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hodyreva [135]
3 years ago
9

if 120 ml of oxygen is collected over water at 27 degrees celsius & 740 mm pressure what will the volume of the dry gas be a

t STP
Chemistry
1 answer:
ZanzabumX [31]3 years ago
8 0

Answer:

                       V₂ = 106.33 ml

Explanation:

Data Given:

                 Initial Temperature  =  T₁  =  27 °C + 273.15 = 300.15 K

                 Initial Pressure  =  P₁  =  740 mmHg

                 Initial Volume  =  V₁  =  120 ml

                 Final Temperature  =  T₂  =  0 °C + 273.15 = 273.15 K

                 Final Pressure  =  P₂  =  760 mmHg

                 Fnal Volume  =  V₂  =  ?

Formula Used:

Let's assume that the Oxygen gas is acting as an Ideal gas, the according to Ideal Gas Equation,

                       P₁ V₁ / T₁  =  P₂ V₂ / T₂

Solving for V₂,

                       V₂ = P₁ V₁ T₂/ T₁ P₂

Putting values,

                       V₂ = 740 mmHg × 120 ml × 273.15 K / 300.15 K × 760 mmHg

                       V₂ = 106.33 ml

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To solve this problem, we should recall that the change in enthalpy is calculated by subtracting the total enthalpy of the reactants from the total enthalpy of the products:

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You did not insert the table in this problem, therefore I will find other sources to find for the enthalpies of each compound.

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ΔHf Fe(s) = 0.0 kJ/mol

Since the given enthalpies are still in kJ/mol, we have to multiply that with the number of moles in the formula. Therefore solving for ΔH:

ΔH = [<span>3 mol </span><span>( − </span><span>393.5 </span>kJ/mol<span>) + 1 mol (</span>0.0 kJ/mol)<span>] − [</span><span>3 mol </span><span>( − </span><span>110.5 </span>kJ/mol<span>) + </span><span>2 mol </span><span>( − </span><span>822.1 </span>kJ/mol<span>)]</span>

ΔH = <span>795.2 kJ</span>

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Molecules of each of the three hybridization demonstrate spatial configurations that would maximizes the separation between the electron domains.

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Bond angles are characteristic of the spatial configuration of electron domains and identifies the hybridization of the central carbon atom.

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What mass of solute must be used to prepare 500ml of 0.100M aqueous sodium borate Na2B4O7 from solid hydrated sodium borate Na2B
motikmotik

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Explanation:

First we need to find the number of moles of sodium borate (Na₂B₄O₇) in the solution:

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number of moles = molar concentration × volume (L)

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We know now that we need 0.05 moles of hydrated sodium borate (Na₂B₄O₇ · 10 H₂O) to make the solution.

Now to find the mass of hydrated sodium borate we use the following formula:

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mass of hydrated sodium borate = 0.05 / 381 = 0.000132 g

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molar concentration

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