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alexandr402 [8]
3 years ago
7

Calculate the number of moles of water molecules in 12 dm' of watervapour at STP.​

Physics
1 answer:
Vinvika [58]3 years ago
4 0

Answer:

22.4 \:  {dm}^{3}  \: are \: occupied \: by \: 1 \: mole \\ 12 \:  {dm}^{3}  \: will \: be \: occupied \: by \: (  \frac{12}{22.4} ) \: moles \\  = 0.536 \: moles

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-7.04

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To cite a source is to ______.
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For any molecule, formula unit, or ion, the sum of the average atomic masses of all the atoms represented in a formula is the.
alukav5142 [94]

The mass of a substance is given in atomic mass units and is calculated by adding the average atomic masses of all the atoms in the substance's chemical formula.

<h3>What empirical formula represents the total average atomic mass of every atom?</h3>

The Method The average atomic masses of all the atoms included in a formula's representation are added to get the mass of any molecule, formula unit, or ion. It has no bearing on the number of significant figures because the number of atoms is an exact quantity. One H2O molecule weighs 18.02 amu on average.

<h3>What connection exists between the empirical formula and the molecular formula?</h3>

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5 0
1 year ago
The city council is considering discussing whether or not to put fluoride in the city's water supply. Many other towns add it al
umka2103 [35]
They should look for <span>a report from an independent scientific research firm,
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8 0
3 years ago
Read 2 more answers
Friction provides the force needed for a car to travel around a flat, circular race track. What is the maximum speed at which a
blagie [28]

Answer:

The maximum speed at which the car can safety travel around the track is 18.6m/s.

Explanation:

Since the car is in circular motion, there has to be a centripetal force F_c. In this case, the only force that applies for that is the static frictional force f_sbetween the tires and the track. Then, we can write that:

f_s=F_c

And since f_s\leq \mu N and F_c=\frac{mv^{2}}{r}, we have:

\mu N\geq \frac{mv^{2}}{r}

Now, if we write the vertical equation of motion of the car (in which there are only the weight and the normal force), we obtain:

N-mg=0\\\\\\implies N=mg

Substituting this expression for N and solving for v, we get:

\mu mg\geq \frac{mv^{2}}{r}\\\\v\leq \sqrt{\mu gr}

Finally, plugging in the given values for the coefficient of friction and the radius of the track, we have:

v\leq \sqrt{(0.42)(9.81m/s^{2})(84.0m)}\\\\v\leq 18.6m/s

It means that in its maximum value, the speed of the car is equal to 18.6m/s.

7 0
3 years ago
Read 2 more answers
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