what is the entropy when 1.23 moles of ccl₂f₂ vaporize at 25°c? [∆h(vap) = 17.2 kj/mol at 25°c]

Glyceraldehyde-3-phosphate (G₃P) is produced in the stroma of chloroplasts all of the listed responses are correct.

Option C is correct.

It is a sugar with three carbons. For each three particles of CO₂ diminished in the Calvin cycle, six atoms of G₃P are shaped yet only one of these atoms leaves the cycle to be utilized by the plant cell.

Plant cells use glyceraldehyde-3-phosphate G₃P() to produce sucrose and starch. The Calvin cycle produces the Glyceraldehyde 3-phosphate (G₃P) found in plant cells. G₃P is a 3-C (carbon) sugar that is utilized to make various starches by the plant cells.

In order to become a component of a carbohydrate molecule, one of G₃P's three-carbon molecules leaves the cycle. The remaining G₃P molecules remain in the cycle so that they can be reformed into RuBP, which is prepared to react with additional CO₂. Together with the process of cellular respiration, photosynthesis creates an energy cycle that is balanced.

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The pH of the solution is 1.60, by mixing 20.0 ml of 0.15 m hcl with 20.0 ml of 0.10 m koh.

pH = -log[H+]

Also;

mole of substance = molarity × volume in L

The moles of HCl = 0.15 × 0.02 = 0.003 moles

The moles of KOH = 0.10 × 0.02 = 0.002 moles

The solution will be acidic since the moles of HCl is greater than that of KOH.

Moles of remaining HCl [H⁺] remaining = 0.003 - 0.002 = 0.001 moles

Molarity of remaining HCl [H⁺] remaining = 0.001/0.04 = 0.025 M

Hence, [H⁺] = 0.025 M

pH = - log(0.025)

pH = 1.6

Therefore, the pH of the resulting solution is 1.60.

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Hard water is not typically caused by the concentration of dissolved oslolum (assuming you meant "solutes") but rather by the presence of high concentrations of dissolved minerals, such as calcium and magnesium ions

The net ionic equation for the formation of calcium carbonate (CaCO3) is as follows:

Ca2+ (aq) + CO32- (aq) → CaCO3 (s)

In this equation, the calcium ions (Ca2+) from the dissolved calcium compounds in hard water react with carbonate ions (CO32-) to form insoluble calcium carbonate precipitate (CaCO3), which appears as a white solid. The precipitation occurs because the solubility product of calcium carbonate is exceeded, resulting in the formation of a solid.

The formation of a precipitate is driven by the principle of solubility. When the concentration of a dissolved compound exceeds its solubility limit, the excess ions come together and form a solid. In the case of hard water, when the concentration of calcium ions and carbonate ions surpasses their respective solubility limits, calcium carbonate precipitate forms.

Regarding the replacement of calcium ions with sodium ions, if calcium ions were replaced with sodium ions in the net ionic equation, the reaction would be different. Sodium ions (Na+) do not form insoluble compounds with carbonate ions, and sodium carbonate (Na2CO3) is soluble in water.

In summary, the formation of calcium carbonate precipitate in hard water is driven by the reaction between calcium ions and carbonate ions, leading to the exceeding of the solubility product and subsequent precipitation. If calcium ions are replaced with sodium ions, no precipitate would form as sodium carbonate remains soluble in water.

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The oxidation number of arsenic (As) in the compound H3AsO3 is +3. This is because the oxidation number of hydrogen (H) is always +1, and the oxidation number of oxygen (O) is always -2. Therefore, we can calculate the oxidation number of arsenic (As) by adding up the oxidation numbers of all the atoms in the compound and setting it equal to zero (since the compound is neutral). In general, some guidelines to determine oxidation numbers are:

The oxidation number of an element in its elemental form is 0.

The oxidation number of a monatomic ion is equal to its charge.

In a compound, the sum of the oxidation numbers of all atoms is equal to the charge of the compound.

Fluorine always has an oxidation number of -1 in compounds.

Oxygen usually has an oxidation number of -2 in compounds, except in peroxides (such as H2O2) where its oxidation number is -1.

Hydrogen usually has an oxidation number of +1 in compounds, except in metal hydrides (such as NaH) where its oxidation number is -1.

In this case, we have:
(+1 x 3) + (x) + (-2 x 3) = 0
Solving for x, we get:
x = +3
So the oxidation number of arsenic (As) in H3AsO3 is +3.
In the compound H3AsO3, the oxidation number of arsenic (As) is +3. Here's the breakdown: hydrogen (H) has an oxidation number of +1, and oxygen (O) has an oxidation number of -2. Using the formula H3AsO3, we can calculate the oxidation number of As as follows:
3(+1) + As + 3(-2) = 0
3 - 6 + As = 0
As = +3

So, the oxidation number of arsenic in H3AsO3 is +3.

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Answer:

Explanation:

Isopropyl alcohol has three intermolecular forces: London dispersion forces from its hydrocarbon chain. dipole-dipole forces from the difference in polarities due to the presence of oxygen atom. hydrogen bonding due to the presence of hydrogen bonded to an oxygen atom.

Answer:

1. Lanthanum-139 atom is the stable isotope of lanthanum with relative atomic mass 138.906348, 99.9 atom percent natural abundance and nuclear spin 7/2.

2. In fact, over 80% of electric cars sold globally utilized permanent magnet-based motors in 2019. These magnets are typically made with rare-earth materials such as neodymium and dysprosium, which have a very geographically constrained supply chain.

3. It is the second most reactive of the rare-earth metals after europium. Lanthanum oxidizes in air at room temperature to form La2O3. It slowly reacts with water and quickly dissolves in diluted acids, except hydrofluoric acid (HF) because of formation of a protective fluoride (LaF3) layer on the surface of the metal.

Explanation:

Answer:

Explanation:

True or false 4 questions help please!

Most mixtures are not physical changes.

True

False   FALSE  when I add solid NaCl salt to water I no longer have a solid salt, I have a solution of sodium and chlorid ions in liquid water.  this is  physical change

Chemical Changes do not create a new substance.

True

False  FALSE 0ne of the proofs that you have a chemical change is the presence of a new substance

Fire is an example of a Chemical Change.

True  TRUE  fire is oxygen reacting wiIh mattter to form new compounds

False

Explosions are not chemical changes.

It depends on the explosion  if I blow up a balloon until it POPS!! that is physical chame.  If I light a firecracker and it explodes, that is a chemical change of oxygen reacting with hot gunpowder

A change in color and odor are both examples of chemical changi

I would rewrite the question as  

A change in color and odor  both INDICATE a chemical change

and that is

True  TRUE

False

F

True   Fire is an example of a Chemical Change.

True

False

Explosions are not chemical changes.

True

False

Answer:

Explanation:

hopes it help you a lot

pls. (mark me as brainlist ).......

An electron configuration determines the kinds and numbers of bonds an atom will form with each other  atom . Carbon can form four covalent bonds with a variety of atoms due to its four valence electrons .

Other molecules can bond to carbon. An atom's features are largely determined by its electron configuration . Four covalent bonds can form with carbon. Chemical bonds known as covalent bonds can occur between nonmetals. Two atoms with covalent bonds have equal or identical electronegativity. Two atoms so share four pairs of electrons.

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The climate zones lying between 23.5° and 66.5° north and south latitude are called the Temperate Climate Zones. The regions lying between the latitudes of the Tropics of Cancer and Capricorn (23.5° north and 23.5° south, respectively) are characterized by a tropical climate.

These regions are called the Tropical Climate Zones. The earth is divided into three broad climatic regions based on latitude. The region from the equator to the Tropic of Cancer and the Tropic of Capricorn is referred to as the tropical zone. Because of the proximity to the equator, the tropical zone receives more heat and has higher temperatures throughout the year. The area between the Tropic of Cancer and the Arctic Circle and the Tropic of Capricorn and the Antarctic Circle is known as the temperate zone. In this zone, the seasons are typically more pronounced, and the temperatures vary significantly from summer to winter. The polar zone extends from the Arctic Circle to the North Pole and the Antarctic Circle to the South Pole. It has extremely low temperatures and receives little to no sunlight for half of the year due to the tilt of the Earth's axis.

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Dilute solutions of NaOH, H2SO4, and H3PO4 were used to treat an unknown ionic chemical. and

ba2+ precipitate with precipitates created in each of the three instances.

Pb2+ precipitates with NaOH,

while ba2+ precipitates with h2so4 and h3po4.

However, neither Pb 2+ nor h3po4 k+ precipitate with the aforementioned solutions of NaOH, h2so4, or h3po4, respectively.

When naming an ionic compound, the cation is stated first, then the anion. Charges that are positive and negative must be equal. Roman numerals are used in parentheses to name some anions that have several forms.

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The final temperature of the gas is approximately 40.96 Kelvin.

To determine the final temperature of the gas, we can use the ideal gas law, which states:

PV = nRT

where P is the pressure, V is the volume, n is the number of moles of the gas, R is the ideal gas constant, and T is the temperature in Kelvin.

First, let's convert the given temperatures to Kelvin. We have:

Initial temperature: -125 degrees Celsius = 148 K (approximate)

Final temperature: Unknown

The initial conditions of the gas are as follows:

Initial pressure (P1) = 1.25 atm

Initial volume (V1) = 1250 mL = 1.25 L (since 1 L = 1000 mL)

Initial temperature (T1) = 148 K

The final conditions of the gas are as follows:

Final pressure (P2) = 50.0 atm

Final volume (V2) = 325 mL = 0.325 L

Final temperature (T2) = Unknown

Using the ideal gas law, we can set up the following equation:

(P1 * V1) / T1 = (P2 * V2) / T2

Substituting the known values:

(1.25 atm * 1.25 L) / 148 K = (50.0 atm * 0.325 L) / T2

Simplifying the equation:

T2 = (50.0 atm * 0.325 L * 148 K) / (1.25 atm * 1.25 L)

T2 = 40.96 K

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Answer: They heated the water

Explanation:

The hotter the water was the quicker the sugar dissolved

Answer: At a given temperature and pressure, one mole of any gas occupies the same volume. The molar volume is the volume occupied by one mole of any gas, at room temperature and pressure. The molar volume is equal to 24 dm3 (24,000 cm3).

Explanation: its the internet

A certain metal M forms a soluble nitrate salt MNO₃. Suppose the left half cell of a galvanic cell apparatus is filled with a 1. 50 M solution of MNO₃ and the right half cell with a 15. 0 mM solution of the same substance. The right side of the cell is positive and voltage of the cell is 0.12V.

a) The representation of the galvanic cell with a different concentration of MNO₃ is shown as

M(s)|MNO₃(aq)(1.50M)||MNO₃(aq)(15.0mM)|M(s)

On the left hand side of the cell, the oxidation is occurred which is shown as

M(s)→M⁺(aq)+e⁻

Therefore, this electrode is negative as it shows oxidation reaction.

On the right hand side of the cell, the reduction is occurred and it is shown as

M⁺(aq)+e⁻→M(s)

Therefore, this electrode is positive as it shows reduction reaction.

b) The emf of cell is calculated using the formula

E(cell)=E⁰(cell)-RT/nF×ln([reductant/[oxidant])

The concentration of oxidant is 1.50M and of reductant is 15.0mM which is equal to 0.015M and E⁰(cell) is 0(zero) because both are same solution. Hence, the equation will be

E(cell)=-RT/nF×ln([reductant/[oxidant])

Where, T=20⁰C=20+273=293K

n=1

F=96485C

R=8.314J/molK

Plug all values in the formula

E(cell)=-(8.314×293J×mol×K/1mol×96485C×K)×ln(0.015/1.50)

E(cell)=-(2436.002J/96485C)×ln(0.01)

Ecell=-(0.025247×-4.6052)

Ecell=-(-0.1162)J/C

Ecell=0.1162 J/C=0.12V (1V=1J/C)

Therefore, the voltage of the cell is 0.12V.

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Answer:

40

Correct option is A)

3dm

3

×

24 dm

3

1 mole

​

=0.125 moles of O

2

​

2H

2

​

O

2

​

⟶2H

2

​

O+O

2

​

2 moles of H

2

​

O

2

​

 produces 1 mole of O

2

​

.

0.125 moles of O

2

​

 requires 2×0.125 H

2

​

O

2

​

=0.25 moles of H

2

​

O

2

​

Concentration of H

2

​

O

2

​

=

0.1

0.25

​

=2.5 M

Hence, option A is correct.

Explanation:

Answer

Based on the previous we can say that: The law of conservation of the mass states that the total mass will always stay the same during a chemical reaction. After the reaction, the vessel will have a mass equal to the initial 245 g.

Explanation

The Law of Conservation of Mass states that the mass is neither created nor destroyed in chemical reactions. This means the mass of any one element at the beginning of a reaction will equal the mass of that element at the end of the reaction.

Based on the previous we can say that: The law of conservation of the mass states that the total mass will always stay the same during a chemical reaction.

After the reaction, the vessel will have a mass equal to the initial 245 g.

Answer:

19,199,280 grams

Explanation:

0.062502343837894 grams in one mole of oxygen

The concentration of CO (g) in the equilibrium mixture is 0.020 M. In other words, only a small amount of CO (g) is produced in this reaction at 1000°C. T 0.020 M.

The concentration of CO in an equilibrium mixture (in mol/L) is 5.8 M.

Given that the concentration of H2O (g) is [H2O] = 0.500 M at 1000°C, and  the reaction is:

C(s) + H2O(g) ⇄ CO(g) + H2(g) KC = 0.2 at 1000°C

We need to determine the concentration of CO in an equilibrium mixture (in mol/L)

if a reaction mixture initially contains only reactants.

We can solve this problem using the ICE table method as follows:

Let x be the change in concentration of H2O (g) and CO (g) when they reach equilibrium.

Then the equilibrium concentrations of CO (g) and H2 (g) are equal to x. Hence, the equilibrium concentration of H2O (g) is (0.500 - x) M. Substitute these values in the expression for Kc and solve for x.

Kc = [CO (g)] [H2 (g)] / [H2O (g)] [C (s)]

= 0.2[CO (g)] = Kc [H2O (g)] [C (s)] / [H2 (g)]

= 0.2 × (0.500 - x) / x

We can simplify this expression by cross-multiplication to get:

5x = 0.1 - 0.2xx = 0.02 M

Substituting x = 0.02 M in the expression for [CO (g)], we get:

[CO (g)] = 0.2 × (0.500 - 0.02) / 0.02 = 5.8 M (approx.)

Therefore, the concentration of CO (g) in an equilibrium mixture (in mol/L) is 5.8 M.                                                                                               The problem requires us to find the equilibrium concentration of CO (g) in a mixture that initially contains only reactants.

To solve this problem, we need to use the expression for the equilibrium constant Kc, which is given by:

Kc = [CO (g)] [H2 (g)] / [H2O (g)] [C (s)]

We can also use the ICE table method to solve this problem. In this method, we start with the initial concentration of the reactants and calculate the change in concentration of each species as they reach equilibrium.

We then use the equilibrium concentrations to calculate the value of Kc and solve for the unknowns. Here is how we can set up the ICE table for this problem: Reaction:

C(s) + H2O(g) ⇄ CO(g) + H2(g)

Initial: [C] = [H2]

= 0 M,

[H2O] = 0.500 M

Equilibrium: [C] = [H2] = x,

[H2O] = 0.500 - x,

[CO] = [H2] = x

Change: +x +x -x -x

Substituting the equilibrium concentrations into the expression for Kc, we get:

Kc = [CO] [H2] / [H2O] [C]

= x² / (0.500 - x)

= 0.2

Solving for x, we get: x = 0.020 M

Substituting this value of x into the expression for [CO], we get:

[CO] = x = 0.020 M

Therefore, the concentration of CO (g) in the equilibrium mixture is 0.020 M.

In other words, only a small amount of CO (g) is produced in this reaction at 1000°C. T 0.020 M.

The concentration of CO in an equilibrium mixture (in mol/L) is 5.8 M.

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For conducting chemical reactions, electrochemical methods rely on electrochemical cells to supply the necessary voltage and current.

How are electrochemical cells dependent on them?

Electrochemical cells, which are machines that use electricity to move ions from one electrode to another, are the foundation of electrochemical methods. I

In an electrochemical cell, a voltage source provides the energy required to move the ions while also driving their motion. A chemical reaction between the electrodes and the ions is sparked by this energy, and the result is a current.

The concentration of the electroactive species in the solution is then determined using the current generated. As a result, electrochemical cells are necessary for electrochemical methods. Chemical reactions can also be used to create energy in electrochemical cells.

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Answer:

I think it is c

Explanation:

When water turns into water vapor it goes into the air so it must be lighter

Answer:

5

Explanation:

Answer: IONIC EQUATION.

Explanation:

A chemical equation is defined as the form by which a chemical reaction is represented mathematically. These are written in the form of symbols and chemical formulas of reactants and products which are taking part in the chemical reaction. A chemical equation can be written in two forms, these include:

--> MOLECULAR EQUATION: in this type of equations, the compounds are written and represented in a molecular form. This is sometimes referred to as a balanced equation.

--> IONIC EQUATION: This is a type of chemical equation in which the electrolytes in aqueous solution are expressed as dissociated ions. A typical illustrated example is seen in the reaction between AgNO3(aq) and NaCl(aq) :

Ag+(aq) + NO3-(aq) + Na+(aq) + Cl-(aq) → AgCl(s) + Na+(aq) + NO3-(aq)

The (aq) written in the above equation signifies they are in aqueous solution.

NaHCO3(s) + HC2H3O2(aq) → CO2(g) + H2O(l) + NaC2H3O2(aq)

This equation represents the reaction of baking soda (sodium bicarbonate) with vinegar (acetic acid) to generate carbon dioxide gas, water, and sodium acetate.

The balanced equation for the reaction of NaHCO3(s) with HC2H3O2(aq) including phase symbols is

NaHCO3(s) + HC2H3O2(aq) → CO2(g) + H2O(l) + NaC2H3O2(aq)

Baking soda, also known as sodium bicarbonate, is a white, crystalline powder with the chemical formula NaHCO3. It is an alkaline substance that neutralizes acids.

Vinegar is mostly composed of acetic acid, HC2H3O2, which is a weak acid. Vinegar has a sour flavor and a strong smell due to the presence of acetic acid.

NaHCO3(s) + HC2H3O2(aq) → CO2(g) + H2O(l) + NaC2H3O2(aq)  This equation represents the reaction of baking soda (sodium bicarbonate) with vinegar (acetic acid) to generate carbon dioxide gas, water, and sodium acetate. When the baking soda and vinegar are combined, a chemical reaction occurs, causing carbon dioxide gas bubbles to form. This is due to the reaction between the acid and base in the mixture, which generates carbon dioxide gas as a byproduct. This reaction is commonly used in baking as a leavening agent to make cakes, muffins, and other baked goods rise.

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1 mole of CO2 is 6.022 x 10^23 molecules of CO2. It is also 44.01 g of CO2. Here 44.01 g comes as the sum of individual atoms of CO2 i.e. 12.01 + 16 + 16 i.e. 44.01 g

CO2 is composed of one carbon atom and two oxygen atoms, which have the atomic masses of 12.01 g/mol and 16.00 g/mol respectively. So, the molar mass of CO2 is the sum of the atomic masses of the atoms that make up CO2, which is 12.01 + 16.00 + 16.00 = 44.01 g/mol. Carbon dioxide (CO2) is a naturally occurring compound made up of one carbon atom and two oxygen atoms. It is a colorless, odorless gas and is a byproduct of many biological and industrial processes. CO2 is a greenhouse gas, which means it helps trap heat in the Earth's atmosphere and contributes to global warming.

Therefore, one mole of CO2 weighs 44.01 grams.

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Answer:

Atomic number = 26

Explanation:

We always rely on the proton number as the atomic number, as proton numbers don't change, even if it became an ion. Only electrons decrease or increase.

Answer:

3,4–diethyloctane

Explanation:

To name the compound, we must do the following:

1. Locate the longest continuous carbon chain. This gives the parent name of the compound. In this case, the longest chain is carbon 8 i.e octane.

2. Identify the substituent attached to the compound. In this case, the substituent group attached is ethyl i.e CH2CH3.

Note: there are two ethyl group attached to the compound.

3. Locate the position of the substituent group attached, giving them the lowest possible count. In this, the ethyl group are at carbon 3 and 4.

4. Combine the above to get the name of the compound. Therefore, the name of the compound is:

3,4–diethyloctane.