All categories

3 years ago

What means that the "p" orbital has 3 orientations?

Chemistry

1 answer:

Vlada [557]3 years ago

3 0

Answer:

The p orbital is a lobed region describing where an electron can be found, within a certain degree of probability. All p orbitals have l = 1, with three possible values for m (-1, 0, +1).

Explanation:

You might be interested in

Which solution has a higher percent ionization of the acid, a 0.10M solution of HC2H3O2(aq) or a 0.010M solution of HC2H3O2(aq)

Svetach [21]

Answer:

The solution 0.010 M has a higher percent ionization of the acid.

Explanation:

The percent ionization can be found using the following equation:

$\% I = \frac{[H_{3}O^{+}]}{[CH_{3}COOH]} \times 100$

Since we know the acid concentration in the two cases, we need to find [H₃O⁺].

By using the dissociation of acetic acid in the water we can calculate the concentration of H₃O⁺ in the two cases:

1. Case 1 (0.1 M):

CH₃COOH(aq) + H₂O(l) ⇄ CH₃COO⁻(aq) + H₃O⁺(aq) (1)

0.1 - x x x

$Ka = \frac{[CH_{3}COO^{-}][H_{3}O^{+}]}{[CH_{3}COOH]}$ (2)

Where:

Ka: is the dissociation constant of acetic acid = 1.7x10⁻⁵.

$1.7 \cdot 10^{-5} = \frac{x^{2}}{0.1 - x}$

$1.7 \cdot 10^{-5}*(0.1 - x) - x^{2} = 0$

By solving the above equation for x we have:

x = 1.29x10⁻³ M = [CH₃COO⁻] = [H₃O⁺]

Hence, the percent ionization is:

$\% I = \frac{1.29 \cdot 10^{-3} M}{0.1 M} \times 100 = 1.29 \%$

2. Case 2 (0.01 M):

The dissociation constant from reaction (1) is:

$Ka = \frac{[CH_{3}COO^{-}][H_{3}O^{+}]}{[CH_{3}COOH]}$

With [CH₃COOH] = 0.01 M

$1.7 \cdot 10^{-5} = \frac{x^{2}}{0.01 - x}$

$1.7 \cdot 10^{-5}*(0.01 - x) - x^{2} = 0$

By solving the above equation for x:

x = 4.04x10⁻⁴ M = [CH₃COO⁻] = [H₃O⁺]

Then, the percent ionization for this case is:

$\% I = \frac{4.04 \cdot 10^{-4} M}{0.01 M} \times 100 = 4.04 \%$

As we can see, the solution 0.010 M has a higher percent ionization of the acetic acid.

Therefore, the solution 0.010 M has a higher percent ionization of the acid.

I hope it helps you!

4 0

3 years ago

What is the quantity of heat (in kJ) associated with cooling 185.5 g of water from 25.60°C to ice at -10.70°C?Heat Capacity of S

Cerrena [4.2K]

Taking into account the definition of calorimetry, sensible heat and latent heat, the amount of heat required is 37.88 kJ.

<h3>Calorimetry</h3>

Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.

<h3>Sensible heat</h3>

Sensible heat is defined as the amount of heat that a body absorbs or releases without any changes in its physical state (phase change).

<h3>Latent heat</h3>

Latent heat is defined as the energy required by a quantity of substance to change state.

When this change consists of changing from a solid to a liquid phase, it is called heat of fusion and when the change occurs from a liquid to a gaseous state, it is called heat of vaporization.

<u><em>25.60 °C to 0 °C</em></u>

First of all, you should know that the freezing point of water is 0°C. That is, at 0°C, water freezes and turns into ice.

So, you must lower the temperature from 25.60°C (in liquid state) to 0°C, in order to supply heat without changing state (sensible heat).

The amount of heat a body receives or transmits is determined by:

Q = c× m× ΔT

where Q is the heat exchanged by a body of mass m, made up of a specific heat substance c and where ΔT is the temperature variation.

In this case, you know:

c= Heat Capacity of Liquid= 4.184 $\frac{J}{gC}$
m= 185.5 g
ΔT= Tfinal - Tinitial= 0 °C - 25.60 °C= - 25.6 °C

Replacing:

Q1= 4.184 $\frac{J}{gC}$ × 185.5 g× (- 25.6 °C)

Solving:

<u><em>Change of state</em></u>

The heat Q that is necessary to provide for a mass m of a certain substance to change phase is equal to

Q = m×L

where L is called the latent heat of the substance and depends on the type of phase change.

In this case, you know:

n= 185.5 grams× $\frac{1mol}{18 grams}$ = 10.30 moles, where 18 $\frac{g}{mol}$ is the molar mass of water, that is, the amount of mass that a substance contains in one mole.