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ArbitrLikvidat [17]
3 years ago
7

Which of the following would likely occur as a result of an oil spill in the ocean?

Chemistry
2 answers:
Naddik [55]3 years ago
8 0
The answerrrrrrrr is D
FinnZ [79.3K]3 years ago
6 0

Answer:

D.

Explanation:

The Oil spill effects on environments and habitats can be catastrophic: they can kill plants and animals, disturb salinity/pH levels, pollute air/water and more. Read more about the types of oil pollution.

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What has 39 neutrons which element
Rasek [7]

No of Neutrons =Mass No-No of protons

  • Proton=Neutron+1

\\ \sf\longmapsto x+1=39

\\ \sf\longmapsto x=39-1=38

  • Hence it has atomic no 38

The element is Stroncium(Sr)

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3 years ago
What is the Molecular Mass of CuSo4.5H2O​
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The answer to this question is 159.609 g/mol
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An atom of uranium-238 undergoes radioactive decay to form an atom of thorium-234. Which type of nuclear decay has occurred?
WINSTONCH [101]

Answer:-

Alpha decay

Explanation:-

Uranium 238 has atomic number 92 and mass number 238.

Thorium 234 has atomic number 90 and mass number 234.

So, the change in atomic number as uranium 238 disintegrates into thorium234 = 92 – 90 = 2

So, the change in mass number as uranium 238 disintegrates into thorium234= 238 – 234 = 4

We know that when an alpha particle is emitted, the mass number decreases by 4 and the atomic number decreases by 2.

So when an atom of uranium 238 undergoes radioactive decay to form an atom of thorium-234, alpha decay has occurred.

5 0
3 years ago
What effect does filtration have on water with a ph of 3?
creativ13 [48]

Answer:

None  

Step-by-step explanation:

A high pH is caused by an excess of hydrogen ions over hydroxide ions.

Hydrogen ions are so small that they pass through the pores of an ordinary filter.

Thus, ordinary filtration has no effect on the pH of water.

4 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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