For a) [Ru(NH₃)₅Cl]SO₄
Ru configuration = d⁶s²
In this complex Ru oxidation number is +3
Ru³⁺ configuration = d⁵
number of
electrons = 5
For b) Na₂[Os(CN)₆]
Os configuration = d⁶s²
In this complex Os oxidation number is +4
Os⁴⁺ configuration = d⁴
number of
electrons = 4
<span>A 50-gram sample with a half-life of 12 days will have a remaining mass of 25 grams after its 12-day half-life.
Every cycle of a half-life, the sample will lose half of its mass, so if the half-life, itself, is 12 days and the time period passing is 12 days, one half-life has passed and the material will be halved.</span>
When ΔG° is the change in Gibbs free energy
So according to ΔG° formula:
ΔG° = - R*T*(㏑K)
here when K = [NH3]^2/[N2][H2]^3 = Kc
and Kc = 9
and when T is the temperature in Kelvin = 350 + 273 = 623 K
and R is the universal gas constant = 8.314 1/mol.K
So by substitution in ΔG° formula:
∴ ΔG° = - 8.314 1/ mol.K * 623 K *㏑(9)
= - 4536
Answer:
The correct answers are:
a) 180 g
b) 93.7 cm³
Explanation:
The density of a substance is the mass of the substance per unit of volume. So, it is calculated as follows:
density= mass/volume
From the data provided in the problem:
density = 0.8 g/cm³
a) Given: volume= 225 cm³
mass= density x volume = 0.8 g/cm³ x 225 cm³ = 180 g
b) Given: mass= 75.0 g
volume = mass/density = 75.0 g/(0.8 g/cm³)= 93.75 cm³≅ 93.7 cm³
Answer:
= 459.1 nm
This wavelength corresponds to yellow color and thus gold has warm yellow color.
Explanation:
Given that:- Energy = 2.7 eV
Energy in eV can be converted to energy in J as:
1 eV = 1.602 × 10⁻¹⁹ J
So, Energy = 
Considering:-
Where,
h is Plank's constant having value 
c is the speed of light having value 
is the wavelength of the light
So,
= 459.1 nm
This wavelength corresponds to yellow color and thus gold has warm yellow color.
