PV = nRT. If P = 1 atm, V = 5.0 liter, R = 0.0821 L.atm/mol.K, and T = 293 K; what is the value of n?

Answer:

d) NH4F

Explanation:

Hello,

In this case, the base resulting from mixing a weak acid and a weak base is d) NH4F since ammonium hydroxide is a wear base and hydrofluoric acid is a weak acid.

Ammonium hydroxide is a weak base since it is not completely ionized in ammonium and hydroxyl ions:

[tex]NH_4OH\rightarrow NH_4^++OH^-[/tex]

Moreover, hydrofluoric acid is a weak acid since it is not completely ionized in hydrogen and fluoride ions:

[tex]HF\rightleftharpoons H^++F^-[/tex]

For the both of the substances, the limit is established by the basic and the acid dissociation constant respectively.

Regards.

Answer:

[tex][H^+]=0.000123M[/tex]

[tex]pH=3.91[/tex]

Explanation:

Hello,

In this case, dissociation reaction for acetic acid is:

[tex]CH_3COOH\rightleftharpoons CH_3COO^-+H^+[/tex]

For which the equilibrium expression is:

[tex]Ka=\frac{[CH_3COO^-][H^+]}{[CH_3COOH]}[/tex]

Which in terms of the reaction extent [tex]x[/tex] could be written as:

[tex]1.74x10^{-5}=\frac{x*x}{[CH_3COOH]_0-x}=\frac{x*x}{0.0010M-x}[/tex]

Thus, solving by using a solver or quadratic equation we obtain:

[tex]x_1=0.000123M\\\\x_2=-0.000141M[/tex]

And clearly the result is 0.000123M, which also equals the concentration of hydronium ion in solution:

[tex][H^+]=0.000123M[/tex]

Now, the pH is computed as follows:

[tex]pH=-log([H^+])=-log(0.000123)\\\\pH=3.91[/tex]

Best regards.

Answer:

[tex]T=469.1K\\\\T=195.9\°C[/tex]

Explanation:

Hello,

In this case, for the given decomposition reaction, we can compute the enthalpy of reaction considering the enthalpy of formation of each involved species (products minus reactants):

[tex]\Delta _rH=2\Delta _fH_{Ag}+\frac{1}{2} \Delta _fH_{O_2}-\Delta _fH_{Ag_2O}\\\\\Delta _rH=2*0.00+\frac{1}{2} *0.00-(-31.1)=31.1kJ/mol[/tex]

Next, the entropy of reaction considering the standard entropy for each involved species (products minus reactants):

[tex]\Delta _rS=2S_{Ag}+\frac{1}{2} S_{O_2}-S_{Ag_2O}\\\\\Delta _rS=2(42.55)+\frac{1}{2} (205.0)-(121.3)=66.3J/(mol*K)[/tex]

Next, since the Gibbs free energy of reaction is computed in terms of both the enthalpy and entropy of reaction at the unknown temperature, for such Gibbs energy equaling 0, the temperature (in K and °C) turns out:

[tex]\Delta _rG=\Delta _rH-T\Delta _rS\\\\0=31.1kJ/mol-T(66.3\frac{J}{mol*K}*\frac{1kJ}{1000J} )\\\\T=\frac{31.1kJ/mol}{0.0663kJ/(mol*K)} =469.1K\\\\T=195.9\°C[/tex]

Which is within the given rank.

Best regards.

Answer:

96 mmHg

[tex]h=96mmHg[/tex]

Explanation:

From this question,manometer end is closedw, So we can deduced that the height of the column will not be affected by the atmospheric pressure .

The difference of height of the mercury level is given as,

h=9.60cm

h=9.60(10mm/1cm)

[tex]h=96mm[/tex]

But it is obvious that in this closed end manometer.the pressure of the gas is equal to the height

P(gas)=h

P(gas)=96mmHg

This pressure is as a result of the presence of gas.

Therefore, the pressure of the argon gas in the container is 96mmHg.

The pressure of the argon in the container was 96mmHg.

We were told that the manometer has closed ends which means that the

height will not be affected by atmospheric pressure.

The height which is the difference in mercury level is

h=9.60cm

We can convert it to millimeter by multiplying it by 10

h=9.60 × 10 = 96mm

The pressure of the closed end manometer will be equal to the height

P(gas)=h

P(gas)=96mmHg

The pressure of the argon gas in the container is 96mmHg.

Read more on https://brainly.com/question/17151458

Answer:

Explanation:

In all calorimetric experiment , the calorimeter must be isolated from the surrounding . Otherwise the heat change in the experiment can not be determined with precision .

The reaction is endothermic . Hence, there is lowering of temperature due to absorption of heat in the reaction equal to ΔH°. The value of ΔH° can be calculated by measuring fall in the temperature of the content . The fall in the temperature will be less when heat is allowed to come from the surrounding . Less fall of temperature will result in less ΔH° to be calculated .

Hence in the given experiment , if the student neglects to put lid on the cup , the experiment will give less value of ΔH°.

Answer:

Pb + 2H2O --> PbO2 + 2H2

Explanation:

Products:

Solid metal; PbO2

Hydrogen; H

Reactants:

Metal; Pb

Steam; H2O

Reactants --> Products

Pb + H2O --> PbO2 + H2

Upon balancing we have;

Pb + 2H2O --> PbO2 + 2H2

Answer:2.62 L

Explanation:

A sample of gas occupies a volume of 2.62 liters at 25° C and 1.00 atm. and the volume at 50° C and 2 atm then volume is 2.62 liters.

The equation of state for a fictitious perfect gas is known as the ideal gas law, sometimes known as the generic gas equation. Although it has significant drawbacks, it is a decent approximation of the behavior of many gases under various situations.

An ideal gas is one in which there are no intermolecular attraction forces and all collisions between atoms or molecules are entirely elastic. It may be seen as a group of perfectly hard spheres that collide but do not else interact with one another.

By using ideal gas equation,

P₁ V ₁ ÷ T = P₂V₂ ÷ T

1 × 2.62 ÷ 25 = 2 × V₂ ÷ 50

V₂ = 1 × 2.62 × 50 ÷ 25 × 2

V₂ = 2.62 liters.

Thus, a sample of gas occupies a volume of 2.62 liters at 25° C and 1.00 atm. and the volume at 50° C and 2 atm then volume is 2.62 liters.

To learn more about ideal gas law follow the link below;

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Answer:

an increase in 1-butene was observed when t-butoxide was used

Explanation:

When a base reacts with an alkyl halide, an elimination product is formed. This reaction is an E2 reaction.

Here we are to compare the reaction of two different bases with one substrate; 2-bromobutane. Both reactions occur by the E2 mechanism but follow different transition states due to the size of the base.

The Saytzeff product, 2-butene, is obtained when the methoxide is used while the non Saytzeff product, 1-butene, is obtained when t-butoxide is used.

The Saytzeff rule is reliable in predicting the major products of simple elimination reactions of alkyl halides given the fact that a small/strong bases is used for the elimination reaction. Therefore hydroxide, methoxide and ethoxide bases give similar results for the same alkyl halide substrate. Bulky bases such as tert-butoxide tend to yield a higher percentage of the non Saytzeff product and this is usually attributed to steric hindrance.

Answer:

[tex]0.033g[/tex]

Explanation:

Hello,

In this case, since 11 mg per kilogram of body weight has the given lethality, the mg that turn out lethal for a chicken weighting 3 kg is computed by using a rule of three:

[tex]11mg\longrightarrow 1kg\\\\x\ \ \ \ \ \ \longrightarrow 3kg[/tex]

Thus, we obtain:

[tex]x=\frac{3kg*11mg}{1kg}\\ \\x=33mg[/tex]

That in grams is:

[tex]=33mg*\frac{1g}{1000mg} \\\\=0.033g[/tex]

Regards.

Answer:

Oxidation half equation;

3Cd(s) -------> 3Cd^2+(aq) + 6e

Reduction half equation;

2In^3+(aq) + 6e -----> 2In(s)

Explanation:

Since the reduction potentials of Indium and Cadmium are -0.34 V and - 0.40 V respectively, we can see that cadmium will be oxidized while indium will the reduced.

We arrived at this conclusion by examining the reduction potential of both species. The specie with more negative reduction potential is oxidized in the process.

Oxidation half equation;

3Cd(s) -------> 3Cd^2+(aq) + 6e

Reduction half equation;

2In^3+(aq) + 6e -----> 2In(s)

Answer:

d. HF(g)

Explanation:

Hello,

In this case, the standard entropy S° could be predicted by looking at the amount of bonds the compound has, thus, the fewer the number bonds, the lower the standard entropy, it means that d. HF(g) has lowest value as it has one bond only whereas methane has four bonds, water two bonds and ammonia three bonds.

Best regards.

Answer:

b. 0°C and 1 bar

Explanation:

Hello,

In this case, the STP conditions are standard temperature and pressure sets of conditions for experimental measurements to be established to allow comparisons to be made between different sets of data, it means that a specific pressure and temperature is assigned to analyze the properties of a substance. Such conditions are strictly 0°C and 1 bar because a large number of physical, chemical and thermodynamic properties are measured at them, therefore the standard entropy of a substance refers to its entropy at: b. 0°C and 1 bar.

Best regards.

Answer:

7.4 × 10⁻⁹ M

Explanation:

Step 1: Given data

Solubility product constant (Ksp) for Li₃PO₄: 5.9 × 10⁻¹⁷

Concentration of lithium ion: 0.0020 M

Step 2: Write the reaction for the solution of Li₃PO₄

Li₃PO₄(s) ⇄ 3 Li⁺(aq) + PO₄³⁻(aq)

Step 3: Calculate the phosphate concentration required for a precipitate to occur

The solubility product constant is:

Ksp = 5.9 × 10⁻¹⁷ = [Li⁺]³ × [PO₄³⁻]

[PO₄³⁻] = 5.9 × 10⁻¹⁷ / [Li⁺]³

[PO₄³⁻] = 5.9 × 10⁻¹⁷ / 0.0020³

[PO₄³⁻] = 7.4 × 10⁻⁹ M

Answer:

0.0630

Explanation:

The molar mass of urea = 60 g/mol

we all know that:

[tex]\mathtt{number \ of \ moles = \dfrac{mass }{molar \ mass}}[/tex]

Then; the number of moles of urea

= [tex]\mathtt{\dfrac{4.0 \ g}{60 \ g/mol}}[/tex]

= 0.0667 mol

Similarly; the number of moles of methanol

= [tex]\mathtt{\dfrac{32 \ g}{32.04 \ g/mol}}[/tex]

= 0.9988 mol

The total number of moles = (0.0667 + 0.9988) mol

= 1.0655 mol

Finally,the mole fraction of urea  [tex]\mathtt{(X_{urea})}[/tex] = [tex]\mathtt{\dfrac{ n_{urea}}{(n_{urea}+n_{methanol})}}[/tex]

[tex]\mathtt{(X_{urea})}[/tex] = [tex]\mathtt{\dfrac{0.0667 \ mole}{1.0655 \ mole}}[/tex]

= 0.0630

Answer:

C) SO3

Explanation:

Lewis formula shows the bonding between atoms of a molecule and expresses the lone pair present in the atoms.

SO3 or Sulfur trioxide cannot be adequately described by a single Lewis formula because it has majorly 3 resonance structures because Sulfur does not follow the octet rule and can expand electrons in its outer shell.

Hence, the correct answer is C) SO3

Answer:

a. Rate = k×[A]

b. k = 0.213s⁻¹

Explanation:

a. When you are studying the kinetics of a reaction such as:

A + B → Products.

General rate law must be like:

Rate = k×[A]ᵃ[B]ᵇ

You must make experiments change initial concentrations of A and B trying to find k, a and b parameters.

If you see experiments 1 and 3, concentration of A is doubled and the Rate of the reaction is doubled to. That means a = 1

Rate = k×[A]¹[B]ᵇ

In experiment 1 and to the concentration of B change from 1.50M to 2.50M but rate maintains the same. That is only possible if b = 0. (The kinetics of the reaction is indepent to [B]

Rate = k×[A][B]⁰

b. Replacing with values of experiment 1 (You can do the same with experiment 3 obtaining the same) k is:

Rate = k×[A]

0.320M/s = k×[1.50M]

Answer:

Lime water, [tex]Ca(OH)_{2}_({aq} )[/tex] is formed.

Explanation:

Lime-water is a clear and colourless dilute solution of aqueous calcium hydroxide salt.

Small amounts of calcium hydroxide salt,  [tex]Ca(OH)_{2}_(s)[/tex]  is sparsely soluble at room temperature when dispersed vigorously. if in excess, a white suspension called 'milk of lime'is formed.

I hope this explanation is helpful.

Answer:

0.35 g/mL

Explanation:

Use the formula D = [tex]\frac{m}{v}[/tex], where D is density, m is mass, and v is volume.

D = 4.9/14

D = 0.35

D = 0.35 g/mL

Answer:

The energy required to heat 1.30 kg of water from 22.4°C to 34.2°C is 64,121.2 J

Explanation:

Calorimetry is the measurement of the amount of heat that a body gives up or absorbs in the course of a physical or chemical process.

The sensible heat of a body is the amount of heat received or transferred by a body when undergoing a temperature variation (Δt) without there being a change in physical state. That is, when a system absorbs (or gives up) a certain amount of heat, it may happen that it experiences a change in its temperature, involving sensible heat. Then, the equation for calculating heat exchanges is:

Q = c * m * ΔT

Where Q is the heat or quantity of energy exchanged by a body of mass m, constituted by a substance of specific heat c and where ΔT is the variation in temperature (ΔT=Tfinal - Tinitial).

In this case:

Replacing:

[tex]Q=4.18 \frac{J}{g*K}*1,300 g*11.8 K[/tex]

Q= 64,121.2 J

The energy required to heat 1.30 kg of water from 22.4°C to 34.2°C is 64,121.2 J

Answer:

The gas is Methane at 340K

The words given are not clear, so the clear question is as follows:

Match the words below to the appropriate blanks in the sentences. Make certain each sentence is complete before submitting your answer:

A. β-1,4- and α-1,6-glycosidic

B. α-1,4-glycosidic

C. α-1,4-galactose

D. an unbranched glucose

E. a branched fructose

F. α-1,6-glycosidic

Amylose is ......... polymer of ....... units joined by ........ bonds.

Amylopectin is ....... polymer of .......units joined by ........ bonds.

Answer:

D. an unbranched glucose

C. α-1,4-galactose

B. α-1,4-glycosidic

E. a branched fructose

A. β-1,4- and α-1,6-glycosidic

F. α-1,6-glycosidic

Explanation:

Amylose and amylopectin are two types of polysaccharides that can be found in starch granules.

Amylose is linear or unbranched glucose polymer of α-1,4-galactose units that are joined by α-1,4-glycosidic.

Amylopectin is a branched fructose polymer of β-1,4- and α-1,6-glycosidic units joined by α-1,6-glycosidic bonds.

Hence, the correct answers in the sequential order are:

Amylose:

D. an unbranched glucose

C. α-1,4-galactose

B. α-1,4-glycosidic

Amylopectin:

E. a branched fructose

A. β-1,4- and α-1,6-glycosidic

F. α-1,6-glycosidic

Answer:

1.84 × 10⁻³

Explanation:

Step 1: Write the balanced equation

2 NOBr(g) ⇄ 2 NO(g) + Br₂(g)

Step 2: Calculate the initial concentration of NOBr

0.143 moles of NOBr(g) are introduced into a 1.00 liter container. The molarity is:

M = 0.143 mol / 1.00 L = 0.143 M

Step 3: Make an ICE chart

         2 NOBr(g) ⇄ 2 NO(g) + Br₂(g)

I             0.143               0           0

C              -2x               +2x        +x

E          0.143-2x            2x          x

Step 4: Find the value of x

The equilibrium concentration of NOBr(g) was 0.108 M. Then,

0.143-2x = 0.108

x = 0.0175

Step 5: Calculate the concentrations at equilibrium

[NOBr] = 0.108 M

[NO] = 2x = 0.0350 M

[Br₂] = x = 0.0175 M

Step 6: Calculate the equilibrium constant (Kc)

Kc = [0.0350]² × [0.0175] / [0.108]²

Kc = 1.84 × 10⁻³

Answer:

[tex][NH_2NO_2]=0.0868M[/tex]

Explanation:

Hello,

In this case, for the given chemical reaction, the first-order rate law is:

[tex]r=\frac{d[NH_2NO_2]}{dt} =-k[NH_2NO_2][/tex]

Which integrated is:

[tex][NH_2NO_2]=[NH_2NO_2]_0exp(-kt)[/tex]

Thus, the concentration after 31642.0 s for a 0.384-M solution is:

[tex][NH_2NO_2]=0.384M*exp(-4.70x10^{-5}s^{-1}*31642.0s)\\[/tex]

[tex][NH_2NO_2]=0.0868M[/tex]

Best regards.

Answer:

[A] = 0.0868 M

Explanation:

Rate constant = 4.70×10-5 s-1

First order reaction

Initial concentration, [A]o = 0.384 M

Final concentration, [A] = ?

Time, t = 31642.0 s

All these variables are related by the following equation;

[A] = [A]o e^(-kt)

[A] = 0.384  e^(-4.70×10-5 x  31642.0)

[A] = 0.384 e^(-1.4872)

[A] = 0.384 * 0.2260

[A] = 0.0868 M

Answer:

PLEASE LOOK INN TO THE FILE YOU WILL GET ANSWER AND ALSO SUMMARY THANKS FOR ASKING QUESTION.

Explanation:

Answer:

A : At the end of the hour, isotope B has a greater decay constant λ than isotope A

Explanation:

Firstly, we need to understand that radioactive decay follows a first order rate law.

What this means is that we can calculate the radioactive decay constant using the following formula from the half-life

Mathematically;

[tex]t_{1/2}[/tex]  = 0.693/λ

where λ represents the radioactive decay constant.

Rearranging the equation, we can have

λ = 0.693/[tex]t_{1/2}[/tex]

Now, to have a fair level playing ground, it is best that the half-life of both isotopes are in the same unit of time(seconds)

For A, the half-life = 2.3 minutes which is same as 2.3 × 60 = 138 seconds

For B, the half-life is 24 seconds

Thus, at the end of the hour, the decay constant for isotope A will be;

λ = 0.693/138 = 0.0050 [tex]s^{-1}[/tex]

For isotope B, the decay constant will be;

λ = 0.693/24 = 0.028875  [tex]s^{-1}[/tex]

We can see that the decay constant of isotope B is higher than that of A at the end of the experiment

Answer:12.9e-12g or in short 12.9pg

Explanation:as p=1e-12

Answer:

13.11 g.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below :

N2 + O2 —> 2NO

Next, we shall determine the mass of O2 that reacted and the mass of NO produced from the balanced equation. This is illustrated below:

Molar mass of O2 = 16x2 = 32 g/mol

Mass of O2 from the balanced equation = 1 x 32 = 32 g.

Molar mass of NO = 14 + 16 = 30 g/mol

Mass of NO from the balanced equation = 2 x 30 = 60 g.

From the balanced equation above,

32 g of O2 reacted to produce 60 g of NO.

Finally, we shall determine the theoretical yield of NO as follow:

From the balanced equation above,

32 g of O2 reacted to produce 60 g of NO.

Therefore, 6.99 g of O2 will react to produce = (6.99 x 60)/32 = 13.11 g of NO.

Therefore, the theoretical yield of nitrogen monoxide, NO is 13.11 g.

Answer:

0.101 M

Explanation:

Step 1: Given data

Step 2: Calculate [A]

For a second-order reaction, we can calculate [A] using the following expression.

1/[A] = 1/[A]₀ + k × t

1/[A] = 1/0.250 M + 0.117 M⁻¹.min⁻¹ × 50.7 min

[A] = 0.101 M

Answer:

Au^3+(aq) +3e ------> Au(s). 1.50 V

Explanation:

When we construct the galvanic cell, our intention is to produce energy by spontaneous electrochemical reactions. In order to have a spontaneous electrochemical reaction, E°cell must be positive. The more positive the value of E°cell, the more spontaneous the reaction is.

E°cell= E°cathode - E°anode

If E°cathode= 1.50 V

E°anode= -0.25 V

E°cell= 1.50 -(-0.25)

E°cell= 1.75 V

Hence the process; Au^3+(aq) +3e ------> Au(s) yields the highest standard cell potential