The empirical formula is = C2H4O2
Molecular formula=C2H4O2
<h3>Calculation of
Molecular formula and empirical formula:-</h3>
Vinegar has three elements:
oxygen = 53.29%,
hydrogen = 6.70%,
and carbon = 40.01%.
We can create an equation for the total mass of vinegar with 'a' carbon atom,' b' hydrogen atom, and 'c' oxygen atoms using the molar masses of C, H, and O.
12*a + 1*b + 16*c = 60
We also know that C makes up 24 g/mol, or 40.01 percent, of the overall mass of 60 g/mol.
The formula contains two C atoms because each C atom has a molecular mass of 12 g/mol. You may determine that the formula contains 4 H and 2 O atoms by using the same reasoning for H and O.
The empirical formula is = C2H4O2
Molecular formula= (C2H4O2)n
(4 x 12 + 1 x 4 + 16 x 2)n = 60
(84)n = 60
n=60/84
n=0.71=1
Molecular formula=C2H4O2
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Explanation:
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We have that the original pressure of this gas is mathematically given as
P2=17.854 kPa
<h3>
Boyle's law</h3>
Question Parameters:
AT constant temperature, a gas is compressed from 5.5 L to 1 L.
At 1 L, the pressure of the gas is 98.2 kPa
Generally the equation for Boyle's law is mathematically given as
P1V1=P2V2
Therefore

P2=17.854 kPa
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Answer:
1, 2, and 3 are true.
Explanation:
The Henderson-Hasselbalch equation is:
pH = pka + log₁₀ ![\frac{[A^-]}{[HA]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BA%5E-%5D%7D%7B%5BHA%5D%7D)
- If the pH of the solution is known as is the pKa for the acid, the ratio of conjugate base to acid can be determined. <em>TRUE</em>
pH = pka + log₁₀ ![\frac{[A^-]}{[HA]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BA%5E-%5D%7D%7B%5BHA%5D%7D)
If you know pH and pka:
10^(pH-pka) = ![\frac{[A^-]}{[HA]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BA%5E-%5D%7D%7B%5BHA%5D%7D)
The ratio will be: 10^(pH-pka)
- At pH = pKa for an acid, [conjugate base] = [acid] in solution. <em>TRUE</em>
pH = pka + log₁₀ ![\frac{[A^-]}{[HA]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BA%5E-%5D%7D%7B%5BHA%5D%7D)
0 = log₁₀ ![\frac{[A^-]}{[HA]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BA%5E-%5D%7D%7B%5BHA%5D%7D)
10^0 = ![\frac{[A^-]}{[HA]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BA%5E-%5D%7D%7B%5BHA%5D%7D)
1 = ![\frac{[A^-]}{[HA]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BA%5E-%5D%7D%7B%5BHA%5D%7D)
As ratio is 1, [conjugate base] = [acid] in solution.
- At pH >> pKa for an acid, the acid will be mostly ionized. <em>TRUE</em>
pH = pka + log₁₀ ![\frac{[A^-]}{[HA]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BA%5E-%5D%7D%7B%5BHA%5D%7D)
If pH >> pKa, 10^(pH-pka) will be >> 1, that means that you have more [A⁻] than [HA]
- At pH << pKa for an acid, the acid will be mostly ionized. <em>FALSE</em>
pH = pka + log₁₀ ![\frac{[A^-]}{[HA]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BA%5E-%5D%7D%7B%5BHA%5D%7D)
If pH << pKa, 10^(pH-pka) will be << 1, that means that you have more [HA] than [A⁻]
I hope it helps!
For that you have to use the Aufbau principle
For example, writing the electronic configuration of Fe(28)
I recommend that you memorize the aufbau principle. Once you learn it, you don’t need anything else except the atomic number of the element to write it’s electronic configuration.