Calculate the number of mol corresponding to 18.8 g Na2SO4.

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

Yes, the experimentation with corn and other crops led to the development of new fuels called 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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As per the octet rule, the element tellurium will make 2 covalent bonds to complete it's octet.

What is octet rule?

The octet rule describes an atom's propensity to favor eight electrons in its valence shell. Atoms with fewer than eight electrons in the outermost shell are more likely to interact with one another and create better stable molecules. We ignore d or f electrons while considering the octet rule. The octet rule is useful for main group elements (those not in the transition metal or inner-transition metal blocks) since it only involves the s and p electrons. An octet in these atoms corresponds to an electron configuration ending in s2p6.

According to octet rule, an element require to complete it's octet (i.e. 8 electrons in the outermost shell). So, in the case of tellurium there are 6 electrons in the outermost shell therefore, we require two more electrons to complete the octet for which the tellurium would be expected to make 2 covalent bonds.

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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

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.

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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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.

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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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:

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

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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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.

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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}

E₁ = (molecular weight of Mg/2) = (24÷2) = 12

E₂ = (molecular weight of Cu/2) = (63.5) = 31.7

So, M₁/M₂ = E₁×C₁/E₂×C₂ (C₁=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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Ans6

wer:

b

Explanation:

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

Explanation:

Cation

Type 1

Type 2

Anion

The Roman numeral represents  charge and the oxidation state of the transition metal ion.

One of the example is, iron that forms two ions, Fe2+ and Fe3+. To differentiate, Fe2+ is named iron (II) and Fe3+ is named iron (III).

Those compounds in which the cation has only one charge are Type 1 binary ionic compounds whereas compounds in which the cation can have multiple forms are Type 2 binary ionic compounds.

Type II Binary Ionic Compounds contain Transition metals with non-metal ions.

A monatomic cation derives its name from the name of the element. For example, Na+ is called sodium  whereas monatomic anion is named by taking the root of the element and then adding -ide.

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

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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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:

Answer:

Explanation:

To find the number of entities in a given substance we use the formula

N = n × L

where n is the number of moles

N is the number of entities

L is the Avogadro's constant which is

6.02 × 10²³ entities

However we weren't given the number of moles only the mass of oxygen was given. we can find the number of moles from that by using the formula

[tex]n = \frac{m}{M} \\ [/tex]

m is the mass

M is the molar mass

n is the number of moles

Molar mass of oxygen = 16 g/mol

mass in question = 25 g

We have

[tex]n = \frac{25}{16} = 1.56 \\ [/tex]

number of moles = 1.56 mol

The number of oxygen atoms is equal to

N = 1.56 × 6.02 × 10²³ = 9.3912 × 10²³

We have the final answer as

Hope this helps you

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.

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

Explanations:

Given the chemical reaction

Given the following

Mass of Ag2SO4 = 30grams

Mass of KCl = 10grams

Determine the moles of the reactants

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;

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

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 %.

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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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:

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:

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

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

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

The pressure of the gas must be 40.3 atm

Answer: 40.3

Explanation:

We have to find the molar mass of Au(OH)₃. To do that we have to look for the atomic masses of each element. Also we have to consider that one molecule of

Au(OH)₃ has 1 atom of Au, 3 atoms of O and 3 atoms H.

atomic mass of Au = 196.97 amu

atomic mass of O = 16.00 amu

atomic mass of H = 1.01 amu

molar mass of Au = 196.97 g/mol

molar mass of O = 16.00 g/mol

molar mass of H = 1.01 g/mol

molar mass of Au(OH)₃ = 1 * molar mass of Au + 3 molar mass of O + 3 molar mass of H

molar mass of Au(OH)₃ = 1 * 196.97 g/mol + 3 * 16.00 g/mol + 3 * 1.01 g/mol

molar mass of Au(OH)₃ = 248.0 g/mol

Answer: the molar mass of Au(OH)₃ is 248.0 g/mol.

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).

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

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

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

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

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,

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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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.

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