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wariber [46]
2 years ago
10

How do noble gases such as Xenon(Xe) bond

Chemistry
1 answer:
Hunter-Best [27]2 years ago
7 0

Explanation:

The noble gases are found in group 18 of the periodic table with a complete valence electron shell. The elements of group 18 are chemically stable because their valence shells are filled.

However, Xenon differs from other noble elements because of its size and large number of electrons. The electrons in the outer shell are not tightly bound with the nucleus and thus, are somewhat free to part in reactions, this why xenon has so many compounds

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If you set up an experiment with two different independent variables, then the results would be_____.
MariettaO [177]

Answer: A) Inconclusive; you would not know which of the two variables caused the change.

Explanation:

When you set up an experiment, you must make sure that you control the variables such that only one independent variable changes at a time, while all the remainder conditions (the other independent variables) are controlled (fixed).

By observing (measuring) the dependent variable, while only one independent variable changes you can understandhow such independent variable explains (determines) the dependent variable, leading to a conclusion.

Conversely, if two or more independent variables change at a time, then there is no way that you can tell how the output (dependent variable) is related with one or other of the changes of the indipendent variables. You wolud not be able to discriminate (distinguish) the effect of one or other variable, making the experiment inconclusive

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4 0
3 years ago
What is a catalyst?
luda_lava [24]

a substance that increases the rate of a chemical reaction without itself undergoing any permanent chemical change.

 -  a person or thing that precipitates an event.

7 0
3 years ago
Read 2 more answers
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
Explain why chemistry affects all aspects of life and most natural events
Alecsey [184]

Chemistry affects all aspects of life and most natural events because all living and nonliving things are made of matter.

6 0
3 years 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
Read 2 more answers
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