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3 years ago

What happens when gases change to plasma?

Chemistry

1 answer:

IgorLugansk [536]3 years ago

8 0

Answer:. In the case of neon, it is electrical energy that pulls the electrons off.

explanation-Plasma can be made from a gas if a lot of energy is pushed into the gas. In the case of neon, it is electrical energy that pulls the electrons off. When it is time to become a gas again, just flip the neon light switch off. Without the electricity to energize the atoms, the neon plasma returns to its gaseous state.

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3. Use the chart to answer the following question. Amanda found a stone

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Answer:

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Explanation:

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3 years ago

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2 years ago

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A gas sample occupies a volume of 1 L at 200K. The temperature at which the gas would double its volume is

kicyunya [14]

Answer:

400 K

Explanation:

Please see the step-by-step solution in the picture attached below.

Hope this answer can help you. Have a nice day!

8 0

3 years ago

What is the pOH of a 0.150 M solution of potassium nitrite? (Ka HNO2 = 4.5 x 10−4 )

yanalaym [24]

Answer:

11.9 is the pOH of a 0.150 M solution of potassium nitrite.

Explanation:

Solution : Given,

Concentration (c) = 0.150 M

Acid dissociation constant = $k_a=4.5\times 10^{-4}$

The equilibrium reaction for dissociation of $HNO_2$ (weak acid) is,

$HNO_2+H_2O\rightleftharpoons NO_2^-+H_3O^+$

initially conc. c 0 0

At eqm. $c(1-\alpha)$ $c\alpha$ $c\alpha$

First we have to calculate the concentration of value of dissociation constant $(\alpha )$ .

Formula used :

$k_a=\frac{(c\alpha)(c\alpha)}{c(1-\alpha)}$

Now put all the given values in this formula ,we get the value of dissociation constant $(\alpha )$ .

$4.5\times 10^{-4}=\frac{(0.150\alpha)(0.150\alpha)}{0.150(1-\alpha)}$

$4.5\times 10^{-4} - 4.5\times 10^{-4}\alpha =0.150\alpha ^2$

$0.150\alpha ^2+4.5\times 10^{-4}\alpha-4.5\times 10^{-4}=0$

By solving the terms, we get

$\alpha=0.0533$

No we have to calculate the concentration of hydronium ion or hydrogen ion.

$[H^+]=c\alpha=0.150\times 0.0533=0.007995 M$

Now we have to calculate the pH.

$pH=-\log [H^+]$

$pH=-\log (0.007995 M)$

$pH=2.097\approx 2.1$

pH + pOH = 14

pOH =14 -2.1 = 11.9

Therefore, the pOH of the solution is 11.9

4 0

3 years ago

A 29.7 g sample of iron ore is treated as follows. The iron in the sample is all converted by a series of chemical reactions to

Softa [21]

<u>Answer:</u> The mass of iron in the ore is 10.9 g

<u>Explanation:</u>

We are given:

Mass of iron (III) oxide = 15.6 g

We know that:

Molar mass of Iron (III) oxide = 159.69 g/mol

Molar mass of iron atom = 55.85 g/mol

As, all the iron in the ore is converted to iron (III) oxide. So, the mass of iron in iron (III) oxide will be equal to the mass of iron present in the ore.

To calculate the mass of iron in given mass of iron (III) oxide, we apply unitary method:

In 159.69 g of iron (III) oxide, mass of iron present is $(2\times 55.85)=111.7g$

So, in 15.6 g of iron (III) oxide, mass of iron present will be = $\frac{111.7g}{159.69g}\times 15.6g=10.9g$

Hence, the mass of iron in the ore is 10.9 g

7 0

3 years ago