<h3>Answer:</h3>
The Alkane formed is 5,5-dibromo-2,2,3-trimethylhexane. as shown below in attached scheme (Green Color).
<h3>Explanation:</h3>
Alkynes like Alkenes undergo <em>Electrophillic Addition Reactions</em>. The reaction given is a two step reaction. In step 1, the Alkyne adds first equivalent of HBr obeying <em>Markovnikov's rule</em> (i.e. Bromine will add to carbon containing less number of hydrogen atoms) and forms <em>2-bromo-4,5,5-trimethylhex-1-ene</em>. In step 2, the alkene formed in first step (2-bromo-4,5,5-trimethylhex-1-ene) undergoes addition reaction with the second equivalent of HBr via Markovnikov's rule to produce <em>5,5-dibromo-2,2,3-trimethylhexane</em>.
The scheme is attached below, Blue color is assigned to starting Alkyne, Red color is assigned to intermediate Alkene and Green color is assigned to product Alkane respectively.
Answer:
3.37 × 10²³ molecules
Explanation:
Given data:
Mass of C₆H₁₂O₆ = 100 g
Number of molecules = ?
Solution:
Number of moles of C₆H₁₂O₆:
Number of moles = mass/molar mass
Number of moles = 100 g/ 180.16 g/mol
Number of moles = 0.56 mol
Number of molecules:
1 mole contain 6.022 × 10²³ molecules
0.56 mol × 6.022 × 10²³ molecules /1 mol
3.37 × 10²³ molecules
The phosphate group of one nucleotide bonds covalently with the sugar molecule of the next nucleotide, and so on, forming a long polymer of nucleotide monomers. The sugar–phosphate groups line up in a “backbone” for each single strand of DNA, and the nucleotide bases stick out from this backbone. The carbon atoms of the five-carbon sugar are numbered clockwise from the oxygen as 1′, 2′, 3′, 4′, and 5′ (1′ is read as “one prime”). The phosphate group is attached to the 5′ carbon of one nucleotide and the 3′ carbon of the next nucleotide. In its natural state, each DNA molecule is actually composed of two single strands held together along their length with hydrogen bonds between the bases.
*Answer:
Option A: 59.6
Explanation:
Step 1: Data given
Mass of aluminium = 4.00 kg
The applied emf = 5.00 V
watts = volts * amperes
Step 2: Calculate amperes
equivalent mass of aluminum = 27 / 3 = 9
mass of deposit = (equivalent mass x amperes x seconds) / 96500
4000 grams = (9* amperes * seconds) / 96500
amperes * seconds = 42888888.9
1 hour = 3600 seconds
amperes * hours = 42888888.9 / 3600 = 11913.6
amperes = 11913.6 / hours
Step 3: Calculate kilowatts
watts = 5 * 11913.6 / hours
watts = 59568 (per hour)
kilowatts = 59.6 (per hour)
The number of kilowatt-hours of electricity required to produce 4.00kg of aluminum from electrolysis of compounds from bauxite is 59.6 kWh when the applied emf is 5.00V
