Br
Se
As
Ge
Ga
Rb
Elements in the top right corner of the periodic table have the highest electronegativity. Elements on the right side have a higher electronegativity than those on the left, same with the ones on the top in comparison to those on the bottom.
The chart is attached below and the options are as follow,
A Benzene
B Methanol
C Hexane
D Octane
Answer: Option-B (Methanol) is the correct answer.
Explanation: As we know that the substance given to student is soluble in water. Therefore, those substances which are soluble in water are
polar in nature because water is
polar in nature and
Like Dissolves Like.
Hence, in chart it can easily be seen that
Methanol is also polar in nature due to presence of -O-H (polar) bond. So, the substance will dissolve in it.
Energy = Planck's constant * Frequency
E = (6.62607004 × 10⁻³⁴<span>) * 7 * 10</span>¹⁴
E = 46.38 * 10⁻²⁰
E = 4.638 * 10⁻¹⁹ J
Hope this helps!
Answer:
Percentage of first isotope = 69.152 %
Percentage of second isotope = 30.848 %
Explanation:
The formula for the calculation of the average atomic mass is:
Given that:
For first isotope:
Let % = x %
Mass = 62.9296 amu
For second isotope:
% = 100 - x % (Since, there are only two isotopes)
Mass = 64.9278 amu
Average mass = 63.546 amu
Thus,
Solving,
1.9982 x = 138.18
Thus,
<u>Percentage of first isotope = x = 69.152 %</u>
<u>Percentage of second isotope = 100 - x % = 30.848 %</u>
Answer:
Equilibrium shifts to produce more reactant
Explanation:
- <em>Le Châtelier's principle</em> <em>states that when there is an dynamic equilibrium, and this equilibrium is disturbed by an external factor, the equilibrium will be shifted in the direction that can cancel the effect of the external factor to reattain the equilibrium.</em>
- When more product is added to the solution:
<em>This will increase the concentration of the products side, so the reaction will be shifted to the lift side (reactants side) to suppress the increase in the concentration of Products.</em>
<em />
<em>So, the right choice is: Equilibrium shifts to produce more reactant</em>