If 39.5 g AlCl3 is produced, how many grams of HCl was used in the reaction

The molar entropy from the question that we have here is 85.1kJ/K.mol as shown.

The molar entropy of a substance can be calculated by dividing the total entropy of the substance by the number of moles present. The entropy of a substance can be determined experimentally by measuring its heat capacity or by analyzing its thermodynamic behavior under different conditions.

We know that;

Entropy = ΔH/T

= 1.6 * 10^4 * 10^3J/mol/187.95 K

= 85.1kJ/K.mol

For the bromide ion;

Rate = 5/1 * 2.7 * 10^-3 mol/s

= 1.35 * 10^-2 mol/s

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Answer:

1.525834 (1.53 when accounting for significant figures).

Explanation:

This problem relies on the Ideal Gas Law, PV = nRT, where P is pressure, V is volume, n is moles, R is a specific constant, and T is temperature. In this problem, we are solving for n, moles, so we would rewrite it as n = PV/RT. Since the units here are moles, liters, atmospheres, and kelvin, R would be the value in atmosphere liter per mole kelvin, or 0.0821. From here, you just enter the values in the fraction and calculate.

For the significant figures, I followed the measurement of 34.2 L, giving 3, although an argument could be made for 1 significant figure from 1 atm, I imagine your professor would want something more specific than 2.

The correct answer is a) KBr.

KBr is an ionic compound composed of a metal (K) and a non-metal (Br). When this compound is added to water, the polar water molecules surround the ions in the solid and separate them, which leads to the compound dissolving in water.

What is Ionic Compound?

An ionic compound is a chemical compound composed of ions held together by electrostatic forces called ionic bonds. Ions are atoms or molecules that have gained or lost one or more electrons, giving them a positive or negative charge. In an ionic compound, a positively charged ion (cation) and a negatively charged ion (anion) are attracted to each other to form a stable compound.

CO2, CH4, and O2 are nonpolar molecules, and therefore, do not dissolve well in water. CO2 and O2 are gases at room temperature and pressure, while CH4 is a gas at room temperature but can be liquefied under pressure.

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Both the initial heat transfer when the piston initially contacts the stops and the overall heat transfer throughout the procedure are -11,172 kJ.

h1 = 3279.1 kJ/kg

h2 = 751.6 kJ/kg

Q = ΔU = m(u2 - u1) (u2 - u1)

where m is the steam's mass.

We may use the ideal gas law to determine the mass of the steam:

PV = mRT

PV/(RT) = (3.5 MPa) (0.33 m3)/(0.287 kJ/kg-K) (380.4 K) = 4.47 kg for the formula m.

We can now determine the rate of heat transfer:

Q is equal to m(u2 - u1) = (4.47 kg)(751.6 kJ/kg - 3279.1 kJ/kg) = -11,172 kJ

Heat is leaving the system, as indicated by the negative sign.

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Carolyn's observations suggest that the Elodea plant produced significantly more bubbles in the sunlit window than it did on her desk. This is likely because Elodea, like all plants, undergoes photosynthesis, a process in which they use sunlight to convert carbon dioxide and water into glucose and oxygen.

The oxygen produced during photosynthesis is released into the water as bubbles, which is what Carolyn observed.

When the Elodea was on Carolyn's desk, it may not have been receiving enough light to undergo photosynthesis at its maximum capacity, resulting in fewer bubbles being produced.

However, when Carolyn moved the aquarium to a sunlit window, the Elodea likely received more light, allowing it to undergo photosynthesis more efficiently, resulting in a higher number of bubbles being produced.

Overall, Carolyn's observations demonstrate the importance of light for plants to undergo photosynthesis and produce oxygen.

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In order to convert the volume from 21.0 L to 41.0 L while maintaining the same pressure and particle number, the gas would need to be heated to a temperature of 580.49 K.

According to Boyle's law, the pressure and volume of a gas follow an inverse relationship when the gas's temperature and molecular composition are both constant.

This problem can be solved using the coupled gas law, which connects a gas's pressure, volume, and temperature:

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

In this problem, we are given P1 = P2, V1 = 21.0 L, V2 = 41.0 L, and            T1 = 34°C. We want to find T2 in Kelvin.

First, we need to convert T1 from Celsius to Kelvin:

T1 = 34°C + 273.15

T1 = 307.15 K

Next, we can rearrange the combined gas law to solve for T2:

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

Substituting the given values, we get:

T2 = (1 atm * 41.0 L * 307.15 K) / (1 atm * 21.0 L)

T2 = 580.49 K

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The mass of chlorine in 25kg of [tex]CFCl_3[/tex] is 6,435g where mass is a measure of the amount of matter in an object. It is measured in kilograms (kg) or grams (g).

The mass of [tex]CFCl_3[/tex] = 25kg

The molecular weight of [tex]CFCl_3[/tex] (also known as Freon-11) is = 137.37 g/mol.

It means one mole of [tex]CFCl_3[/tex] = 137.37 g.

Since chlorine is an element, one mole of chlorine is equal to its atomic weight, which is 35.45 g/mol.

Therefore, number of moles 25 kg of [tex]CFCl_3[/tex] contains =

(25 kg) / (137.37 g/mol) = 182.4 mol of [tex]CFCl_3[/tex].

To calculate the mass of chlorine in 25 kg of [tex]CFCl_3[/tex], we need to multiply the number of moles of chlorine by its atomic weight.

The mass of chlorine in 25 kg of [tex]CFCl_3[/tex] is = 182.4 mol * 35.45 g/mol = 6,435 g of chlorine.

Hence, the mass of chlorine in 25 kg of [tex]CFCl_3[/tex] is 6,435 g.

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The blood alcohol level in mass percent is 0.0956% if 8.33 mL of 0.04988 M [tex]K2Cr2O7[/tex] is required for titration of a 9.9950 g sample of blood.

To calculate the blood alcohol level in mass percent, we need to first determine the amount of ethanol present in the blood sample. This can be done by using a redox titration with potassium dichromate [tex](K2Cr2O7)[/tex], which oxidizes the ethanol to acetic acid.

The balanced chemical equation for the reaction is:

[tex]C2H5OH + 2Cr2O7^2- + 16H+ → 2CO2 + 4Cr^3+ + 11H2O[/tex]

From the balanced equation, we can see that the stoichiometric ratio between [tex]K2Cr2O7[/tex] and ethanol is 2:1. Therefore, the moles of ethanol in the blood sample can be calculated as:

moles of ethanol = 0.5 × moles of [tex]K2Cr2O7[/tex]

The moles of [tex]K2Cr2O7[/tex]can be determined from its concentration and volume used in the titration:

[tex]moles of K2Cr2O7 = 0.04988 mol/L × 8.33 × 10^-3 L = 4.15 × 10^-4 mol[/tex]

Substituting this value into the equation above, we get:

moles of ethanol = 0.5 × 4.15 × 10^-4 mol = 2.075 × 10^-4 mol

The mass of ethanol in the blood sample can be calculated from its molar mass:

[tex]mass of ethanol = 2.075 × 10^-4 mol × 46.07 g/mol = 9.551 × 10^-3 g[/tex]

Finally, the blood alcohol level in mass percent can be titration calculated as:

mass percent = mass of ethanol / mass of blood sample × 100%

mass percent = [tex]9.551 × 10^-3 g / 9.9950 g × 100%[/tex]

mass percent = 0.0956%

Therefore, the blood alcohol level in mass percent is 0.0956%.

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There are 5.85 grams of acetic acid (CH3CO2H) in the given solution.

To calculate the grams of solute in the given solution, we first need to calculate the number of moles of solute (acetic acid) present in the solution.

We know the volume of the solution (250 mL) and the molarity of the solution (0.39 M). We can use the following equation to calculate the number of moles of solute:

moles of solute = molarity × volume (in liters)

First, we need to convert the volume from milliliters to liters:

250 mL = 0.250 L

Now we can use the equation to calculate the number of moles of acetic acid:

moles of CH3CO2H = 0.39 M × 0.250 L = 0.0975 moles

Finally, we can use the molar mass of acetic acid to convert the number of moles to grams:

molar mass of CH3CO2H = 60.05 g/mol

grams of CH3CO2H = moles of CH3CO2H × molar mass of CH3CO2H

grams of CH3CO2H = 0.0975 moles × 60.05 g/mol = 5.85 g.

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Answer:

88.55%

Explanation:
4.9995 g of crude material contain 4.4275 g of NaOH, therefore  

4.4275/4.9995 = 88.55%

⋅

Selection is the process by which we filter out irrelevant information and focus on the information that is most relevant to us.

For example, when we walk down a busy street, we may not pay attention to every detail of every building or person we pass by. Instead, we might focus on things that stand out, such as a bright billboard or an interesting conversation nearby.

Regarding stereotypes, one common example is the belief that all individuals from a certain race, gender, or religion share certain characteristics or behaviors. This stereotype is often based on limited or inaccurate information and can lead to biased judgments and discrimination. It's essential to recognize and challenge our own stereotypes to avoid unfairly treating others based on their group membership.

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The heat that is going to be released in the reaction can be obtained as 930.3 kJ.

The heat of reaction, also known as the enthalpy change of a reaction, is the amount of heat energy that is released or absorbed during a chemical reaction. It is denoted by the symbol ΔH and has units of joules per mole (J/mol) or kilojoules per mole (kJ/mol).

Number of moles of butane = 41.06g/58 g/mol = 0.7 moles

If 2 mole release 2658 kJ

0.7 moles will release 0.7 * 2658/2

= 930.3 kJ

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Answer:More purple

Explanation: Its more purple

The IUPAC name of the given compound is 1-bromo-3-ethylpentane.

IUPAC (International Union of Pure and Applied Chemistry) naming is a system used to give standardized names to chemical compounds. The steps involved in doing IUPAC naming are as follows:

1. Identify the longest carbon chain: The parent chain is the longest continuous chain of carbon atoms in the molecule.

2. Number the carbon atoms: The carbon atoms in the parent chain are numbered starting from the end nearest to the substituent, and the substituents are given numbers based on the carbon to which they are attached.

3. Identify and name the substituents: Substituents are groups of atoms that replace hydrogen atoms on the parent chain. They are named according to their functional groups.

4. Write the name: The name of the compound is written by listing the names of the substituents in alphabetical order, along with their position on the parent chain.

5. Add prefixes and suffixes: Prefixes are added to indicate the number of substituents on the parent chain, and suffixes are added to indicate the functional group present.

6. Check the name: The final step is to check the name for accuracy and consistency with IUPAC rules.

It's important to note that the naming of complex organic compounds can involve additional rules and naming conventions beyond these basic steps.

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As may be shown, the molar entropy from the question is 85.1 kJ/K.mol.

By dividing the substance's overall entropy by the quantity of moles present, one can determine a substance's molar entropy. By measuring a substance's heat capacity or by examining its thermodynamic behavior under various circumstances, one can empirically ascertain the entropy of that substance.

Entropy = ΔH/T

= 1.6 * 10^4 * 10^3J/mol/187.95 K= 85.1kJ/K.mol

b) If we have the Br- ion;

Rate = 5/1 * 2.7 * 10^-3 mol/s

= 1.35 * 10^-2 mol/s

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Copper ions  and acetate ions make up the chemical known as solid copper (II) acetate . It will dissolve and split into its component ions when placed in water. Moreover, copper (II) acetate is a strong electrolyte, which means that when dissolved in water, it almost entirely separates into ions.

Pb(OAc) 23H₂O, a colorless or white efflorescent monoclinic crystalline material, is the trihydrate that is created when it reacts with water.

Cu₂(OAc)4(H₂O) is a more bluish-green crystalline solid than anhydrous copper(II) acetate. Copper acetates have been utilized as fungicides and green pigments in some way since antiquity. Copper acetates are currently utilized as reagents for  the synthesis of various inorganic and organic compounds.

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The number of moles of acetic acid which has a molar mass of 60.05 g/mol present in 16.84 grams of acetic acid is 3.56.

Given the molar mass of Acetic acid (M) = 60.05 g/mol.

The mass of acetic acid (m) = 16.84g

Let the number of moles = n

A mole is a unit of measurement used in chemistry to measure amount of substance. It is calculated as molar mass by mass of the given substance. The mole is defined as the amount of a substance that contains as many elementary entities (atoms, molecules, ions, etc.) as there are atoms in exactly 12 grams of carbon-12. One mole of a substance contains this number of particles, and can be used to calculate the mass of a given sample of the substance.

n = 60.05/16.84 = 3.56 moles

Hence the required number of moles are 3.56.

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The partial pressure of IBr after equilibrium is reached is:

P_IBr = 0.029 - x = 0.0265 atm.

What is Equilibrium?

In chemistry, equilibrium refers to a state in which the rates of the forward and reverse reactions are equal, resulting in a stable concentration of products and reactants. In other words, equilibrium occurs when the concentrations of reactants and products remain constant over time. The concept of equilibrium is important in many areas of chemistry, including chemical reactions, acid-base equilibria, and solubility equilibria.

We can start by setting up an ICE table:

2IBr(g) I2(g) Br2(g)

Initial 0.029 atm 0 atm 0 atm

Change -x +x +x

Equil. 0.029-x x x

The equilibrium constant expression for the reaction is:

Kp = (P_I2 * P_Br2) / P_IBr^2

At equilibrium, the partial pressures are:

P_I2 = P_Br2 = x

P_IBr = 0.029 - x

Substituting these expressions into the equilibrium constant expression gives:

8.5×10^-3 = (x * x) / (0.029 - x)^2

Solving for x gives:

x = 0.0025 atm

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(a) Partial pressure of Ne is 470 torr. (b) Partial pressure of of CF₄ is 110 torr. (c) Total pressure is 580 torr. (d) Mole fraction of Ne is 0.621.

(a) The initial pressure of Ne is 470 torr, so the partial pressure of Ne after mixing is also 470 torr.

(b) The initial pressure of CF₄ is 110 torr, so the partial pressure of CF₄ after mixing is also 110 torr.

(c) The total pressure is the sum of the partial pressures of the two gases:

Total pressure = partial pressure of Ne + partial pressure of CF₄

Total pressure = 470 torr + 110 torr

Total pressure = 580 torr

(d) To find the mole fraction of Ne, we need to know the number of moles of Ne and CF₄. We can use the ideal gas law to find the number of moles of each gas:

PV = nRT

n = PV/RT

For Ne:

n = (470 torr x 1.50 L)/(0.0821 L·atm/mol·K x 298 K)

n = 19.25 mol

For CF₄:

n = (110 torr x 2.50 L)/(0.0821 L·atm/mol·K x 298 K)

n = 11.72 mol

The total number of moles is:

nTotal = nNe + nCF₄

nTotal = 19.25 mol + 11.72 mol

nTotal = 30.97 mol

The mole fraction of Ne is:

XNe = nNe/nTotal

XNe = 19.25 mol/30.97 mol

XNe = 0.621

Therefore, the mole fraction of Ne is 0.621.

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6.86 g of hydrochloric acid would be neutralized by 5.50 g of magnesium hydroxide.

The balanced chemical equation for the reaction between magnesium hydroxide and hydrochloric acid is:

Mg(OH)₂(s) + 2HCl(aq) → MgCl₂(aq) + 2H₂O(l)

From the equation, we can see that 1 mole of Mg(OH)₂ reacts with 2 moles of HCl. We can use the molar mass of Mg(OH)₂ to convert its mass to moles:

moles of Mg(OH)₂ = mass / molar mass = 5.50 g / 58.32 g/mol = 0.0942 mol

Since 1 mole of Mg(OH)₂ reacts with 2 moles of HCl, we know that twice as many moles of HCl will be neutralized:

moles of HCl neutralized = 2 × moles of Mg(OH)₂ = 2 × 0.0942 mol = 0.1884 mol

Finally, we can use the molar mass of HCl to convert the moles of HCl to grams:

mass of HCl neutralized = moles of HCl × molar mass = 0.1884 mol × 36.46 g/mol = 6.86 g

Therefore, 6.86 g of hydrochloric acid would be neutralized by 5.50 g of magnesium hydroxide.

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Explanation:

The titration of acetic acid with KOH is a weak acid-strong base titration. At the beginning of the titration (part a), we have only acetic acid in the solution, and its concentration is 0.200 M. As we add KOH, it reacts with acetic acid to form acetate and water:

CH3COOH + KOH → CH3COOK + H2O

The acetate ion is the conjugate base of acetic acid and can be considered a weak base. We can use the following equation to calculate the pH of the resulting solution at each point of the titration:

pH = pKa + log([A^-]/[HA])

where pKa is the acid dissociation constant of acetic acid (1.8 × 10^-5), [A^-] is the concentration of acetate ion, and [HA] is the concentration of undissociated acetic acid.

a) At the beginning of the titration (0.0 mL of KOH added), the solution contains only acetic acid. Therefore, [HA] = 0.200 M and [A^-] = 0 M.

pH = pKa + log([A^-]/[HA])

pH = -log(1.8 × 10^-5) + log(0/0.200)

pH = 2.40

The pH of the solution is 2.40.

b) When 50.0 mL of 0.100 M KOH is added, we have added 5.00 mmol of KOH. This amount of KOH reacts with 5.00 mmol of acetic acid, and the remaining 0.050 mol - 0.005 mol = 0.045 mol of acetic acid remains in the solution. At the same time, 0.005 mol of acetate ion is formed.

[HA] = 0.045 mol / 0.100 L = 0.450 M

[A^-] = 0.005 mol / 0.100 L = 0.050 M

pH = pKa + log([A^-]/[HA])

pH = -log(1.8 × 10^-5) + log(0.050/0.450)

pH = 4.41

The pH of the solution is 4.41.

c) When 100.0 mL of 0.100 M KOH is added, we have added 10.00 mmol of KOH. This amount of KOH reacts with 10.00 mmol of acetic acid, and there is no acetic acid remaining in the solution. At the same time, 0.010 mol of acetate ion is formed.

[HA] = 0 mol / 0.100 L = 0 M

[A^-] = 0.010 mol / 0.100 L = 0.100 M

pH = pKa + log([A^-]/[HA])

pH = -log(1.8 × 10^-5) + log(0.100/0)

pH = 4.74

The pH of the solution is 4.74.

d) When 110.0 mL of 0.100 M KOH is added, we have added 11.00 mmol of KOH. This amount of KOH reacts with 10.00 mmol of acetic acid, and there is an excess of 1.00 mmol of KOH in the solution. This excess KOH completely dissociates to give 1.00 mmol of OH^- ion. At

The cold pack is designed to address the problem of pain and swelling due to injury or inflammation.

A cold pack, also known as a cold compress, is a medical device designed to provide cold therapy to a specific area of the body. It typically consists of a gel-filled plastic bag or pouch that is placed in the freezer for a period of time to lower its temperature.

When applied to the affected area, the cold temperature helps to reduce blood flow, which in turn reduces inflammation, swelling, and pain. Cold packs are commonly used to treat minor injuries such as sprains, strains, and bruises, as well as to alleviate pain and swelling associated with chronic conditions like arthritis.

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Answer:

Green Bean Casserole is a classic.

The family recipe has passed down from one great-aunt to another.
Now it's causing gas and bloating across multiple generations of the family.

Explanation:

Answer:

catalysts do not shift the equilibrium position of a chemical reaction because they do not affect the free energy difference between reactants and products

Explanation:

To understand this better, let's consider an example. The Haber process is used to produce ammonia from nitrogen and hydrogen gas:

N2(g) + 3H2(g) ⇌ 2NH3(g)

This reaction is exothermic, meaning that it releases heat. According to Le Chatelier's principle, adding heat to an exothermic reaction will shift the equilibrium position towards the reactants (N2 and H2). Conversely, removing heat from the system will shift the equilibrium position towards the products (NH3).

Now, let's say we add a catalyst to this reaction. The catalyst will speed up both the forward and reverse reactions equally, without affecting their relative rates. This means that although the reaction will reach equilibrium faster with a catalyst present, it will still reach the same equilibrium position as it would without a catalyst.

Answer:

9 ans

Explanation:

cold 20 -11 =9 ans

summer 20-11=9 ans

Answer:

23205 J or 2.37 × 10⁴ J or 23.7 kJ

Explanation:

The amount of heat required to change the temperature of a substance can be calculated using the formula: q = mcΔT, where q is the heat added, m is the mass of the substance, c is its specific heat capacity, and ΔT is the change in temperature.

The specific heat capacity of iron is 0.44 J/g K1. So, to raise the temperature of 235 g of iron from 25.0°C to 250.0°C (a change of 225°C), you would need to add:

q = (235 g) × (0.44 J/g K) × (225 K) = 23205 J

So you would need to add 23205 joules of heat to raise the temperature of 235 g of iron from 25.0°C to 250.0°C.

Answer: 2 L or 2000 mL

Explanation:

A 1 molar (1 M) solution is equal to 1 mole of the solute dissolved in 1 L of solution.

For HCl (mw 36.5) 1 mole = 36.5 g

1 molar (1 M) = 1 mole/1 liter (or 36.5 g/L)

So

0,5 M = 1 mole/x

(x is the volume we are solving for)

Multiply both sides by x and you get

0.5x=1

Now multiple both sides by 2

X=2

So it’s 2L volume

Answer:

Atomic emission is more sensitive to flame instability than atomic absorption because during atomic emission, the intensity of the emitted light is proportional to the concentration of the element being measured. If the flame is unstable, it can cause fluctuations in the intensity of the light being emitted, which can lead to errors in the measurement of the element's concentration. In contrast, in atomic absorption, the intensity of the absorbed light is measured, which is less sensitive to flame instability since the amount of light absorbed by the element is proportional to its concentration regardless of the flame's stability.

Answer:

Atomic emission is more sensitive to flame instability than atomic absorption because atomic emission is based on the analysis of light emitted from excited atoms in the flame. In contrast, atomic absorption is based on light absorption by the flame's particles.

Flame instability can lead to changes in the temperature and pressure of the flame, which can affect the excited states of the atoms in the flame. When the flame is unstable, it can cause fluctuations in the number of excited atoms and the length of time they stay excited. This, in turn, can lead to fluctuations in the amount of light emitted by the excited atoms, making it more difficult to accurately measure the analyte concentrations in the sample using atomic emission spectroscopy.

On the other hand, atomic absorption spectroscopy is less sensitive to flame instability because the light absorption by the atoms in the flame is not as dependent on the excitation states of the atoms. The atoms in the flame absorb light at specific wavelengths regardless of their excited states. Hence, fluctuations in the excited state populations have less of an impact on the absorption signal. However, atomic absorption spectroscopy can still be affected by other factors, such as changes in the temperature and pressure of the flame and the presence of other interfering species in the sample.

1. (a) Consider the following table .

which of the element :

a) is a halogen?

Ans. Element J is halogen.

b) is most likely to be attracted by a magnet?

Ans. Element Y is most likely to be attracted by a magnet.

c) belongs to group I.

Ans. Element X belongs to group 1.

d) would readily form an ion with a double negative charge?

Ans. Element R would readily form an ion with a double negative charge.

e) what type of bond would exist between J and X when they combine?

Ans. Ionic bond would exist between J and X when they combine.

f) how many neutrons are there in Q?

Ans. Atomic number of Q = 13

Mass Number = 27.

Number of neutrons = Mass Number - Atomic number

Number of neutrons = 27 - 13.

Hence, There are 14 neutrons in Element Q.

g) write the formula of the compound formed when R combines with X.

Ans. When R combines with X it forms X₂R.

h) state the element which exist as a diatomic molecule

Ans. Element J exist as a diatomic molecule.

I) select the element which belongs to the d- block of the periodic table

Ans. Element Y belongs to the d- block of the periodic table.

J) write the electronic configuration of the element Y

Ans. Electronic configuration of the element Y is 1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹ 3d⁵ or [Ar] 4s¹ 3d⁵.