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Neko [114]
3 years ago
5

What is the organism that is harmed in parasitism called? (3 points)

Chemistry
1 answer:
SVEN [57.7K]3 years ago
5 0

Answer:

B Host

Explanation:

this is due to the fact that the organism is hosting the parasite

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What volume and mass of steam at 100 Celsius and 760. torr (or 1 atm) would release the same amount of energy as heat during con
erastova [34]
  
<span>As we know that
1 cu cm H2O = 1 mL H2O = 1g H2O 
now

Heat of fusion of water = 79.8 cal/g 
and

Heat of vaporization of water = 540 cal/g 

Atomic weight of water : H=1 O=16 H2O=18 
now by calculating and putting values

65.5gH2O x 79.8cal/gH2O x 1gH2O/540cal = 9.68g H2O (steam) 

9.68gH2O x 1molH2O/18gH2O x 22.4LH2O/1molH2O = 12.0 L H2O
hope it helps</span>
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3 years ago
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What is the volume of a gas at 756.14 Torr. If the volume was 588.56 mL at 588.56 torr.
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Answer:

Explanation:

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nuclear energy is created through the process of splitting apart atoms and releasing large amounts of heat energy that can be co
shutvik [7]
Nuclear fission is a process by which the nucleus of an atom is split into two or more smaller nuclei, known as fission products. The fission of heavy elements is an exothermic reaction, and huge amounts of energy are released in the process.
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3 years ago
An 80.0-gram sample of water at 10.0°C absorbs 1680 Joules of heat energy. What is the final temperature of the water? a 50.0°C
ICE Princess25 [194]

Answer:

b)15.0°C

Explanation:

Specific Heat of Water=4.2 J/g°C

This means, that 1 g of Water will take 4.2 J of energy to increase its temperature by 1°C.

∴80 g Water will take 80×4.2 J of energy to increase its temperature by 1°C.

80×4.2 J=336 J

Total Energy Provided=1680 J

The temperature increase=\frac{\textrm{Total energy required}}{\textrm{energy required to increase temperature by one degree}}

Temperature increase=\frac{1680}{336}

=5°C

Initial Temperature =10°C

Final Temperature=Initial + Increase in Temperature

=10+5=15°C

7 0
3 years ago
A sample of phosphorus-32 has a half-life of 14.28 days. If 55 g of this radioisotope remain unchanged after approximately 57 da
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55= No (1/2)^55/57
55= No (1/2)^3.9
55= No (1/2)^4
55= No (1/16)
No= 880 g
7 0
3 years ago
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