Answer:
Qsp > Ksp, BaCO3 will precipitate
Explanation:
The equation of the reaction is;
Na2CO3 + BaBr2 -------> 2NaBr + BaCO3
Since BaCO3 may form a precipitate we can determine the Qsp of the system.
Number of moles of Na2CO3 = 0.96g/106 g/mol = 9.1 * 10^-3 moles
concentration of NaCO3 = number of moles/volume of solution = 9.1 * 10^-3 moles/10 L = 9.1 * 10^-4 M
Number of moles of BaBr2 = 0.20g/297 g/mol = 6.7 * 10^-4 moles
concentration of BaBr2 = 6.7 * 10^-4 moles/10 L = 6.7 * 10^-5 M
Hence;
[Ba^2+] = 6.7 * 10^-5 M
[CO3^2-] = 9.1 * 10^-4 M
Qsp = [6.7 * 10^-5] [9.1 * 10^-4]
Qsp = 6.1 * 10^-8
But, Ksp for BaCO3 is 5.1*10^-9.
Since Qsp > Ksp, BaCO3 will precipitate
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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solution:
Calculate the molar concentration of the polycyclic aromatic hydrocarbon(PHA)(178.23g/mol),that was found in a well water sample at a concentration of 6.21ppb.Assume the density of the water is 1.00mg/mL
Answer:
aluminium is all of isotope with 14 neutrons .
Test it i guess i dont completely understand the question
