The heat of vaporization delta Hv of dichloromethane (Ch2CL2) is 28.0 kJ/mol . Calculate the change in entropy delta S when 473 g of dichloromethane boils at 39.8 degree.

Answers

Answer 1

Answer:

16 J/K.mol

Explanation:

From the question,

ΔS = ΔH/T............... Equation 1

Where ΔH = Heat change, T = Temperature

But,

ΔH = n(Hv).................. Equation 2

Where n = number of  mole, Hv = heat of vaporization.

Given: Hv = 28.0 kJ/mol, n = 473/85 = 5.59 mole.

Substitute these values into equation 2

ΔH = 28/5.59

ΔH = 5.01 kJ.

Also: T = 273+39.8 = 312.8 J

Substitute into equation 1

ΔS = 5.01/312.8

ΔS = 0.016 kJ/K

ΔS = 16 J/K.mol


Related Questions

Identify some other substances (besides KCl) that might give a positive test for chloride upon addition of AgNO3. do you think it is reasonable to exclude these types of substances as contaminants that would give a false positive when you tested your reaction residue to verify that it is KCl?

Answers

Answer:

-The other substances that give a positive test with AgNO3 are other chlorides present, iodides and bromide.

-It is reasonable to exclude iodides and bromides but it is not reasonable to exclude other chlorides

Explanation:

In the qualitative determination of halogen ions, silver nitrate solution(AgNO3) is usually used. Now, various halide ions will give various colours of precipitate when mixed with with silver nitrate. For example, chlorides(Cl-) normally yield a white precipitate, bromides(Br-) normally yield a cream precipitate while iodides (I-) normally yield a yellow precipitate. Thus, all these ions or some of them may be present in the system.

With that being said, if other chlorides are present, they will also yield a white precipitate just like KCl leading to a false positive test for KCl. However, since other halogen ions yield precipitates of different colours, they don't lead to a false test for KCl. Thus, we can exclude other halides from the tendency to give us a false positive test for KCl but not other chlorides.

What is a major product of the reaction in the box?​

Answers

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!

The combustion of propane (C 3H 8) in the presence of excess oxygen yields CO 2 and H 2O: C 3H 8 (g) + 5O 2 (g) → 3CO 2 (g) + 4H 2O (g) When 2.5 mol of O 2 are consumed in their reaction, ________ mol of CO 2 are produced.

Answers

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.

Two moles of neon gas enclosed in a constant volume system receive 4250 J of heat. If the gas was initially at 293 K, what is the final temperature of the neon

Answers

Answer:

=355.5K

Explanation:

Specific heat, Q = mcΔT

where

Q= 4250JΔT= change in temp = final temp - initial tempc = specific heat capacity = 1.7m = mass of substance in grams

[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.

You are given 10.00 mL of a solution of an unknown acid. The pH of this solution is exactly 2.18. You determine that the concentration of the unknown acid was 0.2230 M. You also determined that the acid was monoprotic (HA). What is the pKa of your unknown acid

Answers

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.

Write the net ionic equation for any precipitation reaction that may be predicted when aqueous solutions of manganese(II) nitrate and sodium hydroxide are combined.

Answers

Answer:

Explanation:

Mn( NO₃ )₂ + 2Na OH = Mn( OH)₂ (s) ↓ +  2Na NO₃

Converting into ions

Mn⁺ + 2 NO₃⁻ + 2 Na⁺ + 2 OH⁻ = Mn( OH)₂ + 2 Na⁻ + 2 NO₃⁻

Cancelling out common terms

Mn⁺ + 2 OH⁻ = Mn( OH)₂

this is net ionic equation required.

D-Fructose is the sweetest monosaccharide. How does the Fischer projection of D-fructose differ from that of D-glucose? Match the words in the left column to the appropriate blanks in the sentences on the right. Fill in the blanks.
a ketone
carbon 3
carbon 2
carbon 1
an aldehyde
carbon 4
In D-glucose, there is__________ functional group, and the carbonyl group is at___________ when looking at the Fischer projection.
In D-tructose. there is functional group, and the carbonyl group is at when looking at______ the Fischer projection.

Answers

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.

which statement describes the use of a flowchart?

Answers

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:

What is an anode? Explain.

Answers

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.

For each reaction, write the chemical formulae of the oxidized reactants in the space provided. Write the chemical formulae of the reduced reactants.
reactants oxidized _____
reactants reduced _____
a. 2Fe(s)+3Pb(NO3)2(aq)→3Pb(s)+2Fe(NO3)3(aq)
b. AgNO3(aq)+Cu(s)→2Ag(s)+CuNO)2(a)
c. 3AgNO(aq)+Al()→3Ags)+Al(NO3)3(aq)

Answers

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.

below are three reactions showing how chlorine from CFCs (chlorofluorocarbons) destroy ozone (O3) in the stratosphere. Ozone blocks harmful ultraviolet radiation from reaching earth’s surface. Show how these 3 equations sum to produce the net equation for the decomposition of two moles of ozone to make three moles of diatomic oxygen (2 O3→ 3 O2), and calculate the enthalpy change. (6 points) R1 O2 (g) → 2 O (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

Answers

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

ΔH = -338.8kJ

An aqueous solution of cobalt(II) fluoride, , is made by dissolving 6.04 grams of cobalt(II) fluoride in sufficient water in a 200. mL volumetric flask, and then adding enough water to fill the flask to the mark. What is the weight/volume percentage of cobalt(II) fluoride in the solution

Answers

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.

Use bond energies provided to estimate 2Br2

Answers

An your use the energies for the estimate or 2br2 oh 2b3

The initial concentrations of I2 and I− in the reaction below are each 0.0401 M. If the initial concentration of I−3 is 0.0 M and the equilibrium constant is Kc=0.25 under certain conditions, what is the equilibrium concentration (in molarity) of I−? I−3(aq)↽−−⇀I2(aq)+I−(aq)

Answers

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

[I⁻] = 0.0352M

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

Calculation of the  equilibrium concentration:

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.

Learn more about molarity here: https://brainly.com/question/14782192

Match each property of a liquid to what it indicates about the relative strength of the intermolecular forces in that liquid.

Strong intermolecular forces

Weak intermolecular forces

Answers

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:-

Dispersion forcesDipole-dipole interactionsHydrogen bondingion-dipole forces.

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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11. (2 pts) Sodium Hydroxide, is also known as lye and was a critical component in
homemade soap. Now it is a commonly used drain cleaner because it chemically reacts
with fats (the typical cause of a clog) to form a soap that can be swept down the drain.
What is the molarity of 5.00 g Sodium Hydroxide in 750.0 mL of solution?

Answers

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

The rate law for the reaction 2NO2 + O3 → N2O5 + O2 is rate = K[NO2][O3].
Which one of the following mechanisms Is consistent with this rate law?
A. NO2 + NO2 → N2O2 (fast)
N2O4 + O3 → N2O5 + O2 (slow)
B. NO2 + O3 → NO5 (fast)
NO5 + NO5 → N2O5 + (5/2)O2 (slow)
C. NO2 + O3 → NO3 + O2 (slow)
NO3 + NO2 → N2O5 (fast)
D. NO2 + NO2 → N2O2 + O2 (slow)
N2O2 + O3 → N2O5 (fast)

Answers

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]

This mechanism is consistent with rate law.

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)

Anita heats a beaker containing water. As the temperature of the
water increases, which change to the water molecules occurs?
Select one:
The molecules separate into atoms of hydrogen and oxygen.
The molecules move at a faster rate.
The molecules become more massive.
The molecules expand and become wider.

Answers

Answer:

The molecules expand and become wider.

Explanation:

Answer:

The molecules move at a faster rate.

Explanation:

Temperature is directly proportional to the average kinetic energy of molecules, so when the temperature of water molecules increases, they have a higher average kinetic energy.

They move at a faster rate. The distance between the molecules only increases when they become gas where they break apart from the bondings with the other water molecules.

How many moles of aqueous magnesium ions and chloride ions are formed when 0.250 mol of magnesium chloride dissolves in water

Answers

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:

“Denitrifying” bacteria return molecular nitrogen gas (N2) back into the biosystem by a series of reductions. Identify the correct sequence. Select the correct answer below: A. NO−3→NO−2→N2O→N2 B. N2O→NO−3→NO−2→N2 C. N2O→NO−2→NO−3→N2 D. NO−3→N2O→NO−2→N2

Answers

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.

If a diatomic molecule has a vibrational force constant of k=240 kg s-2 and a reduced mass of 1.627x10-27 kg, its vibrational frequency should be (in cm-1):
A. 2040
B. 4079
C. 2885
D. 5770
E. 1443

Answers

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

how salt solution can be determined by using hydrometer​

Answers

Answer:

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

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

formic acid buffer containing 0.50 M HCOOH and 0.50 M HCOONa has a pH of 3.77. What will the pH be after 0.010 mol of NaOH has been added to 100.0 mL of the buffer

Answers

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]

pH = 3.95

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

What is Formic Acid?

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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Mass of the condensed unknown liquid: 0.3175 g Temperature of the water bath: 99.00 oC Pressure of the gas: 748.2 mmHg Volume of the flask (volume of the gas): 145.0 mL Given : Kelvin = t oC + 273.15 1 L = 1000 mL 1 atm = 760 mmHg Gas constant: R = 0.08206 atm  L / mole  K; Ideal Gas Law: PV = nRT 1. What is the pressure of the gas in atm? (1 points) 2.

Answers

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.

An astronomy research team has been studying the atmosphere of a moon orbiting a newly discovered exoplanet. The team has determined that the moon has an average temperature of 95K on the surface, with an average pressure of 1.6atm. Remote analysis of this moon's atmosphere has revealed it has a molar mass of 28.6 g/mol. Calculate the density (g/L) of 1 mole of the moon's atmosphere under the given conditions.

Answers

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

Determine whether each of the following salts will form a solution that is acidic, basic, or pH-neutral. Drag the appropriate items to their respective bins.
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AI(NO3)3 CH3NH3CN NaCIO
CH3NH3CI NaNO3
Acidic Basic pH-neutral
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Answers

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:

Al(NO₃)₃  →  Al³⁺  +  3NO₃⁻

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.

CH₃NH₃CN  →  CH₃NH₃⁺  +  CN⁻

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

NaCIO  → Na⁺  + ClO⁻

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.

CH₃NH₃CI → CH₃NH₃⁺  +  Cl⁻

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.

How many molecules of ATP are formed during the complete catabolism of a saturated fatty acid with the chemical formula CH3(CH2)20CO2H?

Answers

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!

Identify a homogeneous catalyst:

a. SO2 over vanadium (V) oxide
b. H2SO4 with concentrated HCl
c. Pd in H2 gas
d. N2 and H2 catalyzed by Fe
e. Pt with methane

Answers

Answer:

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

A crystal lattice formed by positive and negative ions is called a

Answers

Answer:

Ionic Crystal

Explanation:

For a reaction, what generally happens if the temperature is increased? a) A decrease in k occurs, which results in a faster rate. b) A decrease in k occurs, which results in a slower rate. c) An increase in k occurs, which results in a faster rate.

Answers

Answer:

an increase in K occurs,which results in a faster rate

if the temperature is increased for a reaction, An increase in k occurs, which results in a faster rate of reaction. Hence, Option (D) is correct.

What is Rate constant ?

A coefficient of proportionality relating the rate of a chemical reaction at a given temperature to the concentration of reactant (in a unimolecular reaction) or to the product of the concentrations of reactants.

It is represented as 'K'

The negative exponential relationship between k and the temperature indicates that as temperature increases, the value of k also increases.

Since the rate constant can be determined experimentally over a range of temperatures, the activation energy can be calculated using the Arrhenius equation.

Therefore, if the temperature is increased for a reaction, An increase in k occurs, which results in a faster rate of reaction. Hence, Option (D) is correct.

Learn more about Chemical kinetics here ;

https://brainly.com/question/24188785

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