Answer:
1.32 moles.
Explanation:
From the question given above, the following data were obtained:
Density of Al = 2.70 g/cm³
Volume of Al = 13.2 cm³
Number of mole of Al =.?
Next, we shall determine the mass of Al.
This can be obtained as follow:
Density of Al = 2.70 g/cm³
Volume of Al = 13.2 cm³
Mass of Al =?
Density = mass / volume
2.7 = mass of Al / 13.2
Cross multiply
Mass of Al = 2.7 × 13.2
Mass of Al = 35.64 g
Finally, we shall determine the number of mole of Al. This can be obtained as follow:
Mass of Al = 35.64 g
Molar mass of Al = 27 g/mol
Number of mole of Al =?
Mole = mass / molar mass
Number of mole of Al = 35.64 / 27
Number of mole of Al = 1.32 moles
Thus, 1.32 moles of aluminum are present in the block of the metal.
A. to push each other away
Answer:
true
Explanation:
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Explanation:
Let us assume that the given data is as follows.
mass of barium acetate = 2.19 g
volume = 150 ml = 0.150 L (as 1 L = 1000 ml)
concentration of the aqueous solution = 0.10 M
Therefore, the reaction equation will be as follows.
Hence, moles of 
= 
.......... (1)
As, No. of moles = 
Hence, moles of 
will be calculated as follows.
No. of moles =
= 
(molar mass of is 255.415 g/mol)
= 
Moles of = 
= 0.01715 mol
Hence, final molarity will be as follows.
Molarity = 
= 
= 0.114 M
Thus, we can conclude that final molarity of barium cation in the solution is 0.114 M.
The answer is B) sliding friction
