1.00 mole of oxygen at 60 degrees celcius and 5 bar is flowing through a 20cm diameter pipe (inner diameter). what is the velocity of the gas in the pipe? can use relation u = v/a = m / p x a
where u = velocity in m/x, v = volumetric florate, a = cross sectional area, m = mass flow rate, p = density

u = 0.176 m/x

Explanation:

∴ n O2 = 1 mol

∴ mass O2 = 1 mol * 32 g/mol = 32 g

∴ T = 60°C = 333 K

∴ P = 5 bar

⇒ V = RTn/P = (83.14 bar.cm³/mol.K)*(333 K)*(1 mol))/(5 bar)

⇒ V = 5537.124 cm³

∴ d = 20cm

⇒ A = (1/4π)*d² = 314.16 cm²

velocity of the gas (m/x):

∴ ρ = 32 g / 5537.124 cm³ = 5.78 E-3 g/cm³

∴ mass flow rate (m) = 32g / x

⇒ u = (32 g/x) / (( 5.78 E-3 g/cm³)*(314.16cm²))

⇒ u = 17.625 cm/x * ( m/100cm) = 0.176 m/x

answer: the enthalpy of the overall chemical reaction is +87.88 kj or +88 kj.

according to hess's law, the overall enthalpy of a reaction is the difference between the sum of enthalpy values of all the products and the sum of enthalpy values of all the reactants.

δh(reaction) = ∑nδh(products) - ∑nδh(reactants)

δh(reaction) = [δh +δh] - [δh]

δh(reaction) = [1(-287.02) + 1(0) ] - [1(-374.9)] kj = +87.88 kj

here, δh (pcl3) = -287.02 kj, δh(cl2) = 0, δh(pcl5) = -374.9 kj

therefore, enthalpy of the overall reaction is +87.88 kj or +88 kj.

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explanation:

the marks go by 4s