Assuming that you start with 21.4 g of ammonia gas and 18.0 g of sodium metal and assuming that the reaction goes to completion, determine the mass (in grams) of each product.

Answer:

8.05 moles

Explanation:

5.50 / 2.954 = x / 4.325

x = 8.05

According to ideal gas equation, if the temperature and pressure stay constant during the process 0.520 moles have been added  so that the balloon expands to 4.325 L.

The ideal gas equation is a equation which is applicable in a hypothetical state of an ideal gas.It is a combination of Boyle's law, Charle's law,Avogadro's law and Gay-Lussac's law . It is given as, PV=nRT where R= gas constant whose value is 8.314.The law has several limitations.The law was proposed by Benoit Paul Emile Clapeyron in 1834.

In the given example if pressure and temperature are constant then V=nR substituting V=4.325 l and R=8.314  so n=V/R=4.325/8.314=0.520 moles.

Thus, 0.520 moles of helium are added if the temperature and pressure stay constant during this process.

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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:

0.9718M

Explanation:

Rate constant, k =  0.255 M-1s-1

time, t = 8.00 s

Initial concentration, [A]o = 1.33 M

Final concentration, [A] = ?

These quantities are represented by the equation;

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

1 / [A] = 1 /1.33 + (0.255 * 8)

1 / [A] = 0.7519 + 2.04

[A] = 1 / 2.7919 = 0.3582 M

How much of NO2 decomposed is obtained from the change in concentration;

Change in concentration = Initial - Final

Change = 1.33 - 0.3582 = 0.9718M

Answer:

A. Rank the following substances in order of decreasing standard molar entropy (S∘).

Rank the gases from largest to smallest standard molar entropy

I2(g)>Br2(g)>Cl2(g)>F2(g)

B. Rank the gases from largest to smallest standard molar entropy. To rank items as equivalent, overlap them.

H2O2(g)>H2S(g) >H2O(g)

C. Rank the gases from largest to smallest standard molar entropy. To rank items as equivalent, overlap them.

C(s, amorphous) >C(s, graphite)>C(s, diamond)

Explanation:

Hello,

In this case, we can apply the following principles to explain the order:

- The greater the molar mass, the larger the standard molar entropy.

- The greater the molar mass and the structural complexity, the larger the standard molar entropy.

- The greater the structural complexity, the larger the standard molar entropy.

A. Rank the following substances in order of decreasing standard molar entropy (S∘).

Rank the gases from largest to smallest standard molar entropy

I2(g)>Br2(g)>Cl2(g)>F2(g)

This is due to the fact that the greater the molar mass, the larger the standard molar entropy.

B. Rank the gases from largest to smallest standard molar entropy. To rank items as equivalent, overlap them.

H2O2(g)>H2S(g) >H2O(g)

This is due to the fact that the greater the molar mass and the structural complexity, the larger the standard molar entropy as the hydrogen peroxide has four bonds and weights 34 g/mol as well as hydrogen sulfide that has two bonds only.

C. Rank the gases from largest to smallest standard molar entropy. To rank items as equivalent, overlap them.

C(s, amorphous) >C(s, graphite)>C(s, diamond)

Since the molecular complexity is greater in the amorphous carbon (messy arrangement), mid in the graphite and lower in the diamond (well organized).

Regards.

Answer:

a. 121 ml, b. 253 ml and c. 2.57 L.

Explanation:

The new volume can be calculated by using the Boyle's law equation:  

P1V1 = P2V2

In the equation, P1 and P2 are the initial and final pressures and V1 and V2 are the initial and final volumes for a real gas at constant temperature.  

a) Based on the given information, P1 = 755 mmHg, V1 = 125 ml, P2 = 780 mm Hg and V2 will be,  

V2 = P1V1/P2

V2 = 755 mmHg × 125 ml/780 mmHg

V2 = 121 ml

b) Based on the given information, P1 = 1.08 atm, V1 = 223 ml, P2 = 0.951 atm and V2 will be,  

V2 = P1V1/P2

V2 = 1.08 atm × 223 ml/0.951 atm

V2 = 253 ml

c) Based on the given information, P1 = 103 kPa, V1 = 3.02 L, P2 = 121 kPa and V2 will be,  

V2 = P1V1/P2

V2 = 103 kPa × 3.02 L/121kPa

V2 = 2.57 L

Answer:

1.12 × 10⁻⁴ M

Explanation:

Step 1: Write the reaction for the solution of Mg(OH)₂

Mg(OH)₂(s) ⇄ Mg²⁺(aq) + 2 OH⁻(aq)

Step 2: Make an ICE chart

We can relate the solubility product constant (Ksp) with the solubility (S) through an ICE chart.

       Mg(OH)₂(s) ⇄ Mg²⁺(aq) + 2 OH⁻(aq)

I                                0                    0

C                              +S                +2S

E                                S                  2S

The solubility product constant is:

Ksp = 5.61 × 10⁻¹² = [Mg²⁺] × [OH⁻]² = S × (2S)² = 4S³

S = 1.12 × 10⁻⁴ M

Answer:The concentration of the unknown HNO3 solution = 0.01809 M

Explanation:

For the acid-base reaction,  HNO3 + NaOH-----> NaN03 + H20

we have that

C1 V1 = C2 V2

Where ,

C1 = concentration of HNO3=?

V1 = volume of HNO3 = 25.00 mL,

V2 = volume of NaOH = 22.62 mL,  

C2 = concentration of NaOH = 0.02000 M

Therefore ,

25.00 mL x C1 = 22.62 mL x 0.02000 M    

 = (22.62 mL / 25.00 mL) x 0.02000 M = 0.01809 M

The concentration of the unknown HNO3 solution = 0.01809 M

Answer:

A precipitate will form.

[Ag⁺] = 2.8x10⁻⁵M

Explanation:

When Ag⁺ and CrO₄²⁻ are in solution, Ag₂CrO₄(s) is produced thus:

Ag₂CrO₄(s) ⇄ 2 Ag⁺(aq) + CrO₄²⁻(aq)

Ksp is defined as:

Ksp = 1.1x10⁻¹² = [Ag⁺]² [CrO₄²⁻]

Where the concentrations [] are in equilibrium

Reaction quotient, Q, is defined as:

Q = [Ag⁺]² [CrO₄²⁻]

Where the concentrations [] are the actual concentrations

If Q < Ksp, no precipitate will form, if Q >= Ksp, a precipitate will form,

The actual concentrations are -Where 500mL is the total volume of the solution-:

[Ag⁺] = [AgNO₃] = 0.12M ₓ (250mL / 500mL) = 0.06M

[CrO₄²⁻] = [Na₂CrO₄] = 0.33M × (250mL / 500mL) = 0.165M

And Q = [0.06M]² [0.165M] = 5.94x10⁻⁴

As Q > Ksp; a precipitate will form

In equilibrium, some Ag⁺ and some CrO₄⁻ reacts decreasing its concentration until the system reaches equilibrium. Equilibrium concentrations will be:

[Ag⁺] = 0.06M - 2X

[CrO₄²⁻] = 0.165M - X

Where X is defined as the reaction coordinate

Replacing in Ksp expression:

1.1x10⁻¹² = [0.06M - 2X]² [0.165M - X]

Solving for X:

X = 0.165M → False solution. Produce negative concentrations.

X = 0.0299986M

Replacing, equilibrium concentrations are:

[Ag⁺] = 0.06M - 2(0.0299986M)

[CrO₄²⁻] = 0.165M - 0.0299986M

[CrO₄²⁻] = 0.135M

Answer:

1. the siren has a lower pitch to Jane

2. the siren has a higher pitch to John

3. sound different due to moving away from Jane making the sound wave lengths longer and moving toward John making the wave lengths shorter

Explanation:

The Doppler effect expresses that sound is comparative with the spectator or observer. This is demonstrated valid by the model given with Jane and John. To one individual it could sound low and to someone else it could sound high, in light of where they are tuning in from. To John, the police alarm playing is a higher pitch. Be that as it may, to Jane this equivalent alarm is a totally extraordinary pitch and is heard lower than in comparison to the john.

This is a prime case of the Doppler Effect. They sound distinctive on the grounds that the sound is moving far from Jane making the sound frequencies longer and it is advancing toward John making the frequencies shorter. This impacts how the sound is heard by the human ear.

Answer:

2Cl——>Cl2+2e-

Explanation:

It shows an electron loss or gain

Answer: NaCl would give the higher pressure

Explanation:

Osmotic pressure depends only on the number of ions.

NaCl dissociates as Na+ and Cl- ; CsBr  dissociates as Cs+ and Br-

But the concentration of the solutions are different.  

Concentration (morality ) of NaCl = Moles /Litre = (1 g /58.44g/mol)/0.01L

Total number of ions in NaCl solution = 2 x (1 g /58.44g/mol)/0.01L ( 1 mol NaCl gives 2 moles ions, 1 mol Na+ and 1 mol Cl-)  

= 1.71×2RT

Similarly total number of ions in CsBr solution = 2 x (1 g /212.80 g/mol)/0.01L

= 0.47×2RT

Therefore osmotic pressure is higher in NaCl solution.

Answer:

Amount of salt in 1 L seawater = 34 g

Explanation:

According to Archimedes' principle, mass of freshwater and cup = mass of equal volume of seawater

mass of freshwater = density * volume

1 cm³ = 1 mL

mass of freshwater = 0.999 g/cm³ * 735 cm³ = 734.265 g

mass of freshwater + cup = 734.265 + 25 = 759.265 g

Therefore,  mass of equal volume of seawater = 759.265 g

Volume of seawater displaced = 735 mL = 0.735 L (assuming the cup volume is negligible)

1 liter = 1000 cm³ = 1000 mL;

Density of seawater = mass / volume

Density of seawater = 759.265 g / 0.735 L = 1033.01 g/L

Density of freshwater in g/L = 0.999 g/ (1/1000) L = 999 g/L

mass of 1 Liter seawater = 1033.01 g

mass of 1 Liter freshwater = 999 g

mass of salt dissolved in 1 L of seawater = 1033.01 g - 999 g = 34.01 g

Therefore, amount of salt in 1 L seawater = 34 g

Answer:

There are required 70.1 years for the activity of a sample of cesium-137 to fall to 20.0 percent of its original value

Explanation:

The radioactive decay follows always first-order kinetics where its general law is:

Ln[A] = -Kt + ln[A]₀

Where [A] is actual concentration of the atom, k is rate constant, t is time and [A]₀ is initial concentration.

We can find rate constant from half-life as follows:

Rate constant:

t(1/2) = ln 2 / K

As half-life of Cesium-137 is 30.2 years:

30.2 years = ln 2 / K

K = 0.02295 years⁻¹

Replacing this result and with the given data of the problem:

Ln[A] = -Kt + ln[A]₀

Ln[A] = -0.02295 years⁻¹* t + ln[A]₀

Ln ([A] / [A₀]) = -0.02295 years⁻¹* t

As you want time when [A] is 20% of [A]₀, [A] / [A]₀ = 0.2:

Ln (0.2) = -0.02295 years⁻¹* t

70.1 years = t

Answer:

Explanation:

Given that:

The argon atoms are excited into an excited state before emitting the 488.0 nm laser.

the energy of the first ionization energy of argon is 1520 kJ mol-1.

SInce 1 eV = 96.49 kJ/mol

Therefore, the energy of the first ionization energy of argon in eV is = ( 1520/ 96.49) eV

=  15.75 eV

To find where  the energy level of the excited state lies below the vacuum energy level, let's first determine, the energy liberated by using planck expression.

[tex]E = \dfrac{hc}{\lambda}[/tex]

[tex]E = \dfrac{6.6 \times 10^{-34} \times 3 \times 10^8}{488 \times 10^{-9}}[/tex]

[tex]E = \dfrac{1.98 \times 10^{-25}}{488 \times 10^{-9}}[/tex]

[tex]E = \dfrac{1.98 \times 10^{-25}}{488 \times 10^{-9}}[/tex]

[tex]E =4.057 \times 10^{-19} \ J[/tex]

Converting Joules (J) to eV ; we get,

[tex]E =\dfrac{4.057 \times 10^{-19}}{1.6 \times 10^{-19}}[/tex]

E = 2.53 eV

The energy levels of the first exited state = -13.223 eV

Answer:

Ka = 6.15x10⁻⁷

Explanation:

Ka is defined as dissociation constant in the equilibrium of a weak acid with water. The general reaction is:

HA(aq) + H₂O(l) ⇆ H₃O⁺(aq) + A⁻(aq)

And Ka is defined as the ratio between molar concentrations in equilibrium of products over reactants as follows:

Ka = [H₃O⁺] [A⁻] / [HA]

You don't take water in the equilibrium beacuse is a pure liquid

Replacing with the concentrations of the problem:

Ka = [H₃O⁺] [A⁻] / [HA]

Ka = [4.00x10⁻⁴] [4.00x10⁻⁴] / [0.260]

Answer:

Depends on the reaction.

Explanation:

Hello,

In this case, the answer is depends on the reaction since the ratios between the rates of both consumption and formation depend upon the stoichiometric coefficients in the chemical reaction. For instance, for the reaction:

A -> 2B

The relationship is:

[tex]\frac{1}{-1}r_A =\frac{1}{2} r_B[/tex]

Therefore, we can see that the rate of consumption of A half the rate of formation of B, but is we consider the following chemical reaction:

2A -> B

The relationship is:

[tex]\frac{1}{-2}r_A =\frac{1}{1} r_B[/tex]

Therefore we can see that the rate of consumption of A doubles the rate of consumption of B.

Best regards.

A 1.89 pints of blood would contain 873 grams of platelets.

To calculate the amount of platelets present in 1.89 pints, it is first necessary to transform this unit of volume into liters:

                                  [tex]1.89 \times 473.2 = 894.3 mL[/tex]

                                         [tex]\frac{894.3}{1000}= 0.84L[/tex]

Now, just calculate the amount of platelets present in 0.84L:

                                    [tex]\frac{1.04\times10^{3}g}{xg}=\frac{1L}{0.84L}[/tex]

                                       x = 873 grams

So, a 1.89 pints of blood would contain 873 grams of platelets.

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Answer:

One carbon atom and the nitrogen atom have nonzero formal charges.

Explanation:

The compound Acetonitrile has sixteen valence electrons as is easily San from its structure. It contains a carbon nitrogen triple bond with a lone pair of electrons on nitrogen. All atoms satisfy the octet rule and there is no hyper valent atom in the molecule.

The formal charge an carbon and nitrogen is calculated as follows;

No. of valence electron on atom - [non bonded electrons + no. of bonds]

Therefore, for carbon and nitrogen, we have;

formal charge on carbon = 4 - (0 + 4) = 0

formal charge on nitrogen = 5 - (2 + 3) = 0

Hence carbon and nitrogen both possess zero formal charges.

Answer:

B) The complex ion is favored over solid silver chloride

C) The free Ag+ ion is unstable.

Explanation:

Hello,

In this case, since the dissociation of solid silver chloride occurs at equilibrium with a neglectable solubility product (very small Ksp), which means that the solid tends to remain undissolved:

[tex]AgCl(s)\rightleftharpoons Ag^+(aq)+Cl^-(aq)[/tex]

By the addition of ammonia, the following reaction is favored:

[tex]Ag^+(aq)+2NH_3(aq)\rightleftharpoons [Ag(NH_3)_2]^+(aq)[/tex]

Which has a large equilibrium constant, which means that the formation of the complex is assured. In such a way, by addition of more ammonia, more complex will be formed, therefore B) The complex ion is favored over solid silver chloride is true. Moreover, C) The free Ag+ ion is unstable, since they tend to form the complex once they are formed by the solid silver chloride so it readily reacts.

Best regards.

Answer:

Solid sodium hydroxide dissolves in water to form an aqueous solution of ions.

NaOH(s) ⇌ Na+(aq) + OH–(aq) ΔH1 = ?

Solid sodium hydroxide reacts with aqueous hydrochloric acid to form water and an aqueous solution of sodium chloride.

NaOH(s) + H+(aq) + Cl–(aq) ⇌ H2O(l) + Na+(aq) + Cl–(aq) ΔH2 = ?

Solutions of aqueous sodium hydroxide and hydrochloric acid react to form water and aqueous sodium chloride.

Na+(aq) + OH–(aq) + H+(aq) + Cl–(aq) ⇌ H2O(l) + Na+(aq) + Cl–(aq) ΔH3 = ?

Answer:

0.27 V

Explanation:

Given that the both half cells contain 1.0 molar solutions of their respective electrolytes.

E°Pb= -0.13 V

E°Cd = -0.40 V

Since it is a galvanic cell, the electrode having a more negative electrode potential will serve as the anode and the electrode having a less negative electrode potential will serve as the cathode.

Hence cadmium will serve as the anode and lead will serve as the cathode.

E°cell = E°cathode - E°anode

E°cell = -0.13 - (-0.40)

E°cell = 0.27 V

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:Cobb

Explanation:What y'all

Answer:

2,2,3,3-tetrapropyloxirane

Explanation:

In this case, we have to know first the alkene that will react with the peroxyacid. So:

What do we know about the unknown alkene?

We know the product of the ozonolysis reaction (see figure 1). This reaction is an oxidative rupture reaction. Therefore, the double bond will be broken and we have to replace the carbons on each side of the double bond by oxygens. If [tex](CH_3CH_2CH_2)_2C=O[/tex] is the only product we will have a symmetric molecule in this case 4,5-dipropyloct-4-ene.

What is the product with the peroxyacid?

This compound in the presence of alkenes will produce peroxides. Therefore we have to put a peroxide group in the carbons where the double bond was placed. So, we will have as product 2,2,3,3-tetrapropyloxirane. (see figure 2)

Answer:

D. The Ca[OH]2 solution may have been unsaturated

Explanation:

The solubility product constant Ksp of any given chemical compound is a term used to describe the equilibrium between a solid and the ions it contains solution. The value of the Ksp indicates the extent to which any compound can dissociate into ions in water. A higher the Ksp, implies more greater solubility of the compound in water.

If the Ksp is more than the value in literature, this false value must have arisen from the fact that the solution was unsaturated hence it appears to be more soluble than it should normally be when saturated.

Answer:

The correct answer is 1332 KJ.

Explanation:

Based on the given information,  

ΔH°f of C2H4 is 51.9 KJ/mol, ΔH°O2 is 0.0 KJ/mol, ΔH°f of CO2 is -394 KJ/mol, and ΔH°f of H2O is -286 KJ/mol.  

Now the balanced equation is:  

C2H4 (g) + 3O2 (g) ⇔ 2CO2 (g) + 2H2O (l)

ΔH°rxn = 2 × ΔH°f CO2 + 2 × ΔH°fH2O - 1 × ΔH°fC2H4 - 3×ΔH°fO2

ΔH°rxn = 2 (-394) + 2(-286) - 1(51.9) - 3(0)

ΔH°rxn = -1411.9 KJ

Now, the given ΔS°f of C2H4 is 219.8 J/mol.K, ΔS°f of O2 is 205 J/mol.K, ΔS°f of CO2 is 213.6 J/mol.K, and ΔS°f of H2O is 69.96 J/mol.K.  

Now based on the balanced chemical reaction,  

ΔS°rxn = 2 × ΔS°fCO2 + 2 ΔS°fH2O - 1 × ΔS°f C2H4 - 3 ΔS°fO2

ΔS°rxn = 2 (213.6) + 2(69.96) - 1(219.8) -3(205)

ΔS°rxn = -267.68 J/K or -0.26768 KJ/K

T = 25 °C or 298 K

Now putting the values of ΔH, ΔS and T in the equation ΔG = ΔH-TΔS, we get

ΔG = -1411.9 - 298.0 × (-0.2677)

ΔG = -1332 KJ.  

Thus, the maximum work, which can obtained is 1332 kJ.  

Answer:

the chemical equation of ethanol as fuel is

C2H5OH(l)+3 02(g)------2CO2(g)+3H2O(g)

the preparation process of ethanol from cassava is

cassava flour---liquification---saccharification---cooling---fermentation---distillation---ethanol

Answer:

2 Na(s) + C(s) + 3/2 O₂(g) → Na₂CO₃(s)

Explanation:

The molar enthalpy of formation of a chemical is defined as the change in enthalpy during the formation of 1 mole of the substance from its constituent elements (Constituent elements are pure elements you have in the periodic table)

For Na₂CO₃ constituent elements are Na(s), C(s) and O₂(g) and the chemical equation that represents the molar enthalpy is:

Answer:

A) CH3CH2CH2CH2CH2CH2OH

Explanation:

For this question, we have the following answer options:

A) CH3CH2CH2CH2CH2CH2OH

B) (CH3CH2)2CH(OH)CH2CH3

C) (CH3CH2)2CHOHCH3

D) (CH3CH2)3COH

E) (CH3CH2)2C(CH3)OH

We have to remember the reaction mechanism of the substitution reaction with [tex]PBr_3[/tex]. The idea is to generate a better leaving group in order to add a "Br" atom.

The [tex]PBr_3[/tex] attacks the "OH" generation new a bond to P (O-P bonds are very strong), due to this new bond we will have a better leaving group that can remove the oxygen an allow the attack of the Br atom to generating a new C-Br bond. This is made by an Sn2 reaction. Therefore we will have a faster reaction with primary substrates. In this case, the only primary substrate is molecule A. So, "CH3CH2CH2CH2CH2CH2OH" will react faster.

See figure 1

I hope it helps!

Answer:

Animals whose certain organs closely match those of humans or have similar genetic makeup are used in vaccine tests because the results can closely resemble those same results on humans.

Explanation:

Answer:

they use them to test the effectiveness of the vaccine.

Explanation: