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

Molecular formula of a compound whose molecular mass is 60.00 g/mol and has an empirical formula of C₂H₈N₂.

We know the empirical formula and thus the molar mass of the empirical formula, we simply need to find out how many of these fit into the molar mass of the molecular formula.

In this problem, we have an empirical formula of CH₄N

so the molar mass is 12 + 4 + 14 = 30 g/mol.

Molecular formula mass/Empirical formula mass=60 g/mol/30 g/mol=2

The molecular formula is TWICE that of the empirical formula.

Molecular formula = 2XCH₄N = C₂H₈N₂

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Answer:

C2H8N2

Explanation:

edge 2023

The molar mass of the acid if 28. 5 ml of the koh solution is required to neutralize the sample is 90.23g/mol.

Molarity = (moles of solute/volume of solution) × 1000

Given,

Molarity of KOH solution = 0.180 M

Volume of solution = 28.5 mL

0.180 = (moles of KOH/ 28.5) × 1000

Moles = (0.18× 28.5)/1000

= 0.00513 mol

HA + KOH ------- KA + H2O

1 moles of KOH reacts with 1 moles of HA.

So, 0.00513 moles of KOH react with 0.00513 moles of HA.

Number of moles= given mass/ Molar mass

Given,

Mass of HA = 0.462 g

Moles of HA = 0.00512 mol

0.00512 = 0.462/ Molar mass

Molar mass = 90.23 g/ mol.

Thus the molar mass of HA required to neutralize the 28.5 mL of KOH is 90.23g/mol.

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Answer:

C. C-14 decays but C-12 stays the same after death.

Explanation:

C-12 is stable and won't change.  C-14 is slightly radioactive.  It decays over time to C12.  A living plant or animal has a fairly constant C14/C12 ratio during it's lifetime.  Sources of food have a fairly consistent C-14/C-12 ratio.  Once it dies, the C-14 is no longer replenished, so the ratio of C-14 to C-12 drops over time, giving paleontologists the ability to carbon-date the sample.  The less C-14, the older the sample.

The stock solution should have a molarity of 8.24M.

Proportional sets. C₁V₁=C₂V₂ is used to calculate an unknown quantity where two solutions/mixtures are proportional.

It follows that the initial concentration times the initial volume will equal the final concentration times the final volume because the sum does not vary.

Using the formula as follows;

                       C₁V₁ = C₂V₂

Where;

C₁ = initial concentration of the stock solution

C₂ = final concentration of the stock solution

V₁ = initial volume of the stock solution

V₂ = final volume of the stock solution

According to the information given in this question,

V₁ = 25mL = 25/1000 = 0.025L

V₂ = 2L

C₁ =?

C₂ = 0.103M

Using the formula; C₁V₁ = C₂V₂

C₁ = C₂V₂/V₁

C₁ = (0.103 × 2) / 0.025

C₁ = 0.206 / 0.025

C₁ = 8.24 M

Therefore, the stock solution should have a molarity of 8.24M.

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27.94 g/mole is the molar mass of the unknown gas.

Use Graham's Law, which indicates that the square root of a gas's molar mass determines its rate of effusion.

The following equation compares two distinct gases, (a) and (b), and expresses this:

sqrt(M(b)/M) = R(a)/R(b) (a)

R is the effusion rate.

square root = sqrt

the molar mass, M

In response to your query, if gas (a) is the unknown gas and Xe(M = 131 g/mol) is the second gas, we have:

sqrt(M(Xe)/M(a) = 2.165, where R(a)/R(Xe)

Square both sides, and you get:

(2.165)² = 131 g/mol/ M (a)

M(a)(2.165)² = 131

M(a) is equal to 131/(2.165)² = 27.94 g/mol.

Having a molar mass of 27.94 g/mol, the unknown gas.

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The reaction of tetrahedral intermediate and the two leaving groups, methoxide and amide anion favour product.

A leaving group is a group of atoms or an atom which is able to break away from a molecule as a stable species or with a lone pair, breaking the bond between the molecule and itself.

A stage is come before forming product. This stage is termed as transition stage and the reactant at this stage is termed as intermediate.

According to Le Châtelier's Principle, a reactions always tend towards the equilibrium, the reaction produces more products from the excess reactant, that's why it causing the system to shift to the left which allows the system to reach equilibrium.

Due to this reason, this reaction shift towards product.

Thus, we concluded that the reaction of tetrahedral intermediate and the two leaving groups, methoxide and amide anion favour product.

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Answer: Most polyatomic ions are -ate, related ions that have 1 fewer oxygen atom are -ite (must have at least 1 oxygen tho) and nonmetals are ide.

Explanation:

The valence electrons are present in the lewis formula for the perchlorate ion will be 32.

The electrons in such an atom's outermost atomic shell known as the valence electrons may be those that orbit the nucleus. Core electrons are mostly in filled orbitals some of which are located closest to the nucleus.

The formula of perchlorate ion is [tex]ClO^{-} _{4}[/tex].

The valence electrons of Cl is 7 and the valence electrons of oxygen is 6 , one negative charge . By adding the total valence electron, it will be 32.

The valence electrons are present in the lewis formula for the perchlorate ion will be 32.

Hence, the correct answer will be option (e).

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A student compares the boiling point of substances having different intermolecular forces. Boiling points of various substances is the dependent variable that student most likely use.

The boiling point of a substance is proportional to the strength of its intermolecular forces, the higher the boiling point, the stronger the intermolecular forces. We can compare the strengths of intermolecular forces by comparing the boiling points of different substances.

Intermolecular forces are measured by boiling points.

Intermolecular forces increase as bond polarization increases.

Ionic > hydrogen bonding > dipole dipole > dispersion is the order of the strength of intermolecular forces (and thus their impact on boiling points).

Substances with strong intermolecular forces are very attracted to one another and are held together tightly. These substances require a great deal of energy to separate, whereas substances with weak intermolecular forces are held together very loosely and have weak interactions.

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A combustion reaction will always be associated with a change in entropy that will be positive.

The entropy of a system is the measurement of the disorder of the system which happens in the thermodynamic reaction.

During association the value of entropy  will be positive and during dissociation the value will be negative.

During association the number of molecule increases due to which the entropy of the combustion increases and the value will be positive .

Therefore , combustion reaction will always be associated with a change in entropy that will be positive.

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Correct option is C

Latent heat of vaporization is the amount of heat required to bring about the phase transition from liquid to gaseous state, at its boiling point.

The substance is Propane :

10140000 J of heat is used to boil 28.47 kg at 298 K. Here, the important thing is, the liquid is changing to vapor and the temperature is not changing means it's a phase change. So, the formula used for this is:

                        q=m× Δ[tex]H_{vap}[/tex]

where, q is the heat energy, m is mass and  Δ[tex]H_{vapour}[/tex]  is the enthalpy of vaporization.

Let's rearrange the formula for  as:

Given mass of the liquid = 28.47kg

Heat required to boil the liquid = 10.14 ×[tex]10^{2} J[/tex]

Let's plug in the values: [tex]\frac{10.14*10^{2}J }{28.47Kg} =356164\frac{J}{Kg}[/tex]

Rounding off 356164 will turn into 356000 J[tex]Kg^{-1}[/tex]

Hence, the right choice is C. propane 356000 j[tex]Kg^{-1}[/tex]

What is latent heat ?

Latent heat, energy absorbed or released by a substance during a change in its physical state (phase) that occurs without changing its temperature.

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Heat change at constant volume for the combustion of carbon monoxide = -67.71 Kcal

CO(g) + 1/2O2(g) ⟶ CO2(g)

From the above reaction,

∆ng = nP - nR = 1 - ( 1/2 + 1 )

∆ng = - 1/2

Temperature (T) = 17°C (17+ 273) K = 290K (Given)

Now from first law of thermodynamics,

∆H = ∆E + ∆ng RT

ΔΗ = -67.71 + (-1/2) x 2 x 10^-3 x 290

∆H = -67.71 - 0.29 = -68 Kcal

Hence the heat of reaction at constant pressure will be -68 Kcal

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The amount of CO that would be required to generate 635 g of CO2 will be 404.14 g

First, let us get the equation of the reaction:

[tex]2CO + O_2 -- > 2CO_2[/tex]

From the equation, we can see that the mole ratio of CO to that of CO2 is 1:1.

635 g of CO2 is to be generated.

Mole of 635 g CO2 = mass/molar mass = 635/44.01 = 14.43 moles

Thus, the equivalent mole of CO required will also be 14.43 moles.

Mass of 14.43 moles CO = moles x molar mass = 14.43 x 28.01 = 404.14 g

Hence, 404.14 g of CO will be required to produce 635 g of CO2

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

The transformation of water vapour into liquid is known as condensation. The process is the opposite of evaporation, in which liquid water turns into a vapour. Either the air is chilled to its dew point or it gets too saturated with water vapour to retain any more water, causing condensation to occur.

Attraction - Gas Particles move continuously in a straight path in a gas. They are significantly further apart and move independently of one another because the kinetic energy of the molecule is larger than the attraction force between them. There are often almost no attractive forces between particles.

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The two common by-products of sugar fermentation are acid ( like lactic acid) or hydrogen and gas(carbon dioxide). During fermentation , any gas that is produced by the microbes get trapped in the inverted Durham tube and bubbles will appear.

The acid production is indicated by the color change of medium from normal red to yellow and remains red for no acid production.

The respiration process can be aerobic or anaerobic. In both process glycolysis, which occur in absence of oxygen is common. In this process one molecule of glucose is converted into two molecules of pyruvic acid . Gycolysis occur in cell cytoplasm. In anaerobic respiration is a cytosolic process, which results in production of lactic acid ( lactic acid fermentation)   or ethanol and carbon dioxide (alcoholic fermentation)

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The end (final) value of n in a hydrogen atom transition is 2.

Using the Rydberg formula ;

1/lamda = R ( 1/[tex]n_{f} ^{2}[/tex] - 1/[tex]n_{i} ^{2}[/tex] )

where , lamda = wavelength = 486 nm = 486 ×[tex]10^{-9} m[/tex]

[tex]n_{f}[/tex] in a hydrogen atom transition = final state = ?

[tex]n_{i}[/tex] in a hydrogen atom transition = initial state = 4

R = Rydberg constant = 1.097 ×[tex]10^{7} m^{-1}[/tex]

1/ 486 ×[tex]10^{-9} m[/tex] = 1.097 ×[tex]10^{7} m^{-1}[/tex] (  1/[tex]n_{f} ^{2}[/tex] - 1/[tex]4^{2}[/tex] )

[tex]n_{f}[/tex] = 2

Missing parts :

Determine the end (final) value of n in a hydrogen atom transition, if the electron starts in n=4 and the atom emits a photon of light with a wavelength of 486 nm.

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Answer:

Mg+2HCL-Magnesium Chloride +Hygrogen

[tex] > [/tex]

Explanation:

N.O of moles=Mass\Molar Mass

192÷24=8

1:1

8×2=16

[tex]\huge\underline{\underline{\boxed{\mathbb {SOLUTION:}}}}[/tex]

[tex]\longrightarrow \sf{Mg+2HCI \Longrightarrow MgCl_2+H_2}[/tex]

[tex]\longrightarrow \sf{molMg=129gMg \times \dfrac{1molMg}{24.305gMg=} }[/tex]

[tex]\small \sf{\longrightarrow molH_2=7.90molMg \times \dfrac{1molH_2}{1molMg}= 7.90molH_2 }[/tex]

[tex]\small \sf{\longrightarrow massH_2=7.90H_2 \times \dfrac{2.016g}{1molH_2}= 7.90molH_2= 15.9 \: gH_2}[/tex]

[tex]\huge\underline{\underline{\boxed{\mathbb {ANSWER:}}}}[/tex]

[tex] \small\bm{ The \: mass \: of \:H_2 \: produce \: will \: be \: 15.9 }[/tex] [tex]\bm{grams.}[/tex]

The pH of a buffer solution containing acid of pKa 7.5 will be 8.1.

The pH of a solution indicates its acidity or basicity. A pH below 7 is acidic, while pH above 7 is basic. 7 is considered as the neutral value.

A buffer is a solution that has the ability to resist the change in pH when an acid or a base is added to it. The natural example of buffer is blood.

According to the question, pKa of acid = 7.5

If concentration of base is x then, concentration of acid will be x/4.

According to the Henderson–Hasselbalch equation,

pH = pKa + log [A⁻] / [HA]

pH = 7.5 + log [[tex]\frac{x}{x/4}[/tex]]

pH = 7.5 + log 4 ⇒ 7.5 + 0.602

pH = 8.1

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At 2000 K, if the partial pressures of an equilibrium mixture of H2S, H2, and S are 0.0151 , 0.0472 , and 0.0352 atm, respectively, then the value of the equilibrium constant Kp at 2000 K is 9.089.

The pressure that is exerted by one among the mixture of gases if it occupies the same volume on its own is known as Partial pressure.

As we know that,

Kp is the constant of a certain reaction at equilibrium.

For the given chemical reaction:

H2 +S -------H2S

The expression of Kp for above equation follows:

Kp = p(H2S) /{p(H2) ×p(S)}

where,

p(H2S) is the partial pressure of H2S = 0.0151atm

p(H2) is the partial pressure of H2 = 0.0472atm

p(S) is the partial pressure of S = 0.0352 atm

Kp = 0.0151/(0.0472 × 0.0352)

Kp = 0.0151/0.00166144

Kp = 9.089

Thus, we find that the value of equilibrium constant Kp at 2000K is 9.089.

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option (B) KOH is the correct answer

KOH is the substance that, when added to the solution, is most likely to raise the pH .

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n oxidative phosphorylation, ATP production is coupled to the events in the electron-transport chain

Abiotic factors are the physical and chemical compounds of an environment, being able to influence the living beings present in the ecosystem so that plants and animals have to adapt to survive and develop.

The biotic components are all the living things that live in a certain place, while the abiotic components are all the physical, chemical and geological factors of the environment such as

With this information, we can conclude that An abiotic factor is a non-living part of an ecosystem that shapes its environment.

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The molarity of the NaOH is  0.238 M when  1. 477 g sample of KHP required 30. 27 ml of NaOH solution to reach endpoint.

Calculation ,

Mass of  KHP  = 1. 477 g

Number of moles of  KHP = given mass / molar mass of  KHP

Number of moles of  KHP = 1. 477 g × 1 mol / 204.23 g  KHP = 0.007 mol

Since , one mole  KHP  requires one mole of NaOH .

Number of moles of NaOH required by 0.007 mol of KHP = 0.007 mol

Molarity of NaOH  = number of moles / volume in lit = 0.007mol/0.03027ml =  0.238 M

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Volume of hydrogen (in l) is produced is 35.20 L

Given:

moles of hcl = 3.143 mol

To Find:

volume of hydrogen

Solution:

2HCl → H2 + Cl2

STP 2*22.4L 22.4L 22.4L

2 mol 1 mol 1 mol

73 gm 2gm 35.5gm

here we can see that 2 mol hcl is producing 22.4 litres of hydrogen

So, 3.143 mol hcl will produce hydrogen = 22.4/2*3.143 = 35.20 litres

Volume of hydrogen (in l) is produced is 35.20 L

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option (b) freezing water is the correct answer.

Freezing water is an example of an exothermic process.

We can observe that melting of an ice is an example of endothermic process whereas, freezing of water is considered to be an example for exothermic process.

Water releases heat when it solidifies, warming the area around it. As a result, freezing is an exothermic process.

In the above given options,

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Answer:

B.) Freezing water

Explanation:

I got it correct on founders edtell

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Answer:

B) ATP to ADP

Explanation:

Hope this helps! Pls give brainliest!

It is necessary to add zncl 2 to promote the reaction of hcl with some alcohols as this reaction is reversible to make it irreversible ZnCl2 is used.

ZnCl2 is used as Lewis acid catalyst. Since Cl- is weaker nucleophile than Br. The lewis acid ZnCl2 coordinate with oxygen of alcohol and proceed the reaction in forward direction.

HCl is less reactive with alcohols whereas the bond between H and Cl is stronger than HBr and HI.

Hence its reaction with alcohols need a catalyst ZnCl2, which helps in the breaking of bond between H and Cl.

Therefore, Luca's reagent is used in the reaction of HCl and alcohol.

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This is why even though Olive oil and butter both contain fats, olive oil is liquid at room temperature and butter is solid.

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Steps that are required for the correct use of an automatic micropipettor -

1. ensure that the pipette is adjusted to discharge the proper V

2. depress the knob until you reach the point of first resistance.

3. dip the tip of the pipette into the solution

4. gradually depress the plunger to suck solution into the tip

5. Remove the tip from the solution and put it on the inside wall of the container into which the solution to be dispersed.

6. again depress the knob to the first point of resistance

7. After a brief pause, depress the knob as far as it will go.

Pipettes that are automated can eliminate the need for manual work and provide a variety of benefits. The most obvious benefit is increased throughput, since automation frees up time and effort for other activities. Another advantage is increased repeatability.

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