name the elements that have a total of 3 valence electrons

The accurate measurement is 0.51s and precise one is 0.516 s to 0.524 s.

0.51 is accurate measurement because this value is more closer to 0.500 s compared to other given options.

The measurement of 0.516 and 0.524 must have taken from highly precised instrument because it is capable of making measurement upto 3 decimal place.

Hence 0.516s and 0.524s is most precise team.

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Answer:

Dynamic equilibrium is the steady state of a reversible reaction where the rate of the forward reaction is the same as the reaction rate in the backward direction. Static equilibrium, also known as mechanical equilibrium, means the reaction has stopped. In other words, the system is at rest.

Applying an external force to an object causes it to gain mechanical energy, which is measured in Joule.

Some tips to follow when doing lab practical are:

From the table, Column 1 represents the time in half-life cycles, ranging from the initial state to 8 cycles. Column 2 shows the predicted number of radioactive atoms at each time point, based on the assumption that the number of atoms reduces by half in each half-life cycle.

Column 3 represents the simulated number of radioactive atoms at each time point and corresponds to the predicted values of the simulation.

In conclusion, the results as outlined in the lab guide are A=  27 B=  16 C=  9 D=  4 E=  2 F=  2 G=  0 H= 0.

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#complete question:

Do step 3 as outlined in the lab guide. Record your results in the appropriate blanks. A = B = C = D = E = F = G = H = A 3-column table with 9 rows. Column 1 is labeled Time half-life cycles, n with entries Initial, 1, 2, 3, 4, 5, 6, 7, 8. Column 2 is labeled Predicted radioactive atoms with entries 100, 50, 25, 13, 6, 3, 2, 1, 0. Column 3 is labeled Simulated radioactive atoms with entries 100, A, B, C, D, E, F, G, H.

Answer:

b. Ni, Fe, Ti, K

Explanation:

The given elements are present in period 4 of periodic table. Now we will discuss the trend of atomic radius along period.

Atomic radius:

" It is the smallest distance from nucleus to the outer most valance shell of an atom"

When we move from left to right in the periodic table the number of valance electrons in an atom increase. The atomic size goes to decrease in same period because of edition of electron with in the same shell. When the electron are added, at the same time protons are also added in the nucleus. In this way positive charge is also going to increase and this charge is greater in effect than the charge of electrons. This effect causes the greater nuclear attraction. The electrons are pull towards the nucleus and valance shell come closer to the nucleus.

As a result of this greater nuclear attraction atomic radius decreases and ionization energy increases because it is very difficult to remove the electron from atom and more energy is required.

Ni ∠ Fe ∠ Ti ∠ K

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Explanation:

Muscles move bones through contracting and relaxing

The pressure exerted by the vapours over the surface of the liquid a equilibrium at a given temperature is known as equilibrium vapour pressure or saturated vapour pressure.

Answer:

Vapour pressure is a measure of the tendency of a material to change into the gaseous or vapour state, and it increases with temperature. The temperature at which the vapour pressure at the surface of a liquid becomes equal to the pressure exerted by the surroundings is called the boiling point of the liquid

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Answer:

The mechanism for the formation of hexamethylenetetraamine predicts the formation of aminomethanol from the addition of ammonia to formaldehyde. This molecule subsequently undergoes unimolecular decomposition to form methanimine and water.

Explanation:

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Answer:

Explanation:

It is A because the electrice flows through the cord to give energy for the blades to move.

Answer:

Place a glass rod in the solution, if the crystallisation point has been reached, you should notice small crystals formed on it.

Answer:

D.

Explanation:

using stoichiometric factors to solve this one.

The moles of the reactant and the product produced in the reaction is given by the stoichiometric law. The moles of water produced in the reaction is 2.79 mol.

The stoichiometric law states the moles of the reactant and the product in the reaction are equivalent to the stoichiometric coefficient.

For the equation given, stoichiometric law states that:

\(\rm 7\;mol\;O_2=6\;mol\;H_2O\)

The moles of water produced from the 3.25 moles of oxygen are:

\(\rm 7\;mol\;O_2=6\;mol\;H_2O\\\\3.25\;mol\;O_2=\dfrac{6}{7} \;\times\;3.25\;mol\;H_2O\\\\3.25\;mol\;O_2=2.79\;mol\;H_2O\)

The moles of water produced in the reaction are 2.79 mol. Thus, option D is correct.

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Answer:

the rate increases

Explanation:

they are closer to each other, they will collide with each other more frequently and more successful collision

The student needs to add approximately 83.3 mL of 12 M HCl solution to the existing 1 L of 2 M HCl solution to increase the concentration to 3 M. It is important to handle concentrated acids with caution and follow proper safety procedures.

To increase the concentration of a 1 L solution of 2 M HCl to 3 M, the student needs to calculate the volume of concentrated HCl needed and add it to the existing solution. Here's how the calculation can be done:

Given:

Initial concentration of HCl solution = 2 M

Final concentration desired = 3 M

Initial volume of HCl solution = 1 L

Step 1: Calculate the moles of HCl in the initial solution.

Moles of HCl = Initial concentration × Initial volume = 2 M × 1 L = 2 moles

Step 2: Calculate the moles of HCl needed for the desired concentration.

Moles of HCl needed = Final concentration × Final volume = 3 M × 1 L = 3 moles

Step 3: Calculate the moles of HCl to be added.

Moles of HCl to be added = Moles needed - Moles present = 3 moles - 2 moles = 1 mole

Step 4: Convert the moles of HCl to the required volume of concentrated HCl.

To calculate the volume, we need to know the concentration of the concentrated HCl solution. Assuming it is 12 M, we can use the following formula:

Volume of concentrated HCl = Moles of HCl to be added / Concentration of concentrated HCl

Volume of concentrated HCl = 1 mole / 12 M = 0.0833 L or 83.3 mL

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Answer:

K = 3.45

Explanation:

In the reaction:

SO₂(g) + NO₂(g) ⇄ SO₃(g) + NO(g)

And K is:

K = [SO₃] [NO] / [SO₂] [NO₂]

Where [] are concentrations in equilibrium, as volume is 1L, [] could be taken as moles.

Equilibrium concentrations are:

[SO₂] = 2.00 moles - X

[NO₂] = 2.00 moles - X

[SO₃] = X

[NO] = X

AS moles of NO are 1.3 moles, X = 1.3 moles

Replacing:

[SO₂] = 0.7 moles = 0.7M

[NO₂] = 0.7M

[SO₃] = 1.3M

[NO] = 1.3M

K = 1.3M² / 0.7M²

The equilibrium constant is 3.5.

First of all we must obtain the equation of the reaction as follows;

SO₂(g) + NO₂(g) ⇄ SO₃(g) + NO(g)

Next we obtain the initial concentration of each specie;

SO₂(g) =  2.00 moles/1.00-L  = 2 M

NO₂(g)  =  2.00 moles/1.00-L  = 2 M

At equilibrium;

NO =  1.30 moles//1.00-L  = 1.3 M

The we set up the ICE table as follows;

          SO₂(g) + NO₂(g) ⇄ SO₃(g) + NO(g)

I          2             2                0          0

C        -x             -x               +x          +x

E       2 - x         2 - x            0 + x     0 + x

When x = 1.3

Concentration of each specie at equilibrium;

SO₂(g) = 2 - 1.3 = 0.7 M

NO₂(g) = 2 - 1.3 = 0.7 M

SO₃(g)  = 1.3 M

NO(g) M

K = (1.3)^2/(0.7)^2

K = 3.5

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Answer:

528.25 Calories

Explanation:

From the question given above, the following data were obtained:

Energy in Joules = 2210.20 J

Energy in calories =?

We can convert 2210.20 J to calories by doing the following:

4.184 J = 1 cal

Therefore,

2210.20 J = 2210.20 J × 1 cal / 4.184 J

2210.20 J = 528.25 Calories

Thus, 2210.20 J is equivalent to 528.25 Calories.

Answer:

2H2S(g) + 3O2(g) → 2H2O(g) + 2SO2(g) + 248kcal

Explanation:

The reaction of the problem occurs as follows:

H2S(g) + O2(g) → H2O(g) + SO2(g)

To balance the reaction we must balance oxygens:

H2S(g) + 3O2(g) → 2H2O(g) + 2SO2(g)

To balance the complete reaction:

2H2S(g) + 3O2(g) → 2H2O(g) + 2SO2(g)

As the energy is evolved, 124kcal are as product in the reactio per mole of H2S. As the balanced reaction contains 2 moles of H2S, the heat evolved is:

124kcal*2 = 248kcal:

2H2S(g) + 3O2(g) → 2H2O(g) + 2SO2(g) + 248kcal

And this is the balanced equation

Answer:

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Questions & Answers

CBSE

Chemistry

Grade 11

The Gaseous State

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Answer

What volume will a gas occupy at 740mm pressure which at 1480 mm occupies 500 cc? Temperaturebeingconstant.

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Hint: Boyle in the mid 1600’s observed that the product of the pressure and volume of a gas are observed to be nearly constant provided that the temperature is kept constant. This is always true in the case of an ideal gas.

P×V=constant

This relationship between pressure and volume of an ideal gas is known as the Boyle's Law.

Complete answer:

Since it is given in the question that temperature can be assumed as constant, we can apply Boyle’s Law.

According to Boyle's Law we know that:

⇒P×V=constant

We can rewrite this as

⇒P1×V1=P2×V2

Where P1 is the pressure at the volume V1

P2 is the pressure at the volume V2

Thus from the given question we can say that:

P1 = 1480 mm

V1 = 500 cc

P2 = 740 mm

V2 = x

Thus we can rewrite the equation and substitute the above values and obtain the equation given below:

⇒V2=P1×V1P2

⇒V2=1480×500740

⇒V2=1000cc

Thus the volume of gas occupied at a pressure of 740 mm is 1000 cc.

Answer: A. 212

Explanation:

Answer:

a) 1.5 x 10^-3 mol/L

b) 1.35×10^-8

c) decrease

Explanation:

The solubility of lead II iodide is given by the equation;

PbI2(s) -----> Pb^2+(aq) + 2I^-

By looking at the ICE table, I^-=2x= 3.0 x 10^-3 mol/L/2 = 1.5×10^-3 mol/L

Hence molar solubility of PbI2 = 1.5 x 10^-3 mol/L

Ksp= [Pb^2+] [2I^-]^2 =

Let the molar solubility of each ion be x, therefore;

Ksp= 4x^3

Ksp= 4(1.5 x 10^-3 mol/L)^3= 1.35×10^-8

Addition of kI to the saturated solution will shift the equilibrium position to the left thereby decreasing the solubility of the PbI2 in the system due to common ion effect. The concentration of the iodide ion is now excess in the system leading to the reverse reaction being favoured according to Le Chateliers principle.

a) The molar solubility of PbI₂ is  \(1.5 * 10^{-3} mol/L\)

b) The solubility constant is \(1.35*10^{-8}\)

c) The molar solubility of lead (II) will decrease.

The solubility of lead II iodide is given by the equation;

\(PbI_2(s) ----- > Pb^{2+}(aq) + 2I^-\)

By looking at the ICE table,

\(I^-=2x= 3.0 * 10^{-3} mol/L/2 =\)  \(1.5 * 10^{-3} mol/L\)

Hence, molar solubility of PbI2 = \(1.5 * 10^{-3} mol/L\)

\(Ksp= [Pb^{2+}] [2I^-]^2\)

Let the molar solubility of each ion be x, therefore;

\(Ksp= 4x^3\\\\Ksp= 4(1.5 * 10^{-3} mol/L)^3\\\\Ksp= 1.35*10^{-8}\)

The addition of KI to the saturated solution will shift the equilibrium position to the left thereby decreasing the solubility of the PbI₂ in the system due to common ion effect. The concentration of the iodide ion is now excess in the system leading to the reverse reaction being favoured according to Le- Ch-ateliers principle.

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The free energy is 6.128 kJ/mol

Free energy, also known as Gibbs free energy, is a thermodynamic quantity used to describe the energy available to do work in a system

The concept of free energy is important in many areas of chemistry, including chemical thermodynamics, biochemistry, and materials science. It is used to understand and predict the behavior of chemical reactions, phase transitions, and other thermodynamic processes.

ΔG = ΔG° + RTlnQ

ΔG = 3.2 * 10^3 + (8.314 * 295 * ln(3.3)

=   3.2 * 10^3 + 2.928 * 10^3

= 6.128 * 10^3 J/mol

or 6.128 kJ/mol

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The equilibrium expression Kc for the reaction can be written as -

                                                 \(Kc = \frac{O_{3}^{2} }{O_{2} ^{3} }\)

Equilibrium constant of a chemical reaction at equilibrium is defined as the ratio of concentration of products to the concentration of reactants, each raised to their respective stoichiometric coefficients.

The expression that links Kc to the equilibrium concentration of reactants and products and stoichiometry of the equation is called equilibrium expression.

Kc is the equilibrium constant where subscript c refers to the fact that concentrations have been used in the calculations. The units of Kc depend on the equilibrium expression.

Therefore, the equilibrium expression Kc for the reaction can be written as -

                                                 \(Kc = \frac{O_{3}^{2} }{O_{2} ^{3} }\)

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The balanced equation  \(K_{2}Cr_{2}O_{7}\) oxidizes KI in the presence of \(H_{2}SO_{4}\)can be represented as

6 KI + \(K_{2}Cr_{2}O_{7}\) + 7 \(H_{2}SO_{4}\) → 4 \(K_{2}Cr_{2}O_{7}\) + \(Cr_{2}(SO_{4})_{3}\) + 3 \(I_{2}\)\(I_{2}\) + 7 \(H_{2}O\)

The balanced equation for the oxidation of KI (potassium iodide) by \(K_{2}Cr_{2}O_{7}\)(potassium dichromate) in the presence of \(H_{2}SO_{4}\)(sulfuric acid) can be represented as follows:

6 KI + \(K_{2}Cr_{2}O_{7}\) + 7 \(H_{2}SO_{4}\) → 4 \(K_{2}Cr_{2}O_{7}\) + \(Cr_{2}(SO_{4})_{3}\) + 3 \(I_{2}\)\(I_{2}\) + 7 \(H_{2}O\)

In this equation, \(K_{2}Cr_{2}O_{7}\) is the oxidizing agent, and KI is the substance being oxidized. The sulfuric acid (\(H_{2}SO_{4}\)) serves as a catalyst and provides the necessary acidic conditions for the reaction to occur.

The products of the reaction are potassium sulfate (\(K_{2}SO_{4}\)), chromium(III) sulfate (\(Cr_{2}(SO_{4})_{3}\)), iodine (), and water (\(H_{2}O\)).

Note that this equation represents a stoichiometrically balanced equation, ensuring that the number of atoms of each element is the same on both sides of the equation.

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Answer:

2.87

Explanation:

Answer: Melting Point : 42 to 43 °C

Boiling Point : 215 to 216 °C

Answer:

The d-block and transition metal are not interchangeable terms because the d-block elements are a subset of the transition metal elements. The transition metals are defined as the elements that have partially filled d orbitals, which includes the d-block elements as well as other elements that have partially filled d orbitals in other blocks, such as lanthanides and actinides. Therefore, while all d-block elements are transition metals, not all transition metals are d-block elements.

The isotope that has the greater abundance is lithium-7

To know which isotope has the greater abundance, we shall detertmine the abundance of each isotope. This can be obtained as follow:

Average atomic mass = [(Mass of A × A%) / 100] + [(Mass of B × B%) / 100]

6.94 = [(6.02 × A) / 100] + [(7.02 × (100 -A) / 100]

6.94 = 0.0602A + 7.02 + 0.0702A

Collect like terms

6.94 - 7.02 = 0.0602A - 0.0702A

-0.08 = -0.01A

Divide both sides by -0.01

A = -0.08 / -0.01

A = 8%

Thus,

B = 100 - A

B = 100 - 8

B = 92%

From the above calculations, we obtained:

Thus, we can conclude that lithium-7 has the greater abumdance

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