Answer:
the correct option is option c, I and IV
Explanation:
Atomic emission spectroscopy is used to determine quantity of element in a sample.
Atomic emission spectroscopy based on occurring of atomic emission when a valence electron from higher energy orbital comes back to lower energy orbital.
Light intensity emitted by a flame, plasma, arc of particular wavelength are used to excite the valence electron.
The intensity of atomic emission lines are proportional to number of atoms present in the excited state.
Emission intensity is affected by temperature of the excitation source and the efficiency of atomization.
Increase in temperature does not affect the ground state population.
Therefore, statements I and IV are correct.
So, the correct option is c.
Is this a question?
Cause if it is I dont understand it.
No, don't try, it will explode close to 187 kPa
Answer:
³⁸₂₀Ca.
Explanation:
³⁸₁₉K –> __ + ⁰₋₁β
Let ʸₓA represent the unknown.
Thus the equation above can be written as:
³⁸₁₉K –> ʸₓA + ⁰₋₁β
Thus, we can obtain the value of y an x as follow:
38 = y + 0
y = 38
19 = x + (–1)
19 = x – 1
Collect like terms
19 + 1 = x
x = 20
Thus,
ʸₓA => ³⁸₂₀A => ³⁸₂₀Ca
Therefore, the equation is:
³⁸₁₉K –> ³⁸₂₀Ca + ⁰₋₁β
Ionic Equation:
H⁺(aq) + Cl⁻(aq) + Na⁺(aq) + CHO₂⁻(aq) → HCHO₂(aq) + Na⁺(aq) + Cl⁻(aq)
Net ionic equation:
H⁺(aq) + CHO₂⁻(aq) → HCHO₂(aq)
