Answer is
C. 2.07 M
For explanation
M1V1 = M2V2
M2 = (M1V1)/V2
M2 = (1.5M x 345ml) / 250 ml
:. M2 = 2.07 M
Answer:
16.05 amu
Explanation:
12.011 rounds to about 12.01 and 1.008 rounds to about 1.01 so when adding you'd do [12.01 + (1.01×4)]= 16.05
Answer:
3.3167 moles Of AlCl3
Explanation:
We'll begin by writing the balanced equation for the reaction. This is illustrated below:
3Ca + 2AlCl3 —> 3CaCl2 + 2Al
From the balanced equation above,
2 moles of AlCl3 reacted to produce 2 moles of Al.
Finally, we shall obtained the number of moles of AlCl3 that reacted to produce 3.3167 moles of Al as follow:
From the balanced equation above,
2 moles of AlCl3 reacted to produce 2 moles of Al.
Therefore, 3.3167 moles Of AlCl3 will also react to produce 3.3167 moles of Al.
Thus, 3.3167 moles Of AlCl3 is needed for the reaction.