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
They exist in the outer orbitals
Answer:
собак гандонов в гавне))))))))))
Explanation:
hloride,which reactant is in excess?how many moles of aluminium chloride can be produced during the reaction.hloride,which reactant is in excess?how many moles of aluсиськижопыссиськасуксиськипись loride can be produced during the reaction.hloride,which reactant is in excess?how many moles of aluminium chloride can be produced during the reaction.hloride,whichйух reactant is in excess?how many moles of aluminium chloride can be produced during the reaction.
Answer:
P2=0.385atm
Explanation:
step one:
Given that the temperature T1= 60 Celcius
we can convert this to kelvin by adding 273k to 60 Celcius
we have T1= 333k
pressure P1= 0.470 atm
step two:
we know that the standard temperature is T2= 273K
Applying the temperature and pressure relationship we have
P1/T1=P2/T2
substituting our given data we have
0.47/333=P2/273
cross multiply we have
P2= (0.47*273)/333
P2= 128.31/333
P2=0.385 atm
