<span>6.03 moles.
1 molecule of butane contains 4 carbon atoms and ten hydrogen atoms.
The molar mass is 4 times the atomic mass of carbon, 12 g/mol, plus 10 times the atomic weight of hydrogen, 1 g/mol.
Molar mass = 4 * 12 g/mol + 10 * 1 g/mol = 58 g/mol.
This means that 1 mole of butane has a mass of 58 g.
To figure out how many moles are in a sample of butane, divide the mass of sample in grams by 58 grams
Number of moles in sample = 350 g / 58 g/mol = 6.03 moles.</span>
Answer:
1.) To convert between grams and moles, you would use the substance's molar mass. To go from grams to moles, divide the grams by the molar mass. 600 g58.443 g/mol = 10.27 mol of NaCl. It has been found that 1 mol of any gas at STP (Standard Temperature and Pressure = 0 °C and 1 atm) occupies 22.4 L
2-bromo-3,4-dimethylpentane is combined with t-butoxide. The product of this reaction is 3,4 dimethyl - 1- pentene.
The reaction of 2-bromo-3,4-dimethylpentane is combined with t-butoxide forms 2 alkene in the elimination reaction due to steric hindrance. The least stable alkene 3,4 dimethyl - 1- pentene is easy to make. the t-butoxide is (CH₃)₃CO⁻. The reaction involves in this reaction is E2 elimination reaction. This reaction involves the one step reaction. The product will also form that is 3,4 dimethyl - 2 - pentene. so the reaction involve Elimination reaction and the product due to steric hindrance is 3,4 dimethyl - 1- pentene
Thus, 2-bromo-3,4-dimethylpentane is combined with t-butoxide. The product of this reaction is 3,4 dimethyl - 1- pentene.
To learn more about t-butoxide here
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The periodic table increases as you read left to right so it is A.
A redox reaction --> a reaction whereby oxidation & reduction occurs
Reduction:
Charge of Cl2 = 0
Charge of Cl- in NaCl = -1
Hence, since charge of Cl2 decreased from 0 in Cl2 to -1 in NaCl, reduction occured.
Oxidation:
Charge of Na = 0
Charge of Na+ in NaCl = +1
Hence, since charge of Na increased from 0 in Na to +1 in NaCl, oxidation occured.
Since both oxidation & reduction occured in the reaction, it is a redox reaction.
