What is an extensive property that can be calculated?

Answers

Answer 1
Answer:

Answer: The property which depends on the quantity of the substance is called an extensive property. The free energy change for a reaction (Δ G) depends on the quantity of the substance and is therefore an extensive property. It shows the additive nature. The extensive property Δ G is easily calculated from the formula, ΔG = -nFE cell.

Explanation:

Answer 2
Answer:

Final answer:

An extensive property is one that changes when the size of the sample changes. One such property that can be calculated is enthalpy. Enthalpy can be calculated using the formula H = E + PV.

Explanation:

An extensive property is a property that changes when the size of the sample changes. Examples include mass, volume, length, and total charge. One extensive property that can be calculated is enthalpy.

The enthalpy of a system can be calculated using the formula H = E + PV, where H represents the enthalpy, E the internal energy of the system, P the pressure, and V the volume. Like other extensive properties, the enthalpy of a system would change with the quantity or size of the sample.

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Is scandium a transition metal?​

Answers

Answer:no

Explanation:

Answer:

Scandium is a transition metal

Explanation:

Aqueous hydrochloric acid HCl will react with solid sodium hydroxide NaOH to produce aqueous sodium chloride NaCl and liquid water H2O. Suppose 9.84 g of hydrochloric acid is mixed with 3.1 g of sodium hydroxide. Calculate the maximum mass of water that could be produced by the chemical reaction. Round your answer to 2 significant digits.

Answers

Answer:

1.4 g H₂O

Explanation:

In a reaction, the reactants are usually not present in exactstoichiometric amounts, that is, in the proportions indicated by the balanced equation. Frequently a large excess of one reactant is supplied to ensure that the more expensive reactant is completely converted to the desired product. Consequently, some reactant will be left over at the end of the reaction. The reactant used up first in a reaction is called the limiting reagent, because the maximum amount of product formed depends on how much of this reactant was originally present. When this reactant is used up, no more product can be formed.

A chemist has two solutions of H2SO4. One has a 40% concentration and the other has a 25% concentration.How many liters of each solution must be mixed to obtain 78 liters of a 28% solution?

liters of the 40% solution and

liters of the 25% solution must be mixed to obtain a 28% solution of H2SO4.

(Round to the nearest tenth, if necessary.)

Answers

Answer:

16 liters of the solution with 40% concentration must be mixed with 62 liters of the solution with 25% concentration in order to obtain 78 liters, 25% concentration solution.

Explanation:

Let the required volume of solution 1 be represented by x.

The required volume of solution 2 would then be 78-x.

The number of moles of solution 1 that would be required = 0.4x

The number of moles of solution 2 that would be required = 0.25(78-x)

The number of moles of the final mixture = 78 x 0.28 = 21.84

moles of solution 1 + moles of solution 2 = moles of final mixture

0.4x + 0.25(78 - x) = 21.84

  0.4x + 19.5 - 0.25x = 21.84

     0.4x - 0.25x = 21.84 - 19.5

          0.15x = 2.34

            x = 15.6 liters

To the nearest tenth = 16 liters

Liters of 40% solution needed = 16 liters

Liters of 25% solution needed = 78 - 16 = 62 liters.

Hence, 16 liters of the solution with 40% concentration must be mixed with 62 liters of the solution with 25% concentration in order to obtain 78 liters, 25% concentration solution.

You are asked to prepare 500 mL 0.300 M500 mL 0.300 M acetate buffer at pH 4.904.90 using only pure acetic acid ( MW=60.05 g/mol,MW=60.05 g/mol, pKa=4.76), pKa=4.76), 3.00 M NaOH,3.00 M NaOH, and water. Answer the questions regarding the preparation of the buffer. 1. How many grams of acetic acid will be needed to prepare the 500 mL buffer? Note that the given concentration of acetate refers to the concentration of all acetate species in solution.

Answers

The quantity of acetic acid that is needed to prepare the 500 mL buffer is 9.0075 grams.

Given the following data:

  • Volume of acetate buffer = 500 mL to L = 0.5 L.
  • Molarity of acetate buffer = 0.300 M.
  • pH = 4.90.
  • MW = 60.05 g/mol.
  • pKa = 4.76.

How to calculate the mass of acetic acid.

First of all, we would write the equilibrium chemical reaction for acetate-acetic acid as follows:

                                CH_3COOH \rightleftharpoons CH_3COO^(-)+H^+

Next, we would calculate HA by applying Henderson-Hasselbalch equation:

pH =pka+ log_(10) (A^-)/(HA)

Where:

  • HA is acetic acid.
  • A^-  is acetate ion.

Substituting the given parameters into the formula, we have;

4.90 =4.76+ log_(10) (A^-)/(HA)\n\n4.90 -4.76+ log_(10) (A^-)/(HA)\n\n(A^-)/(HA)=1.38\n\nA^- = 1.38[HA]

For the concentration of both acids, we have:

[HA]+[A^-]=0.300M\n\n[HA]+1.38[HA]=0.300M\n\n2.38[HA]=0.300M\n\nHA = 0.126

For acetate ion:

A^- = 1.38[HA] = 1.38 * 0.126\n\nA^- =0.174

At a volume of 0.5 liters, we have:

HA = 0.5 * 0.126\n\nHA = 0.063 \;moles

A^- =  0.5 * 0.174\n\nA^- =0.087 \;moles

By stoichiometry:

Total moles = 0.063 + 0.087 = 0.15 moles.

Mass = number \;of \;moles * molar\;mass\n\nMass =0.15 * 60.05

Mass = 9.0075 grams.

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Answer:

You will need 9,0 g of acetic acid

Explanation:

The equilibrium acetate-acetic acid is:

CH₃COOH ⇄ CH₃COO⁻ + H⁺ pka = 4,76

Using Henderson-Hasselbalch you will obtain:

pH = pka + log₁₀([A^(-)])/([HA])

Where HA is acetic acid and A⁻ is acetate ion

4,90 = 4,76 + log₁₀([A^(-)])/([HA])

1,38 = ([A^(-)])/([HA])(1)

As acetate concentration is 0,300M:

0,300M = [HA] + [A⁻] (2)

Replacing (2) in (1):

[HA] = 0,126 M

And:

[A⁻] = 0,174 M

As you need to produce 500 mL:

0,5 L × 0,126 M = 0,063 moles of acetic acid

0,5 L × 0,174 M = 0,087 moles of acetate

To produce moles of acetate from acetic acid:

CH₃COOH + NaOH → CH₃COO⁻ + Na⁺ + H₂O

Thus, moles of acetate are equivalents to moles of NaOH and all acetates comes from acetic acid, thus:

0,087 moles of acetate + 0,063 moles of acetic acid  ≡ 0,15 moles of acetic acid ×(60,05 g)/(1mol) = 9,0 g of acetic acid

I hope it helps!

Is crumbling a cookie, a physical change? Explain why. My child is having problems with this question, and I'm not so sure how to explain it.

Answers

Sugar, flour, and eggs cannot be separated. The materials' properties have changed, resulting in a chemical change.Therefore, crumbling a cookie is not a physical change.

What is physical change ?

Physical changes affect a chemical substance's form but not its chemical composition. Physical changes can be used to separate mixtures into their component compounds, but not to separate compounds into chemical elements or simpler compounds.

A chemical change is the transformation of one material into another, the formation of new materials with different properties, and the formation of one or more new substances. It occurs when one substance reacts with another to form a new substance.

A physical change is characterized by a change in physical properties. Melting, transition to a gas, change in strength, change in durability, changes in crystal form, textural change, shape, size, colour, volume, and density are all examples of physical properties.

Thus, crumbling a cookie is not a physical change.

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Answer:

Because the sugar flour and eggs can no longer be separated. The properties of the materials have changed so it's a chemical change

According to the valence bond theory the triple bond in ethyne consists of

Answers

Answer:
            According to the valence bond theory the triple bond in ethyne consists of one sigma bond and two pi bonds.

Explanation:
                   Atomic number of carbon is 6. The ground state electronic configuration of carbon is as follow,

                                         1s
², 2s², 2p²

And the excited state electronic configuration of carbon is as follow,

                                         1s², 2s¹, 2px¹, 2py¹, 2pz¹

In ethyne the 2s¹ orbital and 2px¹ orbitals having unpaired electrons form sigma bonds by head to head overlapping with orbitals of hydrogen atom and carbon atom. The remaining 2py¹ and 2pz¹ orbitals of both carbons overlap perpendicular to the existing sigma bond resulting in the formation of two pi bonds.