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3 years ago

8

Polar bears give birth and hunt on sea ice. Which of the following would help polar bears survive during the melting of Arctic i

ce?
Growing another layer of fur during summer
Migrate inland to search for different food sources
Staying put until the ice refreezes
Sticking to the usual diet of seals

2 answers:

galina1969 [7]3 years ago

4 0

i think the answer is the second one

kompoz [17]3 years ago

4 0

Answer:

<em>B </em><em>- Migrate inland to search for different food sources</em>

Explanation:

In Arctic Ice they will Migrate inland to search for different food sources. :D

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Approximately how many particles are in 2 moles?

schepotkina [342]

Answer:

$1.2\times10^2^4$ particles in 2 moles.

Explanation:

The number of particles that are contained in one mole, the international unit of amount of substance: by definition, exactly 6.022×10²³, and it is dimensionless. It is named after the scientist Amedeo Avogadro.

It is also known as Avogadro's constant.

∴ Number of particles in one mole = $6.022\times10^2^3$

∴ Number of particles in 2 mole = 2 times Number of particles in one mole

∴ Number of particles in 2 mole= $2\times6.022\times10^2^3=12.044\times10^2^3\approx1.2\times10^2^4$

Hence there are $1.2\times10^2^4$ particles in 2 moles.

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3 years ago

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Why is frying an egg an example of a chemical change

skelet666 [1.2K]

A Chemical change is any form of chemical reaction changing the state of the matter involved. When you fry an egg, you are making a reaction between the heat and the chemicals that make up the egg. Therefore once you have fried that egg, you have completed a chemical reaction. This is why it's a chemical change when you fry an egg.

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3 years ago

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The gas phase reaction, A2 + B2 ---->2AB, proceeds by bimolecular collisions between A2 and B2 c,o(\'e'9 molecules. The rate

Varvara68 [4.7K]

Answer:

The reaction rate becomes quadruple.

Explanation:

According to the law of mass action:-

The rate of the reaction is directly proportional to the active concentration of the reactant which each are raised to the experimentally determined coefficients which are known as orders. The rate is determined by the slowest step in the reaction mechanics.

Order of in the mass action law is the coefficient which is raised to the active concentration of the reactants. It is experimentally determined and can be zero, positive negative or fractional.

The order of the whole reaction is the sum of the order of each reactant which is raised to its power in the rate law.

Thus,

Given that:- The rate law is:-

$r=k[A_2][B_2]$

Now, $[A'_2]=2[A_2]$ and $[B'_2]=2[B_2]$

So, $r'=k[A'_2][B'_2]=k\times 2[A_2]\times 2[B_2]=4\times k[A_2][B_2]=4r$

<u>The reaction rate becomes quadruple.</u>

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3 years ago

Consider the reaction of metallic copper with iron(!!) to give copper(ll) and ironin 0.77V Fe* (aq) + e-Fe (aq) Cup (aq) + 2e --

frosja888 [35]

Answer :

(a) The anode and cathode will be $E^o_{(Cu^{2+}/Cu)}$

and $E^o_{(Fe^{3+}/Fe^{2+})}$ respectively.

(b) The emf of cell potential is 1.022 V

Explanation :

(a) The standard reduction potentials for iron and copper are:

$E^o_{(Fe^{3+}/Fe^{2+})}=0.77V\\E^o_{(Cu^{2+}/Cu)}=0.34V$

In the voltaic cell, the oxidation occurs at an anode which is a negative electrode and the reduction occurs at the cathode which is a positive electrode.

From the standard reduction potentials we conclude that, the substance having highest positive $E^o$ potential will always get reduced and will undergo reduction reaction.

So, iron will undergo reduction reaction will get reduced. Copper will undergo oxidation reaction and will get oxidized.

The given cell reactions are:

Oxidation half reaction (anode): $Cu\rightarrow Cu^{2+}+2e^-$

Reduction half reaction (cathode): $Fe^{3+}+1e^-\rightarrow Fe^{2+}$

Thus, the anode and cathode will be $E^o_{(Cu^{2+}/Cu)}$

and $E^o_{(Fe^{3+}/Fe^{2+})}$ respectively.

(b) Now we have to calculate the potential of a cell.

Oxidation half reaction (anode): $Cu\rightarrow Cu^{2+}+2e^-$

Reduction half reaction (cathode): $Fe^{3+}+1e^-\rightarrow Fe^{2+}$

In order to balance that electrons, we will multiple the reduction reaction by 2, we get:

Oxidation half reaction (anode): $Cu\rightarrow Cu^{2+}+2e^-$

Reduction half reaction (cathode): $2Fe^{3+}+2e^-\rightarrow 2Fe^{2+}$

The overall cell reaction will be,

$2Fe^{3+}+Cu\rightarrow Cu^{2+}+2Fe^{2+}$

$E^o_{[Fe^{3+}/Fe^{2+}]}=2\times 0.77V=1.54V$

$E^o_{[Cu^{2+}/Cu]}=0.34V$

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

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

$E^o=E^o_{[Fe^{3+}/Fe^{2+}]}-E^o_{[Cu^{2+}/Cu]}$

$E^o=1.54V-(0.34V)=1.20V$

Now we have to calculate the cell potential.

Using Nernest equation :

$E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Fe^{2+}]^2[Cu^{2+}]}{[Fe^{3+}]^2}$

where,

n = number of electrons in oxidation-reduction reaction = 2

$E_{cell}$ = emf of the cell = ?

Now put all the given values in the above equation, we get:

$E_{cell}=1.20-\frac{0.0592}{2}\log \frac{(0.20)^2(0.25)}{(0.0001)^2}$

$E_{cell}=1.022V$

Therefore, the emf of cell potential is 1.022 V

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3 years ago

Which of the following ions has a pseudo-noble-gas electron configuration?

Darya [45]

Answer:A pseudo-noble gas configuration is: 1s2 2s2 2p6 3s2 3p6 3d10. Cu^+ has this electron configuration. The answer is C)

Explanation:

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