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
Answer:
Mg²⁺(aq) + SO₃²⁻(aq) + 2 H⁺(aq) + 2 I⁻(aq) ⇄ Mg²⁺(aq) + 2I⁻(aq) + H₂O(l) + SO₂(g)
Explanation:
<em>Give the complete ionic equation for the reaction (if any) that occurs when aqueous solutions of MgSO₃ and HI are mixed.</em>
When MgSO₃ reacts with HI they experience a double displacement reaction, in which the cations and anions of each compound are exchanged, forming H₂SO₃ and MgI₂. At the same time, H₂SO₃ tends to decompose to H₂O and SO₂. The complete molecular equation is:
MgSO₃(aq) + 2 HI(aq) ⇄ MgI₂(aq) + H₂O(l) + SO₂(g)
In the complete ionic equation, species with ionic bonds dissociate into ions.
Mg²⁺(aq) + SO₃²⁻(aq) + 2 H⁺(aq) + 2 I⁻(aq) ⇄ Mg²⁺(aq) + 2I⁻(aq) + H₂O(l) + SO₂(g)
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<h3>Further explanation</h3>
Given
3000 L of gas at 39°C and 99 kPa to 45.5 kPa and 16°C,
Required
the new volume
Solution
Combined with Boyle's law and Gay Lussac's law

T₁ = 39 + 273 = 312
T₂ = 16 + 273 = 289
Input the value :
V₂ = (P₁V₁.T₂)/(P₂.T₁)
V₂ = (99 x 3000 x 289)/(45.5 x 312)
or we can write it as:
V₂ = 3000 L x (289/312) x (99/45.5)
So, if 1 mole occupies 22.4 L, the imediate conclusion is that a bigger number of moles will occupy more than 22.4 L, and a smaller number of moles will occupy less than 22.4 L. In your case, 3 moles of gas will occupy 3 times more volume than 1 mole of gas
no need to show work i am a teacher at a college