Answer:
32.5g of sodium carbonate
Explanation:
Reaction of sodium carbonate (Na₂CO₃) with Mg²⁺ and Ca²⁺ as follows:
Na₂CO₃(aq) + Ca²⁺(aq) → CaCO₃(s)
Na₂CO₃(aq) + Mg²⁺(aq) → MgCO₃(s)
<em>1 mole of carbonate reacts per mole of the cations.</em>
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To know the mass of sodium carbonate we must know the moles of carbonate we need to add based on the moles of the cations:
<em>Moles Mg²⁺:</em>
2.91L * (0.0661 moles MgCl₂ / 1L) = 0.192 moles MgCl₂ = Moles Mg²⁺
<em>Moles Ca²⁺:</em>
2.91L * (0.0396mol Ca(NO₃)₂ / 1L) = 0.115 moles Ca(NO₃)₂ = Moles Ca²⁺
That means moles of sodium carbonate you must add are:
0.192 moles + 0.115 moles = 0.307 moles sodium carbonate.
In grams (Using molar mass Na₂CO₃ = 105.99g/mol):
0.307 moles Na₂CO₃ * (105.99g / mol) =
<h3>32.5g of sodium carbonate</h3>
From the equation; ΔTf = Kf × m
Where, Kf for water = 1.853 K kg/mole; m is the molarity = number of solute/amount of solvent in kg.
Glucose is the solute whose molecular mass is 180 g/mole and water is the solvent.
Moles of solute = 15.5/180 = 0.0861 moles
Amount of solvent in kg = 245/1000 = 0.245 Kg
Therefore; molarity = 0.0861/0.245 = 0.3515 moles/Kg
Therefore; ΔTf = 1.853 × 0.3515 = 0.6513 K
Hence; the depression in freezing point is 0.6513
The freezing point of solution will therefore be;
= 273 - 0.6513 = 272.3487 K
Answer:
97 000 g Na
Explanation:
The absortion (or liberation) of energy in form of heat is expressed by:
q=m*Cp*ΔT
The information we have:
q=1.30MJ= 1.30*10^6 J
ΔT = 10.0°C = 10.0 K (ΔT is the same in °C than in K)
Cp=30.8 J/(K mol Na)
If you notice, the Cp in the question is in relation with mol of Na. Before using the q equation, we can find the Cp in relation to the grams of Na.
To do so, we use the molar mass of Na= 22.99g/mol

Now, we are able to solve for m:
=97 000 g Na
Williamson synthesis is the most common way for obtaining ethers, called after its developer Alexander Williamson. It is an organic reaction of forming ethers from an organohalide and an alkoxide. The reaction is carried out according to the SN2 mechanism.
On the attached picture it is shown required alkoxide ion, <span>alkyl(aryl)bromide and the ether that forms from the reactants. </span>