Answer: Non polar solvents
Explanation:
Since with increasing the size of alkyl group hydrophobic nature increases and solubility in polar solvents decreases .
Hence Carboxylic acids with more than 10 carbon atoms, solubility is more in non polar solvents.
Answer:
1. Saturated hydrocarbons may be cyclic or acyclic molecules.
2. An unsaturated hydrocarbon molecule contains at least one double bond.
Explanation:
Hello,
In this case, hydrocarbons are defined as the simplest organic compounds containing both carbon and hydrogen only, for that reason we can immediately discard the third statement as ethylenediamine is classified as an amine (organic chain containing NH groups).
Next, as saturated hydrocarbons only show single carbon-to-carbon bonds and carbon-to-hydrogen bonds, they may be cyclic (ring-like-shaped) or acyclic (not forming rings), so first statement is true
Finally, since we can find saturated hydrocarbons which have single carbon-to-carbon and carbon-to-hydrogen bonds only and unsaturated hydrocarbons which could have double or triple bonds between carbons and carbon-to-hydrogen bonds, the presence of at least one double bond makes the hydrocarbon unsaturated.
Therefore, first and second statements are correct.
Best regards.
Answer:
982.5 kg/m³
Explanation:
When the temperature of a fluid increases, it dilates, and because of the variation of the volume, it's density will vary too. The density can be calculated by the expression:
ρ₁ = ρ₀/(1 + β*(t₁ - t₀))
Where ρ₁ is the final density, ρ₀ the initial density, β is the constant coefficient of volume expansion, t₁ the final temperature, and t₀ the initial temperature.
At t₀ = 4°C, the water desity is ρ₀ = 1,000 kg/m³. The value of the constant for water is β = 0.0002 m³/m³ °C, so, for t₁ = 93°C
ρ₁ = 1,000/(1 + 0.0002*(93 - 4))
ρ₁ = 1,000/(1+ 0.0178)
ρ₁ = 982.5 kg/m³
Partial pressure is the amount of pressure or force that is exerted by the atoms into the outer environment. it is dependent on the temperature and pressure of the present surroundings. in this case, we are asked in this problem to determine the partial pressure of oxygen at 16oC and 1 atm. We have to look into a solubility data table commonly found in handbooks and determined via experiments and correlations. According to literature, the value of the partial pressure is equal to 0.617 mM.This is under the assumption that the salinity of the water in which oxygen is dissolved is equal to zero.
<span>Let's assume
that the F</span>₂ gas has ideal gas behavior.
<span>
Then we can use ideal gas formula,
PV = nRT
Where, P is the pressure of the gas (Pa), V is the volume of the gas
(m³), n is the number of moles of gas (mol), R is the universal gas
constant ( 8.314 J mol</span>⁻¹ K⁻<span>¹) and T is temperature in Kelvin.</span>
Moles = mass / molar mass
Molar mass of F₂ = 38 g/mol
Mass of F₂ = 76 g
Hence, moles of F₂ = 76 g / 38 g/mol = 2 mol
<span>
P = ?
V = 1.5 L = 1.5 x 10</span>⁻³ m³
n = 2 mol
R = 8.314 J mol⁻¹ K⁻<span>¹
T = -37 °C = 236 K
By substitution,
</span>
P x 1.5 x 10⁻³ m³ = 2 mol x 8.314 J mol⁻¹ K⁻¹ x 236 K
p = 2616138.67 Pa
p = 25.8 atm = 26 atm
Hence, the pressure of the gas is 26 atm.
Answer is "a".
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