Which types of orbitals are found in the principal energy level n=2?

Answers

Answer 1

The kind of orbitals precisely found in the principal energy level n=2 are the one 2s orbital and three 2p orbitals

Why the n=2 orbital has four orbitals?

It follows that in the electronic configuration of elements describes the distribution of electrons of elements is shared into different energy levels through the orbitals.

In the second orbital, there are actually four different types of orbitals which are found here, these are:

The one 2s orbitalThe three 2p orbitals.

The position of elements in the periodic table is largely based on the shared electrons of atoms which are being technically distributed at various level.

An element is a substance which cannot be split into different forms by ordinary means. Electrons revolve around the orbits of atoms of elements.

So therefore, from the given explanation above, it can be deduced that the the second energy level has four different orbitals.

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Related Questions

yes or no ? experimentation with corn and other crops led to the development of new fuels called biofuels

Answers

Yes, the experimentation with corn and other crops led to the development of new fuels called biofuels.

What are biofuels?

Biofuel is a biodegradable,  inexhaustible, fuel that is produced from Biomass. Biofuel is considered the easiest available and pure fuel on the earth. Biofuels are manufactured from biomass such as wood and straw, which are liberated by direct combustion of dry matter and converted into a gaseous and liquid fuel.

The organic matter such as sludge, sewage, and vegetable oils, can be changed into biofuels by a wet process such as fermentation and digestion. Biofuel is available in all regions of the world, and mainly includes fuels such as Biodiesel, Bioethanol, and Bio methanol.

The two types of biofuels are bioethanol and biodiesel most commonly used these days. Both of these biofuels are the first generation of biofuel technology.

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Calculate the heat of reaction when 25.00 mL of 0.1102 M HCl(aq) at 25.14°C is added to 50.00 mL of 0.1024 M NaOH(aq) at the same temperature in a coffee-cup calorimeter that has a calorimeter constant of 0.001 J/°C. After mixing the temperature of the solution was observed to be 25.93°C.

Answers

The heat of reaction is obtained from the calculation as -0.65kJ/mol.

What is the heat of reaction?

We already know that the heat of the reaction is computed by the use of the information that have been presented in the question. We know that this is a 1:1 reaction thus;

Number of moles of HCl = 0.1102 M * 25/1000 = 2.755 * 10^-3 moles

Number of moles of NaOH = 0.1024 M * 50/1000 L = 5.12 * 10^-3 moles

We can see from above that the HCl is the limiting reactant.

The hat evolved is obtained from;

0.001 J/°C. * (25.93°C - 25.14°C) = 1.79 * 10^-3 J

The heat of the reaction is then;

-( 1.79 * 10^-3 J) * 10^-3 * 1/2.755 * 10^-3 moles

-0.65kJ/mol

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34.8 g of Na₂O are used to form a solution with a volume of 450.0 mL L. What is the
molarity?
34.89

Answers

Answer:

1.25 M

Explanation:

(Step 1)

Convert grams to moles using the molar mass of Na₂O.

Atomic Mass (Na): 22.990 g/mol

Atomic Mass (O): 15.999 g/mol

Molar Mass (Na₂O): 2(22.990 g/mol) + 15.999 g/mol

Molar Mass (Na₂O): 61.979 g/mol

 34.8 g Na₂O               1 mole
----------------------  x  ---------------------  =  0.561 moles Na₂O
                                    61.979 g

(Step 2)

Convert milliliters to liters.

1,000 mL = 1 L

 450.0 mL                 1 L
------------------  x  -------------------  =  0.4500 L
                             1,000 mL

(Step 3)

Calculate the molarity using the molarity ratio.

Molarity = moles / volume (L)

Molairty = 0.561 moles / 0.4500 L

Molarity = 1.25 M

how many atoms are in 6 moles of oxygen

Answers

Each oxygen molecule has two atoms

6 * 2 = 12

12

How many molecules are in 59.73 grams of the theoretical acid borofuric acid, H2B2O2?

remember units and sig figs.

If your answer is a very large number, use scientific notation with an 'E' in this format:

120,000,000 = 1.2E8

Answers

The number of molecules in 59.73 grams of the theoretical acid, borofuric acid, would be 6.44 x [tex]10^2^3[/tex] molecules.

Avogadro's number

According to Avogadro, 1 mole of any substance contains 6.022 x [tex]10^2^3[/tex] molecules.

Recall that: number of moles in a substance = mass of the substance/molar mass of the substance.

Molar mass of borofuric acid, [tex]H_2B_2O_2[/tex]:

H = 1 g/mol

B = 10.8 g/mol

O = 16 g/mol

              (1x2) + (10.8x2) + (16x2) = 55.6 g/mol

Number of moles of 59.73 grams  [tex]H_2B_2O_2[/tex] = 59.73/55.6

        = 1.07 moles

Since 1 mole = 6.022 x [tex]10^2^3[/tex] molecules

1.07 moles of borofuric acid = 1.07 x 6.022 x [tex]10^2^3[/tex] molecules.

                                            = 6.44 x [tex]10^2^3[/tex] molecules

Thus, 59.73 grams of borofuric acid contains 6.44 x [tex]10^2^3[/tex] molecules.

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At STP, exactly 1 mole of any gas occupies 22.4 L. What size container (in Liters) do you need to hold 2.1mol O₂ gas at STP?(Show all work)A) 44.2B) 47.0C) 40.0D) 55.5

Answers

Answer

B) 47.0

Explanation

If at STP, exactly 1 mole of any gas occupies 22.4 L

Then, 2.1 moles of O₂ gas at STP will occupy x L container

To get x L, cross multiply and divide both sides by 1 mole.

[tex]x\text{ }L\text{ }container=\frac{2.1\text{ }mol}{1\text{ }mol}\times22.4\text{ }L=47.0\text{ }L[/tex]

Therefore, the size of the container (in Liters) needed to hold 2.1 mol O₂ gas at STP is 47.0 L

Silver Sulfate reacts with Potassium Chloride according to the following reaction:Ag2SO4 + 2KCl -> 2AgC1 + K2SO4a. If 30.0 grams of Ag2SO4 reacts with 10.0 grams of KCl, what mass of AgCl is produced by the reaction b. the limiting reactant is ______c. how many grams ok K2SO4 can be produced.d. how many grams of excess reactant remain after the reaction e. what is the percent yield if there is 10.0g of AgCl

Answers

Answer:

Explanations:

Given the chemical reaction

[tex]Ag_2SO_4+2KCl\rightarrow2AgCl+K_2SO_4[/tex]

Given the following

Mass of Ag2SO4 = 30grams

Mass of KCl = 10grams

Determine the moles of the reactants

[tex]\begin{gathered} mole\text{ of Ag}_2SO_4=\frac{mass}{molar\text{ mass}}molar\text{ mass} \\ mole\text{ of Ag}_2SO_4=\frac{30g}{311.799} \\ mole\text{ of Ag}_2SO_4=0.0962moles \end{gathered}[/tex][tex]\begin{gathered} mole\text{ of KCl}=\frac{10g}{74.5513} \\ moleof\text{ KCl}=0.1341moles \\ 1mole\text{ of KCl}=\frac{0.1341}{2}=0.06707moles \end{gathered}[/tex]

B) B) Since the 1 moles of KCl is lower than the moles of Ag2SO4, hence KCl willl be the limiting reactant.

A) A) According to stoichiometry, 2 moles of KCl produces 2 moles of AgCl, the mass of AgCl produced will be given as;

[tex]\begin{gathered} mass=mole\times molar\text{ mass} \\ mass\text{ of AgCl}=0.1341\times143.32 \\ mass\text{ of AgCl}=19.22grams \end{gathered}[/tex]

C) According to stoichiometry, 2 moles of KCl produces 1 moles of K2SO4, the mole of K2SO4 produced is;

[tex]\begin{gathered} mole\text{ of }K_2SO_4=\frac{1}{2}\times0.1341 \\ mole\text{ of }K_2SO_4=0.06707moles \end{gathered}[/tex][tex]\begin{gathered} mass\text{ of K}_2SO_4=mole\times molar\text{ mass} \\ mass\text{ of K}_2SO_4=0.06707\times174.259 \\ mass\text{ of K}_2SO_4=11.69grams \end{gathered}[/tex]

Question 1 (3 points)
Listed in the Item Bank are some important labels for sections of the image below.
Select the label for each corresponding area it identifies in


-noble gas
-alkaline earth metals
-transition metals
-alkali metals
-nonmetal
-metalloids
-halogens

Answers

The first column represent the alkali metal, the second column represent alkaline earth metals, the 3rd horizontal middle block is d block, the right 6 columns of the periodic table represent p block and the rightest column is noble gas.

What is noble gas, transition metals and it's places?See figuring out the right place of the elements mentioned in the periodic table is easy because its a predicted and factual thing.The first column we see in the periodic table is alkali metals.The second column we see in the periodic table is alkaline earth metals.The middle horizontal slit of elements called as d block elements starting from scandium.The right six columns belongs to p block elements and the rightest column represents noble gas elements.

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PLEASE HELP ASAP, I will mark the "brainiest"
In the generic reaction: 5A + 2B --> 6C + 3D The molar mass of A is 146.70g/mol while the molar mass of C is 21.31g/mol. If 4.253 g of A reacts with an excess of B, how many grams of C can form? Remember significant figures and units. ​

Answers

The amount, in grams, of C that can form from the reaction, is 0.7414 g.

Stoichiometric problem

From the equation of the reaction, the mole ratio of A to that of C is 5:6.

In other words, for every 5 moles of A that reacts, 6 moles of C is produced.

Molar mass of A = 146.70 g/mol

Molar mass of C = 21.31 g/mol

Recall that: mole = mass/molar mass

Mole of 4.253 g of A = 4.253/146.70

                                  = 0.029 mol

From the mole ratio, the equivalent mol of C that will be produced is:

             0.029 x 6/5 = 0.03479 mol

mass = mole x molar mass

Mass of 0.03479 mol of C = 0.03479 x 21.31

                                            = 0.7414 g (to the appropriate sig. fig.)

In other words, the amount of C that will be produced from 4.253 g of A will be 0.7414 g.

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explain Maxwell equations and maxwell thermodynamics relations; it's significance and application to ideal gases?​

Answers

Answer:

In order to understand the Maxwell equations and relations derived from Maxwell relations, you will have to know Euler's Reciprocity.

 

 It states that for any state thermodynamic quantity (let x and y) having a state function phi, it must satisfy the following condition.

[tex](\frac{ {∂}^{2}\phi }{∂x \cdot∂y}) = (\frac{ {∂}^{2}\phi }{∂y \cdot∂x})[/tex]

Maxwell equations: These are a set of equations derived from the application of Euler's Reciprocity. The four Maxwell equations are as follows:

[tex]dH = TdS + VdP[/tex]

[tex]dG = VdP - SdT[/tex]

[tex]dA = -PdV - SdT[/tex]

[tex]dU = TdS - PdV[/tex]

Let's derive each Maxwell relations step by step,

1) dH = TdS + VdP

Enthalpy is a function of Entropy & Pressure

[tex] \sf \qquad \qquad H = f(S,P)[/tex]

[tex]dH = TdS + VdP[/tex]

[tex]dH = (\frac{∂H}{∂S})_{_P} dS + (\frac{∂H}{∂P})_{_S}dP[/tex]

Comparing both the above equation,

[tex]T = (\frac{∂H}{∂S})_{_P}[/tex]

[tex]V =(\frac{∂H}{∂P})_{_S}[/tex]

now we know that Enthalpy is a state function hence applying cross reciprocality,

[tex](\frac{∂V}{∂S})_{_P}= (\frac{∂T}{∂P})_{_S}[/tex]

This is called the first Maxwell relation,

Similarly,

2) dG = -SdT + VdP

Free energy is a function of Temperature & Pressure,

[tex] \sf \qquad \qquad G = f(T,P)[/tex]

[tex]dG = -SdT + VdP[/tex]

[tex]dG = (\frac{∂G}{∂P})_{_T} dP + (\frac{∂G}{∂T})_{_P}dT[/tex]

Comparing both the above equation,

[tex](\frac{∂G}{∂P})_{_T} = V [/tex]

[tex] (\frac{∂G}{∂T})_{_P}= -S[/tex]

now we know that Free energy is a state function hence applying cross reciprocality,

[tex]-(\frac{∂S}{∂T})_{_P}= (\frac{∂V}{∂P})_{_T}[/tex]

This is called the second Maxwell relation,

3) dA = -PdV - SdT

helmholtz free energy is a function of Temperature & Volume,

[tex] \sf \qquad \qquad A = f(T,V)[/tex]

[tex]dA = -PdV - SdT[/tex]

[tex]dA = (\frac{∂A}{∂V})_{_T} dV + (\frac{∂A}{∂T})_{_V}dT[/tex]

Comparing both the above equation,

[tex](\frac{∂A}{∂V})_{_T} = -P [/tex]

[tex] (\frac{∂A}{∂T})_{_V}= -S[/tex]

now we know that helmholtz free energy is a state function hence applying cross reciprocality,

[tex]-(\frac{∂S}{∂V})_{_T}= -(\frac{∂P}{∂T})_{_V}[/tex]

[tex](\frac{∂S}{∂V})_{_T}= (\frac{∂P}{∂T})_{_V}[/tex]

This is called the third Maxwell relation,

4) dU = TdS - PdV

Internal energy is a function of Entropy & Volume,

[tex] \sf \qquad \qquad A = f(S,V)[/tex]

[tex]dU = TdS - PdV [/tex]

[tex]dU = (\frac{∂U}{∂S})_{_V} dS + (\frac{∂U}{∂V})_{_S}dV[/tex]

Comparing both the above equation,

[tex](\frac{∂U}{∂V})_{_S} = -P [/tex]

[tex] (\frac{∂U}{∂S})_{_V}= T[/tex]

now we know that Internal energy is a state function hence applying cross reciprocality,

[tex](\frac{∂T}{∂V})_{_S}= -(\frac{∂P}{∂S})_{_V}[/tex]

This is called the fourth Maxwell relation,

The main significance of the Maxwell relation is that those thermodynamic quantities which are unmeasurable can be replaced with measurable quantities with the help of the Maxwell relation.

The derivative of the extensive asset in relation to the extensive asset gives the intensive asset; with respect to the intensive asset, the derivative of the extensive asset gives the extensive asset. This is the result of the overall Maxwell relations.

The coefficient of expansion and compression of a gas in thermodynamics is the application of the Maxwell relations.

There are 3 coefficients introduced,

Coefficient of thermal expansion (expansivity) α

[tex]α= \frac{1}{V} (\frac{∂V}{∂T} )_{_P}[/tex]

Coefficient of isothermal compressibility β

[tex]β = -\frac{1}{V} (\frac{∂V}{∂P} )_{_T}[/tex]

Isochoric thermal expansion coefficient γ

[tex] γ= \frac{1}{P} (\frac{∂P}{∂T} )_{_V}[/tex]

For ideal gases,

PV = nRT

For one mole ideal gas (n=1),

PV = RT

Taking derivative with respect to T at constant pressure,

[tex]V(\frac{∂P}{∂T} )_{_P}+ P(\frac{∂V}{∂T} )_{_P}= R(\frac{∂T}{∂T} )_{_P}[/tex]

At constant pressure ∂P = 0, & R.H.S = 1, hence

[tex](\frac{∂V}{∂T} )_{_P}= \frac{R}{P}[/tex]

[tex]α= \frac{1}{V} (\frac{∂V}{∂T} )_{_P}[/tex]

[tex]α= \frac{1}{V} \cdot \frac{R}{P}[/tex]

[tex]\sf Also, PV=RT\\ \frac{R}{PV} = \frac{1}{T}[/tex]

[tex]\boxed{α= \frac{1}{T}}[/tex]

Following the same procedure, by taking derivating w.r.t. pressure at constant temperature.

[tex]V(\frac{∂P}{∂P} )_{_T}+ P(\frac{∂V}{∂P} )_{_T}= R(\frac{∂T}{∂P} )_{_T}[/tex]

[tex]V+ P(\frac{∂V}{∂P} )_{_T}= 0[/tex]

[tex](\frac{∂V}{∂P} )_{_T}= \frac{-V}{P}[/tex]

Substituting the above value in,

[tex]β = -\frac{1}{V} (\frac{∂V}{∂P} )_{_T}[/tex]

[tex]β = -\frac{1}{V} \cdot \frac{-V}{P}[/tex]

[tex] \boxed{β = \frac{1}{P}}[/tex]

Repeating the same procedure again, i.e. derivative w.r.t. T at constant volume

[tex]V(\frac{∂P}{∂T} )_{_V}+ P(\frac{∂V}{∂T} )_{_V}= R(\frac{∂T}{∂T} )_{_V}[/tex]

At constant Volume ∂V = 0, and R.H.S = 1, hence overall equation becomes,

[tex]V(\frac{∂P}{∂T} )_{_V} = R \\ (\frac{∂P}{∂T} )_{_V} = \frac{R}{V}[/tex]

Substituting above value in,

[tex]γ= \frac{1}{P} (\frac{∂P}{∂T} )_{_V}[/tex]

[tex]γ= \frac{1}{P} \cdot \frac{R}{V}[/tex]

[tex] \boxed{γ= \frac{1}{T}}[/tex]

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The Maxwell equation in thermodynamics is very important and useful because the set of relations allows the scientists to change specific unknown quantities.

What are the Maxwell's thermodynamic?

The Maxwell equation in thermodynamics is very useful because this is the set of relations that allows physicists to change certain unknown quantities that are hard to measure in the real world. These quantities need to be replaced by many easily measured quantities. Maxwell's relations are a set of equations in thermodynamics that are derivable from the second derivatives and from the definitions of the thermodynamic potentials. These relations are named for the nineteenth-century physicist James Clerk Maxwell. So entropy and pressure are the natural variables of enthalpy. Maxwell relations are thermodynamic equations that establish the relations between various thermodynamic quantities in equilibrium and other fundamental quantities known as thermodynamical potentials

So we can conclude that Maxwell's thermodynamics are the set of relations allows the scientists to change unknown quantities.

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1. What is the difference between a homogenous and a heterogeneous mixture? 2. Give 3 examples of each mixture and explain its uses in our daily lives.

Answers

Answer:By combining two or more substances, a mixture is produced. A homogeneous solution tends to be identical, no matter how you sample it. Homogeneous mixtures are sources of water, saline solution, some alloys, and bitumen. Sand, oil and water, and chicken noodle soup are examples of heterogeneous mixtures.

Explanation:

6. In order to prepare 50.0 mL of 0.100 M NaCl you will add _____ grams of NaCl to _____ mL of wate

Answers

The first step to solve this problem is to find the number of moles of NaCl in 50.0mL of 0.100M NaCl. To do it, convert the volume of solution to liters and multiply it by the concentration of the solution (M=mol/L).

[tex]50mL\cdot\frac{1L}{1000mL}\cdot\frac{0.100mol}{L}=0.005mol[/tex]

Now, use the molar mass of NaCl to find the mass of 0.005moles of it (MM=58.44g/mol):

[tex]0.005mol\cdot\frac{58.44g}{mol}=0.2922g[/tex]

It means that you have to add 0.2922g to 50mL of water.

Please help me it’s due today!!!!

Answers

The following are the examples of a chemical property of a substance;

1) A banana reacts with orange and turns brown

2) Iron will rust when exposed to oxygen and water

3) Iron will react with acid to form iron chloride.

What is a chemical reaction?

The term chemical reaction has to do with the change in the properties of substances that are combined together in order to yield a product. Recall that the properties of the reactant change as the product is formed.

When a chemical reaction occurs

1: A new color may appear

2) A new gas may be seen

3) The temperature may be changed

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A marine biologist is preparing a deep-sea submersible for a dive. The sub stores breathing air under high pressure in aspherical air tank that measures 73.0 cm wide.The biologist estimates she will need 8200. L of air for the dive. Calculate the pressure to which this volume of air must becompressed in order to fit into the air tank. Write your answer in atmospheres. Round your answer to 3 significant digits.0atm0.0XS ?EoloPEBH

Answers

To solve the problem we will assume the following:

1. Air behaves as an ideal gas during all the process.

2. The initial air equivalent to 8200L is at atmospheric pressure. It means 1 atm.

3. The temperature remains constant.

Taking into account the above, we can apply the Boyle-Marriote Law that relates the change in pressure and volume at constant temperature. The equation that we will use will be:

[tex]P_1V_1=P_2V_2[/tex]

Where,

P1 is the atmospheric pressure. 1atm

V1 is the initial volume of air required, 8200L

P2 is the final pressure we want to find

V2 is the final volume, it means the volume of the spherical air tank. We will calculate this volume using the volume equation for a sphere:

[tex]V_2=\frac{4}{3}\pi r^3[/tex]

r is the radius of the sphere, r=73cm/2=36.5cm

So, the volume of the spherical air tank will be:

[tex]\begin{gathered} V_2=\frac{4}{3}\pi\times(36.5cm)^3=20.4\times10^4cm^3 \\ V_2=20.4\times10^4cm^3\times\frac{1L}{1000cm^3}=204L \end{gathered}[/tex]

No, we clear P2 from the first equation and replace known data:

[tex]\begin{gathered} P_2=\frac{P_1V_1}{V_2} \\ P_2=\frac{1atm\times8200L}{204L} \\ P_2=40.3atm \end{gathered}[/tex]

The pressure of the gas must be 40.3 atm

Answer: 40.3

For the reaction A (g) → 3 B (g), Kp = 67900 at 298 K. When ∆G = -14.2 kJ/mol, what is the partial pressure of A when the partial pressure of B is 2.00 atm for this reaction at 298 K.

Answers

The partial pressure is the individual pressure of a gas. The partial pressure of gas A is 1.11 × 10⁻⁴atm.

What is partial pressure?

In a mixture of gases, the partial pressure of a gas is the individual pressure of that gas in the mixture of gases.

Given the reaction; A (g) → 3 B (g)

Kp = P(B)³/PA

Where Kp  = 67900

PB = 2.00 atm

67900 = (2)³/PA

PA =  (2)³/67900

PA = 1.11 × 10⁻⁴atm

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Consider the following processes: 1. A ─→ B delta H = ─100 kJ 2. B ─→ C + D delta H = ─50 kJ 3. E ─→ D delta H = ─25 kJ What is the delta H for the process: A ─→ C + E?

Answers

The enthalpy change, ΔH of the process:  A → C + E  is -125 kJ.

What is the enthalpy change of a reaction?

The enthalpy change of a reaction is the sum of the heat changes that occur as reactants combine to form products.

The enthalpy change of the reaction is given by the following formula:

Enthalpy change = sum of energy of bonds broken - sum of energy of the bonds formedΔH = ∑H(bonds broken) - ∑H(bonds formed)

The delta H for the process A → C + E is given as follows;

1. A ─→ B ΔH = ─100 kJ

2. B ─→ C + D ΔH = ─50 kJ

3. E ─→ D delta H = ─25 kJ

-3. D → E = +25 kJ

A → C + D = 1 + 2 + (- 3)

A → C + E = (- 100 - 50 + 25) kJ

A → C + E = -125 kJ

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what is combustion? explain to me pls ​

Answers

Answer:

rapid chemical combination of a substance with oxygen, involving the production of heat and light.

Explanation:

Combustion is a chemical process in which a substance reacts rapidly with oxygen and gives off heat. The original substance is called the fuel, and the source of oxygen is called the oxidizer.

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https://www.britannica.com/science/combustion

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Explain how the following reaction demonstrates that matter is neither created or destroyed in a chemical reaction: Ca(OH)2 + 2HCI-> CaCl2 + 2H20

Answers

Answer:

In this reaction, Ca(OH)2 is a reducing agent. It reacts with hydrogen chloride to form calcium chloride and water. Therefore, the following reaction shows that matter is neither created nor destroyed in a chemical reaction: Ca(OH)2 + 2HCI -> CaCl2 + 2H20. The formation of calcium chloride and water from the hydrolysis of calcium hydroxide is not an example of matter being created or destroyed in a chemical reaction because it does not involve the breaking down of any bonds between atoms.

Explanation:

Imagine the movement of gas particles in a closed container. According to the kinetic molecular theory, which statements below are true of the gas particles? Check all that apply.

Answers

Based on the kinetic molecular theory, the statements that are true of the gas particles are as follows:

Gas particles act like tiny, solid spheres; option A.Gas particles are in constant, random motion; option BCollisions are elastic, there is no energy lost as the particle hits the sides of the container; option C.

What is the kinetic molecular theory?

The Kinetic Molecular Theory is the theory that molecules of a gas are in constant random motion colliding with one another and with the wall of their container.

These collisions between the gas molecules are perfectly elastic and are responsible for the pressure of gas molecules.

The higher the kinetic energy of the gas molecules, the higher the temperature of the gas.

The kinetic molecular theory explains the physical state of gases since gas molecules have negligible intermolecular forces.

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Complete question:

Imagine the movement of gas particles in a closed container. According to the kinetic molecular theory, which statements below are true of the gas particles? Check all that apply. Gas particles act like tiny, solid spheres. Gas particles are in constant, random motion. Gas particles at lower temperatures move faster. Collisions are elastic, there is no energy lost as the particle hits the sides of the container. Slower moving particles collide more often and with more force with the container.

The cost of electricity to deposit 10gm of Mg is Rs. 60. How much would it cost to deposit 100gm of copper from CuSO
, ? (At. wt. of Cu = 63.5)​

Answers

The cost of electricity to deposit 100g of copper from CuSO₄ will be Rs227.1

We know, m = zit

1unit current= 1kwh

Energy=Vit

m= (E. weight/F)×it

m proportional to (E. weight)×cost

m₁/m₂ = E₁/E₂ × C₁/C₂

E.weight = (Molecular weight/n-factor)

According to question,

Cu² + 2e- --- Cu

Mg² + 2e- --- Mg { n-factor of both=2}

E = (molecular weight of Mg/2) = (24÷2) = 12

E = (molecular weight of Cu/2) = (63.5) = 31.7

So, M₁/M₂ = E₁×C₁/E₂×C₂ (C=cost given)

10/100 = 12×60/31.7×C₂

C₂= Rs227.1

Therefore the cost of electricity to deposit 100g of copper from CuSO₄ will be Rs227.1

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When a bright light is shined on glow in the dark powder what is happening to the electrons.

a.The electrons are absorbing some of the light and are become excited.

b.The atoms become negatively charged ions.

c.Nothing, the electrons are not affected by light.

d.The electrons move to a lower energy state.

Answers

A they become excited
A. The electrons are absorbing some of the light and are become excited,

Show the conversions required to solve this problem and calculate the grams of Al2O3 .

Answers

This problem is an example of a gram-to-gram stoichiometry problem. You are given the mass of 
Al
Al
 in grams and you are asked to find the mass of 
Al

2
O

3
Al
2


O
3

 in grams. For questions such as this, the strategy is to convert from grams of 
Al
Al
 to moles of 
Al
Al
, then to moles of 
Al

2
O

3
Al
2


O
3

, and finally to grams of 
Al

2
O

3
Al
2


O
3

.

grams Al⟶moles Al⟶moles 
Al

2
O

3
⟶grams 
Al

2
O

3
grams
 
Al

moles
 
Al

moles
 
Al
2


O
3



grams
 
Al
2


O
3


To convert grams of 
Al
Al
 to moles of 
Al
Al
, you use the molar mass of 
Al
Al

26.98 g/mol
26.98
 g/mol
. The balanced chemical equation is used to relate the moles of 
Al
Al
 to the moles of 
Al

2
O

3
Al
2


O
3

. There are 2 moles of 
Al

2
O

3
Al
2


O
3

 formed for every 4 moles of 
Al
Al
 that react. To convert moles of 
Al

2
O

3
Al
2


O
3

 to grams of 
Al

2
O

3
Al
2


O
3

, you use the molar mass of 
Al

2
O

3
Al
2


O
3


101.96 g/mol
101.96
 g/mol
. This can be done one conversion at a time, or you can string the conversions together.

30.6 g Al×
1 mole Al
26.98 g Al

×
2 moles 
Al

2
O

3
4 moles Al

×
101.96 g 
Al

2
O

3
1 mole 
Al

2
O

3

=57.8 g 
Al

2
O

3


Hopefully the picture will show up this time.

Saturated solutions of each of the following compounds are made at 20°C. Circle the letter(s) of the solution(s), which will form a precipitate upon heating.a) NaClb) Na2SO4c) Li2CO3d) Sucrose

Answers

In this question, we need to analyze the solubility of some substances and the changes that can occur to each one upon heating, at 20°C, every substance listed is soluble, NaCl, Na2SO4, Li2CO3 and Sucrose, but when we have a change in temperature, it will affect directly the solubility of the solution, for example, if you increase the temperature, Lithium carbonate and Sodium sulfate, will have a lower solubility in water, therefore if we have a certain amount of these two substances at 20°C and in 100g of water, the solution will be soluble, but if we increase the temperature, the solubility will change and these two compounds will start to precipitate, as the solubility will be lowering down.

Therefore the answers are Na2SO4 and Li2CO3

How many molecules are in 59.73 grams of the theoretical acid borofuric acid, H2B2O2?

remember units and sig figs.

Answers

The number of molecules in 59.73 grams of the theoretical acid borofuric acid, H₂B₂O₂ is 6.47 × 10²³ molecules.

How to calculate number of molecules?

The number of molecules in a substance can be estimated by multiplying the number of moles in the substance by Avogadro's number (6.02 × 10²³) as follows:

no of molecules = no of moles × 6.02 × 10²³

According to this question, there are 59.73 grams of the theoretical acid borofuric acid. The molar mass of this acid is as follows:

H₂B₂O₂ = 1(2) + 10.8(2) + 16(2) = 55.6g/mol

moles = 59.73g ÷ 55.6g/mol = 1.07mol

no of molecules = 1.07 × 6.02 × 10²³

no of molecules = 6.47 × 10²³ molecules

Therefore, 6.47 × 10²³ molecules is the number of molecules in the acid.

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The Ka of a monoprotic weak acid is 0.00469. What is the percent ionization of a 0.141 M solution of this acid?

Use quadratic equation.

Answers

The percent ionization of an acidic solution can be calculated from the H+ concentration. the percent ionization of the monoprotic acid of 0.141 M is 18.23 %.

What is percent ionization?

Percent ionization of an acidic solution is the percent of H+ ions in the solution. Thus, mathematically, it is the ratio of H+ ion concentration to the concentration of solution multiplied by 100.

Let HA be the monoprotic acid when it ionizes, forming equal concentration of H+ and A- let it be x. Thus ionization  constant can be written as follows:

Ka = [x]² /[HA]

0.00469 =[x]²/[0.141 M]

   [X] = 0.025. = [H+]

Percentage ionization  = (0.025 M / 0.141 M)× 100

                                      = 18.23 %

Therefore percentage ionization of the acid is 18.23%.

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Can someone help me with this? I’m not understanding what it’s asking of me

Answers

Answer: You need to research what ever you are working on and find 2 examples.

Explanation:

The partial pressure of oxygen was observed To be 156 torr

Answers

When the partial pressure of oxygen is 156 torr and the atmospheric pressure is 743 torr, the mole fraction of oxygen is 0.210.

What is partial pressure?

Partial pressure is the pressure exerted by an individual gas in a mixture of gases. The partial pressure of a gas depends on its mole fraction.

The relationship between the partial pressure of a gas and the total pressure is given by Dalton's law, which states that the sum of the partial pressures is equal to the total pressure.

We can calculate the partial pressure of oxygen using the mathematical expression of Dalton's law:

pO₂ = P × X(O₂)

X(O₂) = pO₂ / P

X(O₂) = 156 torr / 743 torr = 0.210

where,

pO₂ is the partial pressure of oxygen.P is the total pressure of the mixture.X(O₂) is the mole fraction of oxygen.

The mole fraction of oxygen is 0.210 when its partial pressure is 156 torr and the atmospheric pressure is 743 torr.

The complete question is:

The partial pressure of oxygen was observed To be 156 torr in air with a total atmospheric pressure of 743 torr. Calculate the mole fraction of oxygen present.

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a gaseous product of a chemical reaction is collected at 285k and 1.3atm. what was the molar mass of the gas in grams per mole if 6.2 g of gas occupies 5.4l

Answers

The molar mass of the gas in grams per mole if 6.2 g of gas occupies 5.4L is 336.97 grams.

What is molar mass?

Molar mass is defined as the mass of a sample of a certain chemical divided by the quantity of the material, expressed as the number of moles in the sample.

It can also be defined as the product of the mass of a specific substance and the amount of that substance in the sample.

Given Pressure = 1.3 atm

Temperature = 285 K

Volume = 5.4 L

Gas content = 8.3

So, PV = nRT

n = RT / PV

n = 8.3 x 285 / 1.3 x 5.4

n = 336.97 grams

Thus, the molar mass of the gas in grams per mole if 6.2 g of gas occupies 5.4L is 336.97 grams.

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The volume of a gas is 480 mL at 45.0 oC the temperature is increased to 60.0 oC at a constant pressure what is the new volume

Answers

Answer

The new volume = 0.503 L or 503 mL

Explanation

Given:

Initial volume, V₁ = 480 mL = (480/1000) = 0.480 L

Initial temperature, T₁ = 45.0°C = (45 + 273 K) = 318 K

Final temperature, T₂ = 60.0°C = (60 + 273 K) = 333 K

What to find:

The new volume, V₂ when the temperature increases to 60.0°C.

Step-by-step solution:

Since the pressure is constant, the new volume, V₂ can be calculated using Charle's law formula.

[tex]\begin{gathered} \frac{V_1}{T_1}=\frac{V_2}{T_2} \\ \\ \Rightarrow V_2=\frac{V_1T_2}{T_1} \end{gathered}[/tex]

Plugging the values of the given parameters into the formula, we have

[tex]V_2=\frac{0.480L\times333K}{318K}=0.503\text{ }L[/tex]

The new volume, V₂ when the temperature increased to 60.0°C = 0.503 L or (0.503 x 1000 mL) = 503 mL

An atom with 3 protons in the nucleus and 3 electrons in the orbitals would have what overall charge?

Answers

Ans6

wer:

b

Explanation:

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