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Viefleur [7K]
3 years ago
13

The volume of a gas is always the same as the volume of __________.

Chemistry
2 answers:
Len [333]3 years ago
7 0
The answer would be A: its container.

Gas will expand to fill whatever container it is put in.
den301095 [7]3 years ago
6 0

Answer: its container

Explanation:

There are three states of matter:

Solid state : In this state, the molecules are arranged in regular and repeating pattern. The molecules are closely packed that means they are fixed and thus have fixed shape and volume.

For example : Copper

Liquid state : In this state, the molecules are present in random and irregular pattern. The molecules are closely packed but they can move from one place to another.  They have fixed volume and take the shape of the container.

For example : water

Gaseous state : In this state, the molecules are present in irregular pattern. The molecules are not closely packed and they can move freely from one place to another and spread out.  Thus they do not have any fixed shape and occupy the whole volume of the container.

For example : Helium gas

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C.<br> Calculate the number of moles in 62g of CO2
worty [1.4K]

Answer:

32÷5

I'm just tryna get points I'm sorry

goodluck tho❤

4 0
3 years ago
Read 2 more answers
What is the ratio of Al 3+ ions to S 2- ions in a neutral compound?
iren2701 [21]
In a neutral ionic compound, you can determine its sub-scripts by simply flipping the ionic charges and dropping the signs: so AlS would be Al2S3
4 0
3 years ago
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Extend the aufbau sequence through an element that has not yet been identified, but whose atoms would completely fill 7p orbital
Thepotemich [5.8K]

Answer:

<u>Number of electrons</u> = 118

<u>Electronic configuration</u>: [Rn] 5f¹⁴ 6d¹⁰ 7s² 7p⁶

Explanation:

Given: A chemical element that has completely filled 7p orbital.

According to this, the principal occupied electron shell or the <u>valence shell of such an element is 7p.</u>

⇒ the principal quantum number <u>(n) for the valence shell is 7.</u>

∴ this element belongs to the period 7 of the periodic table.

Also, an element that has completely filled p-orbital belongs to the group 18 of the p-block.

Therefore, an element that belongs to the group 7 and period 18 of the periodic table, should have a <u>completely filled 5f, 6d, 7s and 7p orbitals</u>.

Therefore, the electronic configuration should be: [Rn] 5f¹⁴ 6d¹⁰ 7s² 7p⁶

And, the <u>number of electrons</u> = atomic number of radon (Rn) + 14 + 10 + 2 + 6 = 86 + 32 = <u>118</u>.

<u>Therefore, the given element has atomic number 118 and has the electronic configuration: [Rn] 5f¹⁴ 6d¹⁰ 7s² 7p⁶. Thus the given element can be Oganesson.</u>

7 0
3 years ago
An atom with 38 protons and 48 neutrons is called_____?
rewona [7]

Answer:

It's not atom it is an element Strontium.

4 0
2 years ago
A 5.76 liter sample of a gas at 22.00C mL 748 torr pressure was heated to a final volume of 17.28 liters, with the pressure rema
timama [110]

Answer:

The final temperature was 612 °C

Explanation:

Charles's law relates the volume and temperature of a certain amount of ideal gas, maintained at a constant pressure, using a constant of direct proportionality. In this law, Charles says that at constant pressure, as the temperature increases, the volume of the gas increases and as the temperature decreases, the volume of the gas decreases. That is, Charles's law is a law that says that when the amount of gas and pressure are kept constant, the ratio between volume and temperature will always have the same value:

\frac{V}{T}=k

When you want to study two different states, an initial and a final one of a gas and evaluate the change in volume as a function of temperature or vice versa, you can use the expression:

\frac{V1}{T1} =\frac{V2}{T2}

In this case:

  • V1= 5.76 L
  • T1= 22 °C= 295 °K (Being 0°C=273°K)
  • V2=17.28 L
  • T2=?

Replacing:

\frac{5.76 L}{295 K} =\frac{17.28 L}{T2}

Solving:

T2 =\frac{17.28 L}{\frac{5.76 L}{295 K}}=\frac{17.28 L*295 K}{5.76 L}

T2= 885 °K = 612 °C

<u><em>The final temperature was 612 °C</em></u>

5 0
3 years ago
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