Ml=−2,−1,0,+1,+2.
<span>Since each of these orbitals can hold a maximum of </span>two electrons<span>, one having spin-up and one having spin-down, a total of </span>10 electrons<span> can share the quantum numbers n = 4 and l = 2</span>
Answer:
7430.5 Joules (7.4*10^4 Joules)
Explanation:
Q=mc∆T
where Q is energy in Joules.
Now m=250 g
c= 0.386 J/g°C
∆T = 99 - 22 = 77 °C
plugging the values in gives
Q=250*0.386*77=7430.5 Joules
(7.4*10^4 Joules, if 2 significant figures)
Answer: 0.225 atm
Explanation:
For this problem, we have to use Boyle's Law.
Boyle's Law: P₁V₁=P₂V₂
Since we are asked to find P₂, let's manipulate the equation.
P₂=(P₁V₁)/V₂

With this equation, the liters cancel out and we will be left with atm.
P₂=0.225 atm
Answer:
The equation to show the the correct form to show the standard molar enthalpy of formation:

Explanation:
The standard enthalpy of formation or standard heat of formation of a compound is the change of enthalpy during the formation of 1 mole of the substance from its constituent elements, with all substances in their standard states.
Given, that 1 mole of
gas and 1 mole of
liquid gives 2 moles of HBr gas as a product.The reaction releases 72.58 kJ of heat.

Divide the equation by 2.

The equation to show the the correct form to show the standard molar enthalpy of formation:

Mass of solute ( m1 ) = 50.0 g
mass of solvent ( m2 ) = 150.0 g
Therefore:
m/m = ( m1 / m1 + m2 )
m/m = ( 50.0 / 50.0 + 150.0 )
m/m = ( 50.0 / 200 )
m/m = 0.25