First write out the balanced equation. 3Cu+2Ag(NO3)3=2Ag+3Cu(NO3)2
Then convert copper from grams to moles
15 g*1 mol cu/63.54 g= 15/63.54 mol cu
Then use the mole ratio to convert Moles Cu to Moles Ag
15/63.54 moles Cu* 2 moles Ag/3 moles Cu
The final awnser is (15*2)/(63.54*3) moles Ag =0.157 moles Ag. If the question wants the answer in grams, convert from moles Ag to grams Ag.
0.157 moles Ag*107.87 g Ag/ mol Ag=16.98 g Ag
P2O5 = Phosphorus pentoxide
CuO = Copper (II) oxide
NH4CI = Ammonium Chloride
Mn(OH)2 = Pyrochroite
H2O2 = Hydrogen peroxide
P4S9 = Tetraphosphorus nonasulfide
CIO2 = Chlorine dioxide
NaF = Sodium fluoride
FeSO3 = Iron (II) Sulfite
Fe(NO3)3 = Iron (III) Nitrate
Cr(NO2)3 = Chromium (III) Nitrite
NaHCO3 = Sodium Hydrogen Carbonate
H2PO4 = Dihydrogen Phosphate Ion
NaCN = Sodium Cyanide
IF7 = Iodine Heptafluoride
PCI3 = Phosphorus Trichloride
Answer: The molarity of KBr in the final solution is 1.42M
Explanation:
We can calculate the molarity of the KBr in the final solution by dividing the total number of moles of KBr in the solution by the final volume of the solution.
We will first calculate the number of moles of KBr in the individual sample before mixing together
In the first sample:
Volume (V) = 35.0 mL
Concentration (C) = 1.00M
Number of moles (n) = C × V
n = (35.0mL × 1.00M)
n= 35.0mmol
For the second sample
V = 60.0 mL
C = 0.600 M
n = (60.0 mL × 0.600 M)
n = 36.0mmol
Therefore, we have (35.0 + 36.0)mmol in the final solution
Number of moles of KBr in final solution (n) = 71.0mmol
Now, to get the molarity of the final solution , we will divide the total number of moles of KBr in the solution by the final volume of the solution after evaporation.
Therefore,
Final volume of solution (V) = 50mL
Number of moles of KBr in final solution (n) = 71.0mmol
From
C = n / V
C= 71.0mmol/50mL
C = 1.42M
Therefore, the molarity of KBr in the final solution is 1.42M
Answer:
Explanation:
The moles of H2 and N2 are as follows respectively, 0.3915mol of H2 and 0.1305 mol of N2.
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