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3 years ago

What is the charge on a fluorine ion? explain answer plz

1 answer:

4 0

-1

An ion is an atom which has gained or lost electrons.
Atoms want to have full outer shells of electrons.
Electrons are negative.
Fluorine has 7 electrons in its outer shell. When it gains one to fill its shell it will be made negative.
Thus the charge of a fluorine ion is -1

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What is the volume of 2.5 moles of nitrogen gas (N2)<br>at standard temperature and pressure (STP)?

gregori [183]

Answer:

1 mole of gas = 22.4L

2.5 moles of gas takes up = ( 22.4 L/ 1 mole ) x 2.5 mole

= 56 L

Explanation:

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8 0

3 years ago

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nydimaria [60]

Answer: iss as kfnandn

Explanation:

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3 years ago

Convert 533 cm/s to units of meters per minute. Show the unit analysis by dragging the conversion factors into the unit‑factor s

lukranit [14]

Answer:

319.8 m/min

Explanation:

533 cm/s

We can convert 533 cm/s to m/min by doing the following:

First, we shall convert 533 cm/s to m/s. This can be obtained as illustrated below:

Recall:

100 cm/s = 1 m/s

Therefore,

533 cm/s = 533 cm/s /100 cm/s × 1 m/s

533 cm/s = 5.33 m/s

Finally, we shall convert 5.33 m/s to m/min. This can be obtained as follow:

1 m/s = 60 m/min

Therefore,

5.33 m/s = 5.33 m/s / 1 m/s × 60 m/min

5.33 m/s = 319.8 m/min

Therefore, 533 cm/s is equivalent to 319.8 m/min

8 0

3 years ago

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JulsSmile [24]

Answer:

Either A or D

But i think D

hope this helps =3

sorry if im wrong

7 0

3 years ago

An ideal gas contained in a piston-cylinder assembly is compressed isothermally in an internally reversible process.

Tju [1.3M]

Answer:

a) $\Delta S$

b) entropy of the sistem equal to a), entropy of the universe grater than a).

Explanation:

a) The change of entropy for a reversible process:

$\delta S=\frac{\delta Q}{T}$

$\Delta S=\frac{Q}{T}$

The energy balance:

$\delta U=[tex]\delta Q- \delta W$

If the process is isothermical the U doesn't change:

$0=[tex]\delta Q- \delta W$

$\delta Q= \delta W$

$Q= W$

The work:

$W=\int_{V1}^{V2}P*dV$

If it is an ideal gas:

$P=\frac{n*R*T}{V}$

$W=\int_{V1}^{V2}\frac{n*R*T}{V}*dV$

Solving:

$W=n*R*T*ln(V2/V1)$

Replacing:

$\Delta S=\frac{n*R*T*ln(V2/V1)}{T}$

$\Delta S=n*R*ln(V2/V1)}$

Given that it's a compression: V2<V1 and ln(V2/V1)<0. So:

$\Delta S$

b) The entropy change of the sistem will be equal to the calculated in a), but the change of entropy of the universe will be 0 in a) (reversible process) and in b) has to be positive given that it is an irreversible process.

7 0

3 years ago