Answer:
Formation of sedimentary rocks
Explanation:
I believe this is correct, May I get a brainliest???
Answer:
15.95
Explanation:
This question is a modification of the calculation of the empirical formula of a compound given its percent composition and atomic weights of the elements in the compound.
Here we are given the formula and the percent composition, so we know that there are 4 atoms of E per 2 atoms of N so lets solve using the information given.
In 100 grams of the binary compound we have
30.46 g N
69.54 g E
The number of moles is the mass divided by atomic weight:
mol N = 30.46 g / A.W N = 30.46 g / 14.00 g/mol = 2.18 mol N
mol E = 65.54 g / A.W E
Thus,
4 mol E/ 2 mol N = ( 69.54 g/ A.W E ) / 2.18
2 A.E = 65.54 g / 2.18 ⇒ A.W E = 69.54 g / ( 2 x 2.18 ) = 15.94 g
So the A.W is 15.94 g/mol which is close the atomic weight of O.
By definition, one mole (one gram molecular weight) of any substance, contains Avogadro’s number of particles; atoms if you are discussing an element, or molecules if a compound. Avogadro’s number has been determined by several methods, all of the accepted values lie within a range of +-1% about the value of 6.022045 x 10^23/gm. That is a large number, in this case approximately; 602,204,500,000,000,000,000,000 molecules of glucose.
From the web :v
Answer : The concentration of a solution with an absorbance of 0.460 is, 0.177 M
Explanation :
Using Beer-Lambert's law :
where,
A = absorbance of solution
C = concentration of solution
l = path length
= molar absorptivity coefficient
From this we conclude that absorbance of solution is directly proportional to the concentration of solution at constant path length.
Thus, the relation between absorbance and concentration of solution will be:
Given:
= 0.350
= 0.460
= 0.135 M
= ?
Now put all the given values in the above formula, we get:
Therefore, the concentration of a solution with an absorbance of 0.460 is, 0.177 M
The answer is most likely C
