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The partial pressure of helium to keep the partial pressure of oxygen at 0,21 atm in a scuba-diver tank is
8,09 atmTo solve this question, we can use the
Dalton's Law, which states that the total pressure in a container with a mixture of gases is the sum of the partial pressures o each individual gas. For the case of this mixture the Dalton's Law is as follows:
In this equation, we need to clear for PHe, knowing that the PO₂ should be 0,21 atm, to find the required pressure of Helium:
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Answer:
a) both substances are insoluble in water
b) both substances are soluble in ligroin
c) both substances suffer combustion, octane produces more CO₂ than hexene.
d) both substances are less dense than waterl, with hexene having the lowest density.
e) only hexene would react with bromine
f) only hexene would react with permanganate
Explanation:
a) both substances are non-polar and water is polar
b) both substances are non-polar and lingroin is non-polar
c) C₈H₁₈ + 17.5O₂ → 8CO₂ + 9H₂O
C₆H₁₂ + 9O₂ → 6CO₂ + 6H₂O
d) water = 997 kg/m³
ocatne = 703 kg/m³
hexene = 673 kg/m³
e) bromine test is used to detect unsaturations
f) permanganate test is used to detect unsaturations
Answer: C. ethanol
The enthalpy of combustion is the amount of heat produced when one mole of ethanol undergoes complete combustion at 25 ° C and 1 atmosphere pressure, yielding products also at 25 ° C and 1 atm.
<u>The enthalpy of combustion of the unknown compound is</u>
ΔH = - 320 kJ / 0.25 mol = - 1280 kJ / mol
<u>To choose a probable compound according to this combustion enthalpy, we must evaluate the deviation in relation to the values reported in the literature for the three probable compounds</u> (methane, ethylene and ethanol). The deviation (e%) will be calculated according to the following equation,
e% = ( | ΔHx - ΔH | / ΔHx ) x 100%
where ΔHx is the enthalpy of combustion of the probable compound.
The following table shows the combustion enthalpies of the probable compounds and their deviation in relation to the enthalpy of ΔH = - 1280 kJ / mol
Compound Enthalpy of combustion (kJ/mol) Deviation
Methane - 890.7 43.8%
Ehylene -1411.2 9.3%
Ethanol -1368.6 6.5%
According to the previous table, we can say that the most probable compound is ethanol, since it has the smallest deviation in relation to the experimental enthalpy value of combustion.
Answer:
308 moles of sodium
Explanation:
The balanced equation for the chemical reaction between sodium metal (Na) and water (H₂O) is the following:
2 Na(s) + 2 H₂O → 2 NaOH(aq) + H₂(g)
From the equation, we can see that 2 moles of Na react with 2 moles of H₂O to give 2 moles of NaOH and 1 mol of H₂ (hydrogen gas). So the stoichiometric mole ratio between Na and H₂ is: 2 mol Na/1 mol H₂. Thus, we multiply the mole ratio by the moles of H₂ to be produced to obtain the moles of Na required:
moles of Na required = 2 mol Na/1 mol H₂ x 154 moles H₂ = 308 moles Na
Therefore, 308 moles of sodium are needed to produce 154 moles of hydrogen gas.