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ANEK [815]
3 years ago
6

Which 2 characteristics must and object have in order to be considered matter?

Chemistry
1 answer:
Allisa [31]3 years ago
4 0

Answer:.

Explanation: Matter must have Mass and Volume to be considered matter!

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2 years ago
If an atom has 12 protons 12 neutrons, and 12 electrons, what is the charge of the atom?
Evgesh-ka [11]
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3 years ago
Q5 what's the answer to A?
Snezhnost [94]
The answer is CH3CH2CH2CHO
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3 years ago
At a certain temperature, the solubility of strontium arsenate, sr3(aso4)2, is 0.0560 g/l. what is the ksp of this salt at this
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The solution for this problem is:
Get into moles first. .0560 grams over 540.8 grams per mole = 1.04 x l0^-4 moles 
Sr3(As04)2 = 3 Sr++(aq) plus 2 As04^-3(aq) 
Ksp = (Sr++)^3(As04^-3)^2 
(Sr++) = 3 X 1.04 x l0^-4= 3.11 x l0^-4 
(As04^-3) = 2 x 1.04 x l0^-4= 2.07 x l0^-4 
Ksp = (1.04 x l0^-4)^3 (2.07 x l0^-4)^2 which equals 4.82 x 10^-20
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
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