The enthalpy of the solution is <u>positive </u>and the entropy is <u>positive</u>.
Potassium trioxonitrate (V) KNO₃(s) is a strong oxidizing solid substance that when dissolved in water changes to aqueous solution.
In its aqueous solution state, the randomness of molecules increases as a result of that the entropy will also increase leading to the positive state of the entropy.
Similarly, provided that the solution becomes quite cold to the touch, the enthalpy is also in it positive state.
Therefore, we can conclude that the enthalpy of the solution is <u>positive </u>and the entropy is <u>positive</u>.
Learn more about Potassium trioxonitrate (V) KNO₃(s) here:
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Answer:
A. 0.0655 mol/L.
B. PbBr2.
C. Pb2+(aq) + Br- --> PbBr2(s).
Explanation:
Balanced equation of the reaction:
Pb(NO3)2(aq) + 2NaBr(aq) --> PbBr2(s) + 2NaNO3(aq)
A.
Number of moles
PbBr2
Molar mass = 207 + (80*2)
= 367 g/mol.
Moles = mass/molar mass
= 3.006/367
= 0.00819 mol.
Since 2 moles of NaBr reacted to form 1 mole of PbBr2. Therefore, moles of NaBr = 2*0.00819
= 0.01638 moles of NaBr.
Since, the ionic equation is
NaBr(aq) --> Na+(aq) + Br-(aq)
Since 1 moles of NaBr dissociation in solution to give 1 mole of Br-
Therefore, molar concentration of Br-
= 0.0164/0.25 L
= 0.0655 mol/L.
B.
PbBr2
C.
Pb(NO3)2(aq)--> Pb2+(aq) + 2No3^2-(aq)
2NaBr(aq) --> 2Na+(aq) + 2Br-(aq)
Net ionic equation:
Pb2+(aq) + 2Br- --> PbBr2(s)
Answer: 31.8 g
Explanation:
To calculate the moles :
According to stoichiometry :
1 mole of 
require 3 moles of 
Thus 0.59 moles of will require=
of
Thus is the limiting reagent as it limits the formation of product and is the excess reagent as it is present in more amount than required.
As 1 mole of give = 2 moles of 
Thus 0.59 moles of give =
of
Mass of 
Thus 31.8 g of will be produced from the given masses of both reactants.
Answer:
.137 moles
Explanation:
need to know the molar mass of water which is 18.01528 g/mol
2.46 g of H₂O divided 18.01528 g/mol molar mass H₂O = moles of H₂O
2.46 divided by 18.01528 =
.137 moles
C4H10 is butane
butane is commonly used in cigarette lighters
2C4H10+1302 -> 8CO2 + 10H₂O is the burning of butane
burning butane with oxygen makes carbon dioxide and water
vsumhcwagovau
