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:
B
Explanation:
first find the no. Of moles of H2SO4
= 75/98.1
= 0.7645
Next step
In on mole of H2SO4 the no.of oxygen atom is 4.
1 H2SO4 : 4 O
0.7645 : x x= 0.7645x4= 3.0581
next multiply avogadros constant
3.0581 x 6.022× 10^23 =1.84x10^24
Answer:
130
Explanation:
This is because that 3atm of N2O4 is used up for the 6atm of NO2, so 1 atm N2O4 is left. Resulting in In(1/4).
Answer:
A. 2.139g of KIO3
B. 26.67mL
Explanation:Please see attachment for explanation