4.42 g mass of CCl4 is required to prepare a 0.25 m solution using 115 g of hexane.
It's easy to find the molecular mass of a compound with these steps: Determine the molecular formula of the molecule. Use the periodic table to determine the atomic mass of each element in the molecule. Multiply each element's atomic mass by the number of atoms of that element in the molecule.
The molar mass of any compound can be found out by adding the relative atomic masses of each element present in that particular compound.
Hexane is an organic compound, a straight-chain alkane with six carbon atoms and has the molecular formula C₆H₁₄.
Therefore,
⇒ 0.115 g of Hexane x (0.25 mol CCl4/1 mol hexane) x (153.81 g of CCl4/1 mol CCl4) = 4.42g CCl4.
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Answer:
1= K⁺= (Z=19) = 1s² 2s² 2p⁶ 3s² 3p⁶ 4s⁰
2 = Zn²⁺= (Z = 30) =1s² 2s² 2p⁶ 3s² 3p⁶ 4s⁰, 3d¹⁰
Explanation:
When an atom lose or gain the electron ions are formed. There are two types of ions cation and anion.
Cation are formed when atom lose the electron.
Anion are formed when an atom gain the electron.
In given question potassium loses its valance electron and form K⁺ cation. Thus its electronic configuration will be written as,
₁₉K⁺ = 1s² 2s² 2p⁶ 3s² 3p⁶ 4s⁰
While the electronic configuration of potassium in neutral form is:
₁₉K = 1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹
The atomic number of zinc is 30 and its electronic configuration is:
₃₀Zn= 1s² 2s² 2p⁶ 3s² 3p⁶ 4s², 3d¹⁰
When zinc atom loses its 2 valance electrons the electron configuration will be,
₃₀Zn²⁺= 1s² 2s² 2p⁶ 3s² 3p⁶ 4s⁰, 3d¹⁰
Answer:
Cupric ions
Explanation:
In the single displacement reaction shown, the cupric ions lost two electrons.
Cu²⁺ + 2e⁻ → Cu
The replacement of a metallic ion in solution by a metal atom higher in the activity series than than the metal in solution falls into this category of reactions.
Since Zn higher in the series, it displacements the cupric ions.
Answer:
The answer to the question is;
The equilibrium constant for the reaction is 0.278
Reversibility.
Explanation:
Initial concentration = 0.500 M N₂ and 0.800 M H₂
N₂ (g) + 3·H₂ (g) ⇔ 2·NH₃ (g)
One mole of nitrogen combines with three moles of hydrogen form 2 moles of ammonia
That is 1 mole of ammonia requires 3/2 moles of H₂ and 1/2 moles of N₂
0.150 M of ammonia requires 3/2×0.150 moles of H₂ and 1/2×0.150 moles of N₂
That is 0.150 M of ammonia requires 0.225 moles of H₂ and 0.075 moles of N₂
Therefore at equilibrium we have
Number of moles of Nitrogen = 0.500 M - 0.075 M = 0.425 M
Number of moles of Hydrogen = 0.800 M - 0.225 M = 0.575 M
Number of moles of Ammonia = 0.150 M
K
= ![\frac{[NH_3]^2}{[N_2][H_2]^3} = \frac{(0.15)^2}{(0.425)*(0.575)^3}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BNH_3%5D%5E2%7D%7B%5BN_2%5D%5BH_2%5D%5E3%7D%20%3D%20%5Cfrac%7B%280.15%29%5E2%7D%7B%280.425%29%2A%280.575%29%5E3%7D)
= 0.278
The kind of reaction is a reversible one as the equilibrium constant is greater than 0.01 which as general guide, all components in a reaction with an equilibrium constant between the ranges of 0.01 and 100 will be present when equilibrium is reached and the chemical reaction will be reversible.
