Answer:
4.8 %
Explanation:
We are asked the concentration in % by mass, given the molarity of the solution and its density.
0.8 molar solution means that we have 0.80 moles of acetic acid in 1 liter of solution. If we convert the moles of acetic acid to grams, and the 1 liter solution to grams, since we are given the density of solution, we will have the values necessary to calculate the % by mass:
MW acetic acid = 60.0 g/mol
mass acetic acid (the solute) = 0.80 mol x 60 g / mol = 48.00 g
mass of solution = 1000 cm³ x 1.010 g/ cm³ (1l= 1000 cm³)
= 1010 g
% (by mass) = 48.00 g/ 1010 g x 100 = 4.8 %
Answer:
The molecular formula is C12H18O3
Explanation:
Step 1: Data given
The empirical formula is C4H6O
Molecular weight is 212 g/mol
atomic mass of C = 12 g/mol
atomic mass of H = 1 g/mol
atomic mass of O = 16 g/mol
Step 2: Calculate the molar mass of the empirical formula
Molar mass = 4* 12 + 6*1 +16
Molar mass = 70 g/mol
Step 3: Calculate the molecular formula
We have to multiply the empirical formula by n
n = the molecular weight of the empirical formula / the molecular weight of the molecular formula
n = 70 /212 ≈ 3
We have to multiply the empirical formula by 3
3*(C4H6O- = C12H18O3
The molecular formula is C12H18O3
Answer: E
Explanation:
The lattice energy is the energy change when one mole of a crystal is formed from its components ions in its gaseous sate
Therefore lattice energy = heat of Sublimation+ ionization energy +electron affinity-(heat of formation)
Therefore lattice Energy = 109 +495 -328 +570.
Lattice energy = --923kjmol-1
The correct answer is <span>ball-and-stick model I just take it</span>
<span>Boron has a lot of different isotopes, most of which having a very short half life (ranging from 770 milliseconds for Boron-8 down to 150 yoctoseconds for boron-7). But the two isotopes Boron-10 and Boron-11 are stable with about 80.1% of the naturally occurring boron being boron-11 and the remaining 19.9% being boron-10. The weighted average weight of those 2 isotopes has the value of 10.81.
The reason they use the average mass of an element for it's atomic weight is because elements in nature are rarely single isotopes. The weighted average allows us to easily compare relative number of atoms of one element against relative numbers of atoms of another element assuming that the experimenters are getting isotope ratios close to their natural ratios.</span>
