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Sliva [168]
2 years ago
13

When the pressure and number of particles are kept constant for a sample of gas, which of the following is also constant for the

sample-
A) Sum of the volume and temperature

B) The difference of the volume and temperature

C) The product of the volume and temperature

D) The quotient of the volume and temperature
Chemistry
1 answer:
Leno4ka [110]2 years ago
7 0
The answer is D....................
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When a solution of 0.1 M Mg(NO3)2 was mixed with a limited amount of aqueous ammonia, a light white, wispy solid was observed, i
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<u>Answer:</u> The net ionic equation is written below.

<u>Explanation:</u>

Net ionic equation of any reaction does not include any spectator ions.

Spectator ions are defined as the ions which does not get involved in a chemical equation. They are found on both the sides of the chemical reaction when it is present in ionic form.

The chemical equation for the reaction of magnesium nitrate and aqueous ammonia (ammonium hydroxide) is given as:

Mg(NO_3)_2(aq.)+2NH_4OH(aq.)\rightarrow Mg(OH)_2(s)+2NH_4NO_3(aq.)

A white precipitate of magnesium hydroxide is formed in the above reaction.

Ionic form of the above equation follows:

Mg^{2+}(aq.)+2NO_3^-(aq.)+2NH_4^+(aq.)+2OH^-(aq.)\rightarrow Mg(OH)_2(s)+2NH_4^+(aq.)+2NO_3^-(aq.)

As, ammonium and nitrate ions are present on both the sides of the reaction. Thus, it will not be present in the net ionic equation and are spectator ions.

The net ionic equation for the above reaction follows:

Mg^{2+}(aq.)+2OH^-(aq.)\rightarrow Mg(OH)_2(s)

Hence, the net ionic equation is written above.

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3 years ago
Consider an electron with charge −e−e and mass mmm orbiting in a circle around a hydrogen nucleus (a single proton) with cha
PtichkaEL [24]

Answer:

Explanation:

The net force on electron is electrostatic force between electron and proton in the nucleus .

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This provides the centripetal force for the circular path of electron around the nucleus .

Centripetal force required = \frac{m\times v^2}{r}

So

\frac{m\times v^2}{r}=\frac{1}{4\pi\epsilon} \times \frac{e\times e}{r^2}

v^2=\frac{e^2}{4\pi \epsilon m r}

v=(\frac{e^2}{4\pi \epsilon m r})^{\frac{1}{2} }

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2 years ago
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labwork [276]

Answer:

144.6

Explanation:

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When matter is a gass, a lot of the time it is ____?
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