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Balancing Redox Reactions VO2++MnO4 V(OH)4++Mn2+(acidic)

Natasha_Volkova [10]

Answer:

MnO4- + 5 VO +2 + 11 H2O = 5 V(OH)4+ + Mn+2 + 2 H+

Explanation:

3 0

3 years ago

During action potential, positively charged ________ ions move inside the cell.

Tems11 [23]

During action potential, positively charged sodium ions move inside the cell.

So option D is correct one.

The sodium ion moves inside the cell during a action potential. The stage of action potential is called depolarization . This open voltage gated sodium channel.

Action potentials ( those electrical impulse that send signals around body ) is nothing but more than temporary shift ( from negative to positive ) in the neuron's membrane potential caused by ions suddenly flowing in and out of the neuron.

It consists of phases:

Depolarization
overshoot
repolarization

An active potential propagates along the cell membrane of an axon until it reaches the terminal button.

to known more about action potential

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8 0

2 years ago

The partial pressure of CO2 gas in a bottle of carbonated water is 4.60 atm at 25 ºC. How much CO2 gas (in g) will be released f

frutty [35]

If the partial pressure of CO₂ in a bottle of carbonated water decreases from 4.60 atm to 1.28 atm, the mass of CO₂ released is 0.265 g.

The partial pressure of CO₂ gas in a bottle of carbonated water is 4.60 atm at 25 ºC. We can calculate the concentration of CO₂ using Henry's law.

$C = k \times P = \frac{1.65 \times 10^{-3} M }{atm} \times 4.60 atm = 7.59 \times 10^{-3} M$

We can calculate the mass of CO₂ in 1.1 L considering its molar mass is 44.01 g/mol.

$\frac{7.59 \times 10^{-3} mol}{L} \times 1.1 L \times \frac{44.01 g}{mol} = 0.367 g$

Now, we will repeat the same procedure for a partial pressure of 1.28 atm.

$C = k \times P = \frac{1.65 \times 10^{-3} M }{atm} \times 1.28 atm = 2.11 \times 10^{-3} M$

$\frac{2.11 \times 10^{-3} mol}{L} \times 1.1 L \times \frac{44.01 g}{mol} = 0.102 g$

The mass of CO₂ released will be equal to the difference in the masses at the different pressures.

$m = 0.367 g - 0.102 g = 0.265 g$

If the partial pressure of CO₂ in a bottle of carbonated water decreases from 4.60 atm to 1.28 atm, the mass of CO₂ released is 0.265 g.

Learn more: brainly.com/question/18987224

The partial pressure of CO₂ gas in a bottle of carbonated water is 4.60 atm at 25 ºC. How much CO₂ gas (in g) will be released from 1.1 L of the carbonated water when the partial pressure of CO2 is lowered to 1.28 atm? At 25 ºC, the Henry’s law constant for CO₂ dissolved in water is 1.65 x 10⁻³ M/atm, and the density of water is 1.0 g/cm³.

5 0

2 years ago

What is the change in boiling point for a 0.615m solution of Mgl2 in water?

ivanzaharov [21]

Answer : The change in boiling point is, $0.94^oC$

Explanation :

Formula used :

$\Delta T_b=i\times K_f\times m$

where,

$\Delta T_b$ = change in boiling point = ?

i = Van't Hoff factor = 3 (for MgI₂ electrolyte)

$K_f$ = boiling point constant for water = $0.51^oC/m$

m = molality = 0.615 m

Now put all the given values in this formula, we get

$\Delta T_b=3\times (0.51^oC/m)\times 0.615m$

$\Delta T_b=0.94^oC$

Therefore, the change in boiling point is, $0.94^oC$

4 0

3 years ago

Americans spend up to $100 billion annually for bottled water (41 billion gallons). The only beverages with higher sales are car

grigory [225]

Answer: C) be hypertonic to Tank B.

Explanation:

The ability of an extracellular solution to move water in or out of a cell by osmosis is known as its tonicity. Additionally, the tonicity of a solution is related to its osmolarity, which is the total concentration of all the solutes in the solution.

Three terms (hypothonic, isotonic and hypertonic) are used to compare the osmolarity of a solution with respect to the osmolarity of the liquid that is found after the membrane. When we use these terms, we only take into account solutes that can not cross the membrane, which in this case are minerals.

If the liquid in tank A has a lower osmolarity (lower concentration of solute) than the liquid in tank B, the liquid in tank A would be hypotonic with respect to the latter.
If the liquid in tank A has a greater osmolarity (higher concentration of solute) than the liquid in tank B, the liquid in tank A would be hypertonic with respect to the latter.
If the liquid in tank A has the same osmolarity (equal concentration of solute) as the liquid in tank B, the liquid in tank A would be isotonic with respect to the latter.

In the case of the problem, option A is impossible because the minerals can not cross the membrane, since it is permeable to water only. There is no way that the concentration of minerals decreases in tank A, so the solution in this tank can not be hypotonic with respect to the one in Tank B.

Equally, both solutions can not be isotonic and neither we can say that the solution in tank A has more minerals that the one in tank B because the liquid present in tank B is purified water that should not have minerals. Therefore, options B and D are also not correct.

Finally, the correct option is C, since in the purification procedure the water is extracted from the solution in tank A to obtain a greater quantity of purified water in tank B. In this way, the solution in Tank A would be hypertonic to Tank B.

5 0

3 years ago