If the formula mass of one molecule is X amu, the molar mass is X g/mol. Group of answer choices True False

Non metal are usually poor conductor of hart and electricity . The are non -lustourous,non-sonours,non-malleable and are coloure

vekshin1

Check the solution in the attachment.

7 0

2 years ago

You determine that it takes 26.0 mL of base to neutralize a sample of your unknown acid solution. The pH of the solution was 7.8

Lubov Fominskaja [6]

Complete Question

You determine that it takes 26.0 mL of base to neutralize a sample of your unknown acid solution. The pH of the solution was 7.82 when exactly 13 mL of base had been added, you notice that the concentration of the unknown acid was 0.1 M. What is the pKa of your unknown acid?

Answer:

The pK_a value is $pK_a =7.82$

Explanation:

From the question we are told

The volume of base is $V_B = 26.mL = 0.0260L$

The pH of solution is $pH = 7.82$

The concentration of the acid is $C_A = 0.1M$

From the pH we can see that the titration is between a strong base and a weak acid

Let assume that the the volume of acid is $V_A = 18mL= 0.018L$

Generally the concentration of base

$C_B = \frac{C_AV_A}{C_B}$

Substituting value

$C_B = \frac{0.1 * 0.01800}{0.0260}$

$C_B= 0.0692M$

When 13mL of the base is added a buffer is formed

The chemical equation of the reaction is

$HA_{(aq)} + OH^-_{(aq)} --------> A^{+}_{(aq)} + H_2 O_{(l)}$

Now before the reaction the number of mole of base is

$No \ of \ moles[N_B] = C_B * V_B$

Substituting value

$N_B = 0.01300 * 0.0692$

$= 0.0009 \ moles$

Now before the reaction the number of mole of acid is

$No \ of \ moles = C_B * V_B$

Substituting value

$N_A = 0.01800 *0.1$

$= 0.001800 \ moles$

Now after the reaction the number of moles of base is zero i.e has been used up

this mathematically represented as

$N_B ' = N_B - N_B = 0$

The number of moles of acid is

$N_A ' = N_A - N_B$

$= 0.0009\ moles$

The pH of this reaction can be mathematically represented as

$pH = pK_a + log \frac{[base]}{[acid]}$

Substituting values

$7.82 = pK_a +log \frac{0.0009}{0.0009}$

$pK_a =7.82$

8 0

2 years ago

Which subatomic particle gives off visible light when it drops back down to a lower energy state?

icang [17]

Answer:

Option C = electron

Explanation:

Electrons are responsible for the production of colored light.

Electron:

The electron is subatomic particle that revolve around outside the nucleus and has negligible mass. It has a negative charge.

Symbol= e-

Mass= 9.10938356×10⁻³¹ Kg

It was discovered by j. j. Thomson in 1897 during the study of cathode ray properties.

How electrons produce the colored light:

Excitation:

When the energy is provided to the atom the electrons by absorbing the energy jump to the higher energy levels. This process is called excitation. The amount of energy absorbed by the electron is exactly equal to the energy difference of orbits.

De-excitation:

When the excited electron fall back to the lower energy levels the energy is released in the form of radiations. this energy is exactly equal to the energy difference between the orbits. The characteristics bright colors are due to the these emitted radiations. These emitted radiations can be seen if they are fall in the visible region of spectrum.

Other process may involve,

Fluorescence:

In fluorescence the energy is absorbed by the electron having shorter wavelength and high energy usually of U.V region. The process of absorbing the light occur in a very short period of time i.e. 10 ∧-15 sec. During the fluorescence the spin of electron not changed.

The electron is then de-excited by emitting the light in visible and IR region. This process of de-excitation occur in a time period of 10∧-9 sec.

Phosphorescence:

In phosphorescence the electron also goes to the excitation to the higher level by absorbing the U.V radiations. In case of Phosphorescence the transition back to the lower energy level occur very slowly and the spin pf electron also change.

5 0

2 years ago

Which properties cause chemical reactions?

GREYUIT [131]

D. Chemical and Physical Properties

3 0

3 years ago

Read 2 more answers

Calculate the de Broglie wavelength of a body of mass 1 kg moving with a velocity of

skelet666 [1.2K]

Answer:

4.42 × 10⁻³⁷ m

Explanation:

Step 1: Given and required data

Mass of the body (m): 1 kg

Velocity of the body (v): 1500 m/s

Planck's constant (h): 6.63 × 10⁻³⁴ J.s

Step 2: Calculate the de Broglie wavelenght (λ) of the body

We will use de Broglie's equation.

λ = h / m × v

λ = (6.63 × 10⁻³⁴ J.s) / 1 kg × (1500 m/s) = 4.42 × 10⁻³⁷ m

6 0

2 years ago