Boric acid, H3BO3, in aqueous solution would only give out one H+ ion. As it is also produce OH ion and by hydrolysis it produces one proton. <span>All the boron compounds (BX3) are having only 6 valence electrons in it and should follow the octet rule by taking another electron.</span>
B(OH)3 + 2 H2O → B(OH)4− + H3O
Answer:
B. Cu (s) +Ni(NO3)2 (aq) - 2CuNO3 +Ni (s)
Explanation:
the above reaction is a substitution reaction
Answer:
0.143 g of KCl.
Explanation:
Equation of the reaction:
AgNO3(aq) + KCl(aq) --> AgCl(s) + KNO3(aq)
Molar concentration = mass/volume
= 0.16 * 0.012
= 0.00192 mol AgNO3.
By stoichiometry, 1 mole of AgNO3 reacts with 1 mole of KCl to form a precipitate.
Number of moles of KCl = 0.00192 mol.
Molar mass of KCl = 39 + 35.5
= 74.5 g/mol
Mass = molar mass * number of moles
= 74.5 * 0.00192
= 0.143 g of KCl.
Answer:
This question is incomplete, here's the complete question:
<em><u>"Suppose 0.0842g of potassium sulfate is dissolved in 50.mL of a 52.0mM aqueous solution of sodium chromate. Calculate the final molarity of potassium cation in the solution. You can assume the volume of the solution doesn't change when the potassium sulfate is dissolved in it. Round your answer to 2 significant digits."</u></em>
Explanation:
Reaction :-
K2SO4 + Na2CrO4 ------> K2CrO4 + Na2SO4
Mass of K2SO4 = 0.0842 g, Molar mass of K2SO4 = 174.26 g/mol
Number of moles of K2SO4 = 0.0842 g / 174.26 g/mol = 0.000483 mol
Concentration of Na2CrO4 = 52.0 mM = 52.0 * 10^-3 M = 0.052 mol/L
Volume of Na2CrO4 solution = 50.0 ml = 50 L / 1000 = 0.05 L
Number of moles of Na2CrO4 = 0.05 L * 0.052 mol/L = 0.0026 mol
Since number of moles of K2SO4 is smaller than number of moles Na2CrO4, so 0.000483 mol of K2SO4 will react with 0.000483 mol of Na2CrO4 will produce 0.000483 mol of K2CrO4.
0.000483 mol of K2CrO4 will dissociate into 2* 0.000483 mol of K^+
Final concentration of potassium cation
= (2*0.000483 mol) / 0.05 L = 0.02 mol/L = 0.02 M
Answer:
la particula que queda es h2o
Explanation:
