The balanced equation for photosynthesis is shown below.

What is defined as the basic unit in the metric system that describes a quality

Bingel [31]

Answer:

They are:

Length - meter (m)

Time - second (s)

Amount of substance - mole (mole)

Electric current - ampere (A)

Temperature - kelvin (K)

Luminous intensity - candela (cd)

Mass - kilogram (kg)

3 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

Read 2 more answers

The combustion of 1.5011.501 g of fructose, C6H12O6(s)C6H12O6(s) , in a bomb calorimeter with a heat capacity of 5.205.20 kJ/°C

avanturin [10]

Answer : The internal energy change is -2805.8 kJ/mol

Explanation :

First we have to calculate the heat gained by the calorimeter.

$q=c\times (T_{final}-T_{initial})$

where,

q = heat gained = ?

c = specific heat = $5.20kJ/^oC$

$T_{final}$ = final temperature = $27.43^oC$

$T_{initial}$ = initial temperature = $22.93^oC$

Now put all the given values in the above formula, we get:

$q=5.20kJ/^oC\times (27.43-22.93)^oC$

$q=23.4kJ$

Now we have to calculate the enthalpy change during the reaction.

$\Delta H=-\frac{q}{n}$

where,

$\Delta H$ = enthalpy change = ?

q = heat gained = 23.4 kJ

n = number of moles fructose = $\frac{\text{Mass of fructose}}{\text{Molar mass of fructose}}=\frac{1.501g}{180g/mol}=0.00834mole$

$\Delta H=-\frac{23.4kJ}{0.00834mole}=-2805.8kJ/mole$

Therefore, the enthalpy change during the reaction is -2805.8 kJ/mole

Now we have to calculate the internal energy change for the combustion of 1.501 g of fructose.

Formula used :

$\Delta H=\Delta U+\Delta n_gRT$

or,

$\Delta U=\Delta H-\Delta n_gRT$

where,

$\Delta H$ = change in enthalpy = $-2805.8kJ/mol$

$\Delta U$ = change in internal energy = ?

$\Delta n_g$ = change in moles = 0 (from the reaction)

R = gas constant = 8.314 J/mol.K

T = temperature = $27.43^oC=273+27.43=300.43K$

Now put all the given values in the above formula, we get:

$\Delta U=\Delta H-\Delta n_gRT$

$\Delta U=(-2805.8kJ/mol)-[0mol\times 8.314J/mol.K\times 300.43K$

$\Delta U=-2805.8kJ/mol-0$

$\Delta U=-2805.8kJ/mol$

Therefore, the internal energy change is -2805.8 kJ/mol

5 0

3 years ago

What is the average kinetic energy of 1 mole of a gas that is at 100 degrees Celsius?

Paha777 [63]

The average kinetic energy of an ideal gas is calculated as

KE_avg = 3/2 kT

where T is the temperature in Kelvin and k=R/N_A; R is the universal gas constant and N_A is the number of moles.

Thus, upon substitution we get

KE_avg = 3/2(8.314/1)(100+273)
KE_avg = 3/2(8.314)(373)
KE_avg = 4651.683

The average kinetic energy of 1 mole of a gas at 100 degree Celsius is 4651.683 J.

3 0

3 years ago

Gas is $2.50/gal at Station A and 85 cents/L at Station B. A) Convert the cost of gas at Station B to $/gal. B) At which station

SVETLANKA909090 [29]

Answer:

85 cents/L is equal to 3.2176$/gallon.

gas wiuld be heaper to buy from the station A.

Explanation:

As 85 cents/L is more than $2.5/gallon therefore buying gas from station A would be cheaper than the other one.

8 0

3 years ago