Number of moles:
1 mole ---------- 6.02x10²³ molecules
? moles --------- 9.45x10²⁴ molecules
1 x ( 9.45x10²⁴) / 6.02x10²³ =
9.45x10²⁴ / 6.02x10²³ => 15.69 moles of CH3OH
Therefore:
Molar mass CH3OH = 32.04 g/mol
1 mole ------------ 32.04 g
15.69 moles ----- mass methanol
Mass methanol = 15.69 x 32.04 / 1 => 502.7076 g
Answer:
Niels Bohr, refined the model of an atom by proposing a quantized shell structure atomic model in order to describe how the electrons are able to maintain stable orbits around the nucleus
Based on the predictions of classical mechanics the electron motion of the Rutherford model was unstable as the electrons where expected to have lost some energy during motion and thus having to come rest in the nucleus
According to the modification by Neils Bohr in 1913, electrons move in shells or orbits of fixed energy and emission of electromagnetic radiation takes place only when electrons changes the orbit in which they move
Explanation:
In the reaction of silver nitrate with copper metal, metallic silver comes out of solution, and the solution turns blue. This as a <u>single replacement</u> reaction.
<h3>What is
single replacement reaction?</h3>
A single replacement reaction, also known as a single displacement reaction, occurs when one element in a molecule is swapped out for another. The starting materials are always pure elements, such as a pure zinc metal or hydrogen gas, plus an aqueous compound.
A + BC → B + AC
When A is more reactive than B or when the product AC is more stable than BC, single replacement reactions happen. A and B could either be two halogens or two metals (with hydrogen included) (C is a cation). C functions as a spectator ion when BC and AC are in aqueous solutions.
For example, 2HCl(aq)+Zn(s)→ZnCl₂(aq)+H₂(g)
Learn more about single replacement reactions here:
brainly.com/question/19068047
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Answer:
we can do it again and again and again and again and again and again
Given buffer:
potassium hydrogen tartrate/dipotassium tartrate (KHC4H4O6/K2C4H4O6 )
[KHC4H4O6] = 0.0451 M
[K2C4H4O6] = 0.028 M
Ka1 = 9.2 *10^-4
Ka2 = 4.31*10^-5
Based on Henderson-Hasselbalch equation;
pH = pKa + log [conjugate base]/[acid]
where pka = -logKa
In this case we will use the ka corresponding to the deprotonation of the second proton i.e. ka2
pH = -log Ka2 + log [K2C4H4O6]/[KHC4H4O6]
= -log (4.31*10^-5) + log [0.0451]/[0.028]
pH = 4.15