A threat is a potential risk loss to an asset
To get a result with the best degree of precision, the
number of significant figures should be equal to the smallest number of
significant figures of the given numbers. In this case, the smallest is 3 as
given by the number 9.03 mL.
Therefore density is:
<span>11.50 g / 9.03 mL = 1.27 g/mL</span>
Answer:
The empirical formula for the compound is C3H4O3
Explanation:
The following data were obtained from the question:
Carbon (C) = 40.92%
Hydrogen (H) = 4.58%
Oxygen (O) = 54.50%
The empirical formula for the compound can be obtained as follow:
C = 40.92%
H = 4.58%
O = 54.50%
Divide by their molar mass
C = 40.92/12 = 3.41
H = 4.58/1 = 4.58
O = 54.50/16 = 3.41
Divide by the smallest i.e 3.41
C = 3.41/3.41 = 1
H = 4.58/3.41 = 1.3
O = 3.41/3.41 = 1
Multiply through by 3 to express in whole number
C = 1 x 3 = 3
H = 1.3 x 3 = 4
O = 1 x 3 = 3
The empirical formula for the compound is C3H4O3
Methane is lighter than air, having a specific gravity of 0.554. It is only slightly soluble in water. It burns readily in air, forming carbon dioxide
and water vapour; the flame is pale, slightly luminous, and very hot.
The boiling point of methane is −162 °C (−259.6 °F) and the melting
point is −182.5 °C (−296.5 °F). Methane in general is very stable, but
mixtures of methane and air, with the methane content between 5 and 14
percent by volume, are explosive. Explosions of such mixtures have been
frequent in coal mines and collieries and have been the cause of many
mine disasters.
Answer is: V<span>an't Hoff factor (i) for this solution is 2,26.
</span>Change in freezing point
from pure solvent to solution: ΔT =i · Kf · m.
<span>Kf - molal freezing-point depression constant for water is 1,86°C/m.
</span>m - molality, moles of solute per kilogram of solvent.
n(K₂SO₄) = 16,8 g ÷ 174,25 g/mol
n(K₂SO₄) = 0,096 mol.
m(K₂SO₄) = 0,096 mol/kg.
ΔT = 0,405°C.
i = 0,405 ÷ (1,86 · 0,096)
i = 2,26.
