Answer:
- <u><em>294.307 g/mol</em></u>
Explanation:
The first question for this statment is:
- <em>Calculate the gram-formula-mass of aspartame. </em>
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<h2>Solution</h2>
The chemical formula is:
The <em>gram-formula-mass </em>is calculated adding the masses for all the atoms in the molecular formula:
Atom Number of atoms Atomic mass Total mass
g/mol g/mol
C 14 12.011 14 × 12.011 = 164.154
H 18 1.008 18 × 1.008 = 18.144
N 2 14.007 2 × 14.007 = 28.014
O 5 15.999 5 × 15.999 = 79.995
===================
Total 294.307 g/mol
Answer: 294.307 g/mol
Answer:
one mole of atom of any element contains6.022×1033 atoms regardless of the type of elements the mass of one mole of an element depend on what that element is and is equal to atom mass of that element in gram
Answer:
The total mass of the reactants is <u>equal to </u> the total mass of the products in a chemical reaction.
Explanation:
<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.
