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tekilochka [14]
3 years ago
13

A sample of citric acid, C6H8O7, has a mass of 22.9 g. What amount, in moles, does this mass represent?

Chemistry
1 answer:
aliina [53]3 years ago
3 0

Answer

C6H807 = 192.124 g/mol

n = m/M

    \frac{22.9g}{192.124 g/mol}

n = 0.119 mol

Explanation:

to calculate amount in moles you need to find the molar mass of citric acid then divide the mass of the acid with the molar mass

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3-6. A drum contains 0.16 mº of toluene. If the lid is left open (lid diameter is 0.92 m), deter-
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This problem is providing us with the volume of toluene, 0.16 m³, and the local ventilation rate 28.34 m³/min,  that is inside a drum at 30 °C and 1 atm. Thus, the time to evaporate all of the toluene is required as well as the parts per million in the drum, and found to be 96.2 min and 3.77 ppm respectively.

<h3>Ventilation:</h3>

In mass transfer problems, ventilation involves mass transfer coefficients, cross-section area and saturation pressure, that we can calculate for this problem as shown below:

A=\frac{\pi}{4}*0.92m^2=0.665 m^2\\\\k=0.83\frac{cm}{s} (\frac{18.02g/mol}{92.13g/mol} )^{\frac{1}{3} }=0.482\frac{cm}{s}=0.00482\frac{m}{s}\\ \\ P_{sat}=exp(16.0137-\frac{3096.52}{53.67+303} )mmHg*\frac{1atm}{760mmHg}=2.01 atm

After that, we can calculate the rate at which the toluene is evaporated:

Qm=\frac{M_{tol}*k*A*P_{sat}}{RT}\\ \\Qm=\frac{(92.13\frac{g}{mol})(0.00482\frac{m}{s})(0.665m^2)(2.01atm)*\frac{1000L}{1m^3} }{(0.08206\frac{atm*L}{mol*K} )(303K)}=23.77\frac{g}{s}

Where all mol, m, atm, L and atm are cancelled out. Next, given the volume of toluene and its density at 30 °C, 0.8574 g/mL, one can calculate its mass and hence the time to evaporate it all:

m_T=0.16m^3*\frac{1x10^6 mL}{1m^3} *0.8574\frac{g}{mL}=137,184g

t=\frac{137,184g}{23.77\frac{g}{s} } =5771.3 s*\frac{1min}{60s} \\\\t=96.2min

Next, the concentration of toluene in parts per million can be calculated with the rate at which toluene is evaporated and the local ventilation rate via the following equation:

Cppm=\frac{Qm*R*T}{K*Qv*P*M_T} \\\\Cppm=\frac{(23.77\frac{g}{s}*\frac{1min}{60 s} )(0.08206\frac{atm*L}{mol*K} )(303K)}{(28.34\frac{m^3}{min}*\frac{1000L}{1m^3} )(1atm)(92.13\frac{g}{mol} )} *10^6\\\\Cppm=3.77ppm

Learn more about mass transfer: brainly.com/question/25309236

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