The initial sample has a molecular formula of MnSO₄·H₂O. This molecule is a hydrate as it has a unit of water within its structure for every molecule of MnSO₄. This sample is being dehydrated to remove the water to give.
MnSO₄·H₂O → MnSO₄ + H₂O
MnSO₄·H₂O has a molecular mass of 169.02 g/mol. While MnSO₄ has a molecular mass of 151 g/mol. Water has a molecular mass of 18.02 g/mol. We now can use the ratio of the mass of water to the mass of the initial sample to determine the percentage of each component by mass.
% water by mass:
18.02/169.02 x 100% = 10.7% Water by mass.
% MnO₄ by mass:
151/169.02 x 100% = 89.3% MnSO₄ by mass.
Water makes up 10.7% of the initial mass of MnSO₄·H₂O.
The statement above is FALSE.
Unlabeled atom joined to carbon atoms which are not directly part of a ring structure are assumed to be CARBON ATOMS. In a ring structure, an unlabeled atom at the angle where two lines joined together is always assumed to be a carbon atom<span />
A) the prefix
For example with the prefix prop: 3 carbons
Propane: C3H8
Propene: C3H6
Propyne: C3H4
<em>that the relative atomic mass of the middle element in each triad was close to the average of the relative atomic masses of the other two elements.</em>
- <em>This gave other scientists a clue that relative atomic masses were important when arranging the elements.</em>
<h2><em>hope</em><em> </em><em>it</em><em> </em><em>helps!</em></h2>
<span>Helium atom's Atomic Weight = He = 4.0026 g/mol.
Avagrado's number is = 6.023 x 10^23 molecules / mol.
In order to find the number of atoms in 535kg of helium blimp.
First we need to convert the weight of 535 kg to mol which can be done by multiplying the atomic weight of helium into present atom with respect to grams.
535 kg converted into grams ---> 535000 g
There fore 535000 g X 4.0026 g/mol = 133,663 mol. (Note The grams will get cancel as per multiplication rules)
Multiplying the avagadro's number with the equation we get:
133,663 mol *6.0221415 Ă— 10^23 molecules/mol= 8.04919303 Ă— 10^28 molecules.
Since Helium is having 2 Atoms :
8.04919303 Ă— 10^28 molecules *2= 1.60983861 Ă— 10^29 atoms
The No of Helium atoms in 535 kg of helium is 1.61 x 10^29.</span>
