Answer:
5.3 × 10⁻¹⁷ mol·L⁻¹
Explanation:
Let <em>s</em> = the molar solubility.
Cu₂S(s) ⇌ 2Cu⁺(aq) + S²⁻(aq); K_{sp} = 6.1 × 10⁻⁴⁹
E/mol·L⁻¹: 2<em>s</em> <em>s
</em>
K_{sp} =[Cu⁺]²[S²⁻] = (2<em>s</em>)²×<em>s</em> = 4s^3 = 6.1 × 10⁻⁴⁹
![s = \sqrt[3]{1.52 \times 10^{-49}} \text{ mol/L} = 5.3 \times 10^{-17} \text{ mol/L}](https://tex.z-dn.net/?f=s%20%3D%20%5Csqrt%5B3%5D%7B1.52%20%5Ctimes%2010%5E%7B-49%7D%7D%20%5Ctext%7B%20mol%2FL%7D%20%3D%205.3%20%5Ctimes%2010%5E%7B-17%7D%20%5Ctext%7B%20mol%2FL%7D)
Lithium, sodium, potassium, rubidium, cesium, and francium
Answer:
2.5 L will be the volume of HNO₃
Explanation:
To find out the total volume of nitric acid in liters we begin from molarity.
HNO₃ solution is 0.10 M
This means that 0.10 moles are contained in 1L of solution.
As we used 0.25 moles of nitric, let's determine the volume by a rule of three:
0.10 moles of nitric acid are contained in 1L of solution
0.25 moles of nitric acid will be contained in (0.25 . 1) / 0.1 = 2.5 L
A density of the substance is an intrinsic property. Each substance has its own value of density, and it is constant. Since density is equal to mass over volume, a graph of mass vs volume would have a constant slope equal to density. So, it will be a linear graph. The mass is in the y-axis, and the volume is on the x-axis. Locate V = 16 mL on the x-axis, project it upwards until it intersects with the linear graph, then, move towards the left to determine the corresponding y-value, represented by the mass.
Answer: E
=
1.55
⋅
10
−
19
J
Explanation:
The energy transition will be equal to 1.55
⋅
10
−
1
J
.
So, you know your energy levels to be n = 5 and n = 3. Rydberg's equation will allow you calculate the wavelength of the photon emitted by the electron during this transition
1
λ =
R
⋅
(
1
n
2
final −
1
n
2
initial )
, where
λ
- the wavelength of the emitted photon;
R
- Rydberg's constant - 1.0974
⋅
10
7
m
−
1
;
n
final
- the final energy level - in your case equal to 3;
n
initial
- the initial energy level - in your case equal to 5.
So, you've got all you need to solve for λ
, so
1
λ =
1.0974
⋅10 7
m
−
1
⋅
(....
−152
)
1
λ
=
0.07804
⋅
10
7
m
−
1
⇒
λ
=
1.28
⋅
10
−
6
m
Since
E
=
h
c
λ
, to calculate for the energy of this transition you'll have to multiply Rydberg's equation by
h
⋅
c
, where
h
- Planck's constant -
6.626
⋅
10
−
34
J
⋅
s
c
- the speed of light -
299,792,458 m/s
So, the transition energy for your particular transition (which is part of the Paschen Series) is
E
=
6.626
⋅
10
−
34
J
⋅
s
⋅
299,792,458
m/s
1.28
⋅
10
−
6
m
E
=
1.55
⋅
10
−
19
J
