Explanation:
Given
The enthalpy of formation of RbF (s) is –557.7kJ/mol
The standard enthalpy of formation of RbF (aq, 1 m) is –583.8 kJ/mol
The enthalpy of solution of RbF = Enthalpy of RbF (aq) - Enthalpy of formation of RbF (s)
= -583.8 - (-557.7) kJ/mol
= -26.1 kJ/mol
The enthalpy is negative which means that the temperature will rise when RbF is dissolved.
Answer:
714 nm
Explanation:
Using the equation: nλ=d<em>sin</em>θ
where
n= order of maximum
λ= wavelength
d= distance between lines on diffraction grating
θ= angle
n is 1 because the problem states the light forms 1st order bright band
λ is unknown
d=
or 0.0000014 (meters)
sin(30)= 0.5
so
(1)λ=(0.0000014)(0.5)
=0.000000714m or 714 nm
Answer:
[H₂] = 1.61x10⁻³ M
Explanation:
2H₂S(g) ⇋ 2H₂(g) + S₂(g)
Kc = 9.30x10⁻⁸ = ![\frac{[H_{2}]^2[S_{2}]}{[H_{2}S]^2}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BH_%7B2%7D%5D%5E2%5BS_%7B2%7D%5D%7D%7B%5BH_%7B2%7DS%5D%5E2%7D)
First we <u>calculate the initial concentration</u>:
0.45 molH₂S / 3.0L = 0.15 M
The concentrations at equilibrium would be:
[H₂S] = 0.15 - 2x
[H₂] = 2x
[S₂] = x
We <u>put the data in the Kc expression and solve for x</u>:


We make a simplification because x<<< 0.0225:

x = 8.058x10⁻⁴
[H₂] = 2*x = 1.61x10⁻³ M
Answer:
<h3>Chlorine </h3>
It has 7 electrons in it last shell. it will gain one electron to complete its octet and become Cl-. The ionic bond of chlorine form with potassium is Kcl ( potassium chloride)
- Phosphorus has 5 electrons in its last shell it need 3 more electron to stable .
- Sulphur has 6 electron. it needs 2 electrons
- Argon is already stable as it has 8 electrons in its last shell it will not gain or loose any electrons