Answer:
As
Explanation:
For any element to exhibit the pattern of ionization energy shown in the question, it must possess five electrons in its outermost shell. These five electrons are not lost at once. They are lost progressively until the valence shell becomes empty. The ionization energy increases steadily as more electrons are lost from the valence shell.
The only pentavalent element among the options in arsenic, hence the answer.
Answer:
81°C.
Explanation:
To solve this problem, we can use the relation:
<em>Q = m.c.ΔT,</em>
where, Q is the amount of heat released from water (Q = - 1200 J).
m is the mass of the water (m = 20.0 g).
c is the specific heat capacity of water (c of water = 4.186 J/g.°C).
ΔT is the difference between the initial and final temperature (ΔT = final T - initial T = final T - 95.0°C).
∵ Q = m.c.ΔT
∴ (- 1200 J) = (20.0 g)(4.186 J/g.°C)(final T - 95.0°C ).
(- 1200 J) = 83.72 final T - 7953.
∴ final T = (- 1200 J + 7953)/83.72 = 80.67°C ≅ 81.0°C.
<em>So, the right choice is: 81°C.</em>
Answer : The normality of the solution is, 30.006 N
Explanation :
Normality : It is defined as the number of gram equivalent of solute present in one liter of the solution.
Mathematical expression of normality is:

or,

First we have to calculate the equivalent weight of solute.
Molar mass of solute
= 94.97 g/mole

Now we have to calculate the normality of solution.

Therefore, the normality of the solution is, 30.006 N
First, let us find the corresponding amount of moles H₂ assuming ideal gas behavior.
PV = nRT
Solving for n,
n = PV/RT
n = (6.46 atm)(0.579 L)/(0.0821 L-atm/mol-K)(45 + 273 K)
n = 0.143 mol H₂
The stoichiometric calculations is as follows (MW for XeF₆ = 245.28 g/mol)
Mass XeF₆ = (0.143 mol H₂)(1 mol XeF₆/3 mol H₂)(245.28 g/mol) = <em>11.69 g</em>