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guapka [62]
2 years ago
8

Is it true that a gas occupies volume of any vessel it is put on ?If so, what if we transferred the gas kept in a small containe

r into a large container ??​
Chemistry
1 answer:
SashulF [63]2 years ago
5 0

Answer:

It fills the container. Initially, in the small container, the gas had filled it with some amount of molecular space. When the container is made big, it arranges itself with more amount of molecular space filling the gas. Hence this happens when gas is transferred to a container.

Thanks !!

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Which object forms when a supergiant explodes
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black hole

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3 years ago
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I have no idea how to do this, any help would be appreciated
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Proton number = the atomic number (which is the smaller number
neutron number = the mass number (the bigger number) - the atomic number
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6 0
2 years ago
Acetylene gas (ethyne; HC = CH) burns in an oxyacetylene torch to produce carbon dioxide and water vapor. The heat of reaction f
Yanka [14]

The mass of CO2 produced by 26g of acetylene is 88g.

Given ,

In an oxyacetylene torch, acetylene gas (ethyne; HCCH) burns to produce carbon dioxide and water vapour.

The acetylene combustion reaction is given by,

H2O + HCCH + 5/2 O=O 2CO2

Heat of reaction for acetylene combustion = 1259kj/mol

CO2 has a molecular mass of 44g/mol.

2 moles of CO2 have a molecular mass of 88g.

On combustion, 1 mole of acetylene yields 2 moles of CO2.

Thus, 26g of acetylene produces 88g of CO2 when burned.

As a result, the mass of carbon dioxide produced by 26g of acetylene is 88g.

Learn more about acetylene here :

brainly.com/question/15346128

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6 0
1 year ago
Copper is formed when aluminum reacts with copper (II) sulfate in a single-replacement reaction. How many moles of copper can be
serg [7]

Answer:

The answer to your question is the limiting reactant is CuSO₄ and 0.975 moles of Cu were obtained

Explanation:

moles of Copper = ?

mass of Aluminum = 29 g

mass of CuSO₄ = 156 g

Limiting reactant = ?

Balanced Chemical reaction

                  3 CuSO₄   +  2 Al   ⇒   3 Cu   +  Al₂(SO₄)₃

Calculate the moles of reactants

CuSO₄ = 64 + 32 + (16 x 4) = 160g

Al = 27 g

                160 g of CuSO₄  ----------------- 1 mol

                156 g                   -----------------  x

                      x = (156 x 1) / 160

                      x = 0.975 moles

               27 g of Al -------------------------- 1 mol

               29 g of Al -------------------------- x

                x = (29 x 1)/27

                x = 1.07 moles

Calculate proportions to find the limiting reactant

Theoretical     3 moles CuSO₄/2 moles Al = 1.5 moles

Experimental  0.975 moles CuSO₄/1.07 moles = 0.91

The experimental proportion was lower than the theoretical proportion that means that the limiting reactant is CuSO₄.      

                    3 moles of CuSO₄ ------------------ 3 moles of Cu

                 0.975 moles of CuSO₄ ---------------  x

                         x = (0.975 x 3)/3

                        x = 0.975 moles of Cu were obtained.        

3 0
3 years ago
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Be kind yk? tell him have some respect for your relationship and if he can't then cut him off completely
3 0
3 years ago
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