All categories

2 years ago

Select the conjugate acid-base pair(s). a) HI, I b) HCHO2, SO4^2- c) CO3^2-, HCI d) PO4^3-, HPO4^2-

Chemistry

1 answer:

Semenov [28]2 years ago

6 0

Answer:

d) PO4^3-, HPO4^2-

Explanation:

Basically, an acid and a base which differs only by the presence or absence of proton (hydrogen ion) are called a conjugate acid-base pair.

a) HI, I

This is incorrect. For the acid, HI the conjugate base is I⁻ ion.

b) HCHO2, SO4^2-

This is incorrect, there's no relationship between both entities.

c) CO3^2-, HCI

This is incorrect, there's no relationship between both entities.

d) PO4^3-, HPO4^2-

This is correct. The difference between both entities is the Hydrogen ion. This is the conjugate acid-base pair

You might be interested in

When does absorption occur in infrared spectroscopy

Vladimir79 [104]

Answer: hopefully this is right UwU

Explanation: The infrared spectrum of a sample is recorded by passing a beam of infrared light through the sample. When the frequency of the IR is the same as the vibrational frequency of a bond or collection of bonds, absorption occurs. Examination of the transmitted light reveals how much energy was absorbed at each frequency (or wavelength).

5 0

2 years ago

600 s after initiation of a first order reaction 48.5% of the initial reactant concentration remains present. What is the rate c

Ludmilka [50]

Answer:

$k=1.20x10^{-3} s^{-1}$

Explanation:

For a first order reaction the rate law is:

$v=\frac{-d[A]}{[A]}=k[A]$

Integranting both sides of the equation we get:

$\int\limits^a_b {\frac{d[A]}{[A]}} \, dx =-k\int\limits^t_0 {} \, dt$

where "a" stands for [A] (molar concentration of a given reagent) and "b" is {A]0 (initial molar concentration of a given reagent), "t" is the time in seconds.

From that integral we get the integrated rate law:

$ln\frac{[A]}{[A]_{0} } =-kt$

$[A]=[A]_{0}e^{-kt}$

$ln[A]=ln[A]_{0} -kt$

$k=\frac{ln[A]_{0}-ln[A]}{t}$

therefore k is

$k=\frac{ln1-ln0,485}{600}=1,20x10^{-3}$

8 0

3 years ago

In a sample of oxygen gas at room temperature, the average kinetic energy of all the balls stays constant. Which postulate of ki

Levart [38]

Answer:

Collisions between gas particles are elastic; there is no net gain or loss of kinetic energy.

Explanation:

When a gas is paced in a container, the molecules of the gas have little or no intermolecular interaction between them. There is a lot of space between the molecules of the gas.

The gas molecules move at very high speed and collide with each other and with the walls of container.

The collision of these particles with each other is perfectly elastic hence the kinetic energy of the colliding gas particles do not change.

7 0

2 years ago

An equation thats shows how an objects acceleration relates to its mass

Nady [450]

Answer:

Newton's second law of motion

F = ma

5 0

2 years ago

You have to prepare some 2 M solutions, with 10 g of solute in each. What volume of solution will you prepare, for each solute b

alexdok [17]

Answer:

0.045 L or 45 mL

Explanation:

Moles = Mass/M.Mass

Moles = 10 g / 109.94 g/mol

Moles = 0.09 moles

Also,

Molarity = Moles / Vol in L

Or,

Vol in L = Moles / Molarity

Vol in L = 0.09 mol / 2 mol/L

Vol in L = 0.045 L

8 0

1 year ago