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3 years ago

How do you compare the nuclear membrane and the cell membrane?

Chemistry

2 answers:

Mademuasel [1]3 years ago

8 0

ANSWER:

they're both protective covering, the nuclear membrane is a covering for the nucleas while the cell membrane is the covering for the whole cell the nucleus (which is the control center of the cell) decides what comes in and out of the cell.

~batmans wife dun dun dun...aka ~serenitybella

iragen [17]3 years ago

5 0

A. Both the cell membrane and the nuclear membrane are protective coverings.

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Wich of the following does not directly affect the weather

Drupady [299]

Answer:

Tings that effect weather is basically the following:

Distance from sea
Altitude
Distance to the equator or poles.
Mountains
Jet streams,etc

8 0

3 years ago

How do hydrogen ions (h+) and chloride ions (cl–) get into the lumen of the stomach?

user100 [1]

In the stomach cell, the enzyme carbonic anhydrase converts one molecule of carbon dioxide and one molecule of water indirectly into a bicarbonate ion (HCO3-) and a hydrogen ion (H+). In the stomach ion exchange is used to move H+ ions out the cells and into the lumen of the stomach

8 0

3 years ago

A 13.5g sample of gold is heated, then placed in a calorimeter containing 60g of water. The temperature of water increases from

sweet-ann [11.9K]

Answer:

$T_i~=163.1$ ºC

Explanation:

We have to start with the variables of the problem:

Mass of water = 60 g

Mass of gold = 13.5 g

Initial temperature of water= 19 ºC

Final temperature of water= 20 ºC

Initial temperature of gold= Unknow

Final temperature of gold= 20 ºC

Specific heat of gold = 0.13J/gºC

Specific heat of water = 4.186 J/g°C

Now if we remember the heat equation:

$Q_H_2_O=m_H_2_O*Cp_H_2_O*deltaT$

$Q_A_u=m_A_u*Cp_A_u*deltaT$

We can relate these equations if we take into account that all heat of gold is transfer to the water, so:

$m_H_2_O*Cp_H_2_O*deltaT=~-~m_A_u*Cp_A_u*deltaT$

Now we can put the values into the equation:

$60~g*4.186~J/g{\circ}C*(20-19)~{\circ}C=-(13.5~g*0.13~J/g{\circ}C*(20-T_i)~{\circ}C)$

Now we can solve for the initial temperature of gold, so:

$T_i~=(\frac{60~g*4.186~J/g{\circ}C*(20-19)~{\circ}C}{13.5~g*0.13~J/g{\circ}C})+20$

$T_i~=163.1$ ºC

I hope it helps!

5 0

3 years ago

A student prepares a 0.47mM aqueous solution of acetic acid CH3CO2H. Calculate the fraction of acetic acid that is in the dissoc

Aliun [14]

The fraction of acetic acid that is dissociated is 0.18

Why?

The chemical equation for the dissociation of acetic acid (HAc) is the following:

HAc(aq) + H₂O(l) ⇄ H₃O⁺(aq) + Ac⁻(aq)

To find the fraction of acetic acid that is in the dissociated form (f), we apply the following equation (Ka for acetic acid is 1.76*10⁻⁵). This equation comes from solving the equation of the equilibrium constant for the dissociated fraction of HAc:

$f=\frac{-Ka+\sqrt{Ka^{2} +4KaC} }{2C} = 0.18$

Have a nice day!

#LearnwithBrainly

6 0

3 years ago

How is the AHfusion used to calculate volume of liquid frozen that produces 1

yulyashka [42]

Answer:

option B

Explanation:

3 0

3 years ago

Read 2 more answers