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viva [34]
4 years ago
10

Elements that exceed the octet rule must have an _______________. unoccupied s-orbital unoccupied p-orbital unoccupied d-orbital

unoccupied first energy level
Chemistry
1 answer:
Irina18 [472]4 years ago
7 0
Elements that exceed the octet rule must have an unoccupied d orbital. I believe.
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An ax is used to chop wood. The metal part chops through the wood, pushing it apart into two smaller sections. which simple mach
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5 0
3 years ago
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.

3 0
3 years ago
A solution is made using 148.5 g of dimethyl ether (MM = 46.07 g/mol) and 90.0 g of methanol (MM = 32.04 g/mol). What is the mol
Mariulka [41]

Answer:

0.534

Explanation:

Mole fraction can be calculated using the formula:

Mole fraction = number of moles of solute ÷ number of moles of solvent and solute (solution).

In this question, solute is dimethyl ether while the solvent is methanol.

Mole (n) = mass (M) ÷ molar mass (MM)

Mole of solute (dimethyl ether) = 148.5 ÷ 46.07

= 3.22moles.

Mole of solvent (methanol) = 90 ÷ 32.04

= 2.81moles.

Total moles of solute and solvent = 3.22 + 2.81 = 6.03moles.

Mole fraction of dimethyl ether = number of moles of dimethyl ether ÷ number of moles of solution (dimethyl ether + methanol)

Mole fraction = 3.22/6.03

= 0.534

6 0
3 years ago
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