<u>Answer:</u> The equilibrium constant for 
equation is 
<u>Explanation:</u>
The given chemical equation follows:
The value of equilibrium constant for the above equation is 
Calculating the equilibrium constant for the given equation:
The value of equilibrium constant for the above equation will be:
Hence, the equilibrium constant for equation is
to the proper number of significant figures) to the following? (12.67+19.2)(3.99)/(1.36+ 11.366).
Given reaction represents dissociation of bromine gas to form bromine atoms
Br2(g) ↔ 2Br(g)
The enthalpy of the above reaction is given as:
ΔH = ∑n(products)Δ
- ∑n(reactants)Δ
where n = number of moles
Δ
= enthalpy of formation
ΔH = [2*ΔH(Br(g)) - ΔH(Br2(g))] = 2*111.9 - 30.9 = 192.9 kJ/mol
Thus, enthalpy of dissociation is the bond energy of Br-Br = 192.9 kJ/mol
1) ideal gas law: p·V = n·R·T.
p - pressure of gas.
V -volume of gas.
n - amount of substance.
R - universal gas constant.
T - temperature of gas.
n₁ = 0,04 mol, V₁ = 0,06 l.
n₂ = 0,07 mol, V₂ = 0,06 · 0,07 ÷ 0,04 = 0,105 l.
2) V₁ = 0,06 l, T₁ = 240,00 K.
T₂ = 340,00 K, V₂ = 340 · 0,06 ÷ 240 = 0,05 l.
Hello!
To find the amount of atoms that are in 80.45 grams of magnesium, we will need to know Avogadro's number and the mass of one mole of magnesium.
Avogadro's number is 6.02 x 10^23 atoms, and one mole of magnesium is equal to 24.31 grams.
1. Divide by one mole of magnesium
80.45 / 24.31 = 3.309 moles (rounded to the number of sigfigs)
2. Multiply moles by Avogadro's number
3.309 x (6.02 x 10^23) = 1.99 x 10^24 (rounded to the number of sigfigs)
Therefore, there are 1.99 x 10^24 atoms in 80.45 grams of magnesium.
