Let initially there are 10 molecules of O2 and 3 molecules of C3H8 present
The reaction will be
C3H8(g) + 5O2(g) ----> 3CO2(g) + 4H2O
so here oxygen molecules are limiting as for 3 molecules of C3H8 we need 15 molecules of O2
now the given 10 molecules of O2 will react with only 2 molecules of C3H8 and they will form six molecules of CO2 and 8 molecules of H2O
Hence answer is
molecules of CO2 formed = 6
Molecules of H2O formed = 8
molecules of C3H8 left = 1
molecules of O2 left = 0
All of them are soluble salt.
First one dissociates into two ions.
The second one dissociates into 3 ions.
The third dissociate into 4 ions. therefore, Al(NO3)3
OK, so to answer this question, you will simply use the molality equation which is as follows:
<span>M1V1 = M2V2
In the givens you have:
M1 = 2M
V1 is the unknown
M2 = 0.4M
V2 = 100 ml
</span>plug in the givens in the above equation:
<span>2 x V1 = 0.4 x 100
</span>therefore:
V1 = 20 ml
Based on this: you should take 20 ml of the 2 M solution and make volume exactly 100 ml in a volumetric flask by diluting in water.
Answer:
–0.16 m
Explanation:
From the question given above, the following data were obtained:
Time (t) = 0.18 s
Acceleration due to gravity (g) = –9.81 m/s²
Height (h) =?
We can obtain how far the ruler will fall by using the following equation:
H = ½gt²
H = ½ × –9.81 × 0.18²
H = ½ × –9.81 × 0.0324
H = –0.16 m
Thus, the ruler will fall –0.16 m before you will catch it.
Answer:
B.) trigonal bipyramidal
Explanation:
A.) is incorrect. In octahedral molecules, the central atom is bonded to six other atoms.
B.) is correct. In trigonal bipyramidal structures, the central atom is bonded to five other atoms.
C.) is incorrect. In tetrahedral molecules, the central atom is bonded to four other atoms.
D.) is incorrect. There is not such thing as a pyramidal molecular shape. This term is most likely referring to the shape, trigonal pyramidal. However, this is still incorrect. In trigonal pyramidal molecules, the central atom is bonded to three other atoms and a lone pair of electrons.
