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3 years ago

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A cell must work to maintain a stable internal environment. It is also important for the environment around the cell to be stabl

e. What reasoning explains what happens when the concentration of water inside a cell is lower than the concentration of water outside the cell?
The cell loses nutrients.

The cell splits and creates two new cells.

The cell gains too many lipids and carbohydrates.

The cell either loses water and dries up or gains too much water and bursts.

(science 7th grade)

1 answer:

IRINA_888 [86]3 years ago

3 0

Answer: the last one

Explanation: if a cell has too little water, it will begin to function incorrectly, and if a cell has too much water it will burst.

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The coiling of the protein chain backbone into an alpha helix (like a spring) is referred to as the ________.

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Answer: secondary structure

Explanation:

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A transfer of heat within a liquid or gas that involves warm particles moving in currents is?

myrzilka [38]

It is called convection. When warm air, or current, moves up and disperse outwards as cold air, or current, moves into the warmer region.

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3 years ago

Read 2 more answers

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

5 0

2 years ago

If there is a loss of 5.63 x 10-7kg of mass in a nuclear reaction, how many Joules of energy would be released? Recall that c =

Fudgin [204]

Thank you for posting your question here at brainly.

E = mc^2
<span>where E is the energy in joules, </span>
<span>m is the mass in kilograms, </span>
<span>and c is the speed of light. </span>

<span>E = mc^2 </span>
<span>E = (5.63 x 10^-7 kg)(3 x 10^8 m/s)^2 </span>
<span>E = 5.07 x 10^10 J </span>

5 0

3 years ago

A runner runs 4339 ft in 7.45 minutes. What is the runner's average speed in miles per hou 6.62 mi/hr O 0.110 mi/hr 6.618 mi/h e

slavikrds [6]

<u>Answer:</u> The average speed of the runner is 6.618 miles/hr

<u>Explanation:</u>

Average speed is defined as the ratio of total distance traveled to the total time taken.

To calculate the average speed of the runner, we use the equation:

$\text{Average speed}=\frac{\text{Total distance traveled}}{\text{Total time taken}}$

We are given:

Distance traveled = 4339 ft

Time taken = 7.45 mins

Putting values in above equation, we get:

$\text{Average speed of runner}=\frac{4330ft}{7.45min}=582.42ft/min$

To convert the speed into miles per hour, we use the conversion factors:

1 mile = 5280 ft

1 hr = 60 mins

Converting the speed into miles per hour, we get:

$\Rightarrow \frac{528.42ft}{min}\times (\frac{1miles}{5280ft})\times (\frac{60min}{1hr})\\\\\Rightarrow 6.618mil/hr$

Hence, the average speed of the runner is 6.618 miles/hr

6 0

3 years ago