Lets take 100 g of this compound,
so it is going to be 2.00 g H, 32.7 g S and 65.3 g O.
2.00 g H *1 mol H/1.01 g H ≈ 1.98 mol H
32.7 g S *1 mol S/ 32.1 g S ≈ 1.02 mol S
65.3 g O * 1 mol O/16.0 g O ≈ 4.08 mol O
1.98 mol H : 1.02 mol S : 4.08 mol O = 2 mol H : 1 mol S : 4 mol O
Empirical formula
H2SO4
CH₃CH₂OCH₂ is more soluble in water because it has shorter hydrocarbon chain.
<h3>What is hydrocarbon?</h3>
Hydrocarbon is defined as the compound which contain hydrocarbon and carbon atoms.
The carbon atom attached to each other to form framework and hydrogen atom attach to them in different ways to give different configuration. One of the most popular hydrocarbon compound is diamond.
<h3>Solubility of hydrocarbon in water</h3>
Hydrocarbon is non polar compound whereas water is polar compound. So, hydrocarbon is in soluble in water. But as they have weak intermolecular interactions known as London dispersion forces i.e. Instantaneous dipole-induced dipole interactions.
make them less soluble in water.
Greater the hydrocarbon chain lesser will be the solubility of ketone in water. On the other hand, lesser the hydrocarbon chain greater will be the solubility of ketone in water.
Thus, we concluded that the CH₃CH₂OCH₂ is more soluble in water because it has shorter hydrocarbon chain.
learn more about hydrocarbon:
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Explanation:
Z = atomic mass of the element and , A = atomic mass of the element .
a) Z = 11, A = 23
Element = Sodium
symbol: ²³₁₁Na .
b) Z = 28, A = 64
Element = Nickel
symbol: ⁶⁴₂₈Ni .
c) Z = 50, A = 115
Element = tin
symbol: ¹¹⁵₅₀Sn .
d) Z = 20, A = 42
Element = Calcium
symbol: ⁴²₂₀Ca .
Answer:
3 AU
Explanation:
The distance from the Earth to the Sun is known as 1 AU, or 1 Astronomical Unit. If an asteroid is three times this distance, it is 3 AU away.
Answer:
V₂ → 106.6 mL
Explanation:
We apply the Ideal Gases Law to solve the problem. For the two situations:
P . V = n . R . T
Moles are still the same so → P. V / R. T = n
As R is a constant, the formula to solve this is: P . V / T
P₁ . V₁ / T₁ = P₂ .V₂ / T₂ Let's replace data:
(1.20 atm . 73mL) / 112°C = (0.55 atm . V₂) / 75°C
((87.6 mL.atm) / 112°C) . 75°C = 0.55 atm . V₂
58.66 mL.atm = 0.55 atm . V₂
58.66 mL.atm / 0.55 atm = V₂ → 106.6 mL
