Answer:
Explanation:
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In this case, since the density is computed by dividing the mass of the substance by its occupied volume (d=m/V), we first need to realize that 0.8206 g/mL is the same to 0.8206 kg/L, which means we first need to compute the volume in L:
Then, solving for the mass in d=m/V, we get m=d*V and therefore the mass of gasoline in that full tank turns out:
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Given:
Be - Beryllium - 9,3227
C - Carbon - 11,2603
O - Oxygen - 13,6181
Ne - Neon - 21,5645
B - Boron - 8,298
Li - Lithium - 5,3917
F - Fluorine - 17,4228
N - Nitrogen - 14,5341
Arranged from highest ionization energy to lowest ionization energy.
Ne ; F ; N ; O ; C ; Be ; B ; Li
The volume of a gas that is required yo react with 4.03 g mg at STP is 1856 ml
calculation/
- calculate the moles of Mg used
moles=mass/molar mass
moles of Mg is therefore=4.03 g/ 24.3 g/mol=0.1658 moles
- by use of mole ratio of Mg:O2 from the equation which is 2:1
the moles 02=0.1679 x1/20.0829 moles
- at STP 1 mole of a gas= 22.4 l
0.0895 moles=? L
- =0.0895 moles x22.4 l/ 1 mole=1.8570 L
into Ml = 1.8570 x1000=1856 ml approximately to 1860
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