32L —> 32000g —> 727.116 Moles (rounded)
41.38 % Mg
55.17 % O
3.45 % H
Explanation:
What is the percent composition of magnesium hydroxide Mg(OH)₂?
To find the percent composition we follow the next algorithm.
First we calculate the molar mass of Mg(OH)₂:
molar mass of Mg(OH)₂ = molar mass of Mg × 1 + molar mass of O × 2 + molar mass of H × 2
molar mass of Mg(OH)₂ = 24 × 1 + 16 × 2 + 1 × 2 = 58 g/mole
Now we devise the next reasoning:
if in 58 g of Mg(OH)₂ there are 24 g of Mg, 32 g of O and 2 g of H
then in 100 g of Mg(OH)₂ there are X g of Mg, Y g of O and Z g of H
X = (100 × 24) / 58 = 41.38 % Mg
X = (100 × 32) / 58 = 55.17 % O
X = (100 × 2) / 58 = 3.45 % H
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percent composition
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Answer:
jchb3prhoucheoucheoucheouchoudehocuehcdouhd
Explanation:
Answer:
A. One unpaired electron
B. 5 unpaired electrons
Explanation:
In A ,Fe is in +3 oxidation state and Electronic configuration- [Ar]3d5
And NO2 is a strong field ligand hence it causes pairing in t2g orbitals and results one unpaired electron in dZX orbital.
In B, also Fe is in +3 oxidation state but F is weak field ligand hence causes no pairing of Electrons hence it results 5 unpaired electrons with electronic configuration t2g^3 eg^2
The percent of histidine side chains would be deprotonated at pH 7.5 is 5.77 %.
<h3>What is pKa?</h3>
The term pKa refers to the negative logarithm of the acid dissociation constant (Ka). The pH is the negative logarithm of the hydrogen ion concentration.
Hence;
Ka = Antilog (-6) =
[H^+] = Antilog (-7.5) = 
We now have to use the formula;
α = ![\sqrt{} \frac{Ka}{[H^+] }](https://tex.z-dn.net/?f=%5Csqrt%7B%7D%20%5Cfrac%7BKa%7D%7B%5BH%5E%2B%5D%20%7D)
α = 
α = 5.77 %
Hence, the percent of histidine side chains would be deprotonated at pH 7.5 is 5.77 %.
Learn more about percent dissociation: brainly.com/question/12273293
