Answer:
Most positive = rubidium
Most negative = fluorine
Explanation:
Electron affinity of an element is the energy released when an electron is attached to a neural atom to form an ion in its gaseous state.
X + e⁻ → X⁻
Electron affinity is similar to electronegativity which is the tendency at which an atom accepts an ion towards itself.
Electron affinity increases across the period and decreases down the group in the periodic table.
In the above option,
Fluorine has the highest electron affinity
Rubidium has the lowest electron affinity
Tellurium and then finally Phosphorus
Helium in this case would have the lowest electron affinity because it has filled orbital and does not require any electron to attain stability. Technically, Helium has the lowest or is expected to have the lowest electron affinity which is close to zero according to quantum mechanics.
Most positive = rubidium
Most negative = fluorine.
You can check periodic table for their exact values
The answer for the problem is explained below.
The option for the answer is "D".
<u><em>Therefore the energy of the light is 4.25 × 10^-19 J</em></u>
Explanation:
Given:
wavelength (λ) = 468 nm = 468×10^-9 m
speed of light (c) = 3.00 x 10^8m/s
Planck's constant is 6.626 x 10^-34J·s
To solve:
energy of light (E)
We know,
E =(h×c) ÷ λ
E = ( 6.626 x 10^-34 × 3.00 x 10^8) ÷ 468×10^-9
E = 4.25 × 10^-19 J
<u><em>Therefore the energy of the light is 4.25 × 10^-19 J</em></u>
Answer: The enthalpy of combustion, per mole, of butane is -2657.4 kJ
Explanation:
The balanced chemical reaction is,
The expression for enthalpy change is,
Putting the values we get :
2 moles of butane releases heat = 5314.8 kJ
1 mole of butane release heat = 
Thus enthalpy of combustion per mole of butane is -2657.4 kJ
Answer:
a) Li2CO3
b) NaCLO4
c) Ba(OH)2
d) (NH4)2CO3
e) H2SO4
f) Ca(CH3COO)2
g) Mg3(PO4)2
f) Na2SO3
Explanation:
a) 2Li + CO3 ↔ Li2CO3
b) NaOH * HCLO4 ↔ NaCLO4 + H2O
c) Ba + 2H2O ↔ Ba(OH)2 +
d) 2NH4 + H2CO3 ↔ (NH4)2CO3 + H2O
c) SO2 + NO2 +H2O ↔ H2SO4 + NOx
f) 2CH3COOH + CaO ↔ Ca(CH3COOH)2 + H2O
g) 3MgO + 2H3PO4 ↔ Mg3(PO4)2 + H2O
h) NaOH + H2SO3 ↔ Na2SO3 + H2O