pplication)Using multiple models simultaneously: This FNT refers to a processinvolving three moles of a diatomic gas (which behaves as an ideal gas).The PV curve at right describes this process.Assume, as is typical near room temperature,vibrational modes are frozen out.a)Determine the energy transferred as work, the change in internal energy, and the energy transferred as heat in this process. b)Could youhave stillansweredthe questions in a) if the temperature was not provided on the plot

Explanation:

The given data is:

The mass of hydrogen is 6.54 kg.

The actual yield is 30.4 kg.

The balanced chemical equation of the reaction is:

[tex]N_2(g)+3H_2(g)<=>2NH_3(g)[/tex]

At first the theoretical yield should be calculated by using the balanced chemical equation:

3 mol. of hydrogen forms ---- 2 mol. of ammonia.

The molar mass of hydrogen is 2.0 g/mol.

The molar mass of ammonia is 17.0 g/mol.

Hence, the above statement can be rewritten as:

6g of hydrogen forms --- 34g of ammonia.

Then,

6.54g of hydrogen forms :

[tex]6.54 kg x 34 g / 6 g\\=37.1 kg[/tex]

% yield = (actual yield /theoretical yield )x 100

=(30.4 kg /37.1 kg )x100

=81.9

Hence, % yield is 81.9.

Answer:

The percent abundance of oxygen-18 is 1.9066%.

Explanation:

The average atomic mass of oxygen is given by:

[tex] m_{O} = m_{^{16}O}*\%_{16} + m_{^{17}O}*\%_{17} + m_{^{18}O}*\%_{18} [/tex]

Where:

m: is the atomic mass

%: is the percent abundance

Since the sum of the percent abundance of oxygen isotopes must be equal to 1, we have:  

[tex] 1 = \%_{16} + \%_{17} + \%_{18} [/tex]

[tex] 1 = x + 3.2 \cdot 10^{-4} + \%_{18} [/tex]

[tex] \%_{18} = 1 - x - 3.2 \cdot 10^{-4} [/tex]

Hence, the percent abundance of O-18 is:  

[tex] m_{O} = m_{^{16}O}*\%_{16} + m_{^{17}O}*\%_{17} + m_{^{18}O}*\%_{18} [/tex]  

[tex]15.9982 = 15.972*x + 16.988*3.2 \cdot 10^{-4} + 17.970*(1 - 3.2 \cdot 10^{-4} - x)[/tex]

[tex] x = 0.980614 \times 100 = 98.0614 \% [/tex]                                                              

Hence, the percent abundance of oxygen-18 is:

[tex]\%_{18} = (1 - 3.2 \cdot 10^{-4} - 0.980614) \times 100 = 1.9066 \%[/tex]                      

Therefore, the percent abundance of oxygen-18 is 1.9066%.

I hope it helps you!                                                      

Answer:

The correct answer is "654.54 nm".

Explanation:

According to the question,

⇒ [tex]\frac{1}{\lambda}= Rh(\frac{1}{n1^2} -\frac{1}{n2^2} )[/tex]

By substituting the values, we get

       [tex]=1.1\times 10^7(\frac{1}{4} -\frac{1}{9} )[/tex]

       [tex]=1.1\times 10^7(\frac{9-4}{36} )[/tex]

       [tex]=1.1\times 10^7(\frac{5}{36} )[/tex]

       [tex]=654.54\ nm[/tex]

Thus the above is the right solution.

Answer:

cm3 = 2500.0 g / 10.5 g/cm3 = 238 cm3

Complete Question

A sample of aluminum, which has a specific heat capacity of 0.897 JB loc ! is put into a calorimeter (see sketch at right) that contains 200.0 g of water. The aluminum sample starts off at 85.6 °C and the temperature of the water starts off at 16.0 °C. When the temperature of the water stops changing it's 20.1 °C. The pressure remains constant at 1 atm. Calculate the mass of the aluminum sample.

Answer:

[tex]M=58g[/tex]

Explanation:

From the question we are told that:

Heat Capacity [tex]H=0.897[/tex]

Mass of water [tex]M=200g[/tex]

Initial Temperature of Aluminium [tex]T_a=85.6[/tex]

Initial Temperature of Water [tex]T_{w1}=16.0[/tex]

Final Temperature of Water  [tex]T_{w2}=16.0[/tex]

Generally

Heat loss=Heat Gain

Therefore

[tex]M*0.897*(85.6-20.1) =200*4.184*(20.1-16)[/tex]

[tex]M=58g[/tex]

Explanation:

the reaction is indeed an oxidation reduction reaction

Answer:

d. amine.

It becomes an amine.

Explanation:

With general formular

[tex]{ \bf{primary \: amine :R - NH _{2}}} \\ { \bf{secondary \: amine : R {}^{i} - NH - R}} \\ { \bf{tertiary \: amine :R {}^{ii} - N(R {}^{i} ) - R }}[/tex]

R is the aryl group such as alkane

Answer:

Nitrobenzene < Bromobenzene < Benzene < Phenol

Explanation:

Aromatic compounds undergo electrophilic aromatic substitution reaction in the presence of relevant electrophiles. Certain substituents tend to increase or decrease the tendency of an aromatic compound towards electrophilic aromatic substitution reaction.

Substituents that increase the electron density around the ring such as in phenol tends to make the ring more reactive towards electrophilic substitution. Halogens such as bromine has a -I inductive effect as well as a +M mesomeric effect.

However the -I(electron withdrawing effect) of the halogens supersedes the +M electron donation due to mesomeric effect.

Putting all these together, the order of increasing reactivity of the compounds towards electrophilic aromatic substitution is;

Nitrobenzene < Bromobenzene < Benzene < Phenol

Answer:

groups are based on how many electrons to become stable

Explanation:

Answer:

C

Explanation:

b/c when copper and zinc metal are addedto solution,then the solution will be consider under redox reaction

[tex]Cu^{2+} + Zn[/tex] → [tex]Cu + Zn^{2+}[/tex] is the redox reaction. Hence, option C is correct.

A chemical reaction taking place between an oxidizing substance and a reducing substance.

The oxidizing substance is used to lose electrons in the reaction, and the reducing substance is used to gain electrons.

On the reduction potential chart, zinc is a stronger oxidizing agent than, Copper (Cu), which is a reducing agent as compared to silver

The redox reaction most likely occurs if silver and copper metal were added to a solution that contained silver and copper ions is ;

[tex]Cu^{2+} + Zn[/tex] → [tex]Cu + Zn^{2+}[/tex]

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Answer:

ITS ANSWER IS

OPTION B. ATOMIC NUMBER

HI HAVE A NICE DAY

Answer:

b

Explanation:

Addition of a solute lowers the freezing point of a substance.

Let us recall that the addition of a solute to a solution lowers the freezing point of the pure solution. This is because, freezing point depression is a colligative property. It depends on the number of solute molecules present.

The following are possible errors in the experiment:

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Answer:

2

Explanation:

i did this

Answer:

A white gelatinous precipitate is observed in each case.

Explanation:

Qualitative analysis in chemistry is mostly used to identify the ions present in a sample by adding certain reagents. The observation after adding the reagent often leads to an inference.

When NaOH is added to a solution containing Zn^2+ in drops, a white gelatinous precipitate is observed.

When NH3(aq) is added in drops to a solution containing Zn^2+, a white gelatinous precipitate is also observed.

Answer:

Mg> H> Cu

Explanation:

We can see from the question that hydrochloric acid reacted with magnesium as follows;

Mg(s) + 2HCl(aq) ----> MgCl2(aq) + H2(g)

Copper does not react with HCl which means that copper is less reactive than hydrogen hence it can not displace hydrogen from a dilute acid solution.

The order of reactivity of the elements then is ; Mg> H> Cu

Answer:

K = 3.3

Explanation:

Nitric acid, HNO3, reacts with nitrogen monoxide, NO, to produce nitrogen dioxide, NO2 and water H2O as follows:

2HNO3(g) + NO(g) → 3NO2(g) + H2O(g)

Where equilibrium constant, K, is:

K = [NO2]³[H2O] / [HNO3]²[NO]

[] is the molar concentration of each species at equilibrium.

To solve this question we need to find molarity of each gas and replace these in the equation as follows:

[NO2] -Molar mass NO2-46.0g/mol-

18.6g * (1mol/46.0g) = 0.404mol / 7.7L = 0.0525M

[H2O] -Molar mass:18.01g/mol-

236.7g * (1mol/18.01g) = 13.14 moles / 7.7L = 1.707M

[HNO3] -Molar mass:53.01g/mol-

16.2g * (1mol/53.01g) = 0.3056 moles / 7.7L = 0.0397M

[NO] -Molar mass: 30.0g/mol-

11.0g * (1mol/30.0g) = 0.367 moles / 7.7L = 0.0476M

Replacing:

K = [NO2]³[H2O] / [HNO3]²[NO]

K = [0.0525M]³[1.707M] / [0.0397M]²[0.0476M]

We are given:

Volume of Solution: 5.7 Liters

Concentration: 1.64 Molar

Number of moles required:

We know that the formula for concentration (in molar) is:

M = N / V              

where M is the concentration, N is the number of moles, and V is the volume of solution (in L)

plugging the given values in the equation:

1.64 = N / 5.7

N = 1.64 * 5.7            

N = 9.35 moles

Answer:

See explanation and image attached

Explanation:

The product of the Grignard reaction between methyl benzoate and excess phenylmagnesium bromide is triphenyl methanol.

The reaction proceeds by nucleophillic reaction as the carbonyl moiety is attacked. A tetrahedral intermediate is formed. Loss of the -OMe group is accompanied by the attack of the first molecule of PhMgBr.

Attack by a second PhMgBr molecule yields trimethyl phenoxide. Protonation of this specie yields the final product which is obtained by aqueous workup.

Answer:

+0.151, spontaneous

Explanation:

Given that;

Co^2+(aq) + 2e ---->Co(s) -0.28 V

Pb^2+(aq) + 2e ---->Pb(s). -0.13 V

Hence Co is the anode and Pb is the cathode

E°cell = E°cathode - E°anode

So;

E°cell = -0.13 V - (-0.28 V)

E°cell = 0.15 V

The cell reaction is spontaneous since E°cell is positive.

Answer:

A. 1.9 × 103 g

(next one)

Which metric unit would be the best choice to report the result?  

A. kg

Answer:

1. 2

2. 1.9 × 10^3 g

3. kg

Explanation:

Answer:

[tex]m_{isoborneol }=0.39g\\\\m_{Camphor}=0.38g\\[/tex]

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to infer that the reaction whereby isoborneol goes to camphor occurs in a 1:1 mole ratio, that is why the theoretical yield of the latter is also 2.5 mmol (0.0025 mol) but the masses can be calculated as follows:

[tex]m_{isoborneol }=0.0025mol*\frac{154.25g}{1mol} =0.39g\\\\m_{Camphor}=0.0025mol*\frac{152.23 g}{1mol} =0.38g\\[/tex]

Because of the fact this is a rearrangement reaction whereas the number of atoms is not significantly modified.

Regards!

Answer:

a. 181.5 kcal

Explanation:

Step 1: Calculate the enthalpy of the process (ΔH°).

Let's consider the following process.

CO₂(s) → CO₂(g)

We can calculate the enthalpy of the process using the following expression.

ΔH° = ∑ np × ΔH°f(p) - ∑ nr × ΔH°f(r)

ΔH° = 1 mol × ΔH°f(CO₂(g)) - 1 mol × ΔH°f(CO₂(s))

ΔH° = 1 mol × (-393.5 kJ/mol) - 1 mol × (-427.4 kJ/mol) = 33.9 kJ

According to the balanced equation, 33.9 kJ are required to sublime 1 mole of CO₂.

Step 2: Convert 986 g of CO₂ to moles

The molar mass of CO₂ is 44.01 g/mol.

986 g × 1 mol/44.01 g = 22.4 mol

Step 3: Calculate the enthalpy needed to sublime 22.4 moles of CO₂

22.4 mol × 33.9 kJ/1 mol = 759 kJ

We can convert it to Kcal using the conversion factor 1 kcal = 4.184 kJ.

759 kJ × 1 kcal/4.184 kJ = 181.5 kcal

Answer:

A

Explanation:

The iron corrodes so it oxidized

I think the answer most be d

In chemistry like dissolves like hence hexane will dissolve in gasoline.

Dissolution stems from intermolecular interaction between solute and solvent molecules.

If this interaction can not occur, dissolution of one substance in another is impossible.

Hexane dissolves in gasoline because the both substances are non-polar and can interact with each other effectively.

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Answer: True

Explanation:

Phosphagens are high energy storage compounds that are usually found in the tissue of animals.

Based on the question, the molecules have similar functions in different organisms such as the fact that they can accept phosphoryl groups from ATP in a situation where the ATP is in excess. Also, they donate phosphoryl groups to ADP in order for the regeneration of ATP.

Best practices for fume hoods: work 25 cm inside, organize items to one side, use raised ledge; avoid open sash and quick movements.

Laboratory fume hoods must be used safely. Workers should operate at least 25 cm within the hood to preserve ventilation and avoid dangerous chemicals. Place things on one side of the hood to preserve ventilation and prevent clogging.

A raised ledge on the rear of the hood prevents things from falling in and impeding airflow. Avoid fully opening the sash to maintain ventilation and containment. Fast, rapid motions can interrupt airflow, so prevent them. These practises guarantee the fume hood contains harmful compounds, making the lab safer. Therefore, option (B), (D) and (E) are correct.

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Answer:

It is 1.25 moles

Explanation:

Molar mass of sulphur = 32 g

[tex]{ \bf{moles = \frac{given \: mass}{molar \: mass} }} [/tex]

Substitute:

[tex]{ \sf{moles = \frac{40}{32} }} \\ { \sf{ = 1.25 \: moles}}[/tex]