Answer:
1.32×10²⁵ atoms of sulfate are contained in 22 units of it
Explanation:
1 unit = 1 mol
Al₂(SO₄)₃ → Aluminum sulfate
As 1 unit = 1 mol, 1 unit has 6.02×10²³ atoms of aluminum sulfate.
Let's make a rule of three:
1 unit of Al₂(SO₄)₃ contains 02×10²³ atoms
Then, 22 units of Al₂(SO₄)₃ must contain (22 . 6.02×10²³) / 1 = 1.32×10²⁵ atoms
The intermolecular forces that are responsible for the dissolution of Ethylene glycol in water is hydrogen bonding dipole-dipole forces and dispersion forces.
Both ethylene glycol and water contains the pair of hydrogen and oxygen.
The hydrogen of one atom create a bond with the oxygen of other atom this results in the formation of intra molecular hydrogen bonding.
The electron are non uniformly distributed over the molecule or the atom which results in the fluctuation of the electron density in the atom.
So it creates are dispersion forces which is present all over the molecule this forces helps to increase the strength of the bond formed between the ethylene glycol and water because they have large masses.
Both ethylene glycol and water are polar molecules because of being polar they form dipole and the dipole of both the molecules interact with each other in order to form bond between the atoms which eventually results in the formation dissolution of ethylene glycol in water.
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Compete Question - which intermolecular forces are responsible for the dissolution of ethylene glycol? select all that apply hydrogen bonding, dipole-dipole, dispersion and Ion dipole interaction.
Answer:
see explanations
Explanation:
4NH₃(g) + 5O₂(g) => 4NO(g) + 6H₂O(g)
Ci(NH₃) = 3.5mole/4L = 0.875M
Cf(NH₃) = 1.6mole/4L = 0.400M
Rate-1 => Δ[NH₃]/Δt = |(0.400M - 0.875M)/3min| = 0.158M/s
Rate-2 => 6(Δ[NH₃]/Δt) = 4(Δ[H₂O]/Δt) => 6/4(0.158M/s) = 0.237M/s
Rate-3 => 5(Δ[NH₃]/Δt) = 4(Δ[O₂]/Δt) => 5/4(0.158M/s) = 0.237M/s
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NOTE: When setting up comparative rate expressions for a given reaction, set the rates expressions as equal then swap coefficient values. Then solve for rate of interest and substitute givens.
example: for NH₃ and H₂O
- set rates expressions equal => Δ[NH₃]/Δt = Δ[H₂O]/Δt
- then swap and insert coefficients from given rxn ...
- solve for rate of interest ...
4NH₃(g) + 5O₂(g) => 4NO(g) + 6H₂O(g)
=> 6(Δ[NH₃]/Δt) = 4(Δ[H₂O]/Δt)
=> Δ[H₂O]/Δt = 6/4(Δ[NH₃]/Δt) = 6/4(0.237M/s) = 0.237M/s
Answer: 0.0748 grams of CO will dissolve in 1.00 L of water if the partial pressure of CO is 2.75 atm
Explanation:
Henry's law states that the amount of gas dissolved or molar solubility of gas is directly proportional to the partial pressure of the liquid.
To calculate the molar solubility, we use the equation given by Henry's law, which is:
where,
= Henry's constant = 
= partial pressure of CO = 2.75 atm
Putting values in above equation, we get:
Hence, the solubility of carbon monoxide gas is 0.0748 g/L
<span>The expected results of the Rutherford's gold foil experiment were that the relative massive alpha particles (respect to electrons) could go through the gold foil without being deviated of their trajectory or only small deviations due to the uniformly distributed positive charge of the protons. The real results showed that some particles were significantly deviated of the trajectory (large deviation angles and even some particles bounced back to the source). This lead Rutherford to reject the plum pudding model and propose a new one. The new model proposed by Rutherford was that the atom consisted of a small and every dense nucleus (which contained the positive charge, protons) and a vast region, almost empty, but where the electrons were, surrounding the nucleus.</span><span />
