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tino4ka555 [31]
3 years ago
14

QUESTION 14

Chemistry
1 answer:
AveGali [126]3 years ago
7 0

Within each functional group, the boiling points of members of the group vary according to molecular weight. Therefore, one cannot deduce any functional group from boiling points.

Organic compounds are divided into families called homologous series. Each homologous series has a functional group common to all the members of the series. As a result of this, the chemical properties of all the members of the homologous series are similar.

However, the boiling points of the members of a homologous series vary according the their molecular weights. As such, boiling points can not be used as evidence to classify substances into any particular homologous series.

Summarily, one cannot deduce any functional group from physical properties.

Learn more: brainly.com/question/1078956

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Air at 25°C with a dew point of 10 °C enters a humidifier. The air leaving the unit has an absolute humidity of 0.07 kg of water
madreJ [45]

Explanation:

The given data is as follows.

         Temperature of dry bulb of air = 25^{o}C

          Dew point = 10^{o}C = (10 + 273) K = 283 K

At the dew point temperature, the first drop of water condenses out of air and then,

        Partial pressure of water vapor (P_{a}) = vapor pressure of water at a given temperature (P^{s}_{a})

Using Antoine's equation we get the following.

            ln (P^{s}_{a}) = 16.26205 - \frac{3799.887}{T(K) - 46.854}

            ln (P^{s}_{a}) = 16.26205 - \frac{3799.887}{283 - 46.854}

                               = 0.17079

                   P^{s}_{a} = 1.18624 kPa

As total pressure (P_{t}) = atmospheric pressure = 760 mm Hg

                                   = 101..325 kPa

The absolute humidity of inlet air = \frac{P^{s}_{a}}{P_{t} - P^{s}_{a}} \times \frac{18 kg H_{2}O}{29 \text{kg dry air}}

                  \frac{1.18624}{101.325 - 1.18624} \times \frac{18 kg H_{2}O}{29 \text{kg dry air}}

                 = 0.00735 kg H_{2}O/ kg dry air

Hence, air leaving the humidifier has a has an absolute humidity (%) of 0.07 kg H_{2}O/ kg dry air.

Therefore, amount of water evaporated for every 1 kg dry air entering the humidifier is as follows.

                 0.07 kg - 0.00735 kg

              = 0.06265 kg H_{2}O for every 1 kg dry air

Hence, calculate the amount of water evaporated for every 100 kg of dry air as follows.

                0.06265 kg \times 100

                  = 6.265 kg

Thus, we can conclude that kg of water the must be evaporated into the air for every 100 kg of dry air entering the unit is 6.265 kg.

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what is the minimum mass of ethylene glycol that must be dissolved in 14.5 kg of water to prevent the solution from freexing at
lbvjy [14]

Answer:

The minimum mass of ethylene glycol = 6.641 Kg

Explanation:

\Delta T_f= T_f-T_f'\\T_f'=T_f-\Delta T_f

Where T_f = freezing point of pure solvent water, 0°C

T_f'= Freezing point of solvent after mixture

K_f = Freezing point depression constant = 1.86 °C/m

Moecular weight of ethylene glycol = 60 g/mol

Weight of ethylene glycol = 14.5 Kg= 14.5×10^3 g

molality of ethylene glycol

m = \frac{weight}{mol.wt} \times\frac{1000}{V}

Substitute the values to calculate m

m = \frac{w}{60} \times\frac{1000}{14.5\times1000}

by formula

0-(-14.2) =1.86\frac{w}{60} \times\frac{1000}{14.5\times1000}

calculating we get w = 6641.93 g

Therefore, The minimum mass of ethylene glycol = 6.641 Kg

4 0
2 years ago
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