The correct answer is option B. Dirty water is a mixture of solid particles and liquid. It is both a mixture and pure substance.
The dirty water sample has both gravel and liquid water in it. After filtration the gravel is removed so the water sample looks clearer than before filtration. Liquid water is a pure substance because it is a compound that is made up of elements hydrogen and oxygen. Now the gravel is only physically combined with the liquid water, thus giving the water sample properties of a mixture. And like any mixture, gravel is physically separated through filtration from the liquid water.
Thus the water sample of the chemists is both a mixture and pure substance.
Answer:
0.9715 Fraction of Pu-239 will be remain after 1000 years.
Explanation:
Where:
= decay constant
=concentration left after time t
= Half life of the sample
Half life of Pu-239 = 
[
![\lambda =\frac{0.693}{24,000 y}=2.8875\times 10^{-5} y^{-1]](https://tex.z-dn.net/?f=%5Clambda%20%3D%5Cfrac%7B0.693%7D%7B24%2C000%20y%7D%3D2.8875%5Ctimes%2010%5E%7B-5%7D%20y%5E%7B-1%5D)
Let us say amount present of Pu-239 today = 
A = ?
![A=x\times e^{-2.8875\times 10^{-5} y^{-1]\times 1000 y}](https://tex.z-dn.net/?f=A%3Dx%5Ctimes%20e%5E%7B-2.8875%5Ctimes%2010%5E%7B-5%7D%20y%5E%7B-1%5D%5Ctimes%201000%20y%7D)
0.9715 Fraction of Pu-239 will be remain after 1000 years.
Answer:
The standard enthalpy of formation of this isomer of octane is -220.1 kJ/mol
Explanation:
Step 1: Data given
The combustion reaction of octane produces 5104.1 kJ per mol octane
Step 2: The balanced equation
C8H18(g) + 12.5 O2 ⟶ 8CO2 (g) + 9 H2O (g) ∆H°rxn = -5104.1 kJ/mol
Step 3:
∆H°rxn = ∆H°f of products minus the ∆H° of reactants
∆H°rxn = ∆H°f products - [∆H°f reactants]
-5104.1 kJ/mol = (8*∆H°fCO2 + 9*∆H°fH20) - (∆H°fC8H18 + 12.5∆H°fO2)
∆H°f C8H18 = ∆H°f 8CO2 + ∆H°f 9H2O+ 5104.1 kJ/mol
∆H°f C8H18 = 8 * (-393.5 kJ)/mol + 9 * (-241.8 kJ/mol)] + 5104.1 kJ
/mol
∆H°f C8H18 = -220.1 kJ/mol
The standard enthalpy of formation of this isomer of octane is -220.1 kJ/mol
Zeff = Z - S
Here, Z is the number of protons in the nucleus, that is, atomic number, and S is the number of nonvalence electrons.
For boron, the electronic configuration is 1s₂ 2s₂ 2p₄
Z = 5, S = 2
Zeff = 5-2 = +3
For O, electronic configuration is 1s₂ 2s₂ 2p₄
Z = 8, S = 2
Zeff = 8-2 = +6
Hence, the correct answer is second option, that is, +3 and +6, the Zeff of boron is smaller in comparison to O, thus, boron exhibits a bigger size than O.
