Give an example of homeostasis in your body

What is emitted when an electron returns to its lower energy level?

Yakvenalex [24]

Answer:

When the electron changes levels, it decreases energy and the atom emits photons. The photon is emitted with the electron moving from a higher energy level to a lower energy level. The energy of the photon is the exact energy that is lost by the electron moving to its lower energy level.

Explanation:

4 0

3 years ago

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A standard solution of FeSCN2+ is prepared by combining 9.0 mL of 0.20 M Fe(NO3)3 with 1.0 mL of 0.0020 M KSCN . The standard so

Xelga [282]

Answer : The equilibrium concentration of $SCN^-$ in the trial solution is $4.58\times 10^{-8}M$

Explanation :

First we have to calculate the initial moles of $Fe^{3+}$ and $SCN^-$ .

$\text{Moles of }Fe^{3+}=\text{Concentration of }Fe^{3+}\times \text{Volume of solution}$

$\text{Moles of }Fe^{3+}=0.20M\times 9.0mL=1.8mmol$

and,

$\text{Moles of }SCN^-=\text{Concentration of }SCN^-\times \text{Volume of solution}$

$\text{Moles of }SCN^-=0.0020M\times 1.0mL=0.0020mmol$

The given balanced chemical reaction is,

$Fe^{3+}(aq)+SCN^-(aq)\rightleftharpoons FeSCN^{2+}(aq)$

Since 1 mole of $Fe^{3+}$ reacts with 1 mole of $SCN^-$ to give 1 mole of $FeSCN^{2+}$

The limiting reagent is, $SCN^-$

So, the number of moles of $FeSCN^{2+}$ = 0.0020 mmole

Now we have to calculate the concentration of $FeSCN^{2+}$ .

$\text{Concentration of }FeSCN^{2+}=\frac{0.0020mmol}{9.0mL+1.0mL}=0.00020M$

Using Beer-Lambert's law :

$A=\epsilon \times C\times l$

where,

A = absorbance of solution

C = concentration of solution

l = path length

$\epsilon$ = molar absorptivity coefficient

$\epsilon$ and l are same for stock solution and dilute solution. So,

$\epsilon l=\frac{A}{C}=\frac{0.480}{0.00020M}=2400M^{-1}$

For trial solution:

The equilibrium concentration of $SCN^-$ is,

$[SCN^-]_{eqm}=[SCN^-]_{initial}-[FeSCN^{2+}]$

$[SCN^-]_{initial}$ = 0.00050 M

Now calculate the $[FeSCN^{2+}]$ .

$C=\frac{A}{\epsilon l}=\frac{0.220}{2400M^{-1}}=9.17\times 10^{-5}M$

Now calculate the concentration of $SCN^-$ .

$[SCN^-]_{eqm}=[SCN^-]_{initial}-[FeSCN^{2+}]$

$[SCN^-]_{eqm}=(0.00050M)-(9.17\times 10^{-5}M)$

$[SCN^-]_{eqm}=4.58\times 10^{-8}M$

Therefore, the equilibrium concentration of $SCN^-$ in the trial solution is $4.58\times 10^{-8}M$

5 0

3 years ago

Calculate the pH of each of the following solutions: (a) 0.1000M Propanoic acid (HC3H5O2, Ka= 1.3x10-5 ) (b) 0.1000M sodium prop

jek_recluse [69]

(a) The pH of 0.1000 M propanoic acid (HC3H5O2) is 2.9.

(b) The pH of 0.1000 M sodium propanoate (NaC3H5O2) is 8.9.

(c) The pH of 0.1000 M propanoic acid (HC3H5O2) and 0.1000 M sodium propanoate (NaC3H5O2) is 4.9.

<h3>Further explanation:</h3>

(a)

Given information:

The value of acid ionization constant for propanoic acid is 1.3 x $10^{-5}$ .

The initial concentration of propanoic acid is .

To calculate:

The pH of 0.1000 M propanoic acid solution.

Solution:

Propanoic acid is a weak acid. It ionizes partially in water as follows:

The expression for acid dissociation constant is,

…… (1)

Here,

is ionization constant of propanoic acid.

is the equilibrium concentration of propanoate ion.

is the equilibrium concentration of hydronium ion.

is the equilibrium concentration of propanoic acid.

ICE table (1):

Refer ICE table (1),

Substitute the values form the ICE table (1) in equation (1).

The approximation x is very small is valid. Therefore, the value of x can be neglected. Above equation can be modified as,

Rearrange above equation for x.

…… (2)

Substitute for in equation (2) to calculate the value of x.

Therefore, from the ICE table (1) the concentration of hydronium ion is,

.

The negative logarithm of hydronium ion concentration is defined as the pH of the solution. Mathematically,

…… (3)

Substitute for in equation (3) to calculate the pH of the solution.

(b)

Given information:

The value of acid ionization constant for propanoic acid is .

The initial concentration of sodium propanoate is .

To calculate:

The pH of 0.1000 M sodium propanoate solution.

Solution:

Sodium propanoate is conjugate base of weak propanoic acid. It undergoes hydrolysis in water to yield hydroxide ion in the solution as follows:

…… (4)

Propanoic acid is a weak acid. It ionizes partially in water as follows:

…… (5)

Dissociation reaction for water is written as follows:

…… (6)

From equation (4), (5), and (6) the relationship between and is,

…… (7)

Substitute for and for in equation (7).

ICE table (2):

The expression for base dissociation constant is,

…… (8)

Here,

is base ionization constant.

is the equilibrium concentration of propanoate ion.

is the equilibrium concentration of hydroxide ion.

is the equilibrium concentration of propanoic acid.

From the ICE table (2),

Substitute the values form the ICE table (2) in equation (8).

The approximation y is very small is valid. Therefore, the value of y can be neglected. Above equation can be modified as,

Rearrange above equation for y.

…… (9)

Substitute for in equation (9) to calculate the value of y.

Therefore, from the ICE table (2) the concentration of hydroxide ion is,

The negative logarithm of hydroxide ion concentration is defined as pOH of the solution. Mathematically,

…… (10)

Substitute for in equation (10) to calculate pOH of the solution.

The relation between pH and pOH is as follows:

…… (11)

Substitute 5.057 for pOH in equation (11) to calculate the pH of the solution.

(c)

Given information:

The value of acid ionization constant for propanoic acid is .

The initial concentration of sodium propanoate is .

To calculate:

The pH of 0.1000 M sodium propanoate and 0.1000 M propanoic acid solution.

Solution:

Propanoic acid is a weak acid, and sodium propanoate is salt of the conjugate base of propanoic acid. Thus, propanoic acid and sodium propanoate will form a buffer system.

The pH of the buffer solution can be determined with the help of the Henderson-Hasselbalch equation. Mathematically,

For propanoic acid and sodium propanoate buffer system, the Henderson-Hasselbalch equation can be modified as,

…… (12)

The negative logarithm of acid ionization constant is equal to .

…… (13)

Substitute for in equation (13).

Substitute for , for and 4.9 for in equation (12).

Learn more:

1. About Henderson-Hasselbalch equation brainly.com/question/12999557

2. Learn more about how to calculate moles of the base in given volume brainly.com/question/4283309

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Ionic equilibria

Keywords: ionic equilibrium, propanoic acid, sodium propanoate, ionization constant, weak acid, conjugate base, equilibrium concentration, hydronium ion, hydroxide ion, pH, pOH, ICE table, negative logarithm, buffer solution, Henderson-Hasselbalch equation, 0.1000 M, 4.9, 8.9, 2.9.

3 0

3 years ago

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In h2o, the type of bond that holds one of the hydrogen atoms to the oxygen atom is a

Murrr4er [49]

The hydrogen and oxygen<span> atoms from H</span>₂O are <span>bonded together through covalent </span>bonding.

3 0

3 years ago

g If a small amount of a strong acid is added to a buffer made up of a weak acid, HA, and the sodium salt of its conjugate base,

ikadub [295]

Answer:

Explanation is in the answer

Explanation:

The pH of the buffer solution does not change appreciably because the strong acid (free H⁺) reacts with conjugate base of buffer producing more weak acid. pH formula of buffers is (Henderson-Hasselbalch formula):

pH = pKa + log ( [A⁻] / [HA] )

The addition of strong acid decreases [A⁻] increasing [HA]. pH change just in the log of the ratio of [A⁻] with [HA], that is a real little effect over pH of the buffer solution.

5 0

3 years ago