Using Charles’s law, V1/T1 =V2/T2 and the information below, solve the question. If a balloon takes up 625L at 273K, what will the new volume be when the balloon is heated to 353K.

Answer:

NO−3→NO−2→N2O→N2

Explanation:

Denitrification is the process by which nitrogen is returned to the atmosphere by denitrifying bacteria. The process of denitrification involves a sequence of reduction reactions in the sequence; NO3−→NO2−→N2O→N2.

Nitrogen is usually present in soil in the form of soil nitrates which are soluble in water and can be absorbed by plant roots. These denitrifying bacteria reduce soil nitrates to nitrites, then to nitrogen I oxide and finally to molecular nitrogen as shown in the sequence above.

Denitrification can release N2O, is an ozone-depleting substance and

greenhouse gas into the atmosphere with its attendant consequence on global warming.

Answer:

[tex]148~ATP[/tex]

Explanation:

In case, we can start with the structure of [tex]CH_3(CH_2)_2_0CO_2H[/tex]. When we draw the molecule, we will obtain a total amount of 22 carbons. So, in order to calculate the total amount of ATP we have to make several questions:

How many Acetyl CoA is produced?

To calculate the total Acetyl Coa we have to use the equation:

[tex]Number~of~Acetyl~Coa=\frac{n}{2}[/tex], where n is the amount of carbons, so:

[tex]Number~of~Acetyl~Coa=\frac{22}{2}=11~carbons[/tex]

How many rounds take place?

To calculate the rounds we have to use the equation:

[tex]Number~of~rounds=\frac{n}{2}-1[/tex], where n is the amount of carbons, so:

[tex]Number~of~Acetyl~Coa=\frac{22}{2}-1=10~carbons[/tex]

How many [tex]FADH_2[/tex] and [tex]NADH[/tex] are produced for this fatty acid?

For each round, we will have 1 [tex]FADH_2[/tex] and 1 [tex]NADH[/tex], if we have 10 rounds. In total, we will have 10

How many ATP are formed?

The ATP would be formed in the electron transport chain and each coenzyme will have a different yield of ATP. So for the total calculation, we have to keep in mind the following relationships:

-) [tex]1~FADH_2~=~1.5~ATP [/tex]

-) [tex]1~NADH~=~2.5~ATP [/tex]

-) [tex]Acetyl~CoA~=~10~ATP[/tex]

So, know we can do the total calculation:

[tex](10*1.5)+(10*2.5)+(11*10)=150[/tex]

We have to subtract  "2 ATP" molecules that correspond to the activation of the fatty acid, so:

[tex]150-2=148~ATP[/tex]

In total, we will have 148 ATP.

See figure 1

I hope it helps!

Answer:

AI(NO₃)₃ → Acidic   pH < 7

CH₃NH₃CN  → Neutral  pH = 7

NaCIO  → Basic   pH > 7

CH₃NH₃CI → Acidic   pH < 7

NaNO₃ → Neutral  pH = 7

Explanation:

First of all we dissociate the salts:

Nitrate anion comes from the nitric acid which is strong, so the anion is the conjugate weak base. It does not react to water, but the Al is an special case. Aluminum as a cathion comes from the Al(OH)₃ which is a base but this compound can also react as an acid, it is called amphoterous.

Al³⁺  +  H₂O  ⇄ Al(OH)²⁺  +  H⁺

Aluminium cathion reacts to water in order to produce a complex and to give protons to the medium, so the salt is acid.

Both ions come from a weak base and a strong acid, so both ions are the conjugate strong base and acid, respectively. They can make hydrolysis to water so the salt is neutral.

CH₃NH₃⁺   +  H₂O  ⇄  CH₃NH₂  +  H₃O⁺     Ka

CN⁻  +  H₂O  ⇄  HCN +  OH⁻   Kb

Sodium cathion, comes from the strong base NaOH so it is does not react to water. It is the conjugate weak acid. Hypochlorite comes from the weak acid, so it can hydrolyse to water.

ClO⁻  +  H₂O  ⇄  HClO  +  OH⁻     Kb

Hypochlorous acid is formed giving OH⁻ to medium, so the salt is basic.

Chloride comes from the strong acid HCl. It does not react to water.

Methylammonium comes from the weak base, methylamine so it can react to water in order to make hydrolysis. The salt will be acidic.

CH₃NH₃⁺   +  H₂O  ⇄  CH₃NH₂  +  H₃O⁺     Ka

NaNO₃ → Na⁺  +  NO₃⁻

Both ions come from a strong base and acid, so they are the conjugate base and acid, respectively. As they do not make hydrolisis in water, the salt will be neutral.

Answer:

(1) Before the addition of any HBr, the pH is 12.02

(2) After adding 12.0 mL of HBr, the pH is 10.86

(3) At the titration midpoint, the pH is 10.73

(4) At the equivalence point, the pH is 5.79

(5) After adding 45.1 mL of HBr, the pH is 1.18

Explanation:

First of all, we have a weak base:

(CH₃)₂NH  + H₂O  ⇄  (CH₃)₂NH₂⁺  +  OH⁻            Kb = 5.4×10⁻⁴

0.289 - x                             x                x

Kb = x² / 0.289-x

Kb . 0.289 - Kbx - x²

1.56×10⁻⁴ - 5.4×10⁻⁴x - x²

After the quadratic equation is solved x = 0.01222 → [OH⁻]

- log  [OH⁻] = pOH → 1.91

pH = 12.02   (14 - pOH)

We determine the mmoles of H⁺, we add:

0.286 M . 12 mL = 3.432 mmol

We determine the mmoles of base⁻, we have

27.9 mL . 0.289 M = 8.0631 mmol

When the base, react to the protons, we have the protonated base plus water (neutralization reaction)

(CH₃)₂NH     +      H₃O⁺        ⇄  (CH₃)₂NH₂⁺  +  H₂O

8.0631 mm       3.432 mm                 -

4.6311 mm                                  3.432 mm

We substract to the dimethylamine mmoles, the protons which are the same amount of protonated base.

[(CH₃)₂NH] → 4.6311 mm / Total volume (27.9 mL + 12 mL) = 0.116 M

[(CH₃)₂NH₂⁺] → 3.432 mm / 39.9 mL = 0.0860 M

We have just made a buffer.

pH = pKa + log (CH₃)₂NH  / (CH₃)₂NH₂⁺

pH = 10.73 + log (0.116/0.0860) = 10.86

mmoles of base = mmoles of acid

Let's find out the volume

0.289 M . 27.9 mL = 0.286 M . volume

volume in Eq. point = 28.2 mL

(CH₃)₂NH     +      H₃O⁺        ⇄  (CH₃)₂NH₂⁺  +  H₂O

8.0631 mm       8.0631mm               -

                                                8.0631 mm

We do not have base and protons, we only have the conjugate acid

We calculate the new concentration:

mmoles of conjugated acid / Total volume (initial + eq. point)

[(CH₃)₂NH₂⁺] = 8.0631 mm /(27.9 mL + 28.2 mL)  = 0.144 M

(CH₃)₂NH₂⁺   +  H₂O   ⇄   (CH₃)₂NH  +  H₃O⁻       Ka = 1.85×10⁻¹¹

 0.144 - x                                  x               x

[H₃O⁺] = √ (Ka . 0.144) →  1.63×10⁻⁶ M  

pH = - log [H₃O⁺] = 5.79

This is the point where we add, the half of acid. (14.1 mL)

This is still a buffer area.

mmoles of H₃O⁺ = 4.0326 mmol (0.286M . 14.1mL)

mmoles of base = 8.0631 mmol - 4.0326 mmol

[(CH₃)₂NH] = 4.0305 mm / (27.9 mL + 14.1 mL) = 0.096 M

[(CH₃)₂NH₂⁺] = 4.0326 mm (27.9 mL + 14.1 mL) = 0.096 M

pH = pKa + log (0.096M / 0.096 M)

pH = 10.73 + log 1 =  10.73

Both concentrations are the same, so pH = pKa. This is the  maximum buffering capacity.

mmoles of acid = 45.1 mL . 0.286 M = 12.8986 mmol

mmoles of base = 8.0631 mmoles

This is an excess of H⁺, so, the new [H⁺] = 12.8986 - 8.0631 / Total vol.

(CH₃)₂NH     +      H₃O⁺        ⇄  (CH₃)₂NH₂⁺  +  H₂O

8.0631 mm     12.8986 mm             -

       -               4.8355 mm                        

[H₃O⁺] = 4.8355 mm / (27.9 ml + 45.1 ml)

[H₃O⁺] = 4.8355 mm / 73 mL → 0.0662 M

- log [H₃O⁺] = pH

- log 0.0662 = 1.18 → pH

Answer:

Molecule C

Explanation:

In this case, on the first reaction, we will have the production of a Grignard reagent. This molecule will react with [tex]D_2O[/tex] and a deuterium atom will be transferrred to the benzene ring. Then at the top of the molecule, we will have an acetal structure. This acetal can be broken by the action of the acid [tex]DCl[/tex], In the mechanism at the end, we will obtain a carbonyl group bonded to a hydrogen atom. Therefore we will have in the final product the aldehyde group. See figure 1 to further explanations.

I hope it helps!

Answer:

[I⁻] = 0.0352M

Explanation:

Based on the equilibrium:

I₃⁻(aq) ⇄ I₂(aq) + I⁻(aq)

Kc is defined as:

Kc = 0.25 = [I₂] [I⁻] / [I₃⁻]

The system reaches the equilbrium when the ratio [I₂] [I⁻] / [I₃⁻] is equal to 0.25

In the beginning, you add 0.0401M of both [I₂] [I⁻].  When the reaction reach the equilibrium, xM of both [I₂] [I⁻] is consumed producing xM of  [I₃⁻]. That is written as:

[I₃⁻] = X

[I₂] = 0.0401M - X

[I⁻] = 0.0401M - X

X is known as reaction coordinate.

Replacing in Kc:

0.25 = [I₂] [I⁻] / [I₃⁻]

0.25 = [0.0401M - X] [0.0401M - X] / [X]

0.25X = 0.00160801 - 0.0802X + X²

0 = 0.00160801 - 0.3302X + X²

Solving for X:

X = 0.0049M → Right solution

X = 0.3252M → False solution. Produce negative concentrations

Replacing, equilibrium concentrations will be:

[I₃⁻] = X

[I₂] = 0.0401M - X

[I⁻] = 0.0401M - X

[I₃⁻] = 0.0049M

[I₂] = 0.0352M

The equilibrium concentration (in molarity) of  [I⁻] should be considered as the 0.0352M.

Since

I₃⁻(aq) ⇄ I₂(aq) + I⁻(aq)

Here Kc should be defined

Kc = 0.25 = [I₂] [I⁻] / [I₃⁻]

Also, The system finished the equilibrium at the time when the ratio [I₂] [I⁻] / [I₃⁻] is equivalent to 0.25.

Also,

[I₃⁻] = X

[I₂] = 0.0401M - X

[I⁻] = 0.0401M - X

Also,

0.25 = [I₂] [I⁻] / [I₃⁻]

0.25 = [0.0401M - X] [0.0401M - X] / [X]

0.25X = 0.00160801 - 0.0802X + X²

0 = 0.00160801 - 0.3302X + X²

Now

X = 0.0049M → Right solution

X = 0.3252M → False solution

Now equilibrium concentrations will be:

[I₃⁻] = X

[I₂] = 0.0401M - X

[I⁻] = 0.0401M - X

[I₃⁻] = 0.0049M

[I₂] = 0.0352M

[I⁻] = 0.0352M

Hence, The equilibrium concentration (in molarity) of  [I⁻] should be considered as the 0.0352M.

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Answer:

1. 0.98 atm

Explanation:

The following data were obtained from the question:

Mass of unknown liquid (m) = 0.3175 g

Temperature (T) = 99 °C

Pressure (P) = 748.2 mmHg

Volume (V) = 145.0 mL

Gas constant (R) = 0.08206 atm.L/Kmol

1. Determination of the pressure in atm.

760 mmHg = 1 atm

Therefore,

748.2 mmHg = 748.2/760 = 0.98 atm

Therefore, the pressure in atm is 0.98 atm.

Answer:

ΔH = -338.8kJ

Explanation:

it is possible to sum the enthalpy changes of some reactions to obtain the enthalpy change of the whole reaction (Hess's law).

Using the reactions:

R₁ O₂(g) → 2O(g) ΔH₁°= 449.2 kJ

R₂ O₃(g) + Cl(g) → O₂(g) + ClO(g) ΔH₂° = -126 kJ

R₃ ClO (g) + O (g) → O₂ (g) + Cl (g) ΔH₃°= -268 kJ

By the sum 2R₂ + 2R₃:

(2R₂ + 2R₃) = 2O(g) + 2O₃(g) → 4O₂(g)

ΔH = 2ₓ(-126kJ) + (2ₓ-268kJ) = -788kJ

Now, this reaction + R₁

2O₃(g) → 3O₂(g)

ΔH = -768kJ + 449.2kJ

Answer:

Salt solution may be calculated by measuring the specific gravity of a sample of water using a hydrometer.

Hope this answer correct (^^)....

Answer:

Strong intermolecular forces:  an increase in viscosity of the liquid, increase in surface tension, decrease in vapor pressure, and an increase in the boiling point.

Weak intermolecular forces: a decrease in viscosity, a decrease in surface tension, an increase in vapor pressure and an increase in boiling point.

Explanation:

Intermolecular forces are forces of attraction or repulsion between neighboring molecules in a substance. These intermolecular forces inclde dispersion forces, dipole-dipole interactions, hydrogen bonding, and ion-dipole forces.

The strength of the intermolecular forces in a liquid usually affects the various properties of the liquid such as viscosity, surface tension, vapour pressure and boiling point.

Strong intermolecular forces in a liquid results in the following; an increase in viscosity of the liquid, increase in surface tension, decrease in vapor pressure, and an increase in the boiling point of the liquid.

Weak intermolecular forces in a liquid results in the following; a decrease in viscosity, a decrease in surface tension, an increase in vapor pressure and an increase in boiling point of that liquid.

Strong intermolecular force is defined as the increase in viscosity of the liquid, increase in surface tension, decrease in vapor pressure, and an increase in the boiling point while  weak intermolecular forces define as the decrease in viscosity, a decrease in surface tension, an increase in vapor pressure, and an increase in boiling point.

Intermolecular forces are forces of attraction or repulsion between neighboring molecules in a substance. These intermolecular forces include as follows:-

Strong intermolecular forces in a liquid result in the following; an increase in viscosity of the liquid, increase in surface tension, decrease in vapor pressure, and an increase in the boiling point of the liquid.

Weak intermolecular forces in a liquid result in the following; a decrease in viscosity, a decrease in surface tension, an increase in vapor pressure, and an increase in the boiling point of that liquid.

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Answer:

0.250 mol Mg²⁺

0.500 mol Cl⁻

Explanation:

Magnesium chloride (MgCl₂) dissociates into ions according to the following equilibrium:

MgCl₂  ⇒  Mg²⁺ + 2 Cl⁻

1 mol      1 mol   2 mol

1 mol of Mg²⁺ and 2 moles of Cl⁻ are formed per mole of MgCl₂.  If we have 0.250 mol of MgCl₂, the following amounts of ions will be formed:

0.250 mol MgCl₂ x 1 mol Mg²⁺/mol MgCl₂= 0.250 mol Mg²⁺

0.250 mol MgCl₂ x 2 mol Cl⁻/mol MgCl₂= 0.500 mol Cl⁻

Answer:

HEY THE ANSWER ABOVE ME IS RIGHT!! i defientely misclicked my rating :/

5/5 all the way.

Explanation:

Answer:

The molar concentration of HCl in the aqueous solution is 0.0131 mol/dm3

Explanation:

To get the molar concentration of a solution we will use the formula:

Molar concentration = mass of HCl/ molar mass of HCl

Mass of HCl in the aqueous solution will be 40% of the total mass of the solution.

We can extract the mass of the solution from its density which is 1.2g/mL

We will further perform our analysis by considering only 1 ml of this aqueous solution.

The mass of the substance present in this solution is 1.2g.

The mass of HCl Present is 40% of 1.2 = 0.48 g.

The molar mass of HCl can be obtained from standard tables or by adding the masses of Hydrogen (1 g) and Chlorine (35.46 g) = 36.46g/mol

Therefore, the molar concentration of HCl in the aqueous solution is 0.48/36.46 = 0.0131 mol/dm3

Answer:

pH = 3.95

Explanation:

It is possible to calculate the pH of a buffer using H-H equation.

pH = pka + log₁₀ [HCOONa] / [HCOOH]

If concentration of [HCOONa] = [HCOOH] = 0.50M and pH = 3.77:

3.77 = pka + log₁₀ [0.50] / [0.50]

3.77 = pka

Knowing pKa, the NaOH reacts with HCOOH, thus:

HCOOH + NaOH → HCOONa + H₂O

That means the NaOH you add reacts with HCOOH producing more HCOONa.

Initial moles of 100.0mL = 0.1000L:

[HCOOH] = (0.50mol / L) ₓ 0.1000L = 0.0500moles HCOOH

[HCOONa] = (0.50mol / L) ₓ 0.1000L = 0.0500moles HCOONa

After the reaction, moles of each species is:

0.0500moles HCOOH - 0.010 moles NaOH (Moles added of NaOH) = 0.0400 moles HCOOH

0.0500moles HCOONa + 0.010 moles NaOH (Moles added of NaOH) = 0.0600 moles HCOONa

With these moles of the buffer, you can calculate pH:

pH = 3.77 + log₁₀ [0.0600] / [0.0400]

When the pH be after 0.010 mol of NaOH has been added to 100.0 mL of the buffer pH is = 3.77 + log₁₀ [0.0600] / [0.0400] = 3.95

It is possible to Computation the pH of a buffer using H-H equation.

Then pH is = pka + log₁₀ [HCOONa] / [HCOOH]

Then If concentration of [HCOONa] is = [HCOOH] then = 0.50M and pH = 3.77:

3.77 is = pka + log₁₀ [0.50] / [0.50]

After that, 3.77 = pka

Then, Knowing pKa, the NaOH reacts with HCOOH, thus:

After that,[tex]HCOOH + NaOH \rightarrow HCOONa + H2O[/tex]

Now, That means the NaOH you add reacts with HCOOH producing more HCOONa.

Then, Initial moles of 100.0mL = 0.1000L:

After that, [HCOOH] = (0.50mol / L) ₓ 0.1000L = 0.0500moles HCOOH

Then, [HCOONa] = (0.50mol / L) ₓ 0.1000L = 0.0500moles HCOONa

After that, when the reaction, moles of each species is:

Then, 0.0500moles HCOOH - 0.010 moles NaOH (Moles added of NaOH) = 0.0400 moles HCOOH

Now, 0.0500moles HCOONa + 0.010 moles NaOH (Moles added of NaOH) = 0.0600 moles HCOONa

Then, With these moles of the buffer, you can calculate pH:

pH = 3.77 + log₁₀ [0.0600] / [0.0400]

Therefore, pH = 3.95

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Answer:

a. Oxidized: Fe(s)

Reduced: Pb(NO3)2

b.Oxidized: Cu(s)

Reduced: AgNO3

c. Oxidized: Al(s)

Reduced: AgNO3

Explanation:

In a redox reaction, one reactant is been oxidized whereas the other is reduced.. The reduced reactant is the one that is gaining electrons and the oxidized one is loosing electrons.

In the reactions:

a. 2Fe(s)+3Pb(NO3)2(aq)→3Pb(s)+2Fe(NO3)3(aq)

The Fe is as reactant as Fe(s) (Oxidation state 0) and the product is +3 (Because NO3, nitrate ion, is always -1). That means Fe is oxidized. The Pb as reactant is +2 and as product 0 (Gaining 2 electrons). Pb(NO3)2 is reduced

b. 2AgNO3(aq)+Cu(s)→2Ag(s)+Cu(NO3)2(a)

AgNO3 is +1 and Ag(s) is 0. AgNO3 is reduced. Cu(s) is 0 as reactant and +2 as product. Cu(s) is been oxidized

c. 3AgNO3(aq)+Al(s)→3Ag(s)+Al(NO3)3(aq)

Here, in the same way, AgNO3 is +1 as reactant and 0 as product. AgNO3 is reduced. And Al(s) is 0 as reactant but + 3 as product. Al(s) is oxidized.

Answer:

A flowchart is a type of diagram that represents a workflow or process

Explanation:

Answer: orders in which steps in a process happen

Explanation:

Answer: The equilibrium value of [tex]N_2O_4[/tex] is 0.379 M

Explanation:

Equilibrium constant is the ratio of the concentration of products to the concentration of reactants each term raised to its stochiometric coefficients.

Using ideal gas equation : [tex]PV=nRT[/tex]

P = pressure of gas

V = volume of gas

n = no of moles

R = gas constant

T = Temperature

pressure of [tex]N_2O_4[/tex] = [tex]\frac{1\times 0.0821Latm/Kmol\times 298}{5L}=5atm[/tex]

pressure of [tex]NO_2[/tex] = [tex]\frac{2\times 0.0821Latm/Kmol\times 298}{5L}=10atm[/tex]

[tex]N_2O_4(g)\rightleftharpoons 2NO_2(g)[/tex]

at t= 0    5 atm                                10 atm

at eqm    (5-x) atm                          (10+2x) atm

[tex]K_p=\frac{[p_NO_2]^2}{[p_N_2O_4]}[/tex]

[tex]0.1134=\frac{(10+2x)^2}{(5-x)}[/tex]

[tex]x=-4.48[/tex]

pressure of [tex]N_2O_4[/tex] at equilibrium = (5-(-4.48))= 9.48 atm

pressure of [tex]N_2O_4[/tex] = [tex]\frac{n\times 0.0821Latm/Kmol\times 298}{V}[/tex]

9.48 = [tex]{M\times 0.0821Latm/Kmol\times 298}[/tex]

[tex]M=0.379[/tex]

Thus the equilibrium value of [tex]N_2O_4[/tex] is 0.379 M

Answer:

Anode is the positively charged electrode which has the following characteristics:

1) Electrons leave anode to enter to the cathode by the battery.

2) Negatively charged ions are attracted towards cathode.

3) It is connected to the positive terminal of the battery.

Answer:

[tex]pKa=3.70[/tex]

Explanation:

Hello,

In this case, given the information, we can compute the concentration of hydronium given the pH:

[tex]pH=-log([H^+])\\[/tex]

[tex][H^+]=10^{-pH}=10^{-2.18}=6.61x10^{-3}M[/tex]

Next, given the concentration of the acid and due to the fact it is monoprotic, its dissociation should be:

[tex]HA\rightleftharpoons H^++A^-[/tex]

We can write the law of mass action for equilibrium:

[tex]Ka=\frac{[H^+][A^-]}{[HA]}[/tex]

Thus, due to the stoichiometry, the concentration of hydronium and A⁻ are the same at equilibrium and the concentration of acid is:

[tex][HA]=0.2230M-6.61x10^{-3}M=0.2164M[/tex]

As the concentration of hydronium also equals the reaction extent ([tex]x[/tex]). Thereby, the acid dissociation constant turns out:

[tex]Ka=\frac{(6.61x10^{-3})^2}{0.2164}\\ \\Ka=2.02x10^{-4}[/tex]

And the pKa:

[tex]pKa=-log(Ka)=-log(2.02x10^{-4})\\\\pKa=3.70[/tex]

Regards.

Answer:

aldehyde

carbon-1

ketone

carbon-2

Explanation:

Monosaccharides are colorless crystalline solids that are very soluble in water. Moat have a swwet taste. D-Fructose is the sweetest monosaccharide.

In the open chain form, monosaaccharides have a carbonuyl group in one of their chains. If the carbonyl group is in the form of an aldehyde group, the monosaccharide is an aldose; if the carbonyl group is in the form of a ketone group, the monosaccharide is known as a ketose. glucose is an aldose while fructose is a ketose.

In D-glucose, there is an aldehyde functional group, and the carbonyl group is at carbon-1 when looking at the Fischer projection.

In D-fructose, there is a ketone functional group, and the carbonyl group is at carbon-2 when looking at the Fischer projection.

Answer:

[tex]w/v\%=3.02\frac{g}{mL} \%[/tex]

Explanation:

Hello,

In this case, we first define the formula for the calculation of weight/volume percentage considering cobalt (II) fluoride as the solute, water the solvent and the both of them as the solution:

[tex]w/v\%=\frac{mass_{solute}}{V_{solution}}*100\%[/tex]

In such a way, since the mass of the solute is given as 6.04 g and the final volume of the solution 200 mL, the weight/volume percentage turns out:

[tex]w/v\%=\frac{6.04g}{200mL}*100\%\\\\w/v\%=3.02\frac{g}{mL} \%[/tex]

Regards.

Answer:

2040 cm-1

Explanation:

The vibrations frequency is obtained from;

v=1/2πc √k/μ

Where;

k= force constant = 240kgs-2

μ= reduced mass = 1.627×10^-27 kg

c= speed of light= 3×10^10cms-1

v= 1/2×3.142×3×10^10√240/1.627×10^-27

v= 5.3×10^-12 × 3.84×10^14

v= 20.4×10^2

v= 2040 cm-1

Answer:

C. NO2 + O3 → NO3 + O2 (slow)

NO3 + NO2 → N2O5 (fast)

Explanation:

A reaction mechanism represents an amount of elementary steps that explain how a reaction proceeds. The mechanism must explain the experimental rate law. Also, the slow step is the rate determining step.

This rate law is obtained from the multiplication of the reactants in the slow step, thus:

A. NO2 + NO2 → N2O2 (fast)

N2O4 + O3 → N2O5 + O2 (slow)

Rate law:

rate = k [N2O4] [O3]

This mechanism is not consistent with rate law.

B. NO2 + O3 → NO5 (fast)

NO5 + NO5 → N2O5 + (5/2)O2 (slow)

Rate law:

rate = k [NO5]²

This mechanism is not consistent with rate law.

C. NO2 + O3 → NO3 + O2 (slow)

NO3 + NO2 → N2O5 (fast)

Rate law:

rate = k [NO2] [O3]

D. NO2 + NO2 → N2O2 + O2 (slow)

N2O2 + O3 → N2O5 (fast)

Rate law:

rate = k [NO2]²

This mechanism is not consistent with rate law.

Thus, right solution is:

C. NO2 + O3 → NO3 + O2 (slow)

NO3 + NO2 → N2O5 (fast)

Answer:

b, H2SO4 with HCl, as they are both liquid acids

Answer:

Ionic Crystal

Explanation:

Answer:   3.87M  of HI remains after 27.3 s

Explanation:

Using the Second order decomposition equation of

1/[H]t =K x t +1/[A]o

Given initial concentration ,[A]o = 4.78M

time, t = 27.3 s

rate of constant , k= 1.80 x 10^-3 M-1s-1

1/[H] t= 1/[A] t= concentration after time, t=?

SOLUTION

1/[A] t =kt +1/[A]o

1/[A] t =(1.80 x 10^-3 (27.3)+1/4.78

0.04914+0.2092=0.2583

1/[A] t =0.2583

[A] t =1/0.2583= 3.87M

Answer:

=355.5K

Explanation:

Specific heat, Q = mcΔT

where

[1 mole of Ne = 20g; 2 moles of Ne = 2 × 20 = 40g]

4250 = 40 × 1.7 × (final - 293K)

final - 293k = 4250 / ( 40 × 1.7)

Final temp = 62.5 + 293

=355.5K

I hope this steps are simple to follow and understand.

Answer:

1.5 mol of CO₂

Explanation:

Use the mole ratio to find how many moles of CO₂ are produced from the reaction.

For every 5 moles of O₂, three moles of CO₂ is produced.

2.5 mol O₂ × 3 mol CO₂ ÷ 5 mol O₂

= 2.5 mol O₂ × 0.6

= 1.5 mol CO₂

When 2.5 mol of O₂ is consumed in the reaction, 1.5 mol of CO₂ is produced.

Hope that helps.

Answer:

5.81 g/L

Explanation:

Let's apply the Ideal Gases Law to determine this:

P . V = n . R . T

Pressure = 1.6 atm

Volume = ?

Mol = 1 mol

Temperature = 96 K

In order to find the density, we should know the volume of the atmosphere which is a mixture of gases so, we consider all the atmosphere as a unique ideal gas.

1. 6 atm . V = 1 mol . 0.082 L.atm/mol.K . 96K

V = (1 mol . 0.082 L.atm/mol.K . 96K) / 1.6 atm

V = 4.92 L → As this is the volume for the whole atmosphere and the mass of 1 mol is 28.6 g, density should be:

28.6 g / 4.92L = 5.81 g/L

Density → mass / volume

Answer:

0.167M

Explanation:

Molarity, M, is an unit of concentration in chemistry defined as the ratio between moles of solute (NaOH in this case) and volume of the solution in liters.

To find molarity of 5.00 g Sodium Hydroxide in 750.0 mL of solution we need to convert mass of NaOH to moles (Using its molar mass: 40g/mol) and the mililiters of solution to liters (1L = 1000mL), thus:

Moles NaOH = 5.00g × (1mol/ 40g) = 0.125 moles NaOH = Moles solute

Liters solution = 750.0mL × (1L / 1000mL) = 0.7500L solution

And molariy is:

0.125 moles NaOH  / 0.7500L solution =

0.167M