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vagabundo [1.1K]
3 years ago
11

Which is one feature of the ocean floor found in the open ocean?

Chemistry
2 answers:
kotykmax [81]3 years ago
8 0

Answer:

abyssal plain are found in the open ocean

Explanation:

Artist 52 [7]3 years ago
7 0
Trench. I took the test
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X is dependent Y is independent.... number of students is Y flavor is X
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Which of the following best describes what happens when molecular compounds melt? A.) Molecules separate into their component at
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Which information about a chemical reaction is provided by a potential energy diagram?(1) the oxidation states of the reactants
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D. The energy released or absorbed during the reaction 

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Read 2 more answers
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
Guys please answer this
iren [92.7K]

Answer:

Mass = 96 g

Explanation:

Given data:

Number of moles of C = 8 mol

Mass of C in gram = ?

Solution:

Formula:

Number of moles = mass/molar mass

Molar mass of C = 12 g/mol

8 mol = mass / 12 g/mol

Mass = 8 mol × 12 g/mol

Mass = 96 g

6 0
3 years ago
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