The theoretical yield of a reaction is 75.0 grams of product and the actual yield is 42.0 grams. What is the percent yield?

The coefficient of performance (cop) of a refrigerator is defined as the ratio of " the work necessary to heat or cool something usefully."

The usable heating or cooling delivered to work required ratio, also known as the coefficient of performance, or COP, of a heat pump, refrigerator, as well as air conditioning system. Higher efficiency, less energy (power) usage, and thus reduced operational costs are all related to higher COPs.

Coefficient of performance formula:

[tex]K = Q_{H} / W_{in}[/tex]

where, K = Coefficient of performance, [tex]Q_{H}[/tex] = heat of pumps output, [tex]W_{in}[/tex] = work required by the system.

It is refrigeration's coefficient of performance (COP) will always be greater than 1.

Therefore, the coefficient of performance (cop) of a refrigerator is defined as the ratio of " the work necessary to heat or cool something usefully."

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During the day, due to the existence of sunlight, the procedure of photosynthesis elevations over that of respiration. Photosynthesis uses carbon dioxide to produce carbohydrates removing oxygen as a byproduct. During the night, because there is no sunlight, the rate of photosynthesis decreases. This, therefore, drives the rate of respiration to surpass the speed of photosynthesis. In respiration, oxygen is consumed and carbon dioxide is dismissed.

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A sizable organ that grows during pregnancy is the placenta. It is typically at the top or side of the uterine wall where it is attached. The placenta and your baby are joined by the umbilical cord. Through the placenta's filtering of the mother's blood and the umbilical cord, your baby receives oxygen, glucose, and other nutrients.

A sizable organ that grows during pregnancy is the placenta. It is typically at the top or side of the uterine wall where it is attached. The placenta and your baby are joined by the umbilical cord.

Blood from the mother filters oxygen, glucose, and other nutrients before traveling via the placenta and reaching your baby through the umbilical cord.

Blood from the mother filters oxygen, glucose, and other nutrients before traveling via the placenta and reaching your baby through the umbilical cord. Additionally, the placenta filters out elements that can be dangerous to your unborn child and eliminates waste products and carbon dioxide from your unborn child's blood.

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The Hbr has highest boiling point followed by Kr , [tex]Cl_{2}[/tex] and [tex]F_{2}[/tex] .

The Hbr has  highest boiling point due to presence of intermolecular H-bonding that is hydrogen bonding.

The electronegativity of bromine is greater than hydrogen atom. The electronegativity difference hydrogen atom and bromine atom is high. So, due to this a positive dipole is generated at hydrogen atom and partial negative charge, these partial charge is called dipole. In a HBr molecule one H-atom attracted by Br-atom of another atom by force of attraction and this partial force of attraction is called H-bonding.

Other all given molecules are arrange according to its molecular mass because boiling points is directly proportional to the molecular mass of molecule.

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The given statement is true .

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Bohr's model explained the position of the electron, proton, and neutron in the atom of the element. The energy at the n = 2 level of the atom will be - 3.40 eV.

The principal quantum number (n) has been the distance of the electron of that atom in the nucleus and its energy in the structure. It can also be said to define the size of the atomic orbit.

n = 2 is the first excited state whose energy is calculated as:

Eₙ = − 13.6 ÷ n² eV

E₂ = - 13.6 eV ÷ 2²

= -3.40 eV

Therefore, -3.40 eV is the energy of electron at n = 2.

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Neutrons are required. Atomic Radiation, Nuclear Graphite, and Effects

The fission reaction is often started by an impinging neutron, and each fission produces an average of 2.5 neutrons.

With the exception of the hydrogen nucleus, which has a single proton, neutrons are necessary for the stability of nuclei. Nuclear fission and fusion generate a tonne of neutrons.

Scattering tests and research on nuclear binding energies provide evidence for the nuclear forces' finite range. The neutron-proton (n-p) potential energy map in Figure 44.3a, which was created by scattering neutrons from a target containing hydrogen, illustrates the short range of the nuclear force. The n-p potential energy well is 40 to 50 MeV deep, and the nucleons are prevented from getting much closer than 0.4 fm by a strong repulsive component.

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We need to add one neutron to initiate the fission reaction.

The reaction in which the identity or characteristics of an atomic nucleus change due to the bombarding of it with an energetic particle is known as a nuclear reaction.

The bombarding particle can be an alpha particle, a gamma-ray photon, a neutron, a proton, or a heavy ion.

In nuclear fission, an atom splits into lighter nuclei due to the bombarding of an energetic particle.

The reaction given is

[tex]_{92}^{235}U + .... _{0}^{1}n \rightarrow _{52}^{137}Te + _{40}^{97}Zr +_{0}^{1}n[/tex]

To find the number of neutrons we have to balance the reaction

Atomic mass on reactant side = 235 + 1 = 236

The atomic number on the reactant side = 92

Atomic mass on the products side = 137+97+2 = 236

The atomic number on the products side = 92

As the reaction is already balanced

We need to add one neutron to initiate the fission reaction.

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The kf of Ag(NH₃)₂ for the given equation is 2.786 x 10⁷.

kf value can be calculated by dividing the molal concentration with the freezing point of depression.

The reaction are given:

Ag+(aq) + e- ⇋ Ag(s) E0= 0.81 V

Ag(NH₃)₂ +(aq) + e- ⇋ Ag(s) + 2NH₃(aq) E0= 0.37 V

To get Kf of Ag(NH₃)₂  we will rearrange the above written equations

a. Ag +(aq) + e- ⇋ Ag(s) E0= 0.81 V

b. Ag(s) + 2NH₃(aq) ⇋ Ag(NH₃)₂ +(aq) + e- E0= - 0.37 V

Adding equation (a) and equation (b) we will get

Ag +(aq) + 2NH₃(aq) ⇋ Ag(NH₃)₂  + (aq) Eo = 0.81 -0.37 = 0.44 V

By the Ernst equation,

E = E° - 0.0591 logkf

Now Kf is calculated at equilibrium and at equilibrium E = 0,

The number of transfer of electron is 1 ( 1 e-) , n=1

E° = 0.44 V

Substituting all the values in Ernst equation

0 - 0.44 - 0.0591/ 1 logkf

logkf = 0.44/ 0.0591

kf = 10 (0.44/ 0.0591)

Kf = 2.786 x 10⁷

Hence, Kf of Ag(NH₃)₂ for the given equation is 2.786 x 10⁷.

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Answer:

765.0 grams CuS

Explanation:

The limiting reagent is the reactant which completely reacts before the other reactant(s) is used up. When 9.0 moles Na₂S and 8.0 moles CuSO₄ react, it appears that CuSO₄ is the limiting reagent. You can tell because it results in the production of less product.

You can determine the mass of CuS by multiplying the moles by the molar mass. It is important to arrange the ratio in a way that allows for the cancellation of units.

Molar Mass (CuS): 95.62 g/mol

8.0 moles CuS               95.62 g
-------------------------  x  -----------------------  =  765.0 grams CuS
                                         1 mole

There are  1.2 hr would this current have to be applied to plate out 7. 20 g of iron .

Calculation ,

Given ; Current ( I ) = 5. 68 A

In [tex]Fe(NO_{3} )_{2}[/tex] , the valancy of Fe is +2 .

2 moles of [tex]e^{-}[/tex] are required for the decomposition of 1 mole of Fe .

7. 20 g  of Fe in moles  = 7. 20 g /55.845 g/mol =0.12 mole

x moles of  [tex]e^{-}[/tex] are required for the decomposition of 0.128 mole of Fe .

moles of   [tex]e^{-}[/tex] are required = 0.256 moles

Charge on 1 mole  of   [tex]e^{-}[/tex] = 96500 C

Charge on 0.256  mole  of   [tex]e^{-}[/tex] = 24704 C

Current ( I )= Q/t  

t =Q / I = 24704 C/5. 68 A = 4349 sec = 1.2 hr

Therefore , there are  1.2 hr would this current have to be applied to plate out 7. 20 g of iron .

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solution of potassium chlorate,

K

C

I

O

3

, has 20 grams of the salt dissolved in 100 grams of water at 70 C. Approximately how many more grams of the salt can be added to the solution before reaching the saturation point?

A solution of NaF is added to a solution that is 0.0144 M in Ba²⁺. When the concentration of F- exceeds 0.011 m, BaF₂ will precipitate.

Let's consider the solution of BaF₂.

BaF₂(s) ⇄ Ba²⁺(aq) + 2 F⁻(aq)

We can use the solubility product constant (Ksp) to find the equilibrium concentration of F⁻ when [Ba²⁺] is 0.0144 M.

When [F⁻] exceeds 0.011 M, BaF₂ will precipitate.

A solution of NaF is added to a solution that is 0.0144 M in Ba²⁺. When the concentration of F- exceeds 0.011 M, BaF₂ will precipitate.

What is BaF₂.?

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The grams of silver formed by the passage of 9,604 c through an electrolytic cell that contains a molten silver salt is 10.7g.

Acc. Faraday’s law of electrolysis,

W=(I×t×M) /nF

W=Faraday’s constant=96500

F=time M=molar mass

N=no. Of el.

It=9604C

Given,

Molar mass of silver=108g/mol.

n=1.

Then weight of silver deposited can be given as

W=(9604×108)/(96500×1)

= 10.7g

Thus we concluded that the silver deposited is 10.7g

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The given statement "To use the gas law constant R=0.0821 the unit for temperature should be kelvin and the unit for pressure should be atmospheres" is -----true.

What is the ideal gas law?

The ideal gas law tells us about behavior of ideal gas through a equation and postulates.The ideal gas law is considered as ideal gas equation.

The ideal gas equation is given as.

PV = nRT

where,

The value of R depends upon units of temperature, pressure and volume.

Why is Kelvin used in gas laws?

The Kelvin scale is used in gas law problems because the pressure and volume of a gas depend on the kinetic energy or motion of the particles.

How are pressure and Kelvin temperature related?

It states that, at a constant volume, the pressure of a given amount of a particular gas is directly proportional to its Kelvin temperature. It can be written as: P ∝ T, or. P/T = k where k is a constant,

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To produce a mass of 6. 38 g al using a current of 12. 50 amps the time required would be 7207.4 minutes.

It states that the chemical which is getting decomposed because of the flow of current within an electrolyte is directly proportional to the quantity of electricity passing through it.

                        W= Z × I × t    Z = 1 coulomb

                                               I = current

                                                t = time

substituting the values

                           Z= 96500/Eq.wt

​Al 3+ +3e − →Al n= 3

so, Eq= 27 / 3 = 9

t = W × Z / I

t = 9 × 96500 / 12.50

t = 7207.4 minutes.

Therefore the time required would be, 7207.4 minutes.

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Risk premiums on corporate bonds tend to decreases during business cycle expansions and increases during recessions, everything else held constant.

A bond that is sold for more than it is worth. when the prevailing interest rate is lower than the coupon rate.

Corporate bond risk premiums have become typically anticyclical, meaning they rise during recessions as well as fall throughout expansions of the business cycle.

Bonds that trade above their face value or cost more than their face value are referred to as premium bonds. Since its interest rate has been higher than those of the current market, a bond may trade at a premium. The bond's price may also increase as a result of the company's as well as the bond's credit ratings.

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Answer: C- rearranged

Explanation:

Answer:

C

Explanation:

The molar mass of a protein if a 0. 30 l of solution containing 0. 45 g of the protein has an osmotic pressure of 0. 80 torr at 25∘c is 0.348 g/mol

Here we can apply the formula ∏ = iMRT, where ∏ = osmotic pressure = 0.80 - ( given ). This is only one part of the information we are given / can conclude in this case ....

i = van’t Hoff factor = 1 for a protein molecule,

R = gas constant = 62.36 L torr / K-mol,

T ( temperature in Kelvin ) = 25 + 273 - conversion factor C° + 273 = 298K

( Known initially ) ∏ = osmotic pressure = 0.80 torr

besides the part " M " in the formula, which we have no information on whatsoever, as we have to determine it's value.

_____

Substitute derived / known values to solve for M ( moles / liter ) -

∏ = iMRT

⇒ 0.80 = ( 1 )( M )( 62.36 )( 298 )

⇒ 0.80 = M( 18583.28 )

⇒ M = 0.80 / 18583.28 ≈ 4.3049 ....

_____

We know that M = moles / liter, so we can use this to solve for moles, and hence calculate the molar mass by the formula molar mass = g / mol -

M = mol / l

⇒ 4.3049 = 0.045 / 25 mL ( 0.030 L ),

0.045 / 0.030 = 1.5 g / L

⇒ 1.5 g = 4.3049 moles,

molar mass = 1.5 g = 4.3049 moles = 0.348 g / mol

Thus the concluded that the molar mass of protein is 0.348 g / mol.

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Answer:

1/2     ( the two answers given are not FRACTIONS)

Explanation:

Moles of  CH3OH   = 128 gm / 32 gm / mole   = 4 moles

Moles of water   72 gm / 18 gm /mole = 4

Mole fraction of CH3OH  =   4 / ( 4+4) = 4/8 = 1/2  

Mg (Magnesium) is produced at the cathode during the electrolysis of a mixture of molten calcium bromide, CaBr₂ (l) and molten magnesium iodide, MgI₂ (l).

The anode in the cell has been the negative electrode and attracts the positive ions in the solution. The cathode has been the positive electrode, and attract negative ions.

In electrolysis of Calcium bromide, Calcium will form at the cathode and Bromine will form at the anode.

The reaction is as follows:

2Br⁻ (aq) → Br₂ (g) + 2e⁻

In electrolysis of magnesium Iodide, Magnesium will form at the anode Iodine will form at the cathode.

The reaction is as follows:

2I⁻ (aq) → I₂ (g) + 2e⁻

Thus from the above conclusion we can say that Mg (Magnesium) is produced at the cathode during the electrolysis of a mixture of molten calcium bromide, CaBr₂ (l) and molten magnesium iodide, MgI₂ (l).

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From the question we know that 12 - 11.70 = 0.3  so 0.3 / 12 = 0.025 therefore Simon's percent error is 2.5%.

Percent error compares an estimate to an accurate value and expresses the difference among them as a percent. This statistic lets analysts recognize the scale of the mistake relative to the true cost. it's also referred to as percentage error as well as % mistakes.

Percentage errors Calculation Steps :

Percentage error tells you the way huge your mistakes are when you degree something in an experiment. Smaller values suggest that you are close to the accepted or actual value. for instance, a 1% blunder means that you obtain very near the accepted value, whilst 45% means that you were pretty a long way off from the real value.

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6.32mL of phosphate solution is needed to completely neutralize the lead in this reaction.

A neutralization reaction is defined as the reaction between an strong acid and strong base to produce a salt and water.

The process of the reacting given volume of acids or bases for the determination of the concentration or volume of the acid or base which is required for neutralization is known as titration in volumetric analysis.

The formula used to determine the volume of the acid is given below:

(Ca × Va) /(Cb × Vb) = Na/Nb

where,

where Ca is the concentration of lead nitrate

Cb is the concentration of sodium phosphate

Va is the volume of lead nitrate

Vb is the volume of sodium phosphate

Na is the moles of lead nitrate

Nb is the moles of sodium phosphate

Concentration of lead nitrate = m/V

= 6.931/(0.1 × 331.2)

= 0.0281 M

Ca = 0.222M

Cb = 0.0281M

Na = Nb = 1

Va = ( 0.0281 × 50) /0.222

= 6.32 mL.

Thus, 6.32mL of phosphate solution is needed to completely neutralize the lead in this reaction.

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Magnesium (Mg) metal is the substance produced at the cathode.

As molten calcium bromide is electrolyzed, calcium (Ca) metal is the substance produced at the cathode.

Due to the fact that oxidation always takes place at the anode and reduction always occurs at the cathode,

Consider the two cations that were produced in the two ionisation processes of the two substances: mathrm Ca +2 and mathrm Mg +2. The following is how to express two cations' potential cathode reactions:

[tex]ca { }^{ + 2} + (aq) + 2e {}^{ - } - > \\ ca(s)E {}^{0} _{ca {}^{ + 2} \div ca } = - 2.87v[/tex]

[tex] Mg{ }^{ + 2} + (aq) + 2e {}^{ - } - > \\Mg(s)E {}^{0} _{Mg {}^{ + 2} \div Mg} = - 2.37v[/tex]

[tex] Mg^{ + 2} [/tex]

will be reduced quickly since it has a low electronegative electrode potential.

Thus, magnesium (Mg) metal is the substance produced at the cathode.

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The enthalpy of reaction (ΔH) of the above chemical reaction is equal to -143 kJ/mol.

A chemical reaction can be defined as a chemical process that involves the continuous transformation (rearrangement) of the ionic, atomic or molecular structure of a chemical element by breaking down and forming chemical bonds, in order to produce a new chemical compound while new bonds are formed.

A balanced chemical equation can be defined as a chemical equation wherein the number of atoms on the reactant (left) side is equal to the number of atoms on the product (right) side.

Next, we would write the properly balanced chemical equation for this chemical reaction:

                                    3/2O₂ -----> 2/2O₃

Therefore, the enthalpy of reaction (ΔH) of the above chemical reaction is half the enthalpy of reaction (ΔH) of the initial chemical reaction:

Enthalpy of reaction (ΔH) = ½ × -286 kJ/mol

Enthalpy of reaction (ΔH) = -143 kJ/mol.

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Now , we know that one mole of SO² Contains 2 Moles ( two ) of atoms of oxygen . Hence , 16 Grams of SO² Contains 0.5 Mole(s) of Oxygen

0.5

hope this helps you. :)

Potential advantages of using water as a solvent in organic chemistry are:

Disadvantages of using water as a solvent in organic chemistry

Water is the most common liquefaction solvent. Not only is water an environmentally friendly solvent that is inexpensive and does not pollute the environment, but it also facilitates the recovery and recycling of inorganic matter contained in biomass in ionic form for eventual use as fertilizers.

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Answer:

i think its -91.6

Explanation:

2 ( - 950.8) - (( - 1130.7) + ( - 393.5) + ( - 285.8))

The standard enthalpy change for the given reaction is approximately 91.9 kJ mol⁻¹.

The heat produced during a chemical reaction under specific conditions is known as the standard enthalpy change, abbreviated as H° (zero delta H). Standard conditions often refer to all reactants and products in their standard states, which are their most stable forms at 298 K and 1 bar (or 1 atm) pressure.

Hess's law, which states that the total enthalpy change of a reaction is equal to the sum of the enthalpy changes of its component phases, is used to calculate the standard enthalpy change of the reaction.

The following reaction is given:

[tex]\rm 2 NaHCO_3(s)--- > Na_2CO_3(s)+ CO_2(g) + H_2O(l)[/tex]

The relevant enthalpy changes of formation are:

ΔH°[NaHCO3(s)] = -950.8 kJ mol⁻¹

ΔH°[Na2CO3(s)] = -1130.7 kJ mol⁻¹

ΔH°[CO2(g)] = -393.5 kJ mol⁻¹

ΔH°[H2O(l)] = -285.8 kJ mol⁻¹

By using Hess's Law, the standard enthalpy change (ΔH°) for the reaction can be calculated as follows:

ΔH° = Σ (products) - Σ (reactants)

ΔH° = [ΔH°[[tex]\rm Na_2CO_3[/tex](s)] + ΔH°[[tex]\rm CO_2[/tex](g)] + ΔH°[H2O(l)]] - [2 × ΔH°[[tex]\rm NaHCO_3[/tex](s)]]

ΔH° = [-1130.7 kJ mol⁻¹ + (-393.5 kJ mol⁻¹) + (-285.8 kJ mol⁻¹)] - [2 × (-950.8 kJ mol⁻¹)]

ΔH° = [-1809.7 kJ mol⁻¹] - [-1901.6 kJ mol⁻¹]

ΔH° ≈ -1809.7 kJ mol⁻¹ + 1901.6 kJ mol⁻¹

ΔH° ≈ 91.9 kJ mol⁻¹

Hence, the standard enthalpy change for the given reaction is approximately 91.9 kJ mol⁻¹.

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The correct product from this reaction is the structure in option D.

An amine is a compound that contains an alkyl group and an amino group. We know that we can obtain an amine from ammonia by replacing one or more hydrogen atoms with alkyl groups.

The compound as shown here is a secondary amine because two hydrogen atoms have been replaced with alkyl groups. We know that the reaction between the HCl and the amine produces an ionic compound owing to the fact that the nitrogen and the hydrogen from HCl are involved in a Dative bond.

As such, the correct product from this reaction is the structure in option D.

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3.01× 1024 particles are the number of  particles are there in 5 grams of sodium carbonate.

There are 6.022 × 1023 particles in one gram of a substance according to Avogadro's number. So when we find out for 5 grams, then we multiply 5 with 6.022 × 1023, we get 3.01 × 1024 particles. For one gram atomic weight of hydrogen, one mole of hydrogen contains 6.022 × 1023 hydrogen atoms.

So we can conclude that 3.01× 1024 particles are the number of  particles are there in 5 grams of sodium carbonate.

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H2o is initially present in a state where its molecules are far apart. during a change of state, its molecules slow down. Gaseous state to liquid state has most likely take place.

In the gaseous state, molecules of water are far away. Hence intermolecular forces are weak.

But when water vapours changes into liquid state molecules start coming closer. And hence intermolecular forces become strong.

Due to this entropy of water molecules decreases.

Thus from above we concluded that when water vapour changes into liquid its molecules is slow down.

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