Mole ratio for the reaction is 1:1
no of moles in NaOH that reacted= 1*21.17/1000=0.02117mols
molarity of HCl=0.02117*10/1000
=2.117M
1) Chemical equation
Na2 SiO3 (s) + 8 HF (aq) ---> H2 Si F6 (aq) + 2 Na F (aq) + 3H2O (l)
It is balanced
2) Molar ratios
1 mol Na2 SiO3 : 8 mol HF.
3) Proportion
0.340 mol Na2 SiO3 * 8 mol HF / 1mol Na2SiO3 = 2.72 mol HF.
Answer: 2.72 mol HF
Answer:
[Cl⁻] = 0.016M
Explanation:
First of all, we determine the reaction:
Pb(NO₃)₂ (aq) + MgCl₂ (aq) → PbCl₂ (s) ↓ + Mg(NO₃)₂(aq)
This is a solubility equilibrium, where you have a precipitate formed, lead(II) chloride. This salt can be dissociated as:
PbCl₂(s) ⇄ Pb²⁺ (aq) + 2Cl⁻ (aq) Kps
Initial x
React s
Eq x - s s 2s
As this is an equilibrium, the Kps works as the constant (Solubility product):
Kps = s . (2s)²
Kps = 4s³ = 1.7ₓ10⁻⁵
4s³ = 1.7ₓ10⁻⁵
s = ∛(1.7ₓ10⁻⁵ . 1/4)
s = 0.016 M
<u><em>The process of how we would obtain </em></u><u><em>ethanal</em></u><u><em> </em></u><u><em>free</em></u><u><em> from ethanol is described in the explanations below. </em></u>
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- In Chemistry, Ethanol undergoes oxidation in the presence of sodium dichromate plus sulphuric acid to yield ethanal and water.
The procedure for achieving this in the laboratory is as follows;
- Step 1; Measure a quantity of a solution of sodium dichromate acidified in a dilute sulphuric acid and pour into a test tube.
- Step 2; Add excess <em>ethanol</em>. This is because if we don't do so there will be plenty of oxidizing agent to carry out a second operation which changes the aldehyde to ethanoic acid. However, we need only the aldehyde.
- Step 3; When the aldehyde ethanal begins to form which will be evident by the change in the colour of solution from <em>orange to green</em>, then the mixture should be distilled from the test tube and tbethe aldehyde collevted so that it doesn't undergo additional oxidation into ethanoic acid.
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