To solve this problem, we must assume ideal gas behaviour so
that we can use Graham’s law:
vA / vB = sqrt (MW_B / MW_A)
where,
<span>vA = speed of diffusion of A (HBR)</span>
vB = speed of diffusion of B (unknown)
MW_B = molecular weight of B (unkown)
MW_A = molar weight of HBr = 80.91 amu
We know from the given that:
vA / vB = 1 / 1.49
So,
1/1.49 = sqrt (MW_B / 80.91)
MW_B = 36.44 g/mol
Since this unknown is also hydrogen halide, therefore this
must be in the form of HX.
HX = 36.44 g/mol , therefore:
x = 35.44 g/mol
From the Periodic Table, Chlorine (Cl) has a molar mass of
35.44 g/mol. Therefore the hydrogen halide is:
HCl
Answer:
An ion with 11 protons, 11 neutrons, and 10 electrons would have a charge of 1+, also expressed as a charge of positive one or +1.
Answer:
a.) 22.4 L Ne.
Explanation:
It is known that every 1.0 mol of any gas occupies 22.4 L.
For the options:
<em>It represents </em><em>1.0 mol of Ne.</em>
<em />
using cross multiplication:
1.0 mol occupies → 22.4 L.
??? mol occupies → 20 L.
The no. of moles of (20 L) Ar = (1.0 mol)(20 L)/(22.4 L) = 0.8929 mol.
using cross multiplication:
1.0 mol occupies → 22.4 L.
??? mol occupies → 2.24 L.
<em>The no. of moles of (2.24 L) Xe </em>= (1.0 mol)(2.24 L)/(22.4 L) = <em>0.1 mol.</em>
- So, the gas that has the largest number of moles at STP is: a.) 22.4 L Ne.
Electrons are orbiting around the nucleus in a specific energy level as described in Bohr's atomic model. There are 7 energy levels all in all; 1 being the strongest and nearest to the nucleus, and 7 being the weakest and farthest away from the nucleus. Electron can transfer from one energy level to another. If it increases energy, it absorbs energy. If it goes down an energy level, it emits energy in the form of light. This light can be measure in wavelength through the Rydberg equation:
1/λ =R(1/n₁² -1/n₂²), where
λ is the wavelength
R is the Rydberg constant equal to 1.097 × 10⁻7<span> per meter
n</span>₁ and n₂ are the energy levels such that n₂>n₁
In the Paschen series is an emission spectrum of hydrogen when the energy level is at least n=4. So, this covers n=4 to n=7.
1/λ =(1.097 × 10⁻7)(1/4² -1/7²)
λ = 216.57 ×10⁻⁶ m or 216.57 μm
Answer: The percent ionization of a monoprotic weak acid solution that is 0.188 M is 
Explanation:
Dissociation of weak acid is represented as:
cM 0 0
So dissociation constant will be:
Give c= 0.188 M and 
= ?
Putting in the values we get:
Thus percent ionization of a monoprotic weak acid solution that is 0.188 M is
