Answer:
63.616
Explanation:
DATA
1. first atomic mass;m1=63
- second atomic mass;m2=65
- first percentage;p1= 69.2%
- second percentage me;p2=30.8%
- average mass;avg= ?
SOLUTION
avg=<u> (m1)(p1) + (m2)(</u><u>p2</u><u>)</u>
100
avg= <u>(63)(69.2) + (65)(30.8)</u>
100
avg= <u>4</u><u>3</u><u>5</u><u>9</u><u>.</u><u>6</u><u> </u><u>+</u><u> </u><u>2</u><u>0</u><u>0</u><u>2</u>
100
avg= <u>6361.6</u>
100
avg= 63.616
Answer:
2.87 gram
N2 is the limiting agent
Explanation:
We will find out if there is sufficient N2 and h2 to produce NH3
a) For 2.36 grams of N2
Molar mass of N2 = 28.02
Number of moles of N2 in 2.36 grams = 2.36/28.02
Mass of NH3 = 17.034 g
Now NH3 produced form 2.36 grams of N2 =
2.36/28.02 * 2 * 17.034 = 2.87 g NH3
b) For 1.52 g of H2
NH3 produced = 1.52/2.016 * (2/3) * 17.034 = 8.56
N2 Is not enough to produce 2.87 g of NH3 and also H2 is not enough to make 8.56 g of NH3.
N2 is the limiting agent as it has smaller product mass
Answer:
<em>Acylation can be used to prevent rearrangement reactions that would normally occur in alkylation. </em>
Okay... what are the following
