Answer:
The chemistry will need 2*10⁶ moles of antimony trifluoride.
Explanation:
The balanced reaction is:
3 CCl₄ (g) + 2 SbF₃ (s) → 3 CCl₂F₂(g) + 2 SbCl₃ (s)
By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of each compound participate in the reaction:
- CCl₄: 3 moles
- SbF₃: 2 moles
- CCl₂F₂: 3 moles
- SbCl₃: 2 moles
You can apply the following rule of three: if by reaction stoichiometry 3 moles of freon are produced by 2 moles of antimony trifluoride, 3*10⁶ moles of Freon are produced from how many moles of antimony trifluoride?
moles of antimony trifluoride= 2*10⁶
<u><em>The chemistry will need 2*10⁶ moles of antimony trifluoride.</em></u>
An allylic carbocation is a reactive intermediate in the reaction of 1,3-diene with her, resulting in 1,4-addition.
<h3>What is
carbocation?</h3>
- A molecule called a carbocation has three bonds and a positively charged carbon atom.
- They are essentially carbon cations, to put it simply.
- It was once referred to as carbonium ion.
- Any even-electron cation with a sizable positive charge on the carbon atom is now referred to as a carbocation.
<h3>Why are carbohydrate molecules crucial?</h3>
- Because charge can be exchanged between many atoms when the vacant p orbital of a carbocation overlaps with the p orbitals of another carbon-carbon double or triple bond, carbocations next to other carbon-carbon double or triple bonds are very stable.
Learn more about carbocation here:
brainly.com/question/13164680
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Answer:
5.03 moles
Explanation:
Find the molar mass of C5H12 and you will get 72.17 g/mol
Next to find the number of moles, you divide 362.8 by the molar mass and you get
(362.8 g)/(72.17 g/mol)= 5.03 moles
cis-1-bromo-2-methylcyclohexane on
E2 reaction yield two different cycloalkenes. This is because <span>cis-1-bromo-2-methylcyclohexane contains
two different axial beta hydrogen atoms. Therefore, base can abstract both of them forming two different products.
As shown below, the two axial beta hydrogens are specified as
Ha and Hb.
When base abstracts Ha proton then it produces </span>
3-methylcyclohex-1-ene, which is a
minor product.
And when base abstracts Hb proton the product formed is
1-methylcyclohex-1-ene. This is a
major product because it has
tri-substituted double bond hence more stable according to
Zaitsev rule.
First determine the formal oxidation numbers:
N changes from +2 to +5 going from NO to (NO3)- O remains -2 the whole time Cr changes from +6 to +3
Now write the half reactions, balance the oxygens with the required number of waters and then balance the hydrogens with the required number of protons:
Oxidation half reaction:
NO(aq) + 2 H2O(l) ---> (NO3)-(aq) + 4 H+(aq) + 3 e-
Reduction half reaction:
(Cr2O7)2-(aq) + 14 H+(aq) + 6 e- ---> 2 Cr3+(aq) + 7 H2O(l)
Now balance the number of electrons on both sides and add them together:
2 NO(aq) + 4 H2O(l) ---> 2 (NO3)-(aq) + 8 H+(aq) + 6 e- (Cr2O7)2-(aq) + 14 H+(aq) + 6 e- ---> 2 Cr3+(aq) + 7 H2O(l) --------------------------------------... 2 NO(aq) + (Cr2O7)2-(aq) + 6 H+(aq) ---> 2 (NO3)-(aq) + 2 Cr3+(aq) + 3 H2O(l)
Notice that the charge is the same in both sides, which is an indication that the redox equation has been balanced correctly:
-2 + 6 = -2 + 2(+3) +4 = +4