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2 years ago

How would the 2s and 3p orbitals differ from the 1s and 2p orbitals?.

Chemistry

1 answer:

Gekata [30.6K]2 years ago

8 0

Answer:

The 2s and 3p orbitals would have more nodes than 1s and 2p orbitals.

p orbitals would have same shape, but 3p would be larger

s orbitals would have same shape, but 2s would be larger

Explanation:

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How many sig figs 18.01528

Ulleksa [173]

Answer:7

Explanation:

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4 years ago

Scientist can use the of minerals to tell them apart A.sound B.length C.size D.color

ipn [44]

Answer:

color

Explanation:

Scientists can use the color of minerals to tell them apart.

7 0

4 years ago

ammonia gas is used as refrigerant 0.474 atm. Pressure is required to change 2000 cm3 sample of ammonia initially at 1.0 atm to

Andreas93 [3]

Answer: The pressure required is 0.474 atm

Explanation:

Boyle's Law: This law states that pressure is inversely proportional to the volume of the gas at constant temperature and number of moles.

$P\propto \frac{1}V}$ (At constant temperature and number of moles)

The equation is,

${P_1V_1}={P_2V_2}$

where,

$P_1$ = initial pressure of gas = 1.0 atm

$P_2$ = final pressure of gas = ?

$V_1$ = initial volume of gas = $2000cm^3$

$V_2$ = final volume of gas = $4.22dm^3=4220cm^3$ ( $1dm^3=1000cm^3)$

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

${1.0\times 2000}={P_2\times 4200}$

$P_2=0.474atm$

The pressure required is 0.474 atm

5 0

3 years ago

A mysterious white powder could be powdered sugar (C12H22O11), cocaine (C17H21NO4), codeine (C18H21NO3), norfenefrine (C8H11NO2)

rodikova [14]

Norfenefrine (C₈H₁₁NO₂).

<h3>Further explanation</h3>

We will solve a case related to one of the colligative properties, namely freezing point depression.

The freezing point of the solution is the temperature at which the solution begins to freeze. The difference between the freezing point of the solvent and the freezing point of the solution is called freezing point depression.

$\boxed{ \ \Delta T_f = T_f(solvent) - T_f(solution) \ } \rightarrow \boxed{ \ \Delta T_f = K_f \times molality \ }$

<u>Given:</u>

A mysterious white powder could be,

powdered sugar (C₁₂H₂₂O₁₁) with a molar mass of 342.30 g/moles,
cocaine (C₁₇H₂₁NO₄) with a molar mass of 303.35 g/moles,
codeine (C₁₈H₂₁NO₃) with a molar mass of 299.36 g/moles,
norfenefrine (C₈H₁₁NO₂) with a molar mass of 153.18 g/moles, or
fructose (C₆H₁₂O₆) with a molar mass of 180.16 g/moles.

When 82 mg of the powder is dissolved in 1.50 mL of ethanol (density = 0.789 g/cm³, normal freezing point −114.6°C, Kf = 1.99°C/m), the freezing point is lowered to −115.5°C.

<u>Question: </u>What is the identity of the white powder?

<u>The Process:</u>

Let us identify the solute, the solvent, initial, and final temperatures.

The solute = the powder
The solvent = ethanol
The freezing point of the solvent = −114.6°C
The freezing point of the solution = −115.5°C

Prepare masses of solutes and solvents.

Mass of solute = 82 mg = 0.082 g
Mass of solvent = density x volume, i.e., $\boxed{ \ 0.789 \ \frac{g}{cm^3} \times 1.50 \ cm^3 = 1.1835 \ g = 0.00118 \ kg \ }$

We must prepare the solvent mass unit in kg because the unit of molality is the mole of the solute divided by the mass of the solvent in kg.

The molality formula is as follows:

$\boxed{ \ m = \frac{moles \ of \ solute}{kg \ of \ solvent} \ } \rightarrow \boxed{ \ m = \frac{mass \ of \ solute \ (g)}{molar \ mass \ of \ solute \times kg \ of \ solvent} \ }$

Now we combine it with the formula of freezing point depression.

$\boxed{ \ \Delta T_f = K_f \times \frac{mass \ of \ solute \ (g)}{molar \ mass \ of \ solute \times kg \ of \ solvent} \ }$

It is clear that we will determine the molar mass of the solute (denoted by Mr).

We enter all data into the formula.

$\boxed{ \ -114.6^0C - (-115.5^0C) = 1.99 \frac{^0C}{m} \times \frac{0.082 \ g}{Mr \times 0.00118 \ kg} \ }$

$\boxed{ \ 0.9 = \frac{1.99 \times 0.082}{Mr \times 0.00118} \ }$

$\boxed{ \ Mr = \frac{0.16318}{0.9 \times 0.00118} \ }$

We get $\boxed{ \ Mr = 153.65 \ }$

These results are very close to the molar mass of norfenefrine which is 153.18 g/mol. Thus the white powder is norfenefrine.

<h3>Learn more</h3>

The molality and mole fraction of water brainly.com/question/10861444
About the mass and density of ethylene glycol as an antifreeze brainly.com/question/4053884
About the solution as a homogeneous mixture brainly.com/question/637791

Keywords: a mysterious white powder, sugar, cocaine, codeine, norfenefrine, fructose, the solute, the solvent, dissolved, ethanol, normal freezing point, the freezing point depression, the identity

7 0

3 years ago

Read 2 more answers

Which statements are true?

schepotkina [342]

Explanation:

Evaporation is defined as a process in which liquid state of water is changing into vapor state.

So, we need to break the bonds of liquid substance in order to convert it into vapor state. And, energy is absorbed for breaking of bonds which means that evaporation is an endothermic process.

Hence, the statement evaporation of water is an exothermic process is false.

When a hydrocarbon reacts with oxygen and leads to the formation of carbon dioxide and water then this type of reaction is known as combustion reaction.

A combustion reaction will always release heat energy. Hence, combustion reaction is exothermic in nature.

When energy is transferred as heat from the surroundings to the system then it means energy is being absorbed by the system. And, absorption of heat is an endothermic process for which $\Delta H$ is positive.

Whereas when energy is transferred from system to the surrounding then it means energy is released by the system which is an exothermic process.

Hence, for an exothermic process value of $\Delta H$ is negative.

Thus, we can conclude that statements which are true are as follows.

A combustion reaction is exothermic.

When energy is transferred as heat from the system to the surroundings, $\Delta H$ is negative.

For an endothermic reaction Deta H is positive.

5 0

3 years ago