ize completely 39 mL of 0.137 molar sodium
hydroxide solution?
Answer in units of mL.
The volume of 0.335 molar hydrochloric acid solution would be required to neutralize completely 39 mL of 0.137 molar sodium hydroxide solution is 15.95mL.
The volume of a substance in a neutralization reaction can be calculated using the following formula;
CaVa = CbVb
Where;
According to this question, 0.335 molar hydrochloric acid solution would be required to neutralize completely 39 mL of 0.137 molar sodium hydroxide solution.
0.335 × Va = 39 × 0.137
0.335Va = 5.343
Va = 15.95mL
Learn more about volume at: brainly.com/question/15195026
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True/False
Answer:
True
Explanation:
Liquid hexane burns in oxygen gas to form carbon dioxide gas and water vapor.TRUE.
This is the complete combustion reaction of hexane, which proceeds according to the following equation.
C₆H₁₄(l) + 9.5 O₂(g) → 6 CO₂(g) + 7 H₂O(g)
If the combustion were incomplete, instead of carbon dioxide, the product would be carbon monoxide or carbon.
Answer:
ROYGBIV or Roy G
Explanation:
Thanks for the points!
Alkaline Earth Metals
Alkali metals
Noble Gases
Lanthanides
Answer:
alkali metals- Group 1
Explanation:
they have less valence electrons and therefore are more reactive
O A. flour and water
O B. sand and water
O c.
salt and water
O D.
oil and water
O E. ice and water
An example of a solution is salt and water.
An example of a solution is salt and water. When salt is mixed with water, it dissolves and forms a homogeneous mixture where the salt particles are evenly distributed throughout the water.
Answer: Cis-1-bromo-4-tert-butylcyclohexane would undergo faster elimination reaction.
Explanation:
The two primary requirements for an E-2 elimination reaction are:
1.There must be availability of β-hydrogens that is presence of hydrogen on the carbon next to the leaving group.
2.The hydrogen and leaving group must have a anti-periplanar position .
Any substrate which would follow the above two requirements can give elimination reactions.
For the structure of trans-1-bromo-4-tert-butylcyclohexane and cis-1-bromo-4-tert-butylcyclohexane to be stable it must have the tert-butyl group in the equatorial position as it is a bulky group and at equatorial position it would not repel other groups. If it is kept on the axial position it would undergo 1,3-diaxial interaction and would destabilize the system and that structure would be unstable.
Kindly find the structures of trans-1-bromo-4-tert-butylcyclohexane and cis-1-bromo-4-tert-butylcyclohexane in attachment.
The cis- 1-bromo-4-tert-butylcyclohexane has the leaving group and β hydrogens in anti-periplanar position so they can give the E2 elimination reactions easily.
The trans-1-bromo-4-tert-butylcyclohexane does not have the leaving group and βhydrogen in anti periplanar position so they would not give elimination reaction easily.
so only the cis-1-bromo-4-tert butyl cyclohexane would give elimination reaction.
Trans-1-bromo-4-tert-butylcyclohexane is expected to undergo E2 elimination faster than cis-1-bromo-4-tert-butylcyclohexane due to less steric hindrance.
In determining the rate of E2 elimination, the trans-1-bromo-4-tert-butylcyclohexane would undergo E2 elimination faster than the cis-1-bromo-4-tert-butylcyclohexane. This is due to the larger degree of steric hindrance in the case of the cis isomer.
In trans-1-bromo-4-tert-butylcyclohexane, the bromine is at the equatorial position while the tert-butyl group is axial. It forms a structure that allows the compound to experience less steric hindrance with bromine in a more favorable position for leaving.
In comparison, cis-1-bromo-4-tert-butylcyclohexane has a bromine and tert-butyl group both at equatorial positions. This causes steric hindrance, and in turn, slows down the E2 elimination rate. Despite the more stable conformation, the bromine is not well-oriented for a leaving group in E2 elimination.
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