The following ions contain the same number of electrons. Rank them in order of decreasing ionic radii.?A. Na+B. Mg 2+C. Al 3+D. O 2-E. N 3-F. F -

Answers

Answer 1

Answer:

N3- > O2- > F- > Na+ > Mg2+ > Al3+

Explanation:

The ionic radio can simply be defined as the distance from the center of the nucleus to the electron in the outermost shell.

It should be noticed that all these ions belong to elements in group 3 of the periodic table.

It must be noted that anions I.e negative ions are generally bigger than anions.

Hence, it is expected that the negative ions are bigger. The arrangement goes thus:

N3- > O2- > F- > Na+ > Mg2+ > Al3+

Answer 2
Final answer:

The ions are ranked according to the number of protons in their nucleus, from least to most, which corresponds to the decreasing ionic radii. Therefore, the order of decreasing ionic radii is: N 3-, O 2-, F -, Na+, Mg 2+, Al 3+.

Explanation:

The ions Na+, Mg 2+, Al 3+, O 2-, N 3- and F - all contain the same number of electrons so they are considered isoelectronic. The size of isoelectronic species depends on their nuclear charges, that is, the number of protons in their nucleus. An ion with more protons will have a stronger nuclear charge which attracts the negatively charged electrons more, causing the ion to be smaller.

So, when you put the ions in order of decreasing ionic radii, the ion with the most protons will be the smallest. Therefore, the order of decreasing ionic radii will be: N 3- < O 2- < F - < Na+ < Mg 2+ < Al 3+. This indicates that N 3- has the largest ionic radius while Al 3+ has the smallest ionic radius among these ions.

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Related Questions

Dalton's experiments advanced our understanding of the atom withoutA.carefully observing and measuring events.B.discovering the internal structure of the atom.C.examining the ratios of elements involved in chemical reactions.D.testing a hypothesis.

Answers

Answer:

B: Discovering the internal structure of the atom.

Explanation:

Dalton’s experiments with gases marked the beginning of the modern era of chemistry. The hypotheses about the nature of matter on which Dalton’s atomic theory is based can be

summarized as follows:

Elements are composed of extremely small particles called atoms. All atoms of a given element are identical, having the same size, mass, and  chemical properties. The atoms of one element are different from the atoms of  all other elements. Compounds are composed of atoms of more than one element. In any compound,  the ratio of the numbers of atoms of any two of the elements present is either an  integer or a simple fraction. A chemical reaction involves only the separation, combination, or rearrangement  of atoms; it does not result in their creation or destruction.

Dalton made no attempt to describe the structure or composition  of atom —he had no idea what an atom is really like. But he did realize that the  different properties shown by elements such as hydrogen and oxygen can be explained  by assuming that hydrogen atoms are not the same as oxygen atoms.

According to Newton�s first law of motion, if there is no net force (unbalanced forces) acting on an object that is moving at a constant 30 mph, the object will:


Continue to move at 30 mph

Eventually come to a stop

Change its inertia

Accelerate

Answers

Answer:

Continue to move at 30 mph

Explanation:

Newton's first law of motion :

"Every object persists in its state of rest or uniform motion in a straight line unless it is compelled to change that state by forces exerted on it."

From the question we know that the net forces on the object were zero or that the there were no unbalanced forces on it.

Therefore we can assume that the object is moving along a straight line.

And the object was moving at a constant speed of 30 mph.

So it is clear from the Newton's first law that the object will remain in the state of motion as it was earlier.

That is the object will remain in motion at constant speed of 30 mph.

A researcher is performing a gram stain on a specimen in a medical laboratory. She gets distracted by a phone call and accidentally uses water instead of alcohol to decolorize. What may happen as a result of this mistake?

Answers

Answer:

Explanation:

It wouldn't be able to separate negative gram from positive. it would not wash away some of the crystal violet stain effectively. Gram - Cells will appear to be the same hue. Water can also dilute cells or wash them off the slide

Answer:

Explanation:

Would not be able to differentiate gram - from + it would not efficiently wash some of the crystal violet stain away. gram - cells will still remain the same color. also water may dilute or wash cells off the slide

On Earth, the white colored paper represents areas that are A) covered with snow and reflect sunlight. B) heavily populated and use the sunlight. Eliminate C) covered with water and reflect sunlight. D) covered with concrete and absorb sunlight.

Answers

Option A, covered in snow.

In an ecosystem model of earth , white colored paper represents areas that are covered with snow and reflect sunlight.

What is an ecosystem?

Ecosystem is defined as a system which consists of all living organisms and the physical components with which the living beings interact. The abiotic and biotic components are linked to each other through nutrient cycles and flow of energy.

Energy enters the system through the process of photosynthesis .Animals play an important role in transfer of energy as they feed on each other.As a result of this transfer of matter and energy takes place through the system .Living organisms also influence the quantity of biomass present.By decomposition of dead plants and animals by microbes nutrients are released back in to the soil.

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Consider the following reduction potentials:
2H+ + 2e– → H2 E° = 0.00 V Ni2+ + 2e– → Ni E° = -0.25 V Cd2+ + 2e– → Cd E° = -0.40 V
Based on this information which pairs of substances will react spontaneously under standard state conditions?

Answers

Answer:

H⁺/H₂//Ni/Ni⁺²; H⁺/H₂//Cd/Cd⁺²; Ni⁺²/Ni//Cd/Cd⁺²

Explanation:

The redox reactions are spontaneously when the E° of the substance that is reduced is higher than the E° of the substance that is oxidized. The substance is reduced when gaining electrons and is oxidized when loses electrons.

So, the reactions that occur spontaneously are:

2H⁺ + Ni → H₂ + Ni⁺² (H⁺ is reduced and Ni is oxidized)

2H⁺ + Cd → H₂ + Cd⁺² (H⁺ is reduced and Cd is oxidized)

Ni⁺² + Cd → Ni + Cd⁺² (Ni⁺² is reduced and Cd is oxidized)

Pairs: H⁺/H₂//Ni/Ni⁺²; H⁺/H₂//Cd/Cd⁺²; Ni⁺²/Ni//Cd/Cd⁺²

To determine the moles of ammonium chloride produced, we need to find the limiting reactant. H2O is present in an unlimited amount, so it cannot be the limiting reactant. Let’s then determine the amount of each remaining reactant present (in moles). NaCl is already given in moles (0.250 mol). NH3 and CO2 are not. They are given in liters. How many moles are present in 3.08 L of NH3 and 2.09 L of CO2?

Answers

Answer:

0.094 mols of CO2 are present in 2.09 L.

0.132 mols of NH3 are present in 3.08 L

The moles of ammonium chloride produced are 0.094

Explanation:

First of all, think the reaction

NH3 + 2H2O + NaCl + CO2 → NH4Cl + NaOH + H2CO3

To get the moles with volume, you must need density

Density = mass / volume

Density CO2 = 0,001976 g/ml

Density NH3 =  0,00073 g/ml

Volume of CO2 = 2.09L

2.09 L . 1000 = 2090 mL

0,001976 g/ml = mass / 2090 mL

4.13 g = mass

Molar weight = 44 g/m

mass / molar weight = 4.13 g / 44 g/m = 0.094 moles

Volume of NH3 = 3.08 L

3.08 L . 1000 = 3080 mL

0,00073 g/ml = mass/ 3080 mL

2.25 g = mass

Molar weight = 17 g/m

mass / molar weight = mols → 2.25 g / 17g/m = 0.132 mols

In my reagents, the least amount I have is CO2. This is my limiting reactant. Take account the reaction.

1 mol of NH3 reacts with 1 mol of CO2, so as I have 0.132 moles of NH3 I need 0.132 moles of CO2; I only have 0.094.

Relation between CO2 and NH4Cl is 1:1 so, 1 mol of CO2 is needed to produce 1 mol e chloride, so 0.094 mols are needed to produce the same amount of chloride.

An aqueous mixture containing starch (a colloid), Nacl, glucose, and albumin (a colloid) is placed in a dialyzing bag and in distilled water. Which of the following correctly describes the location of the indicated substance after dialysis?
a) starch outside
b) water inside only
c) albumin inside
d) albumin inside and outside
e) starch inside and outside

Answers

Answer:

c) albumin inside

Explanation:

Let us see look closer at each answer and see why it is wrong/correct.

a) Starch cannot be outside the bag because the bag isn't permeable to starch. The starch stays inside

b) There cannot be water inside only because it would mean starch and albumin diffused out and they cannot pass through the bag

c) This is correct. The bag isn't permeable to albumin so it stays inside

d) Albumin cannot be outside the bag because the bag isn't permeable to albumin. The albumin stays inside

e) Starch cannot be outside the bag because the bag isn't permeable to starch. The starch stays inside

Final answer:

After dialysis, albumin - a colloid, will remain inside the dialyzing bag because the dialysis membrane is designed to let only small molecules and ions pass through, not large colloidal particles. Starch, another colloid, will remain inside for the same reason. Water will reach equilibrium, moving in and out of the membrane.

Explanation:

The question relates to the process of dialysis, which is used to separate different substances in a mixture based upon their size. An aquatic mixture containing both colloids like starch and albumin and small molecules like NaCl (sodium chloride) and glucose is placed in a dialyzing bag to see where the substances will be after the process is complete. A characteristic of dialysis is that the dialyzing membrane allows small molecules and ions to pass through, while retaining larger colloidal particles.

Referring to the given options, the correct descriptions after dialysis are as follows:

Therefore, the accurate statement concerning the location of the substances post-dialysis is

c) albumin inside

Which of the following is not true for an exothermic reaction?
A. the products have a higher enthalpy than reactants
B. the temperature of the surroundings rise
C. the enthalpy change for the reaction is negative heat flows from the rxn system to the surroundings

Answers

Answer:

The false statement regarding an exothermic reaction is: A. the products have a higher enthalpy than reactants

Explanation:

An exothermic reaction is a type of chemical reaction that involves the release of energy from the system to the surroundings. Thus increasing the temperature of the surroundings.

In this reaction, the enthalpy or energy of the reactants is greater than the enthalpy or energy of the products. ([tex]\Delta H_{f} (Products) < \Delta H_{f} (reactants)[/tex])

As the enthalpy change of a reaction: [tex]\Delta H_{r} = \sum \Delta H_{f} (Products) - \sum \Delta H_{f} (reactants)[/tex]

Therefore, the enthalpy change for an exothermic reaction is negative ([tex]\Delta H_{r} < 0[/tex])

The pH of 0.50 M benzoic acid is 2.24. Calculate the change in pH when 2.64 g of C6H5COONa is added to 38 mL of 0.50 M benzoic acid, C6H5COOH. Ignore any changes in volume. The Ka value for C6H5COOH is 6.5 x 10-5.

Answers

Answer:

1.93

Explanation:

Moles of C[tex]_{6}[/tex]H[tex]_{5}[/tex]COOH = 38/1000 × 0.50 = 0.019mol

Moles of C[tex]_{6}[/tex]H[tex]_{5}[/tex]COONa = Mass/Molar mass = 2.64/144.10 = 0.018321mol

Final pH = pKa + log([C[tex]_{6}[/tex]H[tex]_{5}[/tex]COONa]/[C[tex]_{6}[/tex]H[tex]_{5}[/tex]COOH]

= -log Ka  + log(mols of C[tex]_{6}[/tex]H[tex]_{5}[/tex]COONa]/mols of C[tex]_{6}[/tex]H[tex]_{5}[/tex]COOH

= -log(6.5 × 10^(-5)) + log (0.018321/0.019)=4.17

change in pH = final - initial pH

= 4.17 - 2.24

=1.93

pH is the depiction of the acidity and basicity of the solution and can range from 0 to 14 on the pH scale. The change in pH of benzoic acid is 1.93.

What is pH?

pH is the concentration of the hydrogen ion in the solution depicting the acidity and the basicity of the solution.

Calculate moles of Benzoic acid as:

[tex]\begin{aligned}\rm moles &= \dfrac{\rm mass}{\rm molar \;mass}\\\\&=\dfrac{38}{1000}\times 0.50\\\\&= 0.019\;\rm mol\end{aligned}[/tex]

Calculate moles of Sodium benzoate as:

[tex]\begin{aligned}\rm moles &= \dfrac{\rm mass}{\rm molar \;mass}\\\\&=\dfrac{2.64}{144.10}\\\\&= 0.01832\;\rm mol\end{aligned}[/tex]

pH can be calculated as:

[tex]\begin{aligned}\rm pH &= \rm pKa + log\dfrac{([C_{6}H_{5}COONa]}{[C_{6}H_{5}COOH]}\\\\&= \rm -log(6.5 \times 10^{-5}) + log (\dfrac{0.018321}{0.019})\\\\&= 4.17\end{aligned}[/tex]

Change in the pH can be calculated as:

[tex]\begin{aligned} \rm pH &= \rm final - initial \; pH\\\\&= 4.17 - 2.24\\\\&=1.93\end{aligned}[/tex]

Therefore, 1.93 is the change in the pH of benzoic acid.

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Barium can be analyzed by precipitating it as BaS04 and weighing the precipitate. When a 0.713-g sample of a barium compound was treated with excess H2S04, 0.5331 g of BaSO4 formed. What is the percentage of barium in the compound?
Molecular mass of BaSO4 233.39 g/mol
Molecular mass of Ba = 137.327 g/mol

Answers

Answer:

31.37%

Explanation:

For this case, you should consider the following reaction:

Ba⁺²₍aq₎ + H₂SO₄ ₍aq₎ → BaSO₄ ₍s₎ + H₂O

For which you obtain the precipitate of BaSO₄

In order to obtain the mas of Barium on the precipitate, you may use the following formula:

gBa= M₍BaSO₄₎x(M₍Ba₎/M₍BaSO₄₎)

Where:

gBa= mass of Barium

M₍BaSO₄ ₎= mass of BaSO₄ from the precipitate

M₍Ba₎= mass of Barium from the original sample

M₍BaSO₄₎= mass of BaSO₄ from the precipitate

gBa= (0.5331)x(137.327/233.39)= 0.3136 g

Then we ontain the percentage of Barium multiplying by 100:

% Ba on the original sample= 31.36%

Final answer:

The percentage of barium in the compound is calculated by finding the moles of BaSO₄, converting it to moles of Ba, and then finding the mass of Ba before expressing it as a percentage of the initial sample mass. The resulting percentage of barium is approximately 43.99%.

Explanation:

To determine the percentage of barium in the compound, we must calculate the moles of BaSO₄ and then use the molar mass of barium to find the mass of barium in the initial sample. We can calculate the moles of BaSO₄ using the following steps:

First, use the mass of the BaSO₄ precipitate formed (0.5331 g) and the molar mass of BaSO₄ (233.39 g/mol) to find the moles of BaSO₄.

Moles of BaSO₄ = mass of precipitate / molar mass of BaSO₄

= 0.5331 g / 233.39 g/mol

= 0.002284 mol (rounded to six decimal places)

Since the molar ratio of Ba to BaSO₄ is 1:1, the moles of Ba will be the same as the moles of BaSO₄.Next, multiply the moles of Ba by the molar mass of Ba to find the mass of Ba in the sample.Mass of Ba = moles of Ba × molar mass of Ba

= 0.002284 mol × 137.327 g/mol

= 0.3137 g (rounded to four decimal places)

Finally, we calculate the percentage of barium in the sample using the initial sample mass (0.713 g).

Percentage of barium = (mass of Ba / initial sample mass) × 100%

= (0.3137 g / 0.713 g) × 100%

= 43.99%

Therefore, the percentage of barium in the compound is approximately 43.99%.

Which of the following were found to destroy ozone in the upper atmosphere?A. sulfur oxides.B. CFCs.C. nitrogen oxides.D. carbon dioxide.

Answers

Answer:

sulfur oxides

Explanation:

Which of the following processes is endothermic?A. a hot cup of soup (system) cools on a countertop B. the combustion of butaneC. the chemical reaction in a "hot pack" often used to treat sore musclesD. the vaporization of rubbing alcohol E. the freezing of water

Answers

Answer: Option (D) is the correct answer.

Explanation:

A reaction in which heat energy is absorbed by the molecules of a substance is known as an endothermic reaction.

Whereas when energy is released by the molecules of a substance in a reaction then it is known as an exothermic reaction.

So, when a hot cup of soup (system) cools on a countertop then energy is released into the surrounding by the soup. Hence, it is an exothermic reaction.Also, combustion reactions are always exothermic in nature.The chemical reaction in a "hot pack" often used to treat sore muscles because heat is released to the muscles. Therefore, it is an exothermic reaction.When rubbing alcohol vaporizes then heat is absorbed by the molecules of alcohol. Due to which they gain energy and escape into the surrounding. Therefore, it is an endothermic reaction.Freezing of water is an exothermic reaction because during this process energy is released by the molecules of water. Hence, it is an exothermic reaction.

Thus, we can conclude that the vaporization of rubbing alcohol is endothermic.

Final answer:

Among the provided options, the vaporization of rubbing alcohol is the endothermic process, as it involves the absorption of heat from its surroundings.

Explanation:

In the context of chemistry, an endothermic process is one in which heat is absorbed from the surroundings. This results in an overall increase in the internal energy of the system. Given the options, the vaporization of rubbing alcohol is the process that is endothermic. This is because when rubbing alcohol vaporizes, it absorbs heat from its surroundings, causing the surrounding environment to feel cooler. Other examples might be the melting of ice or the evaporation of water, both of which require an external heat source to facilitate the process.

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Hydrogen gas produced in the laboratory by the reaction of zinc and hydrochloric acid was 23) collected over water at 25°C. The barometric pressure at the time was 742.5 mm Hg. What is the pressure of dry hydrogen gas if the vapor pressure of water at 25'C is 23.8 mm Hg?
A) 760.0 mm Hg
B)718.7 mm Hg
C)742.5 mm Hg
D) 766.3 mm Hg
E) none of the above

Answers

Answer:

Explanation:

The balanced chemical reaction is:

[tex]Zn+2HCl\rightarrow ZnCl_2+H_2[/tex]

According to Dalton's law, the total pressure is the sum of individual pressures.

[tex]p_{total}=p_1+p_2[/tex]

As the hydrogen is collected over water, the total pressure will be sum of pressure of water and pressure of dry hydrogen.

[tex]p_{total}=p_{H_2}+p_{H_2O}[/tex]

[tex]p_{total}[/tex] = 742.5 mm Hg

[tex]p_{H_2}[/tex] = ?

[tex]p_{H_2O}[/tex] = 23.8 mm Hg

Putting in the values:

[tex]742.5=p_{H_2}+23.8[/tex]

[tex]742.5-23.8=p_{H_2}[/tex]

[tex]p_{H_2}=718.7mmHg[/tex]

Thus the pressure of dry hydrogen gas is 718.7 mm Hg

Which of the following is true?
a. A reaction in which the entropy of the system increases can be spontaneous only if it is endothermic.
b. A reaction in which the entropy of the system decreases can be spontaneous only if it is endothermic.
c. A reaction in which the entropy of the system decreases can be spontaneous only if it is exothermic.
d. A reaction in which the entropy of the system increases can be spontaneous only if it is exothermic.

Answers

Answer:

a. A reaction in which the entropy of the system increases can be spontaneous only if it is endothermic.

Explanation:

The change in free energy (ΔG) that is, the energy available to do work, of a system for a constant-temperature process is:

[tex]ΔG = ΔH - TΔS[/tex]

When ΔG < 0 the reaction is spontaneous in the forward direction.When ΔG > 0 the reaction is nonspontaneous. The reaction is

spontaneous in the opposite direction.

When ΔG = 0 the system is at equilibrium.

If both ΔH and ΔS are positive, then ΔG will be negative only when the TΔS  term is greater in magnitude than ΔH. This condition is met when T is large.

Photophosphorylation produces ___ , which stores chemical energy that can be used for the Calvin cycle.

Answers

ATP, adenosine triphosphate

Photophosphorylation of sugar moiety produces ATP  (adenosine triphosphate) which stores chemical energy that can be used for the Calvin cycle.

What is Calvin cycle?

The Calvin cycle, often known as the Calvin-Benson cycle after its discoverers, is a series of chemical processes that collect carbon dioxide.

It is also referred to as the C3 cycle alone. C3 plants are defined as those that only fix carbon through the Calvin cycle. In the stroma of chloroplasts, carbon dioxide diffuses and mixes with the five-carbon sugar ribulose 1, 5-biphosphate (RuBP).

Rubisco, a large molecule that may be the most prevalent organic compound on Earth, is the name of the enzyme that catalyzes this reaction.

The three-carbon chemical 3-PGA is changed into another three-carbon compound termed G3P by ATP and NADPH using their stored energy. A reduction reaction is the name for this kind of reaction.

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The atomic number of an element is the same as the number of its

Answers

Answer: The atomic number of an element is the same as the number of its protons.

Explanation: For example, the atomic number of oxygen is 8. So, this would mean that the number of protons would be 8.

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When 0.0400 mol of HCl(aq) is reacted with 0.0400 mol of NaOH(aq) in 50.0 mL of water, the temperature of the solution increases by 9.82°C. What is the enthalpy of reaction for the following thermochemical equation?HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)Assume that the heat capacity of the solution and calorimeter is 227.4 J/°C.

Answers

The enthalpy of reaction for the following thermochemical equation, when the heat capacity of the solution and calorimeter is 227.4 J/°C is 55.7 kJ /mole.

What is thermochemical equation?

Thermochemical equation is defined as a chemical equations that are properly balanced and take into account the physical conditions of all reactants, products, and energy change. A chemical equation is an equation that shows the beginning molecule, reactants, and final products separated by arrows, but a thermochemical equation is a balanced stoichiometric chemical process that also includes the enthalpy change.

Given heat capacity = 227.14 J/°C

Temperature = 9.82 °C

Number of moles of HCl and NaOH is 0.04 moles

q (solution) = -C (solution) x ΔT

= 227.14 J/°C x 9.82 °C

= 2.2305 kJ

For enthalpy of reaction we have to divide it by mole of reactant

= 2.2305 / 0.04

= 55.4 kJ/ mole

Thus, the enthalpy of reaction for the following thermochemical equation, when the heat capacity of the solution and calorimeter is 227.4 J/°C is 55.7 kJ /mole.

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Final answer:

To determine the enthalpy of reaction for HCl reacting with NaOH, the heat released by the reaction is calculated using the mass of the solution, the specific heat capacity, and the change in temperature. The result is then divided by the number of moles reacted to express the enthalpy change in kJ per mole.

Explanation:

The problem here concerns the measurement of an enthalpy change during the reaction of hydrochloric acid (HCl) with sodium hydroxide (NaOH) to form sodium chloride (NaCl) and water (H2O). To find the enthalpy of reaction for the provided thermochemical equation, we use the temperature change and the heat capacity of the solution to calculate the total heat released by the reaction, and then express this heat change in kJ per mole of reactants.

First, calculate the total heat (q) released using the formula q = mc∆T, where m is the mass of the solution, c is the specific heat capacity, and ∆T is the change in temperature. As the reaction takes place in 50.0 mL of water, and assuming the density of water is 1.00 g/mL, the mass of the solution is 50.0 g. Therefore, q = 50.0 g × 227.4 J/°C × 9.82°C.

After calculating q, divide this value by the number of moles of HCl reacted to obtain the enthalpy of reaction per mole. Remember to convert from joules to kilojoules since enthalpy is usually expressed in kJ/mol. Please note that the reaction is exothermic, so the enthalpy change should be a negative value, indicating heat is released.

In which sample of water do the molecules have the highest average kinetic energy 1)20. ML at 100.°C 2)40. ML at 80. °C 3)60. ML at 60.° C 4)80. Ml at 40.° C

Answers

Answer:

1)20. ML at 100.°C

Explanation:

Average kinetic energy is not related with volume.  so increase in temperature will also increase kinetic energy. so this increase means the highest temperature will be a right answer.

Final answer:

The water sample with the highest average kinetic energy is the 20 mL of water at 100.0°C, as average kinetic energy is directly proportional to temperature.

Explanation:

The molecules with the highest average kinetic energy are in the sample of water at the highest temperature. Kinetic energy is directly proportional to the absolute temperature of the substance, meaning that as temperature increases, kinetic energy also increases. Therefore, ignoring the volume of the water samples, the rank from highest kinetic energy to lowest based on temperature would be:

20 mL of water at 100.0°C40 mL of water at 80.0°C60 mL of water at 60.0°C80 mL of water at 40.0°C

The sample at 100.0°C holds the highest average kinetic energy because it has the highest temperature.

The charge of B is +3 and the charge of O is -2. What is its formula?

A) B3O2

B) B2O3

C) O2B3

Answers

The formula is B2O3. The charges need to equal each other to obtain a neutral compound of no charge hence 2 B has a charge of +6 and 3 O has a charge of -6 to cancel out each other, resulting in a stable uncharged compound.
Option B: B2O3

In order to make a compound, it has to be neutral. We have to combine the +3 charged and the -2 charges until they cancel out.

The only possible compound is B2O3, since the positive charges add up to +6 and the negative charges add up to -6.

-T.B.

What is the full name of NI3

Answers

Answer:

Nitrogen triiodide is NI3.

Explanation:

Nitrogen trllodide (compound)

Redox reactions can be written as two half-reactions, focusing on the gain or loss of electrons by one of the chemical substances. One half-reaction shows the oxidation while the other shows the reduction. When the two half-reactions are combined, the overall reaction is obtained.The half-reactions can aid in the balancing of redox equations because the number of each element must be balanced as well as the number of electrons gained and lost. What substance is added to balance the hydrogen in a half-reaction?
A. H₂O₂
B. H₂O
C. H⁺
D. H₂

Answers

Answer:

C) H⁺

Explanation:

When we are balancing the reaction in an acid medium, hydrogen is balanced using the H⁺ species. This is most likely the intended answer of your question.

When the reaction takes place not in an acid medium, but in an alkaline one, then hydrogen is present as the OH⁻ species. However this option is not given in your question.

Thus the answer is option C).

An apparatus consists of a 4.0 dm^3 flask containing nitrogen gas at 25°c and 803 kPa, it is joined by a valve to a 10.0dm^3 flask containing argon gas 25°c and 47.2KPa. The valve is opened and the gases mix , what is the partial pressure of each gas and calculate the total pressure of the gas mixture

Answers

Answer:

so total pressure is = 261.92 kPa

partial pressure of Ar gas=  33.50 Kpa

partial pressure of N2 gas = 227  kPa

Explanation:

PV = nRT  

n = PV / RT

as we know that 1dm3 = 1 Liter  

smaller flask before mixing

ideal gas constant

n = (803 kPa) x (4L) / ((8.3144621 L kPa/K mol) x (25 + 273 K))  

n = (3212 kPa L) / ((8.3144621 L kPa/K mol) x (298 K))  

n = (3212 kPa L) /2477.7

n=  1.29 mol

In the larger flask:

n = (47.2 kPa) x (10 L) / ((8.3144621 L kPa/K mol) x (25 + 273 K)) =  

n = (472 kpaL/ 2477.7

n= 0.19 mol  

PV = nRT  

P = nRT / V  

After mixing:

P = (1.29 mol + 0.19 mol) x (8.3144621 L kPa/K mol) x (25+ 273 K) / (4 L + 10 L)  

    = (1.48) x (2477.7 / (14 L)    

    = 261.92 kPa total pressure  

so total pressure is = 261.92 kPa

a)

(261 kPa) x (0.19 mol Ar) / (1.29 + 0.19mol) =    

= 49.59/1.48 = 33.50 Kpa for Arg

b)

(261 kPa total) - (33.50 kPa Ar) = 227  kPa  for N2

partial pressure of Ar gas=  33.50 Kpa

partial pressure of N2 gas = 227  kPa

The enthalpy of combustion of acetylene C2H2 is described by

C2H2 (g) + (5/2)O2 (g) >>>>>>>CO2 (g) + H2O (l) Heat of Reaction (Rxn) = -1299kJ/mol

Calculate the enthalpy of formation of accetylene, given the following enthalpies of formation
Standard formation [CO2 (g)]= -393.5 kJ/mol
Standard formation [H2O (l)] = -285.8 kj/mol

Answers

Answer: The enthalpy of formation of accetylene is, 226.2 kJ/mol

Explanation:

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

The chemical equation for the combustion of acetylene follows:

(1) [tex]C_2H_2(g)+\frac{5}{2}O_2(g)\rightarrow 2CO_2(g)+H_2O(l)[/tex]

[tex]\Delta H^o_{rxn}=-1299kJ/mol[/tex]

The formation of [tex]CO_2[/tex] will be,

(2) [tex]C(s)+O_2(g)\rightarrow CO_2(g)[/tex]    [tex]\Delta H_f_{(CO_2)}=-393.5kJ/mol[/tex]

The formation of [tex]H_2O[/tex] will be,

(3) [tex]H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(l)[/tex]    [tex]\Delta H_f_{(H_2O)}=-285.8kJ/mol[/tex]

The formation of [tex]C_2H_2[/tex] will be,

(4) [tex]2C(s)+H_2(g)\rightarrow C_2H_2(g)[/tex]    [tex]\Delta H_f_{(C_2H_2)}=?[/tex]

Now we are reversing equation 1, multiplying equation 2 by 2 and then adding equation 1, 2 and 3, we get:

Reaction (1) :

[tex]2CO_2(g)+H_2O(l)\rightarrow C_2H_2(g)+\frac{5}{2}O_2(g)[/tex]

[tex]\Delta H^o_{rxn}=1299kJ/mol[/tex]

Reaction (2) :

[tex]2C(s)+2O_2(g)\rightarrow 2CO_2(g)[/tex]    [tex]\Delta H_f_{(CO_2)}=2\times -393.5kJ/mol=-787kJ/mol[/tex]

Reaction (3) :

[tex]H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(l)[/tex]    [tex]\Delta H_f_{(H_2O)}=-285.8kJ/mol[/tex]

[tex]\Delta H_f_{(C_2H_2)}=\Delta H^o_{rxn}+\Delta H_f_{(CO_2)}+\Delta H_f_{(H_2O)}[/tex]

[tex]\Delta H_f_{(C_2H_2)}=(1299kJ/mol)+(-787kJ/mol)+(-285.8kJ/mol)[/tex]

[tex]\Delta H_f_{(C_2H_2)}=226.2kJ/mol[/tex]

Therefore, the enthalpy of formation of accetylene is, 226.2 kJ/mol

Final answer:

To calculate the enthalpy of formation of acetylene, we can use the enthalpy change of the combustion reaction and the enthalpies of formation of CO2 and H2O. By rearranging the equation and applying Hess's law, we can determine the enthalpy of formation of acetylene.

Explanation:

The enthalpy of formation of a compound can be calculated using Hess's law, which states that the enthalpy change of a reaction is independent of the pathway taken. To calculate the enthalpy of formation of acetylene (C2H2), we can use the equation:

C2H2 (g) + (5/2)O2 (g) → CO2 (g) + H2O (l)

The enthalpy change of this reaction is -1299 kJ/mol. By rearranging the equation, we can compare it to the standard formation equations for CO2 (g) and H2O (l) to determine the enthalpy of formation of acetylene.

The enthalpy of formation of acetylene is:

ΔHf(C2H2) = -1299 kJ/mol - [(-393.5 kJ/mol) + (-285.8 kJ/mol)]

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Given that Δ H ∘ f [ Br ( g ) ] = 111.9 kJ ⋅ mol − 1 Δ H ∘ f [ C ( g ) ] = 716.7 kJ ⋅ mol − 1 Δ H ∘ f [ CBr 4 ( g ) ] = 29.4 kJ ⋅ mol − 1 calculate the average molar bond enthalpy of the carbon‑bromine bond in a CBr 4 molecule.

Answers

Answer:

283.725 kJ ⋅ mol − 1

Explanation:

C(s) + 2Br2(g) ⇒ CBr4(g) , Δ H ∘ = 29.4 kJ ⋅ mol − 1

[tex]\frac{1}{2}[/tex]Br2(g) ⇒ Br(g) ,  Δ H ∘ = 111.9 kJ ⋅ mol − 1

C(s) ⇒ C(g) ,  Δ H ∘ = 716.7 kJ ⋅ mol − 1

4*eqn(2) + eqn(3) ⇒ 2Br2(g) + C(s) ⇒ 4 Br(g) + C(g) , Δ H ∘ = 1164.3 kJ ⋅ mol − 1

eqn(1) - eqn(4) ⇒ 4 Br(g) + C(g) ⇒ CBr4(g) , Δ H ∘ = -1134.9 kJ ⋅ mol − 1

so,

   average bond enthalpy is [tex]\frac{1134.9}{4}[/tex] = 283.725 kJ ⋅ mol − 1

The  average molar bond enthalpy of the carbon‑bromine bond in a CBr 4 molecule is 283.725 kJ ⋅ mol − 1

Calculation of average molar bond:

C(s) + 2Br2(g) ⇒ CBr4(g) , Δ H ∘ = 29.4 kJ ⋅ mol − 1

Br2(g) ⇒ Br(g) ,  Δ H ∘ = 111.9 kJ ⋅ mol − 1

C(s) ⇒ C(g) ,  Δ H ∘ = 716.7 kJ ⋅ mol − 1

4*eqn(2) + eqn(3)

⇒ 2Br2(g) + C(s)

⇒ 4 Br(g) + C(g) , Δ H ∘

= 1164.3 kJ ⋅ mol − 1

eqn(1) - eqn(4)

⇒ 4 Br(g) + C(g)

⇒ CBr4(g) , Δ H ∘

= -1134.9 kJ ⋅ mol − 1

so,

average bond enthalpy is  = 283.725 kJ ⋅ mol − 1

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A 1.50-kilogram ball is attached to the end of a 0.520-meter string and swung in a circle. The velocity of the ball is 9.78 m/s. What is the tension in the string?

Answers

Answer:

F centripetal force (tension) = 275.9  N

Explanation:

Given data:

Mass = 1.50 kg

Radius = 0.520 m

Velocity of ball = 9.78 m/s

Tension = ?

Solution:

F centripetal force (tension) =  m.v² / R

F centripetal force (tension) = 1.50 kg . (9.78 m/s)² / 0.520 m

F centripetal force (tension) = 1.50 kg . 95.65 m²/s² / 0.520 m

F centripetal force (tension) = 143.5 kg. m²/s² / 0.520 m

F centripetal force (tension) = 275.9  N

How many atoms of fluoride are there in one molecule of nitrogen trifluoride?

Answers

Answer:

there are three Fluorine atom. This is because Nitrogen give 3 electron away but fluorine can only take 1. So nitrogen gives 1 neutron to 3 fluorine each which makes it a stable compound

Answer : The number of atoms of fluoride present in one molecule of [tex]NF_3[/tex] are, 3.

Explanation :

Molecule : It is defined as the smallest particle in an element or a compound which has chemical properties of that element or a compound.

Molecules are made up of atoms and the atoms are bonded by the sharing of electrons.

The given molecules is, [tex]NF_3[/tex]

The nitrogen trifluoride molecule is made up of two elements which are nitrogen and fluorine.

In the given molecules [tex]NF_3[/tex], there are 1 nitrogen atom and there are 3 fluorine atoms.

Hence, the number of atoms of fluoride present in one molecule of [tex]NF_3[/tex] are, 3.

A .115 L sample of dry air has a pressure of 1.0895633 atm at 377 K. What is the volume of the sample if the temperature is increased to 483 K when the pressure is kept constant?

Answers

Answer:

V= 0.147 L

Explanation:

This is simply the application of combined gas law twice, to find the unknowns.

Combined gas law states that: [tex]PV = nRT[/tex]

P= pressure of air

V= volume of air

n= moles of air

R= Universal gas constant ( 0.08205 L atm mol⁻¹ K⁻¹)

T= Absolute temperature in kelvin.

[tex]n=\frac{P * 0.115}{R * 377}[/tex]

Now, applying the same gas law at 483K and substituting for n

[tex]V = \frac{P * 0.115}{R * 377} * \frac{R* 483}{P}[/tex]

V= [tex]\frac{0.115 * 483}{377}[/tex]

V= 0.147 L

A quantity of N2 gas originally held at 5.23 atm pressure in a 1.20 −L container at 26 ∘C is transferred to a 14.5 −L container at 20 ∘C. A quantity of O2 gas originally at 5.21 atm and 26 ∘C in a 5.10 −L container is transferred to this same container.

Answers

Final answer:

To find the final pressure of the gas mixture after transferring the gases, we can use the ideal gas law equation. The final pressure is calculated based on the total number of moles of gas and the final volume of the container. The final pressure of the gas mixture is 0.229 atm.

Explanation:

To find the final pressure of the gas mixture after the transfer, we can use the ideal gas law equation, which states that PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.

We can calculate the number of moles of N2 gas using the information given:

n = PV/RT = (5.23 atm)(1.20 L)/(0.0821 atm·L/mol·K)(26 + 273 K) = 0.287 mol

Similarly, we can calculate the number of moles of O2 gas:

n = PV/RT = (5.21 atm)(5.10 L)/(0.0821 atm·L/mol·K)(26 + 273 K) = 1.05 mol

To find the final pressure, we need to find the total number of moles of gas and the final volume of the container:

Total moles of gas = moles of N2 + moles of O2 = 0.287 mol + 1.05 mol = 1.34 mol

Final volume of the container = sum of the initial volumes = 1.20 L + 14.5 L = 15.7 L

Now we can use the ideal gas law again to calculate the final pressure:

P = nRT/V = (1.34 mol)(0.0821 atm·L/mol·K)(20 + 273 K)/(15.7 L) = 0.229 atm

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One difference between mixtures and pure substances is that

Answers

Answer:

The mixture is made up of different atoms and pure substance is made up of same type of atom.

The main difference is that mixture can be separated into its component by physical mean while pure substances can not be separated by physical process

Explanation:

Mixture:

The properties of the mixture are not same and contains the properties of all those component present in it.it is a combination of one or more Pure substances and can be separated by simple physical methods.it have varying boiling and melting point

Examples are:

mixture of salt and sandSalt water is mixture of water and NaCl and can be separated by physical mean.Alloys: its a mixture of different metalAir: mixture of gases

Pure Substance:

Pure substances are those made of same type of atoms all elements and compounds are pure substances.

it can not be separated by simple physical meanit have very constant and consistent melting and boiling point

Examples are:

Water : contain only water moleculeAll elements: all elements are pure substance made of same atomsAll compounds: can not be separated by physical mean.

A new potential heart medicine, code-named X-281, is being tested by a pharmaceutical company, Pharma-pill. As a research technician at Pharma-pill, you are told that X-281 is a monoprotic weak acid, but because of security concerns, the actual chemical formula must remain top secret. The company is interested in the drug's Ka value because only the dissociated form of the chemical is active in preventing cholesterol buildup in arteries.
To find the pKa of X-281, you prepare a 0.089 M test solution of X-281 at 25.0 ∘C. The pH of the solution is determined to be 2.40.
a. What is the pKa of X-281? Express your answer numerically.
At 25∘C, for any conjugate acid-base pair
pKa + pKb = 14.00
b. What is pKb of the conjugate base of X-281? (Assume 25 ∘C.) Express your answer numerically.

Answers

Answer:

a. pka = 3,73.

b. pkb = 10,27.

Explanation:

a. Supposing the chemical formula of X-281 is HX, the dissociation in water is:

HX + H₂O ⇄ H₃O⁺ + X⁻

Where ka is defined as:

[tex]ka = \frac{[H_3O^+][X^-]}{[HX]}[/tex]

In equilibrium, molar concentrations are:

[HX] = 0,089M - x

[H₃O⁺] = x

[X⁻] = x

pH is defined as -log[H₃O⁺]], thus, [H₃O⁺] is:

[tex][H_3O^+]} = 10^{-2,40}[/tex]

[H₃O⁺] = 0,004M

Thus:

[X⁻] = 0,004M

And:

[HX] = 0,089M - 0,004M = 0,085M

[tex]ka = \frac{[0,004][0,004]}{[0,085]}[/tex]

ka = 1,88x10⁻⁴

And pka = 3,73

b. As pka + pkb = 14,00

pkb = 14,00 - 3,73

pkb = 10,27

I hope it helps!

The correct answers are:

a. The pKa of X-281 is 2.40.

b. The pKb of the conjugate base of X-281 is 11.60.

a. To find the pKa of X-281, we can use the Henderson-Hasselbalch equation, which relates the pH of a solution to the pKa of the acid and the concentration of the acid and its conjugate base. For a monoprotic weak acid HA, the equation is:

[tex]\[ \text{pH} = \text{pKa} + \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right) \][/tex]

Given that the pH of the solution is 2.40 and the concentration of X-281 (HA) is 0.089 M, we can assume that the concentration of the conjugate base A^- is negligible compared to the concentration of the acid HA because the pH is close to the pKa. This means that the ratio[tex]\([\text{A}^-] / [\text{HA}]\)[/tex]is approximately zero, and thus the log term is approximately zero. Therefore, the pH is approximately equal to the pKa of the acid:

[tex]\[ \text{pH} \approx \text{pKa} \] \[ \text{pKa} \approx 2.40 \][/tex]

b. At 25°C, the sum of pKa and pKb for any conjugate acid-base pair is 14.00, which is the pKw of water:

[tex]\[ \text{pKa} + \text{pKb} = 14.00 \][/tex]

We have already determined that the pKa of X-281 is 2.40. Therefore, we can solve for the pKb of the conjugate base of X-281:

[tex]\[ 2.40 + \text{pKb} = 14.00 \] \[ \text{pKb} = 14.00 - 2.40 \] \[ \text{pKb} = 11.60 \][/tex]

Thus, the pKb of the conjugate base of X-281 is 11.60.

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