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4 years ago

Atomic size generally _____.

Chemistry

2 answers:

Lunna [17]4 years ago

4 0

A. Increases as you move from left to right across a period

mafiozo [28]3 years ago

4 0

Atomic size generally

D. decreases as you move from left to right across a period

elements are classified into periods based on the number of energy shells.

elements that have the same number of energy shells fall into the same period.

as you go from left to right across a period, the atomic number increases. Atomic number is the number of protons. In ground state atoms the protons and electrons are the same.

so as you go across a period, the number of protons and electrons increase.

protons are positively charged and located in the nucleus. Electrons are negatively charged and are in energy shells. With higher number of protons in the nucleus, higher the positive charge in the nucleus. Then the force of attraction from the nucleus towards the electrons in the energy shells are higher.It will pull the energy shells more towards the nucleus making the atomic size smaller.

therefore atomic size decreases as you move from left to right across a period

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A water bottling company is looking to achieve maximum production from the following reversible reaction. If the reaction has re

dybincka [34]

Answer:

Remove the already produced water and allow the reaction to reach equilibrium again.

Explanation:

According to Le Chatelier principle, when a reaction is in equilibrium and one of the factors that influence the rate of reaction is altered, the equilibrium will shift so as to annul the effects of the change.

If the product is continuously being removed from a reaction that is in equilibrium, more product will continued to be formed in another to annul the effect of reduction in the concentration of product.

Hence, in order to maximize production of water in the reaction, the product (water) needs to be removed and the reaction allowed to reach equilibrium again.

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4 years ago

White moths were found in huge numbers in the city of Birmingham these moths could be easy camouflage by the trees because of st

natima [27]

Answer:

It declined drastically

Explanation:

The moths were active at night and rested by day on white birch trees, where they were effectively camouflaged from predators.

However, as the city became more industrialized, the trees were covered with black soot from the factories.

The white moths lost their camouflage, so their population declined drastically, as their predators could now find them more easily.

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3 years ago

The physical matter of which living or nonliving things are composed is called a ___

kolezko [41]

Answer:

Substance

Explanation:

Any physical matter despite of living or non living things are composed by a set of molecules or atoms commonly known as substance

On specific living things are composed of biomolecules or cells where as non living things are composed of molecules and atoms

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2 years ago

A 32 year old immigrant from a patriarchal country is giving birth. As she is delivering the baby, she tearfully confesses to he

Wittaler [7]

She's 32. She can vote. She is entitled to privacy and it would take almost an act of Nature to disclose what she has asked of her doctor.

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If this question is anything but a school question (I'm assuming that is all it is), and somebody needs to act on the answer, for goodness sake consult someone who really knows and is governed by HIPAA. This is nothing to fool around with.

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3 years ago

The standard reduction potentials of the following half-reactions are given in Appendix E in the textbook:

german

Answer:

For 1: The largest positive cell potential is of cell having 1st and 4th half reactions.

For 2: The standard electrode potential of the cell is 1.539 V

For 3: The smallest positive cell potential is of cell having 3rd and 4th half reactions. The standard electrode potential of the cell is 0.46 V

Explanation:

The substance having highest positive $E^o$ potential will always get reduced and will undergo reduction reaction.

We are given:

$Ag^++(aq.)+e^-\rightarrow Ag(s);E^o_{Ag^+/Ag}=0.799V\\\\Cu^{2+}+(aq.)+2e^-\rightarrow Cu(s);E^o_{Cu^{2+}/Cu}=0.337V\\\\Ni^{2+}(aq.)+2e^-\rightarrow Ni(s);E^o_{Ni^{2+}/Ni}=-0.28V\\\\Cr^{3+}(aq.)+3e^-\rightarrow Cr(s);E^o_{Cr^{3+}/Cr}=-0.74V$

Substance getting oxidized always act as anode and the one getting reduced always act as cathode.

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

Cell having 1st and 2nd half reactions:

Silver has higher electrode potential. So, this will undergo reduction reaction and act as anode. Copper will undergo oxidation reaction and act as cathode.

$E^o_{cell}=0.799-0.337=0.462V$

Cell having 1st and 3rd half reactions:

Silver has higher electrode potential. So, this will undergo reduction reaction and act as anode. Nickel will undergo oxidation reaction and act as cathode.

$E^o_{cell}=0.799-(-0.28)=1.079V$

Cell having 1st and 4th half reactions:

Silver has higher electrode potential. So, this will undergo reduction reaction and act as anode. Chromium will undergo oxidation reaction and act as cathode.

$E^o_{cell}=0.799-(-0.74)=1.539V$

Cell having 2nd and 3rd half reactions:

Copper has higher electrode potential. So, this will undergo reduction reaction and act as anode. Nickel will undergo oxidation reaction and act as cathode.

$E^o_{cell}=0.337-(-0.28)=0.617V$

Cell having 3rd and 4th half reactions:

Nickel has higher electrode potential. So, this will undergo reduction reaction and act as anode. Chromium will undergo oxidation reaction and act as cathode.

$E^o_{cell}=-0.28-(-0.74)=0.46V$

Hence,

For 1: The largest positive cell potential is of cell having 1st and 4th half reactions.

For 2: The standard electrode potential of the cell is 1.539 V

For 3: The smallest positive cell potential is of cell having 3rd and 4th half reactions. The standard electrode potential of the cell is 0.46 V

8 0

4 years ago