The empirical formula of the compound is C. NiF₂.
<em>Step 1</em>. Calculate the <em>moles of each element</em>
The empirical formula is the simplest whole-number ratio of atoms in a compound.
The ratio of atoms is the same as the ratio of moles.
So, our job is to calculate the molar ratio of Ni to F.
Moles of Ni = 9.11 g Ni × (1 mol Ni /(58.69 g Ni) = 0.1552 mol Ni
Moles of F = 5.89 g F × (1 mol F/19.00 g F) = 0.3100 mol F
<em>Step 2</em>. Calculate the <em>molar ratio</em> of the elements
Divide each number by the smallest number of moles
Ni:F = 0.1552:0.3100 = 1:1.997 ≈ 1:2
<em>Step 3</em>: Write the <em>empirical formula</em>
EF = NiF₂
Answer:
The upper and lower limits for the room-temperature thermal conductivity of a magnesium oxide material having a volume fraction of 0.10 of pores that are filled with still air are
Ku = 38.252 W/mK
K lower = 0.199 W/mK
Explanation:
As we know
Ku = Vp * Kair + Vmagnesium * K metal
Ku = 0.10 *0.02 + (1-0.25) * 51
Ku = 38.252 W/mK
The lower limit
K lower = Kmetal* Kair/( Vp * Kmetal + Vmetal * K air)
K lower = (0.02*51)/(0.10*51 + 0.90 * 0.02)
K lower = 0.199 W/mK
Given, half life of a certain radioactive element = 800 years.
Amount of substance remaining at time t = 12.5%
Lets consider the initial amount of the radioactive substance = 100%
Using the half life equation:
A = A₀(1/2)^t/t₁/₂
where A₀ is the amount of radioactive substance at time zero and A is the amount of radioactive substance at time t, and t₁/₂ is the half-life of the radioactive substance.
Plugging the given data into the half life equation we have,
12.5 = 100 . (1/2)^t/800
12.5/100 = (1/2)^t/800
0.125 = (0.5)^t/800
(0.5)^3 = (0.5)^t/800
3 = t/800
t = 2400 years
Thus the object is 2400 years old.
