A test tube of zinc oxide
Methane CH4 CH4 1 hexane C6H14 CH3(CH2)4CH3 5
ethane C2H6 CH3CH3 1 heptane C7H16 CH3(CH2)5CH3 9
propane C3H8 CH3CH2CH3 1 octane C8H18 CH3(CH2)6CH3 18
butane C4H10 CH3CH2CH2CH3 2 nonane C9H20 CH3(CH2)7CH3 35
pentane C5H12 CH3(CH2)3CH3 3 decane C10H22 CH3(CH2)8CH3 75
Answer:
0.3192 M
Explanation:
From the question given above, the following data were obtained:
Volume of stock solution (V1) = 5.32 mL Molarity of stock solution (M1) = 6 M
Volume of diluted solution (V2) = 100 mL
Molarity of diluted solution (M2) =?
We can obtain the molarity of the diluted solution by using the dilution formula as shown follow:
M1V1 = M2V2
6 × 5.32 = M2 ×100
31.92 = M2 × 100
Divide both side by 100
M2 = 31.92 / 100
M2 = 0.3192 M
Therefore, the molarity of the diluted solution is 0.3192 M.
1. its temperature will rise continuously until it melts
I don't believe that any of the other answers are correct because it can not stay at a certain temperature if it is melting