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valina [46]
3 years ago
14

In a resting state, sodium (Na ) is at a higher concentration outside the cell and potassium (K ) is more concentrated inside th

e cell. During an action potential, the sodium levels ________ inside the cell. Group of answer choices
Chemistry
1 answer:
larisa [96]3 years ago
8 0

Answer:

The correct answer is: INCREASES

Explanation:

The membrane potential of a cell is defined as the difference in the electric potential between the outside and inside of the cell.

Ions such as sodium (Na⁺) and potassium (K⁺) ions, have a concentration gradient across the cell membrane.<em> In the </em><em>resting state</em><em>, the intracellular spaces (inside) has higher concentration of K⁺ ions and the extracellular spaces (outside) has high concentrations of Na⁺ ions.</em>

The rapid changes in the membrane potential of a cell gives rise to the action potential.

<em>During an </em><em>action potential</em><em>, the Na⁺ ions move inside the cell into the intracellular spaces, thus increasing the concentration of Na⁺ ions inside the cell.</em>

Therefore, the sodium levels <u>increases</u> inside the cell, during an action potential.

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What is the pOH of 0.50 molar H3BO3?
Crazy boy [7]

<u>Answer:</u>

<em>A. 10.25</em>

<em></em>

<u>Explanation:</u>

Pkb =4.77

So pka = 14 - pka = 9.23

Ka =10^{-pka}

H_3 BO_3 (aq)+ H_2 O(l) H_2 BO_3^- (aq)+H_3 O^+ (aq)

Initial                0.50M                                 0                                 0

Change                  -x                                 +x                               +x

Equilibrium    0.50M-x                               +x                               +x

Ka =\frac {((x)(x))}{(0.50M-x)}

5.88\times10^{-10}= \frac {x^2}{(0.50M-x)}

(-x is neglected) so we get

5.88\times10^{-10}\times0.50=x^2\\\\x^2=2.94\times10^{-10}

x=\sqrt{x^2}=1.72\times10^{-5} M=H^3 O^{+}

pH=-log[H^3 O^+]\\\\pH=-log[1.72\times10^{-5}]\\\\pH=4.76

pOH = 14 - pH

= 14 - 4.76

pOH = 9.24 is the answer

Option A - 10.25 is the answer which is close to 9.24

4 0
2 years ago
How many kilojoules of energy would be required to heat a 225g block of aluminum from 23.0 C to 73.5 C?
gulaghasi [49]

Answer:

\boxed {\boxed {\sf 10.2 \ kJ}}

Explanation:

We are asked to find how many kilojoules of energy would be required to heat a block of aluminum.

We will use the following formula to calculate heat energy.

q=mc \Delta T

The mass (m) of the aluminum block is 225 grams and the specific heat (c) is 0.897 Joules per gram degree Celsius. The change in temperature (ΔT) is the difference between the final temperature and the initial temperature.

  • ΔT = final temperature - inital temperature

The aluminum block was heated from 23.0 °C to 73.5 °C.

  • ΔT= 73.5 °C - 23.0 °C = 50.5 °C

Now we know all three variables and can substitute them into the formula.

  • m= 225 g
  • c= 0.897 J/g° C
  • ΔT= 50.5 °C

q= (225 \ g )(0.897 \ J/g \textdegree C)(50.5 \textdegree C)

Multiply the first two numbers. The units of grams cancel.

q= (225 \ g  * 0.897 \ J/g \textdegree C)(50.5 \textdegree C)

q= (225   * 0.897 \ J / \textdegree C)(50.5 \textdegree C)

q= (201.825\ J / \textdegree C)(50.5 \textdegree C)

Multiply again. This time, the units of degrees Celsius cancel.

q= 201.825 \ J * 50.5

q= 10192.1625 \ J

The answer asks for the energy in kilojoules, so we must convert our answer. Remember that 1 kilojoule contains 1000 joules.

\frac { 1  \ kJ}{ 1000 \ J}

Multiply by the answer we found in Joules.

10192.1625 \ J * \frac{ 1 \ kJ}{ 1000 \ J}

10192.1625  * \frac{ 1 \ kJ}{ 1000 }

\frac {10192. 1625}{1000} \ kJ

10.1921625 \ kJ

The original values of mass, temperature, and specific heat all have 3 significant figures, so our answer must have the same. For the number we found, that is the tneths place. The 9 in the hundredth place tells us to round the 1 up to a 2.

10.2 \ kJ

Approximately <u>10.2 kilojoules</u> of energy would be required.

3 0
2 years ago
Given 1 cm3 = 1 mL<br> A box has dimensions 2.0 cm x 4.0 cm x 8.0 cm.
svp [43]

Answer: 64

Explanation:

you just multiple the 3 numbers to get the answer (i’m in chemistry and just did this question lol)

5 0
3 years ago
Why don’t we include the mass of an atoms electrons in the atomic mass?
Daniel [21]

Because Electrons have a negative charge

8 0
3 years ago
What is the temperature of 1.485 moles of N₂ gas at a pressure of 1.072 atm and a volume of 20 L?
morpeh [17]

Answer:

178.67K

Explanation:

PV=nRT

T=PV/nR

= 1.072atm*20L/1.485mol*0.0821LatmK^-1

=178.67K

8 0
3 years ago
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