PLEASE HELP ASAP.


What conditions are being experienced
by an observer in Florida when the moon
is at Position 2?

PLEASE HELP ASAP. What Conditions Are Being Experiencedby An Observer In Florida When The Moonis At Position

Answers

Answer 1

Answer:

the conditions are that when the moon is at point 2 is that the moon is covering Florida so that means that part of the earth is dark whilst others is light

Explanation:

Answer 2
At position 2 the moon is covering florida.. so half of the earth is covered by sun and while the moon is covering florida it means the observer is experiencing dark/night

Related Questions

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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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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Calcium nitrate and potassium fluoride solutions react to form a precipitate. Classify this reaction.

Answers

When calcium nitrate and potassium fluoride solutions react to form a precipitate, the type of reaction involved is the process is double replacement

Calcium nitrate and potassium fluoride solutions react to form calcium fluoride and potassium nitrate. Calcium fluoride would precipitate out whereas potassium nitrate would be in aqueous form.

Ca (NO₃)₂ (aq) + 2 KF (aq) → CaF₂ (s) + 2 KNO₃ (aq)

In double replacement reaction, the ionic compounds would exchange their respective ions to form a new compound. Here two ionic compounds, nitrate and fluoride from calcium and potassium respectively are exchanged to form calcium fluoride and potassium nitrate.

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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

Determine the numerical ages of rock samples that contain a parent isotope with a half-life of 100 million
years and have the following percentages of original parent isotope:
50%: Age =
25%: Age =
6%: Age =

Answers

50% = 100 million years
25% = 200 million years
6% = 400 million years
I’m not really sure if this is right

Calculate the molar mass of gold (III) hydroxide, Au(OH)3

Answers

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.

3. The density of C₂H4 (OH)₂ is 1.09 g/me. How Many gram of C₂H4 (OH)2 Should be Mixed With 375 ml of Water to make a 7.50% by Mixture?​

Answers

33.136 grams of ethylene glycol should be mixed with  375 ml of Water to make a 7.50% by Mixture.

Density is the measure of how much “stuff” is in a given amount of space.

DENSITY = MASS / VOLUME

We have :-

→ Density of ethylene glycol = 1.09 g/mL

→ Volume of water = 375 mL

→ Concentration (v/v %) = 7.50 %

Let the volume of ethylene glycol (solute) be 'x' mL .

So, volume of the solution = Volume of solute + Volume of solvent

                                              = (x + 375) mL

Concentration (v/v %) = Vol. of solute/Vol. of solution × 100

⇒ x/(x + 375) × 100 = 7.5

⇒ 100x = 7.5(x + 375)

⇒ 100x = 7.5x + 2812.5

⇒ 100x - 7.5x = 2812.5

⇒ 92.5x = 2812.5

⇒ x = 2812.5/92.5

⇒ x = 30.4 mL

Mass of ethylene glycol = Volume × Density

                                          = 30.4 × 1.09 = 33.136 g

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Calculate the following round to the proper number of significant figures and write in standard scientific notation.2.51x104-1.5x103

Answers

Answer ans explanation

2.51x10^4 - 1.5x10^3 = 2.36^4

Answer

2.36^4

Why are Roman numerals needed in the names of ionic compounds containing a metal that forms more than one type of ion? Type one contains a metal with a invariant charge-One that does not vary from one compound to another. Type two contains a metal with a charge that can differ in different compounds. Match the items in the left column to the appropriate blanks in the sentences on the right

Answers

Cation

Type 1

Type 2

Anion

What are Type 1 and Type 2 compounds?

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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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 metric prefix m would be presented as 10 to the power of:

Answers

Answer:

[tex]-3[/tex]

Explanation:

Here, we want to get the metric prefix m value

This means we want to get power to which it would be raised

Mathematically,we have this as the milli

The milli refers to thousandth

From what we have here, this is the power of -3

So the prefix m represents :

[tex]10^{-3}[/tex]

Which of the following substances dissolves most readily in water?a. CHb. NH3c. BaSO4d. CaCO3

Answers

Answer:

BaSO₄. Option C is correct

Explanations

The substances that dissolve readily in water are ionic compounds and polar covalent compounds. Examples of ionic compounds that dissolve in water are salts, oxides, hydroxides, sulfides, and the majority of inorganic compounds.

The molecules in a polar solvent have a dipole, like water, one side is more negative and one is more positive.

Ionic compounds are composed of a positive ion, normally a metal, and a negative ion, normally a nonmetal, so their forces are attracted to their charge difference.

Thus, a polar solvent dissolves each ion with its corresponding parts, dissociating the two ions of the ionic compound.

Since sulfides are ionic compounds hence the substance that will dissolve most readily in water is BaSO₄. The molecule is formed by one barium cation Ba2+ and one sulfide anion S2-. The two ions are bound through an ionic bond.

What mass of glycerin (C3H8O3) must be dissolved in 169.8 g water to give a solution with a freezing point of -3.81°C? Kf for water = 1.86°C*kg/mol.

Answers

2.67 grams of Glycerin should be added to water of mass 169.8 grams to give a solution with freezing point -3.81°.

When glycerin will be added to water, its freezing point will decrease. This phenomena is given a name "Depression in Freezing Point". This is called a colligative property.

We can find the depression in freezing point by using the formula,

ΔT = i.m.K

Where,

i is the Vant Hoff's factor.

Kf is cryoscopic constant

m is the molality of the solution,

Molality m can be defined as,

m = moles of solute/mass of solvent(in KG)

First, let s find molality of solution,

m = Moles of Glycerin/mass of water

m = (Added mass of glycerin(W)/Molecular mass of glycerin)/mass of water.

Molecular weight of glycerin = 92 g/mole

Mass of water = 169.8 g

In kilograms,

mass of water = 0.1698 Kg.

Now,

m = W/92 x 0.1698

Now, putting all the values in the formula,

ΔT = i.m.Kf

Assuming 100% disassociation, we can take i = 1,

But first, ΔT = 0-(-3.81) °C

ΔT  = 3.81  °C.

So, now we can write,

3.81 = 1.86 x W/(92 x 0.1698)

W = 3.81 x 92 x .01698 / 1.86

W = 2.67 grams.

Hence, of is required to add 2.67 grams of glycerin in water.

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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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how many atoms are in 6 moles of oxygen

Answers

Each oxygen molecule has two atoms

6 * 2 = 12

12

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

Answers

Ans6

wer:

b

Explanation:

Sulfur hexafluoride gas is collected at -4.0 °C in an evacuated flask with a measured volume of 5.0 L. When all the gas has been collected, the pressure in the
flask is measured to be 0.220 atm.
Calculate the mass and number of moles of sulfur hexafluoride gas that were collected.

answer needs has the correct number of significant digits

Answers

Mole of sulfur hexafluoride gas that were collected is 20.081 mol

Mass of  sulfur hexafluoride gas that were collected is 0.137 gram

Sulphur hexafluoride gas is used as the electrical insulating material in circuit and breaker, cables and capacitor and sulfur hexafluoride gas is the nontoxic gas and it is an inorganic compound it is colorless and odorless and non flammable

Here given data is

Pressure = 0.220 atm.

Temprature = -4.0 °C = -4.0 °C +273 .15 K = 269.15 K

Volume =  5.0 L

Using ideal gas quation

PV = nRT

P is the pressure

V is the volume

n is the number of moles

T is the temperature  

R = gas constant =  0.0821 L.atm/K.mol

Applying the equation as

0.220 atm ×  5.0 L = n × 0.0821 L.atm/K.mol ×  269.15 K

n = 22.09/1.1

n = 20.081

Molar mass of sulfur hexafluoride gas =  146.06 g/mol

The formula for calculations of mole

Moles = mass taken/molar mass

20.081 = mass/146.06 g/mol

Mass = 0.137 gram

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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.


What is the IUPAC name for the compound shown?

Answers

The IUPAC name for the compound shown is 3-ethyl-2,2-dimethylpentane.

International Union of Pure and Applied Chemistry is referred to as IUPAC. The terminology for naming organic compounds has been provided by IUPAC. The root name, prefix, and suffix are the three components that make up an IUPAC name.

There are five carbon atoms in the longest chain. Consequently, pent is the structure's root name. Choose the longest chain where the substituents are represented by the fewest numbers.

On the longest chain, three substituents are present. It consists of one ethyl group and two methyl groups. One ethyl group and two methyl groups are substituted at C-2 and C-3, respectively. Therefore, 3-ethyl-2,2-dimethyl will be the prefix. Alkane makes up the functional group. Therefore, the suffix is ane.

This ends up naming the compound as 3-ethyl-2,2-dimethylpentane.

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15.00g of hydrated zinc sulfate loses 6.579 H2O during heating what is the formula for the hydrate

Answers

15.00g of hydrated zinc sulfate loses 6.579 H₂O during heating what is the formula for the hydrate is ZnSO₄.7H₂O

Mass of anhydrous ZnSO₄ = 15.00 g - 6.579 g = 8.421 g

8.412 g of anhydrous ZnSO₄ = 6.579 g

molar mass of anhydrous ZnSO₄ = 161.47 g/mol

161.47 g of  ZnSO₄  = (6.579 × 161.47) / 8.421

                                 = 126 g

no. of moles of H₂O = mass / molar mass

                                 = 126 / 18

                                 = 7 molecules of H₂O

the formula for hydrate is ZnSO₄.7H₂O

Thus, 15.00g of hydrated zinc sulfate loses 6.579 H₂O during heating what is the formula for the hydrate is ZnSO₄.7H₂O

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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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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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The element tellurium would be expected to form covalent bond(s) in order to obey the octet rule.

Answers

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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For a species to survive it must be within its ________ for all ___________ factors.

Answers

For a species to survive it must be within its geographical zone for all the genetic factors.

What is a species and how it can be within its geographical area?A species is a group of organisms with most number of characters in common.A species is the smallest unit of taxanomic heirarchy , in which one finds the highest number of characters in common.We can predict there are more than thousand of species living on earth.One species depends on another species either for food or survival also .For all the genetic factors that can be taken into account like the character , and all species needs to live in its geographical area only , if crossed the geographical zone then can be killed.

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Will Argon, Neon, and Krypton react the same or differently? Explain. (make it clear and simple. thank u)

-need help asap. thank u so much :)

Answers

Will Argon, Neon, and Krypton react same

Reactivity is the relative capacity of an atom or molecule or radical to undergo a chemical reaction with another atom or molecule or compound called as reactivity

In the noble gases only helium and neon are inert and the other noble gases will react with a limited scale under very specific conditions and krypton will form solid with fluorine and xenon will form a variety of compounds with oxygen and fluorine and the name comes from the fact that these elements are virtually unreactive towards other elements and they do not react with other element

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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

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]

[tex]\sf \small \pink{Thanks }\: \green{for} \: \blue{joining} \: \orange{brainly } \: \red{community}![/tex]

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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8. How many oxygen atoms are in 25 g of oxygen?
1.9 × 1024 atoms
2.4 × 1026 atoms
9.4 × 1023 atoms
1.5 × 1025 atoms

Answers

Answer:

9.4 × 10²³ atoms

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

9.4 × 10²³ oxygen atoms

Hope this helps you

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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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

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