Answer:
109.7178g of H2O
Explanation:
First let us generate a balanced equation for the reaction. This is illustrated below:
2C3H8O + 9O2 —> 6CO2 + 8H2O
Next we will calculate the molar mass and masses of C3H8O and H20. This is illustrated below:
Molar Mass of C3H8O = (3x12.011) + (8x1.00794) + 15.9994 = 36.033 + 8.06352 + 15.9994 = 60.09592g/mol.
Mass of C3H8O from the balanced equation = 2 x 60.09592 = 120.19184g
Molar Mass of H2O = (2x1.00794) + 15.9994 = 2.01588 + 15.9994 = 18.01528g/mol
Mass of H2O from the balanced equation = 8 x 18.01528 = 144.12224g
From the equation,
120.19184g of C3H8O produced 144.12224g of H20.
Therefore, 91.5g of C3H8O will produce = (91.5 x 144.12224) /120.19184 = 109.7178g of H2O
Noble Gas. Metals have 1 or 2 Valence Electrons. Halogens have 7 Valence Electrons. Semi-Metals can have different amounts.
Answer:

Explanation:
Hello.
In this case, given the volume (1cm³=1mL) and density of the bromine we are to firstly compute the mass since it will allow us to compute the representative particles:


Next, since the mass of one mole of diatomic bromine is 159.82 g (one bromine weights 78.91), we can next compute the moles in that sample:

Finally, via the Avogadro's number we can compute the representative particles of bromine as follows:

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