The solubility of a substance is 12.0 g per 100. g of water at 20.0°C. It is 18.0 g per 100. g of water at 60.0°C. Howmany grams of the substance can crystallize from a saturated solution that contains 200. g of water at 60.0°C if the solution is cooled to 20.0°C?

The refrigerator would use 180 kilowatt-hours (kWh) of electric energy over the course of 30 days, assuming it runs for 12 hours each day.

To calculate the kilowatt-hours (kWh) of electric energy used by the refrigerator in 30 days, we need to multiply the power rating by the total running time.

Given:

Power rating of the refrigerator = 500 W

Running time per day = 12 hours

Number of days = 30

First, we need to convert the power rating from watts to kilowatts:

Power rating = 500 W / 1000 = 0.5 kW

Next, we calculate the total energy used in kilowatt-hours (kWh) over the 30-day period:

Energy used = Power rating × Running time × Number of days

Energy used = 0.5 kW × 12 hours/day × 30 days

Energy used = 180 kWh

Therefore, the refrigerator would use 180 kilowatt-hours (kWh) of electric energy over the course of 30 days, assuming it runs for 12 hours each day.

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Answer:

the last one

Explanation:

ydyeusueueueueudhs

Because hydrogen forms compounds with oxidation numbers of both +1 and -1, many periodic tables include this element in both Group IA (with Li, Na, K, Rb, Cs, and Fr) and Group VIIA (with F, Cl, Br, I, and At).

There are many reasons for including hydrogen among the elements in Group IA. It forms compounds (such as HCl and HNO3) that are analogs of alkali metal compounds (such as NaCl and KNO3). Under conditions of very high pressure, it has the properties of a metal. (It has been argued, for example, that any hydrogen present at the center of the planet Jupiter is likely to be a metallic solid.) Finally, hydrogen combines with a handful of metals, such as scandium, titanium, chromium, nickel, or palladium, to form materials that behave as if they were alloys of two metals.

There are equally valid arguments for placing hydrogen in Group VIIA. It forms compounds (such as NaH and CaH2) that are analogs of halogen compounds (such as NaF and CaCl2). It also combines with other nonmetals to form covalent compounds (such as H2O, CH4, and NH3), the way a nonmetal should. Finally, the element is a gas at room temperature and atmospheric pressure, like other nonmetals (such as O2 and N2).

It is difficult to decide where hydrogen belongs in the periodic table because of the physical properties of the element. The first ionization energy of hydrogen (1312 kJ/mol), for example, is roughly halfway between the elements with the largest (2372 kJ/mol) and smallest (376 kJ/mol) ionization energies. Hydrogen also has an electronegativity (EN = 2.20) halfway between the extremes of the most electronegative (EN = 3.98) and least electronegative (EN = 0.7) elements. On the basis of electronegativity, it is tempting to classify hydrogen as a semimeta

Answer:

1 respuesta. fórmula empírica = CH2

Explanation:

Translatation: 1 Answer. empirical formula = CH2

1. The mass of 1.33×10²² mole of Sb is 1.62×10²⁴ g

2. The mass of 4.75×10¹⁴ mole of Pt is 9.26×10¹⁶ g

3. The mass of 1.22×10²³ mole of Ag is 1.32×10²⁵ g

4. The mass of 9.85×10²⁴ mole of Cr is 5.12×10²⁶ g

Mass = mole × molar mass

Mass of Sb = 1.33×10²² × 122

Mass of Sb = 1.62×10²⁴ g

Mass = mole × molar mass

Mass of Pt = 4.75×10¹⁴ × 195

Mass of Pt = 9.26×10¹⁶ g

Mass = mole × molar mass

Mass of Ag = 1.22×10²³ × 108

Mass of Ag = 1.32×10²⁵ g

Mass = mole × molar mass

Mass of Cr = 9.85×10²⁴ × 52

Mass of Cr = 5.12×10²⁶ g

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260grams

The formula for calculating the moles of NaOH is expressed as

Determine the required mass of NaOH

Hence the grams of sodium hydroxide (NaOH) you need to make 6.5 L of 1.0M solution is 260grams

Answer:

1-Genetics

2-Alleles

3-Organism's genotype

4-Phenotypes and Genotypes

5-Allele

6-Gregor Mendel

7-Genotype

8-Probability

9-Hybrid

10-Genotypic Ratio

11-Phenotypic ratio

12-Heterozygous

Explanation:

Answer:

the answer is : c

hope this will help you

Some of the most toxic organic compounds, widely used in plastics, pesticides, and solvents, are the dioxins.

The toxic organic compounds widely used in solvents are volatile organic compounds (VOCs) such as benzene, toluene, xylene, and chlorinated solvents like trichloroethylene (TCE) and perchloroethylene (PCE). These solvents can pose a risk to human health and the environment due to their potential for leaching into soil and groundwater, as well as their ability to contribute to air pollution.

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The process in which the reactants get converted into products is defined as the chemical reaction. The nature and identity of products will be totally different from the reactants.

The replacement reaction also called the displacement reaction in which an atom or groups of atoms are displaced by another atom in a molecule. In a single replacement reaction one element is replaced by other in the compound.

The reaction between chlorine and sodium bromide to form sodium chloride and bromine is an example for replacement reaction.

Cl₂ (aq) + 2NaBr (aq) → 2NaCl (aq) + Br₂ (aq)

Here 'Cl' is more reactive than 'Br', so it replaces 'Br' in NaCl.

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The two likely mechanisms for the reaction of 2-iodohexane with sodium ethoxide are SN2 (substitution) and E2 (elimination). SN2 replaces iodine with ethoxy, while E2 forms a double bond and eliminates hydrogen iodide. Mechanism choice depends on steric hindrance and reactant concentrations.

The two most likely mechanisms for the reaction of 2-iodohexane with sodium ethoxide are nucleophilic substitution (SN2) and elimination (E2) reactions.

1. Nucleophilic Substitution (SN2):

In this mechanism, the nucleophile (sodium ethoxide) attacks the carbon atom bonded to the iodine in a concerted manner, while the iodide ion leaves. This leads to the substitution of the iodine atom with the ethoxy group.

2-iodohexane + Sodium ethoxide → 2-hexanol + Sodium iodide

2. Elimination (E2):

In this mechanism, the strong base (sodium ethoxide) abstracts a proton from a neighboring carbon, causing the formation of a carbon-carbon double bond and the elimination of a molecule of hydrogen iodide.

2-iodohexane + Sodium ethoxide → 1-hexene + Sodium iodide + Ethanol

Both of these mechanisms are common reactions for alkyl halides with strong nucleophiles or bases. The specific mechanism that occurs depends on factors such as the steric hindrance around the reaction center and the concentration of the nucleophile or base.

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Answer:

44.625862500000004 grams are in 0.376 mol Kbr! Plz mark as brainliest, hope this helped you!

Answer: As one atomic dipole nears another atom it will affect the electron density distribution, so for example if the slightly positive end of the atom is located next to another atom, it will attract the electron(s) in the other atom. A covalent bond happens when atoms share an electron between them. The other kind of bond is called ionic, and it happens when one atom gives an electron to another atom. These bonds are held together with electric forces. Two atoms of the same element can be different if their electrons are in different states. If one copper atom has an electron in an excited state and another copper atom has all of its electrons in the ground state, then the two atoms are different.

Answer:

there are relatively few atoms but they can make many molecules

Explanation:

We must first convert the amount of atoms of helium to moles in order to solve this issue. This much helium gas can be used by Ngawang to fill about 14 balloons at STP.

Along with liquids, solids, and plasmas, gases are among the four basic states of matter. Gases are made up of atoms or molecules that are constantly moving and spaced far apart from one another. Gases, in contrast to solids and liquids, do not have a set shape or volume and always fill their container to the top.

In addition to other characteristics like density, viscosity, and compressibility, gases can be classified according to their volume, pressure, and temperature. Additionally, they can change physically and chemically through processes including expansion, compression.

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Reduction occurs at the cathode in an electrochemical cell. To determine which transformation could take place at the cathode, we need to compare the standard reduction potentials (E°) of the given half-reactions.

The half-reactions are:

\(a. H2O + 1 e- → H2O2, E° = +0.68 V\\b. 2 Br- → Br2 + 2 e-, E° = +1.09 V\\c. 2 H2O + 2 e- → O2 + 4 H+ , E° = +1.23 V\\d. SO42- + 2 e- → SO32-, E° = +0.17 V\\e. Fe2+ + 1 e- → Fe3+, E° = -0.77 V\)

In order for a reduction reaction to occur at the cathode, the reduction potential of the half-reaction must be more positive than the reduction potential of the half-reaction at the anode. The half-reaction with the more positive reduction potential will be reduced (gain electrons), while the half-reaction with the less positive reduction potential will be oxidized (lose electrons).

Therefore, from the given half-reactions, the half-reaction that could take place at the cathode is:

b. 2 Br- → Br2 + 2 e-, E° = +1.09 V

This is because the reduction potential of this half-reaction is the highest among the given options, indicating that it is the most favorable reduction reaction.

Therefore, Br- will be reduced to form Br2 at the cathode, while the corresponding oxidation reaction (Br2 → 2 Br-) will take place at the anode.

Option (b) is the correct choice.

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Answer:

Diamond, graphite and fullerness

Explanation:

Diamond is clear and transparent but graphite is black and opaque

The balanced redox reaction in acidic solution:

6H₂O₂(aq) + 2Cr₂O₇⁻²(aq) + 14H⁺(aq) → 3O₂ + 4Cr³⁺(aq) + 20H₂O(l)

To balance the redox reaction in acidic solution:

H₂O₂(aq) + Cr₂O₇⁻²(aq) → O₂(g) + Cr³⁺(aq)

We will follow the steps for balancing redox reactions in acidic solution:

Step 1: Assign oxidation numbers to all elements in the equation:

H₂O₂(aq): H has an oxidation state of +1, O has an oxidation state of -1.

Cr₂O₇⁻²(aq): Cr has an oxidation state of +6, O has an oxidation state of -2.

O₂(g): O has an oxidation state of 0.

Cr³⁺(aq): Cr has an oxidation state of +3.

Step 2: Identify the elements that are being oxidized and reduced:

Oxidation: Cr is being reduced from +6 to +3.

Reduction: H₂O₂ is being oxidized from -1 to 0.

Step 3: Write the half-reactions for oxidation and reduction:

Oxidation half-reaction: H₂O₂(aq) → O2(g)

Reduction half-reaction: Cr₂O₇⁻²(aq) → Cr³⁺(aq)

Step 4: Balance the atoms other than H and O in each half-reaction:

Oxidation half-reaction: 2H₂O₂(aq) → O₂(g)

Reduction half-reaction: Cr₂O₇⁻²(aq) → 2Cr³⁺(aq)

Step 5: Balance the oxygen atoms by adding water molecules (H2O) to the side that lacks oxygen:

Oxidation half-reaction: 2H₂O₂(aq) → O₂(g) + 2H₂O(l)

Reduction half-reaction: Cr₂O₇⁻²(aq) → 2Cr³⁺(aq)

Step 6: Balance the hydrogen atoms by adding hydrogen ions (H+) to the side that lacks hydrogen:

Oxidation half-reaction: 2H₂O₂(aq) → O₂(g) + 2H₂O(l)

Reduction half-reaction: Cr₂O₇⁻²(aq) + 14H+(aq) → 2Cr³⁺(aq) + 7H₂O(l)

Step 7: Balance the charges by adding electrons (e-) to the appropriate side of each half-reaction:

Oxidation half-reaction: 2H₂O₂(aq) → O₂(g) + 2H₂O(l) + 4e-

Reduction half-reaction: Cr₂O₇⁻²(aq) + 14H+(aq) + 6e- → 2Cr³⁺(aq) + 7H₂O(l)

Step 8: Multiply each half-reaction by a factor that will equalize the number of electrons in both half-reactions:

Multiply the oxidation half-reaction by 3 and the reduction half-reaction by 2:

6H₂O₂(aq) → 3O₂(g) + 6H₂O(l) + 12e-

Cr₂O₇⁻²(aq) + 14H+(aq) + 12e- → 4Cr³⁺(aq) + 14HvO(l)

Step 9: Combine the two half-reactions, canceling out the electrons on both sides:

6H₂O₂(aq) + 2Cr₂O₇⁻²(aq) + 14H⁺(aq) → 3O₂ + 4Cr³⁺(aq) + 20H

Step 10: Combine all the species to form the balanced redox reaction:

6H₂O₂(aq) + 2Cr₂O₇⁻²(aq) + 14H⁺(aq) → 3O₂ + 4Cr³⁺(aq) + 20H₂O(l)

Step 11: Simplify the equation by canceling out common species:

6H₂O₂(aq) + 2Cr₂O₇⁻²(aq) + 14H⁺(aq) → 3O₂ + 4Cr³⁺(aq) + 20H₂O(l)

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Answer:

124g  

Explanation:

The concept stoichiometry is used here to determine the mass of Na₂O. The grams of Na₂O required to produce 1.60 x 10² grams of NaOH is  124 g.

The stoichiometry is an important concept of chemistry which help us to use the balanced chemical equation to calculate the amounts of reactants and products. The stoichiometric coefficient is the number of molecules that can participate in the reaction.

The balanced equation is:

Na₂O + H₂O → 2NaOH

1 mole of Na₂O is required to react with 1 mole of  H₂O to give 2 moles of NaOH.

62 g of Na₂O produces 80 g of NaOH.

1 g of NaOH is produced from 62/80 g of Na₂O .

1.6 × 10² g of NaOH will require 62 × 1.6 × 10² g / 80 g of Na₂O = 124 g Na₂O

Thus the grams of Na₂O required to produce 1.6 × 10² g of NaOH is 124 g.

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Answer:

The mantle is divided into several layers: the upper mantle, the transition zone, the lower mantle, and D” (D double-prime), the strange region where the mantle meets the outer core. The upper mantle extends from the crust to a depth of about 410 kilometers (255 miles).

Explanation:

The static friction is the type of the friction in which it keeps an object at rest. Like paper on a table.

The friction is the force that resist the motion. The types of friction is given as:

When an individual try to move the stationary object on the surface and without triggering the relative motion between the surface and the body. The examples of the static friction is given as :

1) Paper on table

2) A towel hanging on rack

3) A car parked on hall

4) book mark in book

The question is incomplete, I gave the general answer according to my knowledge.

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The optimum pH of formic acid - formate buffer is 3.75

A substance or a solution which resists any changes in pH, when acid or alkali is added to it.

pH = pKa + log[base] / [acid]

Considering equimolar concentration of acid and base

pH = 3.75 + log(x)/(x)

pH = 3.75 + log (1)

pH = 3.75 + 0

pH = 3.75

Hence,

The optimum pH of formic acid - formate buffer is 3.75

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Answer:

they represent their own atoms

Answer:

Elements

Explanation:

1. ENDOTHERMIC
2. EXOTHERMIC
3. ENDOTHERMIC
4. EXOTHERMIC
5. INPUT OF HEAT ENERGY
6. ENDOTHERMIC
7. EXOTHERMIC

Answer to question 3:
Q = (200g) x (1 cal/g°C) x (100.0°C - 24.0°C) = 15200 calories

Answer to question 6:
The can filled with water should have the greatest change in temperature after five minutes.

Answer to question 7:
Cork would be the best choice for an insulator as it has low thermal conductivity and can trap air pockets, which helps to slow down the transfer of heat.
1. Endothermic processes:
- Dry ice (CO2) subliming
- The input of heat energy
- Vaporizing a block of iron
- Ice melting

2. Exothermic processes:
- Condensing of H2O on the side of a glass
- Water freezing
- Soot being deposited on the walls inside a chimney
- The release of heat energy

Question 5:
To calculate the heat required, use the formula q = m x C x ΔT.
q = 200g x 1 cal/g x (100°C - 24°C) = 200g x 1 cal/g x 76°C = 15,200 calories. So, the answer is 15200 calories.

Question 6:
The can with iron filings should have the greatest change in temperature after five minutes, as metals generally have a higher conductivity and lower specific heat capacity.

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Answer:

thank u for the points

Explanation:

The pH of a solution that results from mixing 25. 0 mL of 0. 200 M HA with 12. 5 mL of 0. 400 M NaOH. is 1.70.

To answer your question, we need to use the principles of acid-base chemistry and stoichiometry. HA represents a weak acid, and NaOH represents a strong base. When these two substances are mixed together, they will undergo a neutralization reaction. The balanced chemical equation for this reaction is: HA + NaOH → \(Na^{+}\)+ H2O. Where NaA represents the salt that is formed from the acid and base.
To determine the pH of the resulting solution, we first need to calculate the moles of each reactant that are present in the mixture. We can use the formula: moles = concentration x volume (in liters)
For HA:
moles HA = 0.200 M x 0.0250 L = 0.00500 mol
For NaOH:
moles NaOH = 0.400 M x 0.0125 L = 0.00500 mol
Since the reaction between HA and NaOH is a 1:1 reaction, we can see that they react completely. Therefore, the number of moles of the excess reactant will be equal to the difference between the moles of the two reactants: moles excess = |moles HA - moles NaOH| = |0.00500 mol - 0.00500 mol| = 0 mol
Since there is no excess reactant, we can assume that all of the acid and base have reacted to form the salt \(Na^{+}\) and water. The moles of the salt Na that are formed will be equal to the moles of the acid or base that are consumed in the reaction: moles \(Na^{+}\) = moles HA (or moles NaOH) = 0.00500 mol

We can now use the volume of the mixture to calculate the concentration of the salt \(Na^{+}\): concentration \(Na^{+}\)= moles \(Na^{+}\)/ volume (in liters) = 0.00500 mol / (0.0250 L + 0.0125 L) = 0.100 M
Finally, we can use the principles of weak acid-base chemistry to determine the pH of the solution. Since \(Na^{+}\) is the conjugate base of the weak acid HA, it will hydrolyze in water to produce OH- ions:
\(Na^{+}\) + \(H_{2}0\) → HA + \(OH^{-}\)
The equilibrium constant for this reaction is given by:
Kb = [HA][\(OH^{-}\)] / [\(Na^{+}\)]
We can assume that the concentration of \(OH^{-}\) ions is equal to the concentration of NaOH that was added to the mixture (since NaOH is a strong base and completely dissociates in water). Therefore: Kb = [HA][0.400 M] / [0.100 M] = 1.60 x 10^-13
The relationship between Ka (the acid dissociation constant) and Kb for a weak acid and its conjugate base is: Ka x Kb = Kw. Where Kw is the ion product constant for water (1.00 x 10^-14 at 25°C). Therefore: Ka = Kw / Kb = 6.25 x 10^-2
Since we know the value of Ka for HA, we can use the Henderson-Hasselbalch equation to calculate the pH of the solution: pH = pKa + log([A-]/[HA]). where pKa = -log(Ka) = 1.20 (for HA), and [A-]/[HA] = [\(Na^{+}\)]/[HA] = 0.100 M / 0.200 M = 0.500. Therefore: pH = 1.20 + log(0.500) = 1.70

In conclusion, the pH of the solution that results from mixing 25.0 mL of 0.200 M HA with 12.5 mL of 0.400 M NaOH is 1.70.

Scientific notation is a technique for composing exceptionally huge or small numbers in a smaller and simple to-utilize structure, while ordinary notation is the standard approach to composing numbers.

In scientific notation, a number is communicated as the result of a number somewhere in the range of 1 and 10 and a force of 10. For instance, the number 5,000 can be composed as 5 x \(10^3\) in scientific notation.

Then again, ordinary notation composes a number utilizing the standard decimal spot esteem system. For example, 5,000 can be composed as 5000 in ordinary notation.

Scientific notation is usually utilized in numerous scientific and specialized fields, like physical science, science, and engineering since it considers simpler estimations and examinations of enormous and small numbers.

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The idea that only one electron in an atom can exist in a given quantum energy state is known as the Pauli exclusive principle .

The quantum state of a particle is defined bu the values of the quantum numbers , so what this means is that no two electrons in the same atoms can have the same set of quantum number . This is known as the  Pauli exclusive principle , named after the German physicist Wolfgang Pauli .

All transition emit one quantum ( one photon ) of energy , but an energy to lower energy is released .

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