I think the answer is thermal!
Hope this helps:)
Brainiest please:)
The value of ΔS° for reaction is - 22.2 J/K.mol
→ 
Calculation,
Given value of S°(J/K.mol) for
= 248.5
= 240.5
= 210.6
= 256.2
Formula used:
ΔS° (Reaction) = ∑S°(Product) - ∑S°(Reactant)
ΔS° = (256.2 + 210.6 ) - ( 248.5 + 240.5) = 466.8 - 489 = - 22.2 J/K.mol
The change in stander entropy of reaction is - 22.2 J/K.mol. The negative sign indicates the that entropy of reaction is decreases when reactant converted into product.
learn about reaction
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Answer:
3.50*10^-11 mol3 dm-9
Explanation:
A silver rod and a SHE are dipped into a saturated aqueous solution of silver oxalate, Ag2C2O4, at 25°C. The measured potential difference between the rod and the SHE is 0.5812 V, the rod being positive. Calculate the solubility product constant for silver oxalate.
Ag2C2O4 --> 2Ag+ + C2O4 2-
So Ksp = [Ag+]^2 * [C2O42-]
In 1 L, 2.06*10^-4 mol of silver oxalate dissolve, giving, the same number of mol of oxalate ions, and twice the number of mol (4.12*10^-4) of silver ions.
So Ksp = (4.12*10^-4)^2 * (2.06*10^-4)
= 3.50*10^-11 mol3 dm-9
Please give me brainleist. :)
Answer:
2a. If the temperature is increased, the reaction will shift to the right in an attempt to release some of the heat. As the forward reaction loses heat while the reverse would create more heat.
2b. If the pressure is increased, it would shift to the left to counteract the increase in pressure as the left side will have fewer molecules.
2c. If Cl2 is added the reaction will shift to the left in order to remove the stress of the extra Cl2 and favor the production of more reactant.
2d. If PCl3 is removed, the reaction will shift to the right. When part of the equation is removed the reaction learns to adapt to the loss by trying to make more Pcl3 and counteract the effects of losing the PCl3.
3a. The reaction will shift to the right to produce more heat and counter the negative effects of losing the heat.
3b. It will shift to the left to get rid of the excess HCl being produced and form more reactant from the breakdown of the HCl.
3c. It would shift to the right in order to get rid of the excess form products from it.
3d. If pressure is decreased there will be no effect on the shift of the reaction because there is an even amount of moles of gas on each side.
4a. K=[N2O4(g0] / [NO2(g)]2
4b. (Below)
K=[N2O4(g)] / [NO2(g)]2
0.4 / 0.5(2)
0.4/0.25 = 1.6
Keq= 1.6
Answer:
Explanation:
3: Given data:
Number of moles of strontium nitrate = 3.00×10⁻³ mol
Number of atoms = ?
Solution:
There are 9 moles of atoms in 1 mole of Sr(NO₃)₂.
In 3.00×10⁻³ moles,
9 mol × 3.00×10⁻³
27.00×10⁻³ mol
Number of atoms in 3.00×10⁻³ mol of Sr(NO₃)₂:
27.00×10⁻³ mol × 6.022×10²³ atoms / 1mol
162.59×10²⁰ atoms
4)Given data:
Mass of calcium hydroxide = 4500 Kg (4500/1000 = 4.5 g)
Number of moles = ?
Solution:
Number of moles = mass in g/molar mass
by putting values,
Number of moles = 4.5 g/ 74.1 g/mol
Number of moles = 0.06 mol
5) Given data:
Number of atoms of silver nitrate = 1.06×10²³
Number of moles = ?
Solution:
1 mole of any substance contain 6.022×10²³ atoms .
1.06×10²³ atoms × 1 mol / 6.022×10²³ atoms
0.176 moles of silver nitrate
