A sample of gas has a volume of 2.36 L at a temperature of 53.00 °C. The gas sample is heated to a temperature of 139.00 °C (assume pressure and amount of gas are held constant). Predict whether the new volume is greater or less than the original volume, and calculate the new volume.

the correct order of the ionization was Cu > Ag < Au.

define ionization ?

Ionization, or Ionisation is the process by which an atom or a molecule acquires a negative or positive charge by gaining or losing electrons, often in conjunction with other chemical changes. The resulting electrically charged atom or molecule is called an ion. Ionization can result from the loss of an electron after collisions with subatomic particles, collisions with other atoms, molecules and ions, or through the interaction with electromagnetic radiation. Heterolytic bond cleavage and heterolytic substitution reactions can result in the formation of ion pairs. Ionization can occur through radioactive decay by the internal conversion process, in which an excited nucleus transfers its energy to one of the inner-shell electrons causing it to be ejected.

the correct order of the ionization was Cu > Ag < Au.

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Answer:

conservation of mass

Answer:

Relative organisms

Explanation:

Ecology is a branch of biology that deals with the relations of organisms to one another and to their physical surroundings and conditions.

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Answer: (i) \(C_2H_3O_2H\) and (ii) \(NaC_2H_3O_2\)

Explanation:

Buffer is a solution which resists changes in pH when small quantities of acid or alkali are added to it.

Acidic buffer is formed by combination of weak acid and salt of weak acid with strong base. Example: \(CH_3COOH\) and \(CH_3COONa\)

Basic buffer is formed by combination of weak base and salt of weak base with strong acid. Example: \(NH_4OH\) and \(NH_4Cl\)

Thus pair of substance can be dissolved together (in the right ratio) to prepare a buffer solution is \(C_2H_3O_2H\) and \(NaC_2H_3O_2\)

Answer:

A. Hydrogen cans oxygen are reactants

Explanation:

Oxygen has a charge of -2. Hydrogen has a charge of +1. So it takes two hydrogens to balange the one oxygen hence the H20. So the reaction is balanced, and H20 is indeed made up of 1 oxygen and 2 hydrogen atoms. The equation 2 hydrogen + 1 oxygen (<---reactants) = Water (<--product) So hence Hydrogen cans oxygen is not a true statement. If cans is a misspelling then C would be your answer.

Hope that helped

The correct order of the particles that give texture to the soil from smallest to largest is clay, silt, and sand; option C.

Soil texture refers to a physical classification of the component and types of soils based on their physical texture either as coarse or fine particles.

There are several types of soils and these various types of soils have different textures.

The types of soils and their arrangement based on increasing particle size are as follows:

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Answer:

Element

Explanation:

Iron is an element

16,066.56Joules

The amount of heat energy required is expressed according to the formula;

where:

• m is the mass of water = 48grams

• c is the specific heat capacity of water = 4.184 J/g°C

• △,t ,is the change in temperature, = 120 - 4,0 = 80°C

Substitute the given parameters into the formula

Hence the amount of energy required to heat 48 grams of water from 40 °C to 120 °C is 16,066.56Joules

Answer:

Is this math? Cause as a fourth grader, I can do Algebra, but not this.

Explanation:

x in  Na₂CO₃.xH₂0  is the number of water molecule that is attached to sodium carbonate called water of crystallization. Therefore the value of x is 10

Titration is a technique by which we know the concentration of unknown solution using titration of this solution with solution whose concentration is known. To know the end point we use phenolphthalein as indicator. End point is a point where completion of reaction happen

The balanced equation is

Na₂CO₃ + H₂SO₄ → Na₂SO₄ + H₂O + CO₂

Concentration of H₂SO₄= 0.05 mol/L

Volume of H₂SO₄ = 20 ml = 0.02 L

number of moles of H₂SO₄= 0.05× 0.02 = 0.001mol

number of moles of H₂SO₄ = n(Na2CO3) = 0.001 mol

Molar mass of Na₂CO₃= 106 g/mol

mass of Na₂CO₃ = n× M = 0.001 × 106 = 0.106g

change in mass =∆m = 0.276 – 0.106 = 0.17 g

Molar mass of (H₂O = 18 g/mol

n umber of moles of water = mass÷ Molar mass=0.17 g÷18=0.0094mol

x= number of moles of H₂O÷ number of moles of sodium carbonate

 = 0.0094mol÷0.001

  =10

Therefore the value of x is 10

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Answer:

LICl + K

Explanation:

Li + KCl ---> LiCl + K

reason: Potassium gives one chlorine to Li as in a single displacement reaction but K remains alone.

C₃H₈ has an empirical formula mass of 44 g/mol. Hence, C₃H₈ is also the molecular formula.

When twο οr mοre distinct elements are chemically mixed in a specific ratiο, the resultant substance is knοwn as a cοmpοund.  Chemical bοnds, such as cοvalent bοnds, iοnic bοnds, οr metallic bοnds, hοld the cοnstituent parts οf a cοmpοund tοgether.

We must first cοmpute the mοles οf each element present in the cοmpοund in οrder tο derive its empirical fοrmula.

We can figure οut hοw many mοles οf carbοn and hydrοgen there are in 100 g οf the chemical by dοing the fοllοwing calculatiοn:

81.71 g / 12.011 g/mol = 6.80 moles of carbon

18.29 g / 1.008 g/mol of hydrogen = 18.15 moles.

The simplest whole number ratio of carbon to hydrogen atoms must next be determined. In order to do this, we divide both values by the lesser amount, in this case 6.80 moles:

Carbon: 6.80 moles / 6.80 moles = 1.00 Hydrogen: 18.15 moles / 6.80 moles = 2.67

To get whole numbers, we must multiply both values by a factor. By multiplying each by 3, the smallest whole number ratio is obtained:

1.00 x 3 = 3 carbon

2.67 x 3 = 8.01 x 8 for hydrogen.

As a result, the compound's empirical formula is C₃H₈.

Knowing the compound's molecular weight is necessary to determine its molecular formula. The compound's MWt (molecular weight) is 44, as stated.

C₃H₈ has an empirical formula mass of 44 g/mol. Hence, C₃H₈ is also the molecular formula.

Any molecule with the empirical formula C₃H₈, like propane, might be the structural formula of the chemical (CH₃CH₂CH₃).

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A 5 carat diamond has a mass of 1.000 g. If the gemstone displaces a liquid from 1.00 mL to 1.30 mL, and the density of the diamond is 3.33 g/mL

To calculate the density of the diamond, we can use the formula:

Density = Mass / Volume

Given that the mass of the diamond is 1.000 g and it displaces a liquid from 1.00 mL to 1.30 mL, we can substitute these values into the formula:

Density = 1.000 g / (1.30 mL - 1.00 mL)

By subtracting the initial volume from the final volume, we find:

Density = 1.000 g / 0.30 mL

Performing the division, we obtain:

Density = 3.33 g/mL

Therefore, the density of the diamond is 3.33 g/mL.

It's important to note that the units for density are grams per milliliter (g/mL) because the mass is given in grams and the volume is given in milliliters.

Understanding and applying the formula for density allows us to determine the mass per unit volume of a substance. In this case, the density provides information about how much mass is contained within a given volume for the diamond.

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Answer:

Explanation:

The crabs cannot see the plankton they eat near the ocean floor. For the crabs to see the plankton, some color of visible light would need to reach the plankton so that it can be reflected into the crabs' eyes.

Crabs cannot see the plankton they eat near the ocean floor.

Plankton are a collection of tiny organisms that live at and beneath the surface of lakes, rivers, ponds, and oceans across the planet.

Some crabs have colour coding capabilities , but the crabs to see the Planktons the light must reach the plankton , reflect and reach to the eyes of crabs , but the light does not reach inside the ocean and so crabs see the plankton they eat near the ocean floor.

Hence crabs cannot see the plankton they eat near the ocean floor.

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The pH of the 0.15 M \(CH_3COOH\) solution is approximately 2.38. and around 2.9% of the \(CH_3COOH\) molecules in the 0.15 M solution are deprotonated.

Acetic acid (\(CH_3COOH\)) is a weak acid that only partially dissociates in water to form \(H^+\) ions and \(CH_3COO^-\) ions. To estimate the pH and fraction of \(CH_3COOH\)molecules deprotonated in a 0.15 M \(CH_3COOH\)solution, we can use the following equations and approximations:

The dissociation constant for acetic acid (Ka) is 1.8 x 10^-5.

The initial concentration of \(CH_3COOH\) is equal to its concentration at equilibrium, since it only partially dissociates.

The concentration of \(H^+\) ions is equal to the concentration of \(CH_3COO^-\) ions at equilibrium, since the dissociation reaction involves a 1:1 ratio of \(H^+\) ions to \(CH_3COO^-\) ions.

Using these approximations, we can set up an equilibrium expression for the dissociation of \(CH_3COOH\) :

\(Ka = [H^+][CH_3COO^-]/[CH_3COOH]\)

We also know that the initial concentration of \(CH_3COOH\) is 0.15 M. Let x be the concentration of \(H^+\) ions and \(CH_3COO^-\) ions at equilibrium. Then:

[\(H^+\)] = x

[\(CH_3COO^-\)] = x

[\(CH_3COOH\)] = 0.15 - x

Substituting these values into the equilibrium expression and solving for x, we get:

Ka = x^2 / (0.15 - x)

1.8 x 10^-5 = x^2 / (0.15 - x)

x = 0.0042 M

The pH can be calculated using the formula:

pH = -log[\(H^+\)]

pH = -log(0.0042)

pH = 2.38

Therefore, the pH of the 0.15 M \(CH_3COOH\) solution is approximately 2.38.

To estimate the fraction of \(CH_3COOH\) molecules that are deprotonated, we can use the equation:

Fraction deprotonated = \([CH_3COO^-] / [CH_3COOH]\) x 100%

At equilibrium, the concentration of \(CH_3COO^-\) ions is equal to the concentration of \(H^+\) ions, which we calculated to be 0.0042 M. The concentration of \(CH_3COOH\) at equilibrium is 0.15 - 0.0042 = 0.1458 M. Substituting these values into the equation, we get:

Fraction deprotonated = 0.0042 / 0.1458 x 100%

Fraction deprotonated = 2.9%

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Answer:

using ideal gas equation =12.4576atm to 4.significant figure

using vander Waals equation = 12.3504

The differences is 0.10atm

The chemical names of the given chemical compounds include the following:

4.)Lead(II) nitrate

5.)Aluminium sulphide

6.)Nickel(II) sulfate

7.)Strontium carbonate

8.)lithium acetate

9.)Calcium Phosphate

10.)Ammonium Bromate

11.)Tin(II) oxide

12.) Zinc Chloride

13.) Barium nitride

14.)Gold(III) hydroxide

15.)Copper(I) Chlorate.

A chemical name of a compound is defined as the name given to a chemical compound that shows the presence of each of the elements that makes up the compound.

The rules that should be used to write chemical compound names include the following:

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Answer:

Mass is conserverd

Explanation:

Any chemical reaction must follow the law of conservation of mass. Meaning that mass cannot be created nor destroyed. Since all chemical reactions must follow this law, no matter what reaction happens, mass is conserved.

Answer:

The one cell of a unicellular organism must be able to perform all the functions necessary for life. These functions include metabolism, homeostasis and reproduction. Specifically, these single cells must transport materials, obtain and use energy, dispose of wastes, and continuously respond to their environment.

Answer:

A single celled Organism is able to get energy from complex molecules, to move, and to sense their environment.

Explanation:

The ability to perform these and other functions is part of their organization.

1. The amount of water produced by the reaction is 0.359 g.

2. The amount of water produced by the reaction is 0.219 g.

To calculate the number of moles of acid neutralized by the tablet, subtract the number of moles of acid neutralized in the titration from the initial solution's moles of acid. Understand and explain standardization in the context of acidic and basic solutions used as reagents in experiments.

1. The neutralization reaction,

HCl (aq) + NaOH (aq) → NaCl (aq) + H2O (l)

we have to calculate the number of moles of HCl that react,

moles of HCl = volume of HCl x concentration of HCl

= 7.98 mL x 2.50 mol/L / 1000 mL/L

= 0.01995 mol

Since NaOH is in excess,

As a result, the amount of water produced will be equal to the amount of HCl that reacts:

moles of water = moles of HCl = 0.01995 mol

we can use the molar mass of water (18.015 g/mol)

mass of water = moles of water x molar mass of water

= 0.01995 mol x 18.015 g/mol

= 0.359 g

2. The neutralization reaction between H2SO4 and NaOH is:

H2SO4 (aq) + 2 NaOH (aq) → Na2SO4 (aq) + 2 H2O (l)

we have to calculate the number of moles of H2SO4 that react,

moles of H2SO4 = volume of H2SO4 x concentration of H2SO4

= 2.43 mL x 2.50 mol/L / 1000 mL/L

= 0.00608 mol

Now, we have to calculate the number of moles of NaOH that react:

moles of NaOH = volume of NaOH x concentration of NaOH

= 2.51 mL x 3.00 mol/L / 1000 mL/L

= 0.00753 mol

we need to use the stoichiometry of the balanced equation,

moles of water = moles of H2SO4 x (2 moles of water / 1 mole of H2SO4)

= 0.00608 mol x 2

= 0.01216 mol

we can use the molar mass of water (18.015 g/mol)

mass of water = moles of water x molar mass of water

= 0.01216 mol x 18.015 g/mol

= 0.219 g

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Answer:  A and C are the right answer.

Explanation:

From the given choice of substances, the one that is not a molecule is Mn.

Molecules are defined as the smallest unit of a substance that can exist and still retain the properties of that substance.

Molecules of substance are usually formed from the atom of the substances when two or more atoms of the same elements or different elements combine.

Molecules of substances are capable of existing alone but atoms of an element must combine to form molecules.

Considering the given substance to determine whether they are molecules or not:

H₂S is a molecule because it contains more than one atom of elements.

O₃ is a molecule because it contains more than one atom of elements.

KOH is a molecule because it contains more than one atom of elements.

Mn is a not molecule because it contains only one atom of elements.

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The energy of combustion for one mole of butane to be approximately 2888.81 kJ/mol.

To calculate the energy of combustion for one mole of butane (C4H10), we need to use the information provided and apply the principle of calorimetry.

First, we need to convert the mass of the butane sample from grams to moles. The molar mass of butane (C4H10) can be calculated as follows:

C: 12.01 g/mol

H: 1.01 g/mol

Molar mass of C4H10 = (12.01 * 4) + (1.01 * 10) = 58.12 g/mol

Next, we calculate the moles of butane in the sample:

moles of butane = mass of butane sample / molar mass of butane

moles of butane = 0.367 g / 58.12 g/mol ≈ 0.00631 mol

Now, we can calculate the heat released by the combustion of the butane sample using the equation:

q = C * ΔT

where q is the heat released, C is the heat capacity of the calorimeter, and ΔT is the change in temperature.

Given that the heat capacity of the bomb calorimeter is 2.36 kJ/°C and the change in temperature is 7.73 °C, we can substitute these values into the equation:

q = (2.36 kJ/°C) * 7.73 °C = 18.2078 kJ

Since the heat released by the combustion of the butane sample is equal to the heat absorbed by the calorimeter, we can equate this value to the energy of combustion for one mole of butane.

Energy of combustion for one mole of butane = q / moles of butane

Energy of combustion for one mole of butane = 18.2078 kJ / 0.00631 mol ≈ 2888.81 kJ/mol

Therefore, the energy of combustion for one mole of butane is approximately 2888.81 kJ/mol.

In conclusion, by applying the principles of calorimetry and using the given data, we have calculated the energy of combustion for one mole of butane to be approximately 2888.81 kJ/mol.

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Cations are formed through lose of electrons and thus, possess positive charge. Cations has smaller ionic radius than the neutral atom, because, they have are fewer  electrons and thus experience more nuclear attractive pull.

Cations are charged particles or ions formed by the lose of electrons from the neutral atom. Atoms acquires positive charge when they lose electrons and acquire negative charge by gaining electrons  and the negative ions are called anions.

For example, Na metal loses one electrons to form the cation Na+. Similarly Mg loses two electrons, to form Mg²⁺  ion. When these atoms loses electrons, the outermost shell can be emptied and the atomic radius shrinks to the penultimate shell.

Similarly, as the number of electrons reduces, the screening of electron from neighboring electrons reduces results in greater nuclear attractive pull and thereby the atomic radius shrinks to smaller than the neutral atom.

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An example of heterogeneous equilibrium is:

Heterogeneous equilibrium refers to an equilibrium state in a chemical reaction where the reactants and products exist in different physical states or phases. It occurs when substances in different phases, such as solids, liquids, and gases, are involved in a chemical reaction.

Considering the given equations:

The equation I: FeO(s) + CO(g) ⇌ Fe(s) + CO₂(g) represents a heterogeneous equilibrium.

This is because the reactants and products involve different phases (solid and gas). FeO is a solid (s), CO is a gas (g), Fe is a solid (s), and CO₂ is a gas (g). The reaction involves the conversion of a solid and a gas to another solid and a gas, and the equilibrium is established between these different phases.

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Daria had some beach sand with her. The sand has a 72 gramme mass. She calculated the volume using the graduated cylinder below. The graduated cylinder contains 15 mL of sand.

The volume of the sand is calculated using the graduated cylinder below. The sand's bulk is specified as 72 grammes.

We can use the water displacement method to calculate the volume of the sand. Following is a description of how to estimate the amount of sand using the water displacement method:

The graduated cylinder of water should first be measured for volume.

The graduated cylinder's water volume should then be measured after adding the sand to it. The volume of water increases by the same amount.

Let's use the provided problem to implement this approach.

In the beginning, there is 10 mL of water in the graduated cylinder. The graduated cylinder contains 25 mL of water once the sand has been added.

The amount of sand is therefore equal to the difference between the two volumes, which is: Sand volume equals final water volume minus initial water volume (25 - 10 = 15 mL).

As a result, there are 15 mL of sand in the graduated cylinder.

Answer : 15

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In testing for β-galactosidase, the observed low activity at the 20-minute time interval and the higher activity at the 70-minute time interval when lactose is present can be attributed to several factors. Firstly, it takes time for the bacteria to be induced into making the enzyme. The presence of lactose initiates the expression of β-galactosidase, but this process is not instantaneous.

Secondly, lactose permease, a protein responsible for transporting lactose into the cell, also requires time to bring lactose into the cell. Only after lactose has entered the cell can β-galactosidase come into contact with the substrate and cleave it. This time delay contributes to the low enzyme activity observed at the 20-minute time interval.

Additionally, it takes time for the bacteria to replicate and increase their cell density. As cell density increases, more enzyme is produced, resulting in the higher activity observed at the 70-minute time interval.

Lastly, the lac operon, which controls the expression of β-galactosidase and lactose permease, is known to be leaky. This means that even in the absence of lactose, there is a low level of β-galactosidase activity. However, this leakiness alone does not account for the significant increase in activity observed at the 70-minute time interval.

Overall, the difference in enzyme activity between the 20-minute and 70-minute time intervals can be explained by the time required for bacterial induction, lactose permease function, and bacterial replication, as well as the leakiness of the lac operon.

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250 L of 5.0% w/v glucose solution can be prepared using 125 g of glucose.

w/v = [ mass of solute (g) / volume of solution (mL) ] × 100

5.0 % w/v = [ 125 g / volume of solution (mL) ] × 100

volume of solution (mL) = [ 125 g / 5.0 % w/v ] x 100

volume of solution (mL) = [ 125 g / (5.0 / 100) ] x 100

volume of solution (mL) = [ 125 g / 0.05 ] x 100