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
I would believe it to be 55g. A -> B YIELDS AB. So, 10g + 45g = 55g.
Answer:
Answer is given below.
Explanation:
Anode is that electrode where oxidation occurs. Cathode is that electrode where reduction occurs.
In cell representation, half cell present left to salt-bridge notation
is anodic system and another half cell present right to salt-bridge notation is cathodic system.
So anode is Cu and cathode is Ag.
oxidation: 
[reduction: 
]
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chemical equation: 
Oxidizing agent is that species which takes electron from another species. Here 
takes electron from Cu. Hence is the oxidizing agent.
Reducing agent is that species which gives electron to another species. Here Cu gives electron to . Hence Cu is the reducing agent.
Answer:
Two
Explanation:
Elements in group 16 wants to bond with elements in group IIA, the group of alkaline earth metals.
- The bonding will make it easier for them complete their octet.
- Elements in group 16 has 6 valence electrons.
- To have a complete octet, they require 2 more electrons.
- Group II elements are willing donors as they are metals.
- For Group II elements to fill their octets, they must lose two electrons.
- So the willingness of group II elements to lose two electrons and the readiness for group 16 elements to gain the electrons makes the desire one another.
