Answer:
Q= 245 =2.5 * 10^2
Explanation:
ΔG = ΔGº + RTLnQ, so also ΔGº= - RTLnK
R= 8,314 J/molK, T=298K
ΔGº= - RTLnK = - 6659.3 J/mol = - 6.7 KJ/mol
ΔG = ΔGº + RTLnQ → -20.5KJ/mol = - 6.7 KJ/mol + 2.5KJ/mol* LnQ
→ 5.5 = LnQ → Q= 245 =2.5 * 10^2
Answer:
The product of the reaction between a ketone and an alcohol is initially a hemiketal which yields a ketal on further reaction with another alcohol molecule.
The structure is found in the attachment.
Explanation:
This reaction is a <em>nucleophilic addition to the carbonyl group</em>. In organic chemistry, <em>a nucleophilic addition reaction is an addition reaction where a chemical compound with an electron-deficient or </em><em>electrophilic</em><em> double or triple bond, a pi (π) bond, reacts with electron-rich reactant, termed a </em><em>nucleophile</em><em>, with the elimination of the double bond and creation of two new single, or sigma (σ), bonds.</em>
In the reaction between a ketone and an alcohol, the <em>carbonyl</em> group of the ketone serves as the <em>electrophile</em> while the <em>hydroxyl</em> group of the alcohol is the <em>nucleophile</em>. The first product is known as a hemiketal because a single alcohol group has been aded to the carbonyl group of the ketone. Further nucleophilic additon of an alcohol group initiated by the presence of an acid e.g hydrochloric acid, results in the formation of a ketal which has two alcohol group added to the original ketone.
Hello!
To know how many moles of iron can be recovered from 100 kg of Fe₃O₄ we'll need to use the
molar mass of Fe₃O₄ and apply the conversion factor to go from kg of Fe₃O₄ to moles of Fe in the following way:
![100 kg Fe_3O_4* \frac{1000 g}{1 kg}* \frac{1 mol Fe_3O_4}{231,533 g Fe_3O_4}* \frac{3 mol Fe}{1 mol Fe_3O_4}=1192,68 moles Fe](https://tex.z-dn.net/?f=100%20kg%20Fe_3O_4%2A%20%5Cfrac%7B1000%20g%7D%7B1%20kg%7D%2A%20%5Cfrac%7B1%20mol%20Fe_3O_4%7D%7B231%2C533%20g%20Fe_3O_4%7D%2A%20%5Cfrac%7B3%20mol%20Fe%7D%7B1%20mol%20Fe_3O_4%7D%3D1192%2C68%20moles%20Fe%20)
So, theoretically, one could recover
1192,68 moles of Fe from 100 kg of Fe₃O₄
Have a nice day!
Answer: 125 g
Explanation:
To calculate the moles :
![\text{Moles of} B_2H_6=\frac{36.1g}{17}=1.30moles](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%7D%20B_2H_6%3D%5Cfrac%7B36.1g%7D%7B17%7D%3D1.30moles)
The balanced reaction is:
According to stoichiometry :
1 mole of
require = 3 moles of ![O_2](https://tex.z-dn.net/?f=O_2)
Thus 1.30 moles of
will require=
of ![O_2](https://tex.z-dn.net/?f=O_2)
Mass of ![O_2=moles\times {\text {Molar mass}}=3.90moles\times 32g/mol=125g](https://tex.z-dn.net/?f=O_2%3Dmoles%5Ctimes%20%7B%5Ctext%20%7BMolar%20mass%7D%7D%3D3.90moles%5Ctimes%2032g%2Fmol%3D125g)
Thus 125 g of
will be needed to burn 36.1 g of ![B_2H_6](https://tex.z-dn.net/?f=B_2H_6)