Answer:
V₂ = 0.6 V.
Explanation:
- We can use the general law of ideal gas: <em>PV = nRT.</em>
where, P is the pressure of the gas in atm.
V is the volume of the gas in L.
n is the no. of moles of the gas in mol.
R is the general gas constant,
T is the temperature of the gas in K.
- If n is constant, and have different values of P, V and T:
<em>(P₁V₁T₂) = (P₂V₂T₁).</em>
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V₁ = V, P₁ = P, T₁ = T.
V₂ = ??? V, P₂ = 1.25 P, T₂ = 0.75 T.
<em>∴ V₂ = (P₁V₁T₂)/(P₂T₁) =</em> (P)(V)(0.75 T)/(1.25 P)(T)<em> = 0.6 V.</em>
Answer:
That means Cu2O is limiting reagent and C is excess reagent
Explanation:
Based on the reaction, 1 mole of Cu2O reacts per mole of C. The ratio of reaction is 1:1.
To solve this question we need to convert the mass of each reactant to moles. The reactant with the lower amount of moles is limiting reactant and the excess reactant is the reactant with the higher number of moles.
<em>Moles Cu2O -Molar mass: 143.09 g/mol-</em>
114.2g Cu2O * (1mol / 143.09g) = 0.798 moles Cu2O
<em>Moles C -Molar mass: 12.01g/mol-</em>
11.1g C * (1mol / 12.01g) = 0.924 moles C
<h3>That means Cu2O is limiting reagent and C is excess reagent</h3>
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Answer:
The effect is the increasing of the molar concentration.
Explanation:
When you standarize a solution of NaOH with KHP you are establish its molar concentration (That is the amount of moles of NaOH per liter of solution).
If you evaporated some water of the solution, you are increasing its concentration because volume is decreasing doing the amount of moles per liter increasing.
