space
x moles : 25L of space
to get the x moles, cross multiply
(0.0714 x 25)/1.25
1.785/1.25 = 1.428 moles
