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4 years ago

11

What's a difference between mercury and oxygen? (They both are elements)

1 answer:

borishaifa [10]4 years ago

8 0

-- There are 80 protons in the nucleus of every atom of Mercury,
but only 8 of them in the nucleus of an atom of Oxygen.

-- Mercury must be warmer than 357°C in order to boil, but Oxygen
must only be warmer than -183°C.

-- Mercury must be colder than -39°C in order to freeze, but Oxygen
must be colder than -219°C.

-- Oxygen is required for human life. Mercury is a deadly poison.

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If a 70-kg swimmer pushed off a pool wall with a force of 250N at what rate will the swimmer accelerate from the wall

Vlad [161]

F = ma
250 = 70 x a
a = 250/70
a = 3.57

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3 years ago

What is another name for the breathing mechanism?

aleksandrvk [35]

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3 years ago

A potential difference of 107 mV exists between the inner and outer surfaces of a cell membrane. The inner surface is negative r

sergij07 [2.7K]

Answer:

The workdone is $W = 1.712 *10^{-20 } \ J$

Explanation:

From the question we are told that

The potential difference is $V = 107 mV = 107 *10^{-3} \ V$

Generally the charge on $Na^{+}$ is $Q_{Na^{+}} = 1.60 *10^{-19 } \ C$

Generally the workdone is mathematically represented as

$W = Q_{Na^{+}}V$

=> $W = 1.60 *10^{-19 } * 107 *10^{-3}$

=> $W = 1.712 *10^{-20 } \ J$

8 0

3 years ago

A car is traveling down the highway at speed 5 m/s when the driver slams on the brakes and skids to a stop in a distance 80 m. A

Murljashka [212]

The car is very incredibly fast

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3 years ago

A gas at a pressure of 2.10 atm undergoes a quasi static isobaric expansion from 3.70 to 5.40 L. How much work is done by the ga

BartSMP [9]

Answer:

Total work done in expansion will be $3.60\times 10^5J$

Explanation:

We have given pressure P = 2.10 atm

We know that 1 atm $=1.01\times 10^5Pa$

So 2.10 atm $=2.10\times 1.01\times 10^5=2.121\times 10^5Pa$

Volume is increases from 3370 liter to 5.40 liter

So initial volume $V_1=3.70liter$

And final volume $V_2=5.40liter$

So change in volume $dV=5.40-3.70=1.70liter$

For isobaric process work done is equal to $W=PdV=2.121\times 10^5\times 1.70=3.60\times 10^5J$

So total work done in expansion will be $3.60\times 10^5J$

5 0

3 years ago