Answer:
64.0 g/mol.
Explanation:
- Thomas Graham found that, at a constant temperature and pressure the rates of effusion of various gases are inversely proportional to the square root of their masses.
<em>∨ ∝ 1/√M.</em>
where, ∨ is the rate of diffusion of the gas.
M is the molar mass of the gas.
<em>∨₁/∨₂ = √(M₂/M₁)</em>
∨₁ is the rate of effusion of the unknown gas.
∨₂ is the rate of effusion of He gas.
M₁ is the molar mass of the unknown gas.
M₂ is the molar mass of He gas (M₂ = 4.0 g/mol).
<em>∨₁/∨₂ = 0.25.</em>
∵ ∨₁/∨₂ = √(M₂/M₁)
∴ (0.25) =√(4.0 g/mol)/(M₁)
<u><em>By squaring the both sides:</em></u>
∴ (0.25)² = (4.0 g/mol)/(M₁)
∴ M₁ = (4.0 g/mol)/(0.25)² = 64.0 g/mol.
Answer:
A model or simulation is only as good as the rules used to create it. It is very difficult to create an entirely realistic model or simulation because the rules are based on research and past events. The main disadvantage of simulations is that they aren't the real thing.
Explanation:
Answer:
Explanation:
Utilizing Rydber's equation:
ΔE = Z²Rh ( 1/n₁² - 1/n₂²) and substituting the values given ( using the Rydbers constant value in Joules ), we have
n=1 to n= infinity
ΔE = 3² x (1/1² - 0) x 2.18 x 10⁻^18 J = 2.0 x 10⁻¹⁷ J (1/infinity is zero)
n= 3 to n= infinity
ΔE = 3² x (1/3² - 0) x 2.18 x 10⁻^18 J = 2.28 x 10^-18 J
b. The wavelength of the emitted can be obtained again by using Rydberg's equation but this time use the constant value 1.097 x 10⁷ m⁻¹ given in the problem .
1/λ = Z²Rh (1/n₁² - 1/n₂²) 10 ⁻¹ = 3² x 1.097 x 10⁷ m⁻¹ x (1/1² - 1/3²) m⁻¹
1/λ =8.8 x 10⁷ m⁻¹ ⇒ λ =1.1 x 10^-8 m
λ = 1.1 x 10^-8 m x 1 x 10⁹ nm/m = 11 nm
When to positive atoms connect
C. Decomposition
Since one reactant turns (decomposes) into two products, it is a decomposition reaction
