The molecular formula is D. C_8H_20O_4Si.
<em>Step 1</em>.Calculate the <em>empirical formula
</em>
a) Calculate the moles of each element
Moles of C= 196.01 g C × (1 mol C/12.01 g C) = 16.325 mol C
Moles of H = 41.14 g H × (1 mol H/1.008 g H) = 40.813 mol H
Moles of O = 130.56 g O × (1 mol O/16.00 g O) = 8.1650 mol O
Moles of Si = 57.29 g Si × (1 mol Si/28.085 g Si) = 2.0399 mol Si
b) Calculate the molar ratio of each element
Divide each number by the smallest number of moles and round off to an integer
C:H:O:Si = 8.0027:20.008:4.0027:1 ≈ 8:20:4:1
c) Write the empirical formula
EF = C_8H_20O_4Si
<em>Step </em>2. Calculate the <em>molecular formula</em>
EF Mass = 208.33 u
MF mass = 208.329 u
MF = (EF)_n
n = MF Mass/EF Mass = 208.329 u/208.33 u = 1.0000 ≈ 1
MF = C_8H_20O_4Si
Ok so the way I do it is as simple as possible.
Firstly look at the reactants and products ( there can be one reactant and one product or more ) you will usually be given the moles of the reactant or products, if you are given grams you can convert into moles by this convertion ( grams/R.M.M ) where R.M.M is the relative atomic mass of your substance ( the mass number of all of the elements in your substance).
Ok when you have moles now look at the ratio between the products and reactants. Usually you will won't know the moles of one substance therefore you will be asked to find moles or mass of that substance.
For example:
When 16 grams of oxygen and 1 gram of hydrogen gas react to produce water. Find the number of grams of water being produced.
O2 + 2H2 -> 2H2O
16g 2g xg
Here we're told the mass of the reactants. In stoichiometry we need to work with moles therefore you need to calculate moles of the reactants.
Firstly find the R.M.M of each reactant.
R.M.M of O2 is 16+16=32 since it's diatomic we add atomic masses of two oxygen atoms.
R.M.M of H2 is 1+1=2, it's also diatomic. (Diatomic two atoms of the same element are joined together). (Ignore the number 2 in front of H2, this number shows us the ratio relationship between reactans or products, i.e when we balance an equation.)
Ok so now find moles:
We have 16 grams of O2
16/R.M.M
16/32 = 0.5 moles
We have 2 grams of H2
1/R.M.M
2/2 = 1 mole
Now back to the equation.
O2 + 2H2 -> 2H2O
0.5 moles 1mole xmoles (it's xmoles because we don't know molarity of water that's what we have find firstly in order to find grams.)
Now look at the ratio between any reactant and product i.e you can choose which reactant to compare to the product, it doesn't make a different ( I will do two or you can do two at the same time)
1st method:
Look at the ratio between O2 and H2O from the reaction above we see the ratio is 1:2 therefore for every 0.5 moles of O2 you get 1 mole of H2O.
1:2
0.5 : x
0.5*2 = 1
2nd method;
Look at the ratio betweem H2 and H2O from the reaction above we see the ratio is 2:2 or 1:1. We have 1 mole of H2 there we must have 1 mole of H2O. We see this is true as both methods give us 1 mole of H2O.
3rd method ( combined):
Look at the ratio between O2, H2 and H2O.
We see that the ratio is 1:2:2
So we have 0.5:1:x
If we multiply 0.5 *2 it equals 1 mole
If we multiply 1*1 we get 1 moles.
Any method is correct and it's up to you to find a comfortable way.
We're not finished in the question we are asked for the mass of water.
So just multiply the number of moles (1mole) by R.M.M of H2O.
1 * R.M.M
R.M.M of H2O = 1+1+16=18
1*18= 18 grams.
And you're finished.
I am sorry if this is so long I want you to understand as much as possible.
In stoichiometry you can also be asked about the empirical formula of a substance. I can show you how do it. If you have any question just tell me.
Hope this helps :).
Answer:
Genotypes: Hh, Hh, hh, and hh
Phenotypes: Short Hair, Short Hair, Long Hair, Long Hair
Have a nice day! ^-^
A.
H₃C-CH₃
this is called ethane
B.
H₃C-CH₂-CH₂-CH₃
this is called butane
C.H₃C-CH₂-CH₂-CH₂-CH₂-CH₃
this is called hexane
D.
H₃C-CH₂-CH₂-CH₂-CH₂-CH₂-CH₃
this is called heptane
Answer:
- <u>Alkaline or basic solution </u>(alkaline and basic means the same)
Explanation:
According to the <em>pH</em>, solutions may be classified as neutral, acidic, or alkaline (basic).
This table shows such classification:
pH classification
7 neutral
> 7 alkaline or basic
< 7 acidic
Thus, since the pH of the solution is 8.3, which is greater than 7, the solution is classified as basic (alkaline).
Additionally, you must learn that pH is a logarithmic scale for the concentration of hydronium ions in the solution.
You can calculate the concentration of hydronium ions using antilogarithm properties:
![pH=-log[H_3O^+]\\ \\ {[H_3O^+]}=10^{-pH}\\ \\ {[H_3O^+]}=10^{-8.3}=0.00000000501](https://tex.z-dn.net/?f=pH%3D-log%5BH_3O%5E%2B%5D%5C%5C%20%5C%5C%20%7B%5BH_3O%5E%2B%5D%7D%3D10%5E%7B-pH%7D%5C%5C%20%5C%5C%20%7B%5BH_3O%5E%2B%5D%7D%3D10%5E%7B-8.3%7D%3D0.00000000501)
NaOH solutions are alkaline solutions, bases, according to Arrhenius model, because they contain OH⁻ ions and release them when ionize in water.
