Sodium has a density of 0.968 g/cm3 and crystallizes with a body-centered cubic unit cell. Part A Calculate the radius of a sodium atom.

Answer:

single displacement

Explanation:

the metal replaces the hydrogen of the acid to for a salt and hydrogen gas

When solutions of sodium sulfide and copper(ii) sulfate are mixed, a precipitate of copper(ii) sulfide is formed. The net ionic equation for this reaction is Cu⁺² (aq) + S⁻² (aq) → CuS (s).

The balanced chemical equation is the equation in which the number of atoms on the reactant side is equal to the number of atoms on the product side in an equation.

Now first write the balanced chemical equation

CuSO₄ (aq) + Na₂S (aq)  → CuS(s) + Na₂SO₄ (aq)

Now write the net ionic equation

Cu⁺² (aq) + SO₄⁻² (aq) + 2Na⁺ (aq) + S⁻² (aq) → CuS (s) + 2Na⁺ + SO₄⁻² (aq)

So the net ionic equation is

Cu⁺² (aq) + S⁻² (aq) → CuS (s)

Thus from the above conclusion we can say that When solutions of sodium sulfide and copper(ii) sulfate are mixed, a precipitate of copper(ii) sulfide is formed. The net ionic equation for this reaction is Cu⁺² (aq) + S⁻² (aq) → CuS (s).

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Answer:

A. 2.74g is the correct answer

Explanation:

[tex]average \: mass = \frac{(mass \: pre - 1982 \times abundance) + (mass \: post - 1982 \times abundance}{100 } \\ ave = \frac{ (3.1g \times 40.0) + (2.5g \times 60.0)}{100} \\ ave = \frac{124 + 150}{100} \\ ave = \frac{274}{10} = 2.74g[/tex]

where, the sum of abundance always have to be 100%

Thus, 40.0 +60.0= 100.0

The addition of the product of volume and pressure along with internal energy is called enthalpy. The change in the enthalpy at 1 atm is 24.3 kJ. Thus, option d is correct.

Enthalpy is the measurement of the heat or the energy present in a system that has a constant pressure in the thermodynamic system.

The balanced equation is shown as,

Sn (s) + SnO₂ (s) → 2SnO (s) , ΔH° = 16.2 kJ

The enthalpy change for the reaction is for 2 moles of Tin(II) oxide. So, for 3 moles of Tin(II) oxide produced in the reaction the enthalpy change will be given as,

= (16.2 kJ × 3 mol) ÷ 2 mol

= 24.3 kJ

Therefore, option d. 24.3 kJ is the enthalpy change.

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Your question is incomplete, but most probably your full question was, What Is The Change In Enthalpy At 25°C And 1 Atm For The Production Of 3.00 Mol SnO(S)? Sn (s) + SnO₂ (s) → 2SnO (s) , ΔH° = 16.2 kJ.

The molar solubility is 7.4×[tex]10^{-3}[/tex] M and the solubility is  7.4×[tex]10^{-3}[/tex] g/L .

Calculation ,

The dissociation of silver bromide is given as ,

[tex]AgBr[/tex] → [tex]Ag ^{+}[/tex] + [tex]Br^{-}[/tex]

S  

 -          S        S        

Ksp =  [[tex]Ag ^{+}[/tex] ] [ [tex]Br^{-}[/tex] ]  =  [S] [ S ] = [tex]S^{2}[/tex]

S = √ Ksp = √ 5. 5×[tex]10^{-5}[/tex] = 7.4×[tex]10^{-3}[/tex]

The solubility =7.4×[tex]10^{-3}[/tex] g/L

The molar solubility is the solubility of one mole of the substance.

Since ,  one mole of [tex]AgBr[/tex] is dissociates and form one mole of each  [tex]Ag ^{+}[/tex] and [tex]Br^{-}[/tex] ion . So, solubility is equal to molar solubility but unit is different.

Molar solubility = 7.4×[tex]10^{-3}[/tex] mol/L = 7.4×[tex]10^{-3}[/tex] M

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The molality of a solution containing 3. 0 moles of NaCl and 100. 0 moles of water is 30 mol/kg.

The number of moles of solute in a solution equal to 1 kg or 1000 g of solvent is referred to as its molality. Mole per kilogram of solvent is the SI unit for molality.

Given:

3.0 moles of NaCl in 100 moles of water.

To find:

The molality of the solution

The moles of solute (NaCl) = 3.0 moles

The mass of solvent (water) = 100 moles (0.1 kg/mol)

Molality of a solution = Number of Moles of solute/ Mass of solvent(kg)

                                 = 3.0 moles/0.1 kg/mol

                                 = 30 mol/kg

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Answer:

Reaction rate is increased by an increase in kinetic energy

of the reacting particles when their

is _temperature_

increased

A . Collision theory

B . Surface Area

C . Temperature

D. Concentration

A sample 0. 100 moles of a gas is collected at at STP . 2.24 is the volume of the gas in liters.

The STP means standard temperature and pressure.

At STP,

Temperature = 0 °C =273 K

Pressure = 1 atm

We get value of volume by using ideal gas equation,

PV = nRT

Calculation,

Since, one mole of a gas occupy 22.4 L volume at STP

So, for 0.1 mole volume occupy = 22.4L × 0.1 mole/1 mole =  2.24L

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There are  1.554 moles are in the sample of gas when a  gas has a volume of 27. 5 L at 302 K and 1. 40 atm.

Calculation,

The given question is solved by the using of ideal gas equation which is shown below.

PV = nRT                               .... ( i )

where , P is he pressure of the gas = 1. 40 atm

V is the volume occupied by the gas = 27.5 L

R is the universal gas constant = 0.082 L atm/K mol

T is the temperature =  302 K

n is he number of moles of the gas = ?

By putting the value of pressure  P , volume V , Universal gas constant and temperature T in the equation ( i ) we get the number of moles.

1. 40 atm × 27.5 L = n × 0.082 L atm/K mol×  302 K

n = 1. 40 atm × 27.5 L/0.082 L atm/K mol×  302 K = 1.554 mole

There are  1.554 moles are in the sample of gas

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If we takes 11. 2 kj of energy to raise the temperature of 145 g of benzene from 23. 0°c to 68. 0°c, then the specific heat of benzene is

1716.475 J/kg.°C.

Specific Heat capacity is given as

Q = cm(t₂-t₁) ................[tex]equation 1[/tex]

Where Q = quantity of heat, c = specific heat of benzene, m = mass of benzene, t₁  = initial temperature, t₂ = final temperature.

[tex]C[/tex] = [tex]\frac{Q}{m_{(t2-t1)} } ...........equation 2[/tex]

Given:

[tex]Q = 11.2 kJ = 11200 J \\\\m= 145\\\\g = 0.145 kg\\\\t_{2} = 68.0 °C\\\\t_{1} = 23.0 °C[/tex]

Substitute into equation 2.

[tex]C = \frac{11200}{[0.145(68-23)]}\\\\ C= \frac{11200}{6.525}\\\\ C= 1716.475 J/kg °C[/tex]

Hence the specific heat of benzene = [tex]1716.475 J/kg °C[/tex]

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Answer: 5kg

Explanation: formula K=1/2mv^2

50=10m

Answer:

C

Explanation:

KE = 1/2 m v^2

KE *2 / v^2   = m

50 (2)/ 10^2   = m =  1 kg

The reaction is at equilibrium. Option C

The term equilibrium constant refers to a value that shows the extent to which reactants are converted into products in the reaction.

We know that;

∆G =∆G'° + RT In Keq

∆G'°  = −4.40 kJ/mol

R = 8.314 J/K/mol

T = 25 + 273 = 298 K

Thus;

∆G = −4.40 * 10^3 + (8.314 *  298  * ln[2.90 * 10^-3/0.490  * 10^-3]

∆G =  −4.40 * 10^3 + 4.40 * 10^3

∆G = 0

Hence, the reaction is at equilibrium.

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1) Consider the following metabolic reaction:

2-Phosphoglycerate ® 3-Phosphoglycerate DG°' = −4.40 kJ/mol

What can be said about this reaction when the concentration of 2-phosphoglycerate is 0.490 mM and the concentration of 3-phosphoglycerate is 2.90 mM at 25°C?

I. This reaction is endergonic under these conditions.

II. This reaction is exergonic under these conditions.

III. This reaction is at equilibrium under these conditions.

IV. This reaction is not favorable under standard conditions.

A) I, IV

B) II only

C) III only

D) IV only

E) None of the above

2) Consider the following metabolic reaction:

3.6 moles [tex]NH_{3}[/tex] required when each quantity of reactant completely reacts.

The ratio of the mole quantities of any two compounds present in a balanced chemical reaction is known as the mole ratio. A reaction's product yield can be predicted using mole ratios, as can the amount of reactant required to produce a specific amount of result.

The balanced chemical equation is :

[tex]3N_{2} H_{4}[/tex](l) → [tex]4NH_{3}[/tex](g) + [tex]N_{2}[/tex](g)

Given is = 2.6 mol [tex]N_{2} H_{4}[/tex]

As we see, 3 moles of [tex]N_{2} H_{4}[/tex] reacts with  4 moles of [tex]NH_{3}[/tex].

Therefore,

n( [tex]NH_{3}[/tex]) = n(  [tex]N_{2} H_{4}[/tex] ) × [tex]\frac{4}{3}[/tex]

n( [tex]NH_{3}[/tex]) = [tex]2.6mol^{*} \frac{4mol}{3mol}[/tex]

n( [tex]NH_{3}[/tex]) = 3.6 moles.

Therefores, 3.6 moles [tex]NH_{3}[/tex] required when each quantity of reactant completely reacts.

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The electrons are transferred in the given redox reaction are 2 electrons.

When oxidation and reduction that is gaining of electron and loosing of electrons takes place simultaneously in a chemical reaction then it would be known as redox reaction.

In terms of oxygen the gaining of oxygen is oxidation and loosing of oxygen is reduction.

In the given chemical reaction the transfer of two electrons will be as follows

Zn₂ (2e) + AgNo₃ → 2Ag (2e) + Zn(NO₃)₂

It is clear from the reaction their is transfer of two electrons zinc is loosing 2 electrons undergoes to oxidation while silver gets reduced.

Therefore, transfer of only two electrons.

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Single displacement

Reactant

Double displacement

Product

option (c) emulsion is the right answer

Emulsion is a classification of milk.

In physical chemistry, an emulsion is a combination of two or more liquids in which one of the liquids is present as microscopic or ultramicroscopic droplets dispersed throughout the other.

Emulsions form as a result of the cleansing action of soaps.

(ii) The emulsification process is how lipids are broken down in the intestines.

(iii) Disinfectants and antiseptics combine with water to generate emulsions.

(iv) The emulsification technique is utilized to create medications.

When two immiscible liquids, such as oil and water, are stirred together with an emulsifier—which might be a protein, phospholipid, or even a nanoparticle—emulsion is created.

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Answer:

25 gm would be left

Explanation:

45 hours is 3 half lives

200 gm  * 1/2 * 1/2 * 1/2 = 25 gm

According to the concept of half life , as 45 hours are equal to three half- lives after 3 half-lives 25 g of sodium-24 remains.

Half -life of a substance is defined as the time which is required for half of the quantity of a radioactive substance to get decayed.It is a term which is used in nuclear chemistry for describing how quickly unstable atoms undergo radioactive decay into other nuclear species by emitting particles or the time which is required for number of disintegrations per second of radioactive material to decrease by one half of its initial value.

In the given problem time is=45 hours and as 1 half  life=15 hours , for 45 hours=45/15=3 half lives .

Amount of sodium-24 after 3 half lives= 200×1/2×1/2×1/2=25 grams.

Therefore, after 45 hours 25 g of sodium-24 remains.

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Synthetic fertilizer

It is a process developed by two scientists - Kristian Birkeland and  Sam Eyde - to fix atmospheric nitrogen for the production of fertilizer.

The nitrogen oxide produced is then used in the production of nitric acid and eventually, the production of synthetic fertilizer.

First, nitrogen is fixed into nitrogen oxide:

[tex]N_2 + O_2 --- > 2NO[/tex]

Then, the NO is oxidized:

[tex]2NO + O_2 -- > 2NO_2[/tex]

The nitrous oxide produced is then dissolved in water to form nitric acid:

[tex]3NO_2 + H_2O --- > 2HNO_3 + NO[/tex]

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Considering the definition of molar volume and density,the density of copper at 25°C is 8.96 [tex]\frac{g}{cm^{3} }[/tex].

Molar volume is an intensive property that indicates how much space one mole of a certain substance or compound occupies. In other words, the molar volume is the space occupied by one mole of a substance.

The expression to calculate the molar volume is:

Vm= V÷ n

where:

Density is defined as the property that matter, whether solid, liquid or gas, has to compress into a given space.

In other words, density is a quantity that allows us to measure the amount of mass in a certain volume of a substance.

Then, the expression for the calculation of density is the quotient between the mass of a body and the volume it occupies:

density= mass÷ volume

The relationship between molar volume and density is reflected in the following expression, where the molar volume is calculated by dividing the molar mass (M) by the mass density (ρ):

Vm= M÷ρ

In this case, you know:

Replacing in the relationship between molar volume and density:

7.09 [tex]\frac{cm^{3} }{mol}[/tex]= 63.5 [tex]\frac{g}{mol}[/tex]÷ρ

Solving:

7.09 [tex]\frac{cm^{3} }{mol}[/tex]×ρ= 63.5 [tex]\frac{g}{mol}[/tex]

ρ= 63.5 [tex]\frac{g}{mol}[/tex]÷7.09 [tex]\frac{cm^{3} }{mol}[/tex]

ρ= 8.96 [tex]\frac{g}{cm^{3} }[/tex]

Finally, the density of copper at 25°C is 8.96 [tex]\frac{g}{cm^{3} }[/tex].

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The work done on the surroundings by the expansion of a gas is w = –pdv .The statement is true.

The work is the energy required to move a system or body or anything against a force is called work. The word done = force×distance

Work done in an isobaric process is given by

W = PΔV

Where W = Work done

P =pressure

ΔV = change in volume

In a free expansion , the gas is allowed to expand into a vacuum.

This happens so quickly , so there is no heat transferred . No work is done , because the gas does not display anything.

W = PΔV

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The two ways were,

A cathode-ray tube, often known as a CRT, is a vacuum tube that houses one or more electron cannons that produce electron beams that are then bent to produce pictures on a phosphorescent screen. The visuals might be pictures (from a television or computer display), electrical waveforms (from an oscilloscope), radar targets, or other occurrences.

Tomson came to the conclusion that atoms can be divided into their constituent corpuscles (atoms are made up of smaller particles).

The electron was discovered in 1897 by J. J. Thomson.

His "plum pudding" concept from 1904 postulated that the positive charge contains embedded electrons.

With this model, he disproved his first theory (the atom was composed of immaterial vortices).

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The potential of a standard hydrogen electrode (s. h. e. ) be under the given conditions is - 0.029 V

Calculation ,

Formula used :

[tex]E_{cell}[/tex] = [tex]E_{cell}[/tex] ° - 0.059/n ㏒[tex]H_{2}[/tex]/ [tex][H^{+}]^{2}[/tex]

As we know that ,

[[tex]H^{+}[/tex]] = 0. 84 M

[tex]P_{H_{2}[/tex] = 2. 2 atm

Temperature ( T ) = 298  K

From reaction at electrode number of electrons involve ( n ) = 1

Standard electrode potential of  standard hydrogen electrode [tex]E_{cell}[/tex] ° = 0

Putting the value of all data in equation ( i ) , we get

[tex]E_{cell}[/tex] = 0  - 0.059/1 ㏒2. 2 / [tex](0. 84)^{2}[/tex] =  - 0.059 ㏒3.14 =  - 0.059×0.497

[tex]E_{cell}[/tex] = - 0.029 V

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During transpeptidation, nucleophilic attack occurs between the α-amine group in the A site and the esterified carbon in the P site

A chemical process in which an amino acid residue or peptide residue is transferred from one amino compound to another. This process occurs when a protease reversibly converts one peptide to another.

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Number of significant figures will be 1.

Given:

weight of acetylsalicylic acid = 0. 4g

Significant figures are the digits of value which carry meaning towards the resolution of the measurement. They are also called significant figures in chemistry. All the experimental measurements have some kind of uncertainty associated with them.

When we convert 0.4g acetylsalicylic acid to mg we get value 400mg

when we convert 400mg to gram we get value of 0.4 gram

Since 0 before a decimal is not significant so there is only one significant figure that is 4.

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At STP, the volume of water that is produced is 134.4 L.

Now we know that the equation of the reaction is;

2C6H6(g) + 15O2(g) = 12CO2(g) + 6H2O(g)

We can use the stoichiometry of the reaction to determine the volume of the water produced as follows;

2 mole of C6H6 produced 6 moles of water. If 1 mole of water occupies 22.4 L then 6 moles of water will occupy 6 * 22.4 L = 134.4 L

Thus, we can see that at STP, the volume of water that is produced is 134.4 L.

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As a gaseous element condenses, the atoms become close together and they have more attraction for one another.

What is condensation?

Examples:

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Because of the enzyme action of Amylase, the chemical breakdown of starch happens in the mouth.

It is an enzyme, simply a type of protein that aids in breaking down the carbohydrates in your body. The main two glands as the pancreas and salivary glands help in making amylase in your mouth.

A small amount of amylase is also present in your urine and blood. If the levels present in your body are high then it causes health problems.

The amylase test performs to know the amount of amylase present in your body. This test is called as "amy test".

The normal range of Amylase in the human body is 40 to 140 U/L or 0.38 to 1.42 microkat/L.

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The value of ''n" in the hydrate formula is 10

Mass of anhydrous NA2B4O7 = 2.64g

Molar mass of NA2B4O7 = 201.224 g/mol

Number of Moles of anhydrous NA2B4O7 = Mass of NA2B4O7 / Molar mass of NA2B4O7

                                                     = 2.64 / 201.224

                                                     = 0.0131 mol

Mass of water in 5.00g of NA2B4O7 = 5.00 - 2.64

                                                              = 2.36 g

Moles of water = 2.36 / 18.015

                          = 0.1307 mol

The whole number ratio = moles of water / moles of anhydrous sodium tetra borate

                                           = 0.1307 / 0.0131

The whole number ratio = 10

Hence the value of n in the hydrate formula is 10.

NA2B4O7. 10H2O

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Answer:

exothermic

Explanation:

exothermic reactions give off heat.....or the temp of the substances increases