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4 years ago

HELP with these 2 questions asap

Chemistry

2 answers:

Allisa [31]4 years ago

8 0

Answer: For question one...B. Because cells are too small to see without a microscope. It would have been impossible because cells are hard to deal with without a microscope because they are so tiny. So, without it, he wouldn't have been able to get the answer out of his discovery.

For question two...D. The nucleus is the control center of the cell. The cell theory states that, for 1., All living things are made of cells and their product., 2. New cells are created by only old cells dividing into two, and 3. Cells are he basic units of life. Cells having a nucleus is not mentioned and will never be because, cells don't need a nucleus to keep moving or keep the cell in control.

I hope this helps you! I really do hope it does.

katovenus [111]4 years ago

5 0

Answer:

number 1 is b and and number two d

Explanation:

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4 years ago

You have 750 grams of water at 80° Celsius. Which of the following would lower the temperature of the water by 10° Celsius? (1 p

dem82 [27]

Answer:

adding 750 grams of water at 60° Celsius .

Explanation:

We can calculate the amount of heat lost from 750 grams of water at 80°C to be lowered by 10°C using the relation:

Q = m.c.ΔT,

Where, Q is the amount of heat lost by water (Q = ??? J).

m is the mass of water (m = 750.0 g).

c is the specific heat capacity of the water (c = 4.18 J/g.°C).

ΔT is the temperature difference (ΔT = final T - initial T = - 10.0°C, the temperature of water is lowered by 10.0°C).

∴ Q = m.c.ΔT = (750.0 g)(4.18 J/g.°C)(- 10.0°C) = - 31350.0 J = -31.350 kJ.

Now, we can calculate the Q that is gained by the different added amounts of water:

adding 750 grams of water at 50° Celsius :

ΔT = 70.0°C - 50.0°C = 20.0°C,

∴ Q = m.c.ΔT = (750.0 g)(4.18 J/g.°C)(20.0°C) = 62700.0 J = 62.70 kJ.

adding 325 grams of water at 60° Celsius :

ΔT = 70.0°C - 60.0°C = 10.0°C,

∴ Q = m.c.ΔT = (325.0 g)(4.18 J/g.°C)(10.0°C) = 13585.0 J = 13.585 kJ.

adding 750 grams of water at 60° Celsius :

ΔT = 70.0°C - 60.0°C = 10.0°C,

∴ Q = m.c.ΔT = (750.0 g)(4.18 J/g.°C)(10.0°C) = 31350.0 J = 31.350 kJ.

adding 1000 grams of water at 55° Celsius:

ΔT = 70.0°C - 55.0°C = 15.0°C,

∴ Q = m.c.ΔT = (1000.0 g)(4.18 J/g.°C)(15.0°C) = 62700.0 J = 62.70 kJ.

So, the right choice is:

adding 750 grams of water at 60° Celsius

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4 years ago