Answer:
Molecular formula = C₂H₆
Explanation:
Given data:
Percentage of hydrogen = 20.0 %
Percentage of carbon = 80.0 %
Molar mass = 30.0 g
Empirical formula = ?
Solution:
Number of gram atoms of H = 20 / 1.01 = 19.8
Number of gram atoms of C = 80 /12 = 6.7
Atomic ratio:
C : H
6.7/6.7 : 19.8/6.7
1 : 3
C : H = 1 : 3
Empirical formula is CH₃.
Molecular formula:
Molecular formula = n (empirical formula)
n = molar mass of compound / empirical formula mass
Empirical formula = CH₃ = 12×1 +3×1.01 = 15.03 g/mol
n = 30 g/mol / 15.03 g/mol
n = 2
Molecular formula = n (empirical formula)
Molecular formula = 2 (CH₃)
Molecular formula = C₂H₆
Answer:
Energy lost is 7.63×10⁻²⁰J
Explanation:
Hello,
I think what the question is requesting is to calculate the energy difference when an excited electron drops from N = 15 to N = 5
E = hc/λ(1/n₂² - 1/n₁²)
n₁ = 15
n₂ = 5
hc/λ = 2.18×10⁻¹⁸J (according to the data)
E = 2.18×10⁻¹⁸ (1/n₂² - 1/n₁²)
E = 2.18×10⁻¹⁸ (1/15² - 1/5²)
E = 2.18×10⁻¹⁸ ×(-0.035)
E = -7.63×10⁻²⁰J
The energy lost is 7.63×10⁻²⁰J
Note : energy is lost / given off when the excited electron jumps from a higher energy level to a lower energy level
Answer: 2 grams
Explanation:
chemical formula of C6H:
C6H
relative atomic mass of
C:12
H:1
C6:72
H1:12
relative formula mass of
C6H1=73
molar mass of glucose:
73g
the molar mass is the mass of one mole of a substance.
in
26.2g
of C6H there are
26.2 divide by 73g moles present.
there are 0.358 or 35.8*10-1
I gram of NH3 is 17.03052
2 grams of NH3 is 35
Answer:
0.97014 moles KMnO4
Explanation:
1 g KMnO4 = 0.00633 mol
153.26 g x 0.00633 mol/ 1 g KMnO4 = 0.97014 moles KMnO4
