M = 22.8 g
V = 14.7 mL
ρ - ?
ρ = m/V
ρ = 22.8/14.7 = 1.55 g/mL
Answer:
They are 1.204×10^24 atoms of hydrogen present in 18 grams of water. In order to calculate this,it is necessary to compute the number of hydrogen moles present in the sample.
Answer: C(s) + O2(g) --> CO2(g)12g (C) .... 50.8g (O2)................. initial amounts0g(C) .........18.8g(O2) ................. amounts when reaction completeThat means that C was the limiting reactant, and the amount of CO2 is based on the amount of carbon that burned. Covert 12 grams of carbon to moles. The moles of CO2 will be the same, since they are in a 1:1 mole ratio. Then convert the moles of CO2 to grams.12g C x (1 mol C / 12.0 g C) x (1 mol CO2 / 1 mol C) x (44.0g CO2 / 1 mol CO2) =44 g of CO2
450.0 ML because it has more solution and it would be better if the sciences users it
<span>a chemical change results in the formation of one or more new substances</span>
