Answer:
AlN₃O₉
Explanation:
Assume that you have 100 g of the compound.
Then you have 12.7 g Al, 19.7 g N, and 67.6 g O.
1. Calculate the <em>moles</em> of each atom
Moles of Al = 12.7 × 1/26.98 = 0.4707 mol Al
Moles of N = 19.7 × 1/14.01 = 1.406 mol N
Moles of O = 67.6 × 1/16.00 = 4.225 mol O
2. Calculate the <em>molar ratios</em>.
Al: 0.4707/0.4707 = 1
N: 1.406/0.4707 = 2.987
O: 4.225/0.4707 = 8.976
3. Determine the <em>empirical formula</em>
Round off all numbers to the closest integer.
Al: 1
N: 3
O: 9
The empirical formula is AlN₃O₉.
Answer:
Some things that were wrong with Rutherford's model were that the orbiting electrons should give off energy and eventually spiral down into the nucleus, making the atom collapse. Bohr proposed his quantized shell model of the atom to explain how electrons can have stable orbits around the nucleus. To remedy the stability problem, Bohr modified the Rutherford model by requiring that the electrons move in orbits of fixed size and energy.
Explanation:
Answer:
68.6 °C
Explanation:
From conservation of energy, the heat lost by acetone, Q = heat gained by aluminum, Q'
Q = Q'
Q = mL where Q = latent heat of vaporization of acetone, m = mass of acetone = 3.33 g and L = specific latent heat of vaporization of acetone = 518 J/g
Q' = m'c(θ₂ - θ₁) where m' = mass of aluminum = 44.0 g, c = specific heat capacity of aluminum = 0.9 J/g°C, θ₁ = initial temperature of aluminum = 25°C and θ₂ = final temperature of aluminum = unknown
So, mL = m'c(θ₂ - θ₁)
θ₂ - θ₁ = mL/m'c
θ₂ = mL/m'c + θ₁
substituting the values of the variables into the equation, we have
θ₂ = 3.33 g × 518 J/g/(44.0 g × 0.9 J/g°C) + 25 °C
θ₂ = 1724.94 J/(39.6 J/°C) + 25 °C
θ₂ = 43.56 °C + 25 °C
θ₂ = 68.56 °C
θ₂ ≅ 68.6 °C
So, the final temperature (in °C) of the metal block is 68.6 °C.
Answer:
No photoelectric effect is observed for Mercury.
Explanation:
From E= hf
h= Plank's constant
f= frequency of incident light
Threshold Frequency of mercury= 435×10^3/ 6.6×10^-34 × 6.02×10^23
f= 11×10^14 Hz
The highest frequency of visible light is 7.5×10^14. This is clearly less than the threshold frequency of mercury hence no electron is emitted from the mercury surface
The answer is A. The prokaryote...
