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mart [117]
2 years ago
15

How does the appearance of the protist differ from that of the onion sample?

Chemistry
1 answer:
lana [24]2 years ago
8 0

The appearance of the protist differ from that of the onion sample due to the presence of motile structures.

<h3>What is a Protist?</h3>

This is an eukaryote which could be unicellular or multicellular and examples include protozoa etc.

The major difference between a protist and onion sample is that protists have motile structures such as flagella, cilia etc while plant cells such as onions don't have.

Read more about Protist here brainly.com/question/2169979

#SPJ1

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olga2289 [7]

Answer:

MEMEME

Explanation:

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3 years ago
Describe how water is a destructive force to land features on the surface of the Earth and provide at least three examples. Desc
Leni [432]

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flooding and land erosion

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A 237g sample of molybdnum metal is heated to 100.1 0C and then dropped into an insulated cup containing 244 g of water at 10.0
miv72 [106K]

Answer:

The specific heat of molybdenum is 0.254 joules per gram-Celsius.

Explanation:

We consider the system formed by the molybdenum metal and water as our system, a control mass inside an insulated cup, that is, a container that avoids any energy and mass interactions between system and surroundings.

From statement we notice that metal is cooled down whereas water is heated. According to the First Law of Thermodynamics, we know that:

Q_{metal} - Q_{water} = 0

Q_{metal} = Q_{water}

Where:

Q_{water} - Heat received by water, measured in joules.

Q_{metal} - Heat released by metal, measured in joules.

Now we expand this identity by definition of sensible heat:

m_{metal}\cdot c_{metal}\cdot (T_{m,o}-T) = m_{water}\cdot c_{water}\cdot (T-T_{w,o})

The specific heat of the metal is cleared within equation above:

c_{metal} = \frac{m_{water}\cdot c_{water}\cdot (T-T_{w,o})}{m_{metal}\cdot (T_{m,o}-T)}

If we know that m_{water} = 0.237\,kg, m_{metal} = 0.244\,kg, c_{water} = 4186\,\frac{J}{kg\cdot ^{\circ}C}, T_{w,o} = 10\,^{\circ}C, T_{m,o} = 100.10\,^{\circ}C and T = 15.30\,^{\circ}C, the specific heat of molybdenum is:

c_{metal} = \frac{(0.237\,kg)\cdot \left(4186\,\frac{J}{kg\cdot ^{\circ}C} \right)\cdot (15.30\,^{\circ}C-10\,^{\circ}C)}{(0.244\,kg)\cdot (100.10\,^{\circ}C-15.30\,^{\circ}C)}

c_{metal} = 254.119\,\frac{J}{kg\cdot ^{\circ}C}

The specific heat of molybdenum is 0.254 joules per gram-Celsius.

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3 years ago
Which are characteristics of metals that result from their bonding? Check all that apply.
ioda

Answer:

Conduct Electricity and Are malleable

Explanation:

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BabaBlast [244]
It’s called the Margin. So D) Margin
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