Answer:
Mn is the oxidizing agent.
N is the reducing agent.
Explanation:
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In this case, according to the undergoing chemical reaction, it is seen that the manganese in KMnO4 has an oxidation state of 7+, in MnSO4 of 2+ and nitrogen in KNO2 is 3+ and in KNO3 is 5+; thus we have the following half-reactions:
Thus, since manganese is undergoing a decrease in the oxidation state, we infer it is the oxidizing agent whereas nitrogen, undergoing an increase in the oxidation state is the reducing agent.
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Answer:
1132.8 ml of water
Explanation:
you have an aqueous solution contains 158.2 g KOH per liter
so concentration =158.2/56 = 2.825M
Molarity =2.825
that means you have 2.825 moles of KOH in 1.00L solution
Mass of Soluet(KOH)= 152.8g
Volume of solution= 1.00L
density of solution= 1.13g/cm3 =1.13g/ml
therefore mass of solution = VolumeX density = 1000mL X 1.13g/ml.=1130g
Mass of solvent(water)= mass of solution- mass of solute(KOH)=1130-152.8= 997.2g
Molality= moles of solute/mass of solvent(Kg)
=2.825/(997.2/1000)= 2.832molal
to prepare a 0.250 molal solution of KOH, starting with 100.0ml ofthe orginal solution
0.250*X =2.832 *100
X = 1132.8 ml of water you have to add
I am pretty sure the answer is D. That the patterns of a polarity matched up on both sides.
You're off to a good start, now find the mass of H2O and put it under I mol,
then multiply 1 mol over the mass of H2O by 215 grams
