Answer:
3.4g of Al
Explanation:
you would need to start with 3.4 g of Al
Answer:
Here's what I get
Explanation:
1. Names
I. CH₃-CH₂-COOH = 49. propanoic acid
II. CH₃-CH₂-OH = 46. ethanol
III. CH₃-COO-CH₂-CH₂-CH₃ = 47. propyl ethanoate
IV. H-O-CH₂-CH₂-CH₃ = 48. propan-1-ol
V. H-COO-CH₃ = 51. methyl methanoate
VI. CH₃-COOH = 50. ethanoic acid
2. Precursors
52. methyl propionate ⇒ methanol + propanoic acid
53. ethyl methanoate ⇒ ethanol + methanoic acid
Answer:
d= 14.007 amu
Explanation:
Abundance of N¹⁴ = 99.63%
Abundance of N¹⁵ = 0.37%
Atomic mass of N¹⁴ = 14.003 amu
Atomic mass of N¹⁵ = 15.000 amu
Average atomic mass = ?
Solution:
Average atomic mass = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass) / 100
Average atomic mass = (14.003 × 99.63)+(15.000× 0.37) /100
Average atomic mass = 1395.12 + 5.55 / 100
Average atomic mass = 1400.67/ 100
Average atomic mass = 14.007 amu.
