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2 years ago

7

Based on the Lewis/electron dot representation of the two atoms, predict the ratio of metal cationic (+) atom to nonmetal anioni

c (-) atom in the compound. •Al. N A. 1:1 B. 1:2 C. 2:1 D. 3:1

1 answer:

Aleks [24]2 years ago

8 0

The answer is 1:1

I got it right

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What physical meaning is attributed to ψ², the square of the wave function?

Zigmanuir [339]

Answer:

The probability density (ψ2)

Explanation:

Indicates the probability of finding the electron in a certain region of space when it is squared ψ2.

This means that define2 defines the distribution of electronic density around the nucleus in three-dimensional space; a high density represents a high probability of locating the electron and vice versa.

The atomic orbital, can be considered as the electron wave function of an atom.

APPLICATIONS:

1.- Specify the possible energy states that the electron of the hydrogen atom can occupy and identify the corresponding wave functions medio, by means of a set of quantum numbers, with which an understandable model of the hydrogen atom can be constructed.

2.- It does not work for atoms that have more than one electron, but the problem is solved using approximation methods for polyelectronic atoms.

3 0

3 years ago

Atmospheric pressure at sea level is equal to a column of 760 mm Hg. Oxygen makes up 21 percent of the atmosphere by volume. The

Neporo4naja [7]

<u>Answer:</u> The partial pressure of oxygen is 160 mmHg

<u>Explanation:</u>

We are given:

Percent of oxygen in air = 21 %

Mole fraction of oxygen in air = $\frac{21}{100}=0.21$

To calculate the partial pressure of oxygen, we use the equation given by Raoult's law, which is:

$p_{O_2}=p_T\times \chi_{O_2}$

where,

$p_{O_2}$ = partial pressure of oxygen = ?

$p_T$ = total pressure of air = 760 mmHg

$\chi_{O_2}$ = mole fraction of oxygen = 0.21

Putting values in above equation, we get:

$p_{O_2}=760mmHg\times 0.21\\\\p_{O_2}=160mmHg$

Hence, the partial pressure of oxygen is 160 mmHg

8 0

3 years ago

One part nitrogen gas combines with one part oxygen gas to form how many part(s) dinitrogen monoxide (nitric oxide)?

____ [38]

Answer : The one part nitrogen gas combines with one part oxygen gas to form one part of dinitrogen monoxide.

Explanation :

Balanced chemical reaction : It is defined as the reaction in which the number of atoms of individual elements present on reactant side must be equal to the product side.

When nitrogen gas combines with oxygen gas then it react to give dinitrogen monoxide or nitrous oxide.

The balance chemical reaction will be:

$2N_2(g)+O_2(g)\rightarrow 2N_2O(g)$

By the stoichiometry we can say that, 2 parts of nitrogen gas combines with 1 part of oxygen gas to give 2 parts of dinitrogen monoxide or nitrous oxide.

First we have to determine the limiting reagent.

From the reaction we conclude that,

As, 2 moles of nitrogen gas combine with 1 mole of oxygen gas

So, 1 moles of nitrogen gas combine with 0.5 mole of oxygen gas

It means that, oxygen gas is an excess reagent because the given moles are greater than the required moles and nitrogen gas is a limiting reagent and it limits the formation of product.

Now we have to determine the moles of dinitrogen monoxide.

As, 2 moles of nitrogen gas combine to give 2 mole of dinitrogen monoxide

So, 1 mole of nitrogen gas combine to give 1 mole of dinitrogen monoxide

Thus, the one part nitrogen gas combines with one part oxygen gas to form one part of dinitrogen monoxide.

7 0

3 years ago

When 1.104 grams of iron metal are mixed with 26.023 grams of hydrochloric acid in a coffee cup calorimeter, the temperature ris

Alika [10]

Answer:

a) An exothermic reaction, will release heat. No heat will be absorbed.

b) 903.71 J of heat released

c) reaction is exothermic and ∆H will be negative.

d) ΔHreaction = 818.6 J/g

e) ΔHreaction = 45.6 kJ/mol

f) ΔHreaction = 91.2 kJ

Explanation:

<u>Step 1:</u> Data given

Mass of iron = 1.104 grams

Mass of hydrochloric acid = 26.023 grams

Initial temperature = 25.2°C

Final temperature = 33.5 °C

Temperature change = 8.3 °C

<u>Step 2:</u> The balanced equation

2Fe(s)+6HCl(aq) → 2FeCl3 (aq)+3H2 (g)

(A) Determine the amount of heat (in J) absorbed by the reaction mixture.

Since we have a rise of temperature, this means the reaction is exothermic.

An exothermic reaction, will release heat. No heat will be absorbed.

(B) How much heat (in J) was released by the reaction that occurred?

q = mC∆T

with q = heat released (in J)

with m = the mass (in grams)

with c = the specific heat capacity (in J/g°C)

with ∆T = The change in temperature (in °C)

q = (26.023g)*(4.184 J/g°C)*(8.3 °C) = 903.71 J of heat released

(C) Is this reaction exothermic or endothermic? Is ΔHreaction positive or negative?

Since we have a rise of temperature, this means the reaction is exothermic.

There is heat released so ∆H will be negative.

(D) Under constant pressure conditions (as used in this experiment), the heat released by the reaction equals the reaction enthalpy, qreleased = ΔHreaction. Determine ΔHreaction in Joules per gram of metal used (J/g).

ΔHreaction = 903.71 J/1.104 g = 818.6 J/g

(E) Determine ΔHreaction in kilojoules per mole of metal used (kJ/mol)

Number of moles of iron =1.104 grams / 55.845 g/mol = 0.0198 moles

ΔHreaction = 903.71 J / 0.0198 moles = 45641.9 J/mol = 45.6 kJ/mol

(F) Determine ΔHreaction in kilojoules per mole for the balanced reaction equation provided

Since we have 2moles of Fe in the balanced reaction;

ΔHreaction = 45.6 kJ/mol * 2 mol = 91.2 kJ

7 0

3 years ago

The vapor pressure of ethanol is 30°C at 98.5 mmHg and the heat of vaporization is 39.3 kJ/mol. Determine the normal boiling poi

Gelneren [198K]

Answer : The normal boiling point of ethanol will be, $348.67K$ or $75.67^oC$

Explanation :

The Clausius- Clapeyron equation is :

$\ln (\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}\times (\frac{1}{T_1}-\frac{1}{T_2})$

where,

$P_1$ = vapor pressure of ethanol at $30^oC$ = 98.5 mmHg

$P_2$ = vapor pressure of ethanol at normal boiling point = 1 atm = 760 mmHg

$T_1$ = temperature of ethanol = $30^oC=273+30=303K$

$T_2$ = normal boiling point of ethanol = ?

$\Delta H_{vap}$ = heat of vaporization = 39.3 kJ/mole = 39300 J/mole

R = universal constant = 8.314 J/K.mole

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

$\ln (\frac{760mmHg}{98.5mmHg})=\frac{39300J/mole}{8.314J/K.mole}\times (\frac{1}{303K}-\frac{1}{T_2})$

$T_2=348.67K=348.67-273=75.67^oC$

Hence, the normal boiling point of ethanol will be, $348.67K$ or $75.67^oC$

3 0

3 years ago