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3 years ago

Are there more bones in the arm than in the leg?

Chemistry

2 answers:

hodyreva [135]3 years ago

6 0

They have the same amount but if you count the wrist it was more then the leg and ankle.

Molodets [167]3 years ago

5 0

There's the same amount of bones in both the arm and the leg. But if you start talking about the arm with the hand included and the leg with the foot included, it will get more complicated because there's more bones in the wrist than the ankle.

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Gastric juice is made up of substances secreted from parietal cells, chief cells, and mucous-secreting cells. The cells secrete

neonofarm [45]

Answer:

The amount of energy required to transport hydrogen ions from a cell into the stomach is 37.26KJ/mol.

Explanation:

The free change for the process can be written in terms of its equilibrium constant as:

ΔG° = $-RTInK_(eq)$

where:

R= universal gas constant

T= temperature

$K_eq$ = equilibrum constant for the process

Similarly, free energy change and cell potentia; are related to each other as follows;

ΔG= -nFE°

from above;

F = faraday's constant

n = number of electrons exchanged in the process; and

E = standard cell potential

∴ The amount of energy required for transport of hydrogen ions from a cell into stomach lumen can be calculated as:

ΔG° = $-RTInK_(eq)$

where;

[texK_eq[/tex]= $\frac{[H^+]_(cell)}{[H^+(stomach lumen)]}$

For transport of ions to an internal pH of 7.4, the transport taking place can be given as:

$H^+_{inside}$ ⇒ $H^+_{outside}$

Equilibrum constant for the transport is given as:

$K_{eq}=\frac{[H^+]_{outside}}{[H^+]_{inside}}$

$=\frac{[H^+]_{cell}}{[H^+]_{stomach lumen}}$

$[H^+]_{cell}$ = 10⁻⁷⁴

=3.98 * 10⁻⁸M

$[H^+]_{stomach lumen}$ = 10⁻²¹

=7.94 * 10⁻³M

Hence;

$K_{eq}=\frac{[H^+]_{cell}}{[H^+]_{stomachlumen}}$

= $\frac{3.98*10^{-8}}{7.94*10{-3}}$

= 5.012 × 10⁻⁶

Furthermore, free energy change for this reaction is related to the equilibrium concentration given as:

ΔG° = $-RTInK_(eq)$

If temperature T= 37° C ; in kelvin

=37° C + 273.15K

=310.15K; and

R-= 8.314 j/mol/k

substituting the values into the equation we have;

ΔG₁ = $-(8.314J/mol/K)(310.15)TIn(5.0126*10^{-6})$

= 31467.93Jmol⁻¹

≅ 31.47KJmol⁻¹

If the potential difference across the cell membrane= 60.0mV.

Energy required to cross the cell membrane will be:

ΔG₂ = $-nFE°_{membrane}$

ΔG₂ = $-(1 mol)(96.5KJ/mol/V)(60*10^{-3})$

= 5.79KJ

Therefore, for one mole of electron transfer across the membrane; the energy required is 5.79KJmol⁻¹

Now, we can calculate the total amount of energyy required to transport H⁺ ions across the membrane:

Δ $G_{total} = G_{1}+G_{2}$

= (31.47+5.79) KJmol⁻¹

= 37.26KJmol⁻¹

We can therefore conclude that;

The amount of energy required to transport ions from cell to stomach lumen is 37.26KJmol⁻¹

5 0

3 years ago

What are Metals that are less reactive than alkali metals and alkaline-earth metals called

Margaret [11]

Transition metals are less reactive than alkali metals because of their high ionization potential and high melting point.

On moving from left to right of the periodic table for every period, electrons fill in the same shell or orbital, with the alkali metals having the least filled outermost shells, one electron, which equates to fewer protons in them.

Consequently, they have a lesser attraction power from the nucleus, whereas, the corresponding transition metals of the same period have more protons interacting with electrons at the same distance, far from the nucleus as the alkali metals.

4 0

2 years ago

Read 2 more answers

When two or more magnetic fields overlap, the result is always a _____.

Romashka [77]

Answer:

A. Combined magnetic field

6 0

2 years ago

Read 2 more answers

Using the periodic table, choose the more reactive metal.<br><br> Ta or V

Vera_Pavlovna [14]

V is more reactive, hope this helps!

4 0

3 years ago

Read 2 more answers

What are the coefficients of the reactants and products in the balanced equation above? Remember to include H2O(l)H2O(l) and H+(

poizon [28]

Answer:

The balanced equation is then given as

BrO₃⁻ + 3Sn²⁺ + 6H⁺ → Br⁻ + 3Sn⁴⁺ + 3H₂O

The coefficients written in a bracket separated by commas is then given as

(1, 3, 6, 1, 3, 3)

Explanation:

The equation for the question is

BrO₃⁻ (aq) + Sn²⁺ + __ → Br⁻ + Sn⁴⁺ + __

The H⁺ and H₂O presented in the question shows that this redox reaction takes place in acidic medium.

We first identify which reactant specie is being oxidized or reduced.

Oxidation is the increase in oxidation number of reactant species due to the loss of electrons while Reduction is the reduction in oxidation number of reactant species due to the gain of electrons.

From the reaction given, it is evident that the Tin ion is oxidized as its oxidation number increases from +2 to +4.

And the Br in BrO₃⁻ undergoes reduction to have its oxidation number change from +5 to -1.

Note that +5 was calculated for as thus.

If the oxidation number of Br in BrO₃⁻ is unknown and called x, the oxidation number of the O in BrO₃⁻ is -2.

x + (-2)(3) = -1

x = -1 + 6 = +5

So, back to the question, since we know which reactant species are oxidized and reduced, we can then write the reduction and oxidation half reactions. Balanced reduction and oxidation half reactions at that showing the number of electrons lost or gained.

Reduction half reaction

BrO₃⁻ → Br⁻

This isn't possible, so we add the spectator compounds/elements/ions provided in the form of H⁺ and H₂O

BrO₃⁻ + H⁺ → Br⁻ + H₂O

We then balance this reaction stoichiometrically,

BrO₃⁻ + 6H⁺ → Br⁻ + 3H₂O

Now, we can check the oxidation numbers on both sides to know the number of electrons gained.

-1 + 6 → -1 + 0

+5 → -1

Hence, it is evident that 6 electrons are gained in this reduction half reaction.

So, we rewrite the reduction half reaction finally to be

BrO₃⁻ + 6H⁺ + 6e⁻ → Br⁻ + 3H₂O

We then repeat this for the oxidation half reaction.

Sn²⁺ → Sn⁴⁺

This is possible, So, no need to add the spectator compounds/elements/ions provided in the form of H⁺ and H₂O

The equation is also balance stoichiometrically, So, we just proceed to balance the charges/oxidation numbers.

+2 → +4

This shows that 2 electrons are lost, So, we rewrite our oxidation half reaction as

Sn²⁺ - 2e⁻ → Sn⁴⁺

It is more appropriate to write it in the form

Sn²⁺ → Sn⁴⁺ + 2e⁻

So, we can then write the two balanced half reactions (stoichiometrically and charge balance) on top of each other.

BrO₃⁻ + 6H⁺ + 6e⁻ → Br⁻ + 3H₂O

Sn²⁺ → Sn⁴⁺ + 2e⁻

In order to add, we need the number of electrons on both of these to be the same, so We multiply the reduction half reaction by 1 and multiple the oxidation half reaction by 3

[BrO₃⁻ + 6H⁺ + 6e⁻ → Br⁻ + 3H₂O] × 1

[Sn²⁺ → Sn⁴⁺ + 2e⁻ ] × 3

We then get

BrO₃⁻ + 6H⁺ + 6e⁻ → Br⁻ + 3H₂O

3Sn²⁺ → 3Sn⁴⁺ + 6e⁻

We can then add the two half reactions now

BrO₃⁻ + 6H⁺ + 6e⁻ + 3Sn²⁺ → Br⁻ + 3H₂O + 3Sn⁴⁺ + 6e⁻

Taking out the 6 electrons that appear on both sides, we have

BrO₃⁻ + 6H⁺ + 3Sn²⁺ → Br⁻ + 3H₂O + 3Sn⁴⁺

Written more properly as

BrO₃⁻ + 3Sn²⁺ + 6H⁺ → Br⁻ + 3Sn⁴⁺ + 3H₂O

Hope this Helps!!!

8 0

3 years ago