The nitration of aromatic compounds is a highly exothermic reaction that generally uses catalysts that tend to be corrosive (e.g., HNO3 /H2SO4 ). A less corrosive employs N2 O5 as the nitrating agent as illustrated below: If this reaction is conducted in an adiabatic CSTR, what is the reactor volume and space time necessary to achieve 35 percent conversion of N2O5? The reaction rate is first order in A and second order in B. Data: HRX = -370.1 kJ/mol CpA = 84.5 J/(mol-K) CpB = 137 J/(mol-K) Cpc = 170 J/(mol-K) CpD = 75 J/(mol-K) To = 303 K FAo = 10 mol/min FBo = 30 mol/min v = 1000 L/min CAo =0.01 mol/L

Answers

Answer 1

Answer:

-370.1 kJ/mol CpA = 84.5 J/(mol-K) CpB = 137 J/(mol-K) Cpc = 170 J/(mol-K) CpD = 75 J/(mol-K) To = 303 K FAo = 10 mol/min FBo = 30 mol/min v = 1000 L/min CAo =0.01 mol/L

Explanation:

The nitration of aromatic compounds is a highly exothetion that generally uses catalysts that tend to be corrosive (e.g., HNO3 /H2SO4 ). A less corrosive employs N2 O5 as the nitrating agent as illustrated below: If this reaction is conducted in an adiabatic CSTR, what is the reactor volume and space time necessary to achieve 35 percent conversion of N2O5? The reaction rate is first order in A and second order in B. Data: HRX = -370.1 kJ/mol CpA = 84.5 J/(mol-K) CpB = 137 J/(mol-K) Cpc = 170 J/(mol-K) CpD = 75 J/(mol-K) To = 303 K FAo = 10 mol/min FBo = 30 mol/min v = 1000 L/min CAo =0.01 mol/L

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Related Questions

Searches related to If 0.75 grams of iron (Fe) react according to the following reaction, how many grams of copper (Cu) will be produced? Fe + CuSO4 -> Cu + FeSO4

Answers

Answer:

0.83 g

Explanation:

Step 1: Write the balanced equation

Fe + CuSO₄ ⇒ Cu + FeSO₄

Step 2: Calculate the moles corresponding to 0.75 g of Fe

The molar mass of Fe is 55.85 g/mol.

[tex]0.75g \times \frac{1mol}{55.85g} = 0.013 mol[/tex]

Step 3: Calculate the moles of Cu produced from 0.013 moles of Fe

The molar ratio of Fe to Cu is 1:1. The moles of Cu produced are 1/1 × 0.013 mol = 0.013 mol.

Step 4: Calculate the mass corresponding to 0.013 moles of Cu

The molar mass of Cu is 63.55 g/mol.

[tex]0.013mol \times \frac{63.55g}{mol} = 0.83 g[/tex]

Answer:

If 0.75 grams of iron (Fe) react, 0.85 grams of copper (Cu) will be produced.

Explanation:

You know the following balanced reaction:

Fe + CuSO₄ ⇒ Cu + FeSO₄

By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the following quantities react and are produced:

Fe: 1 moleCuSO₄: 1 moleCu: 1 moleFeSO₄: 1 mole

Being:

Fe: 55.85 g/moleCu: 63.54 g/moleS: 32 g/moleO: 16 g/mole

the molar mass of the compounds participating in the reaction is:

Fe: 55.85 g/moleCuSO₄: 63.54 g/mole + 32 g/mole+ 4* 16 g/mole= 159.54 g/moleCu: 63.54 g/moleFeSO₄: 55.85 g/mole + 32 g/mole+ 4* 16 g/mole= 151.85 g/mole

Then, by stoichiometry of the reaction, the amounts of reagent and product that participate in the reaction are:

Fe: 1 mole*55.85 g/mole= 55.85 gCuSO₄: 1 mole* 159.54 g/mole= 159.54 gCu: 1mole* 63.54 g/mole= 63.54 gFeSO₄: 1 mole* 151.85 g/mole= 151.85 g

Then you can apply a rule of three as follows: if 55.85 grams of Fe produces 63.54 grams of Cu, 0.75 grams of Fe how much mass of Cu does it produce?

[tex]mass of Cu=\frac{0.75 grams of Fe*63.54 grams of Cu}{55.85 grams of Fe}[/tex]

mass of Cu= 0.85 grams

If 0.75 grams of iron (Fe) react, 0.85 grams of copper (Cu) will be produced.

Which short-term environmental change would a very small asteroid or comet impact on Earth most likely cause? flooding extinction surface craters weather pattern changes

Answers

The correct answer is C. Surface craters

Explanation:

Short-term environmental changes involve temporary changes and effects in the ecosystem, which are mainly minor. In the case of a small asteroid or comet, this will likely lead to surface craters or changes in the surface of the impact zone. This is because the craters and asteroids impact the surface at hight speed. Also, because this is a minor event it might lead to the death of some organisms but not the extinction of these and it is not expected this has major effects such as changes in weather. Thus, the short-term effect that this will most likely cause is "surface craters."

Answer:

surface

Explanation:

2.
Name the following compounds:
a. Rb20

Answers

Answer:

Rubidium oxide

Explanation:

The answer is Rubidium oxide the formula is RB O
2

A critical reaction in the production of energy to do work or drive chemical reactions in biological systems is the hydrolysis of adenosine triphosphate, ATP, to adenosine diphosphate, ADP, as described by the reactionATP(aq)+ H2O(l) → ADP(aq)+ HPO4^-2 (aq)for which ΔGrxn = -30.5 kj/mol at 37.0C and pH 7.0. Required:a. Calculate the value of ΔGrxn in a biological cell in which [ATP] = 5.0 mM, [ADP] = 0.30 mM, and HPO4^-2= 5.0mMb. Is the hydrolysis of ATP spontaneous under these conditions?

Answers

Answer:

Δ [tex]G_{rxn}[/tex] = −51. 4 kJ/mol

However, since Δ [tex]G_{rxn}[/tex] is negative. The hydrolysis of ATP for this reaction is said to be spontaneous

Explanation:

From the question; The equation for this reaction can be represented as :

[tex]ATP_{(aq)} + H_2O_{(l)} \to ADP_{(aq)}+ HPO_4^{2-}} _{(aq)}[/tex]

where:

[tex]\Delta G ^0 _{rxn} =[/tex]-30.5 kJ/mol

= -30.5 kJ/mol × 1000 J/ 1 kJ

= -30.5 × 10 ⁻³ J/mol

Temperature T = 37 ° C

= (37+273)

= 310 K

pH = 7.0

[ATP] = 5.0 mM

= 5.0mM × 1M/1000mM

= 0.005 M

[ADP] = 0.30 mM

= 0.30 mM × 1M/1000mM

= 0.0003 M

[tex][HPO_4^{2-}}][/tex] = 5.0 mM

= 5.0mM × 1M/1000mM

= 0.005 M

The objective is to calculate the value for Δ [tex]G_{rxn}[/tex] in the biological cell and to determine if the hydrolysis  of  ATP is spontaneous under these conditions.

Now;

From the equation given; the equilibrium constant [tex]K_{eq}[/tex] can be expressed as:

[tex]K_{eq} = \dfrac{[ADP][ HPO_4^{2-}]} {[ATP]}[/tex]

[tex]K_{eq} = \dfrac{(0.0003 \ M)(0.005 \ M)} {(0.005 \ M)}[/tex]

[tex]K_{eq} = 3*10^{-4}[/tex]

The  Δ [tex]G_{rxn}[/tex] in the biological cell can now be calculated as:

Δ [tex]G_{rxn}[/tex] = [tex](-30.5 * 10 ^3 \ J/mol) + (8.314 \ J/mol.K)(310 K ) In ( 3*10^{-4})[/tex]

Δ [tex]G_{rxn}[/tex] = [tex](-30.5 * 10 ^3 \ J/mol) + (-20906.68126)[/tex]

Δ [tex]G_{rxn}[/tex] = −51406.68 J/mol

Δ [tex]G_{rxn}[/tex] = −51. 4 × 10³ J/mol

Δ [tex]G_{rxn}[/tex] = −51. 4 kJ/mol

Thus since Δ [tex]G_{rxn}[/tex] is negative. The hydrolysis for this reaction is said to be spontaneous

Enter an equation for the formation of CaCO3(s) from its elements in their standard states. Enter any reference to carbon as C(s). Express your answer as a chemical equation. Identify all of the phases in your answer.

Answers

Answer:

CaF2 + CO3- ----> CaCO3 + 2 F-

Explanation:

The chemical compounds found on the left side of the date are the reagents and those found on the right are the products, where calcium carbonate appears.

Calcium carbonate is a quaternary salt

The authors state in the general procedures that the reaction was monitored by TLC. How would this be done? What would you spot in each lane? How would you know the reaction was done?

Answers

Answer:

Thin Layer Chromatography (TLC) can be used to analyze chemical reactions. During this reaction monitoring, a typical TLC plate would have three spots: the reactant lane, the reaction mixture lane, and a "co-spot" where reaction product would be spotted directly on top of reactant.

The co-spot serves as a reference point and is vital for reactions where reactant and product have similar Rfs, and many other variations of eluent tracking.

To indicate completion of the reaction, the disappearance of a spot (usually the starting reactant) is observed.

An aqueous solution of glucose (C6H12O6), called D5W, is used for intravenous injection. D5W contains 54.30 g of glucose per liter of solution. What is the molar concentration of glucose in D5W

Answers

Answer:

The correct answer is 0.30 M

Explanation:

The molar concentration or molarity of a solution is defined as moles of solute per liter of solution. We found the moles of solute (glucose) by dividing the mass (54.30 g) into the molecular weight (MW) of glucose (C₆H₁₂O₆):

MW(C₆H₁₂O₆)= (12 g/mol x 6) + (1 g/mol x 12) + (16 g/mol x 6) = 180 g/mol

Moles of glucose= mass/MW= 54.30 g/(180 g/mol)= 0.30 mol

There is 0.30 mol of solute per liter of solution, thus the molarity is:

M= moles solute/L solution= 0.30 mol/1 L = 0.30 M

When solutions of hydrochloric acid and sodium hydroxide are mixed, a chemical reaction occurs forming aqueous sodium chloride and water. What would you expect to observe if you ran the reaction in the laboratory

Answers

Answer:

a change in temperature would be observed(ΔH is -ve)

Explanation:

Hydrochloric acid react with sodium hydroxide to give salt(sodium chloride) and water

HCl(aq) + NaOH(aq) =====> NaCl(aq) + H2O(l)

There would be no notable change since sodium chloride dissolved in water but there would be a change in temperature.

Since neutralization is exothermic(heat is evolved), therefore ΔH is negative

Order these species by increasing concentration of H30+ in a 1.0 M aqueous solution. (From the
solution with the least hydronium concentration to the solution with the most hydronium concentration)
NO
H2CO3, NH4, OH, HCO3, NH3, H20
Home
ir
H2CO3,NH4+, OH", HCO3, NH3,
H20
Paste
H20, H2CO3, NH4+, OH", HCO3-
NH3
6
con
O
OH", NH3, HCO3, H20, NH4+,
H2CO3
None of the answer choices are
correct.

Answers

Answer:

OH⁻ < NH₃ < HCO₃⁻ < H₂O < NH₄⁺ < H₂CO₃

Explanation:

We can do some rough calculations to find the approximate pH values of these solutions.

H₂CO₃

Kₐ ≈ 10⁻⁶

[tex]\text{H}^{+} = \sqrt{K_{\text{a}}c} = \sqrt{10^{-6} \times 10^{-1}} = \sqrt{10^{-7}} = 10^{-3.5}\\\text{pH} = -\log (10^{-3.5}) = \mathbf{3.5}[/tex]

NH₄⁺

Kb of NH₃ ≈ 10⁻⁵

Kₐ of NH₄⁺ ≈ 10⁻⁹

[tex]\text{H}^{+} = \sqrt{K_{\text{a}}c} = \sqrt{10^{-9} \times 10^{-1}} = \sqrt{10^{-10}} = 10^{-5}\\\text{pH} = -\log (10^{-5}) = \mathbf{5}[/tex]

OH⁻

Strong base

[OH⁻] = 10⁻¹

pOH = 1

pH = 14 - 1 = 13

HCO₃⁻

Salt of dibasic acid

K₁ ≈ 10⁻⁶; K₂ ≈ 10⁻¹⁰

[tex]{\text{H}^{+}} = \sqrt{K_{1}K_{2}} = \sqrt{10^{-6}\times 10^{-10}} = \sqrt{10^{-16}} = 10^{-8}\\\text{pH} = -\log (10^{-8}) = \mathbf{8}[/tex]

NH₃

Kb ≈ 10⁻⁵

[tex]\text{OH}^{-} = \sqrt{K_{\text{b}}c} = \sqrt{10^{-5} \times 10^{-1}} = \sqrt{10^{-6}} = 10^{-3}\\\text{pOH} = -\log (10^{-3}) = 3[/tex]

pOH = 14 - 3 = 11

H₂O

Neutral. pH = 7

Order from lowest [H₃O⁺] to highest [H₃O⁺]:

      OH⁻ < NH₃ < HCO₃⁻ < H₂O < NH₄⁺ < H₂CO₃  

pH   1 3        11          8            7         5           3.5

 

Steam reforming of methane ( ) produces "synthesis gas," a mixture of carbon monoxide gas and hydrogen gas, which is the starting point for many important industrial chemical syntheses. An industrial chemist studying this reaction fills a tank with of methane gas and of water vapor, and when the mixture has come to equilibrium measures the amount of carbon monoxide gas to be .Calculate the concentration equilibrium constant for the steam reforming of methane at the final temperature of the mixture. Round your answer to significant digits.

Answers

The given question is incomplete, the complete question is:

Calculating an equilibrium constant from a partial equilibrium... Steam reforming of methane (CH) produces "synthesis gas," a mixture of carbon monoxide gas and hydrogen gas, which is the starting point for many important industrial chemical syntheses. An industrial chemist studying this reaction fills a 25.0L tank with 8.0 mol of methane gas and 1.9 mol of water vapor, and when the mixture has come to equilibrium measures the amount of carbon monoxide gas to be 1.5 mol. Calculate the concentration equilibrium constant for the steam reforming of methane at the final temperature of the mixture. Round your answer to 2 significant digits.

Answer:

The correct answer is 2.47.

Explanation:

Based on the given information, the equation for the synthesis gas is,  

CH₄ (g) + H₂O (g) ⇔ CO (g) + 3H₂ (g)

Based on the given information, 25.0 L is the volume of the tank, the concentration of CH₄ is 8.0 mol, the concentration of water vapor is 1.9 mol, and the concentration of CO gas is 1.5 mol.  

Therefore, 25 L of the solution comprise 8.0 mole of CH₄. So, 1 L of the solution will comprise 8.0 / 25 mole CH₄,  

= 0.32 mole of CH₄

Thus, the concentration of CH₄ or [CH₄] will be 0.32 mole/L or 0.32 M.  

Similarly, the concentration of H₂O or [H₂O] will be 1.9/25 = 0.076 M

and [CO] is 1.5/25 = 0.06 M

The concentration equilibrium constant for the steam will be,  

Kc = [CO] pH₂ / [CH₄] [H₂O] (Here pH₂ is the partial pressure of H₂)

Now lets us assume that the reaction has taken place in a constant atmospheric pressure, therefore, pH₂ will be equal to 1.  

= 0.06 M/0.32 M × 0.076 M  

= 2.47  

g A laboratory analysis of an unknown compound found the following composition: C 75.68% ; H 8.80% ; O 15.52%. What is the empirical formula of the compound?

Answers

Answer:

THE EMPIRICAL FORMULA FOR THE UNKNOWN COMPOUND IS C7H9O

Explanation:

The empirical formula for the unknown compound can be obtained by following the processes below:

1 . Write out the percentage composition of the individual elements in the compound

C = 75.68 %

H = 8.80 %

O = 15.52 %

2. Divide the percentage composition by the atomic masses of the elements

C = 75 .68 / 12 = 6.3066

H = 8.80 / 1 = 8.8000

O = 15.52 / 16 = 0.9700

3. Divide the individual results by the lowest values

C = 6.3066 / 0.9700 = 6.5016

H = 8.8000 / 0.9700 = 9.0722

O = 0.9700 / 0.9700 = 1

4. Round up the values to the whole number

C = 7

H = 9

O = 1

5 Write out the empirical formula for the compound

C7H90

In conclusion, the empirical formula for the unknown compound is therefore C7H9O

How many grams of H 2O are produced from 28.8 g of O 2? (Molar Mass of H 2O = 18.02 g) (Molar Mass of O 2=32.00 g) 4 NH 3 (g) + 7 O 2 (g) → 4 NO 2 (g) + 6 H 2O (g)

Answers

Answer:  13.9 g of [tex]H_2O[/tex] will be produced from the given mass of oxygen

Explanation:

To calculate the moles :

[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]    

[tex]\text{Moles of} O_2=\frac{28.8g}{32.00g/mol}=0.900moles[/tex]

The balanced chemical reaction is:

[tex]4NIO_2(g)+7O_2(g)\rightarrow 4NO_2(g)+6H_2O(g)[/tex]

According to stoichiometry :

7 moles of [tex]O_2[/tex] produce =  6 moles of [tex]H_2O[/tex]

Thus 0.900 moles of [tex]O_2[/tex] will produce =[tex]\frac{6}{7}\times 0.900=0.771moles[/tex]  of [tex]H_2O[/tex]

Mass of [tex]H_2O=moles\times {\text {Molar mass}}=0.771moles\times 18.02g/mol=13.9g[/tex]

Thus 13.9 g of [tex]H_2O[/tex] will be produced from the given mass of oxygen

a piece of copper weighing 850 grams is placed in a cup with 450 ml of water at 21 C and the Cp of the cup is 47 J/K, how many grams of gasoline would it take to heat the entire system to 110 C?

Answers

Answer:

4.2g of gasoline

Explanation:

In the problem, you need to give energy to the cup from the combustion of gasoline. The energy you need to give is:

Qcup + QWater + QCopper

As you need to increase (110ºC - 21ºC = 89º = Increase 89K) 89K, the Qcup is:

Qcup = 89K × (47J/K) = 4183J.

You can find Qwater using its specific heat, C (4.18Jg⁻¹K⁻¹), its mass (450mL = 450g) and the change of temperature, 89K:

QWater = CₓmₓΔT

QWater = 4.184Jg⁻¹K⁻¹ ₓ 450g×89K

QWater = 167569J

And Q of Copper, QCu, could be obtained in the same way (Specific heat Cu: 0.387 J/g⁻¹K⁻¹:

QCu = CₓmₓΔT

QCu = 0.387 J/g⁻¹K⁻¹ₓ850gₓ89K

QCu = 29277J

Thus, total heat you need is:

Q = Qcup + QWater + QCopper

Q = 4183J + 167569J + 29277J

Q = 201029J = 201kJ

The combustion of gasoline (Octane) produce 47.8kJ/g (Its heat of combustion). that means to produce 201kJ of energy you require:

201kJ × (1g / 47.8kJ) =

4.2g of octane = Gasoline you require

in an endothermic reaction the ____ have more energy than the ____?

Answers

Answer: products; reactants

Explanation: as the endothermic reactions are tye one which absorbs energy

How many moles of gaseous boron trifluoride, BF3, are contained in a 4.3410 L bulb at 788.0 K if the pressure is 1.220 atm What is the complete ground state electron configuration for the neon atom

Answers

Answer:

n= 0.08186

{He}2s^2 2p^6

Explanation:

PV=nRT

n=PV/RT

n= (1.220 atm)(4.3410 L) / (0.0821 atm*L/mol*K)(788.0 K)

n=0.08186

As for the electron configuration:

Ne:

{He} 2s^2 2p^6

or long hang:

1s^2 2s^2 2p^6

The moles of gaseous boron trifluoride is 0.08186 moles.The ground state electronic configuration of neon is:

[tex][Ne]=1s^22s^22p^{10}[/tex]

Given:

A gaseous boron trifluoride in a 4.3410 L bulb at 788.0 K, if the pressure is 1.220 atm.Neon atom.

To find:

The moles of gaseous boron trifluoride in a container.The electronic configuration of neon in the ground state.

1.

The pressure of the gaseous boron trifluoride = P = 1.220 atm

The volume of the gas in bulb = V = 4.3410 L

The moles of the  gaseous boron trifluoride = n

The temperature of gaseous boron trifluoride = T = 788.0 K

Using an ideal gas equation:

[tex]PV = nRT\\\\1.220 atm\times 4.3410 L=n\times 0.0821 atm L/mol K\times 788.0 K\\\\n=\frac{1.220 atm\times 4.3410 L}{0.0821 atm L/mol K\times 788.0 K}\\\\n=0.08186 mol[/tex]

The moles of gaseous boron trifluoride is 0.08186 moles.

2.

The atomic number of neon atom = 10

The electronic configuration in the ground state is the most stable arrangement of the electrons in the lowest energy levels.

The ground state electronic configuration of neon is:

[tex][Ne]=1s^22s^22p^{10}[/tex]

Learn more about the ideal gas equation and electronic configuration here:

brainly.com/question/1056445?referrer=searchResults

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Write the name for the following molecular compounds. Remember to use the correct prefix for each compound.

a. CS2
b. PBr3
c. NO
d. CF4
e. P2O5

Answers

Answer:

Hey there!

CS2) Carbon Disulfide.

PBr3) Phosphorus Tribromide

NO) Nitric Oxide

CF4) Carbon Tetrafluoride

P2O5) Phosphorus Pentoxide

Let me know if this helps :)

Given Ba, Li, Na, Cs, and Be, arrange the group 1 and 2 elements in order of increasing reactivity to water (H2O).
1. Be∠Na∠ Li∠ Ba∠ Cs
2. Be∠ Li∠ Ba∠ Na∠ Cs
3. Cs∠ Na∠ Be∠ Li∠ Ba
4. Li∠ Cs∠ Na∠ Be∠ Ba

I think it's the third option, but the Beryllium and Barium are messing me up! And a brief explanation too. Thanks!

Answers

Answer:

Be∠ Li∠ Ba∠ Na∠ Cs

Explanation:

Beryllium does not react with water. It is the only alkaline earth metal that does not react with water because of its small size and high ionization energy. Beryllium differs considerably from other members of group two, its compounds when anhydrous show a considerable degree of covalent character.

As the atomic number of the group two elements increases, their ionization energies decreases and their electrode potentials become more negative hence their reactivity increases down the group. This implies that barium will have a very negative electrode potential comparable to that of the alkali metals, hence it reacts considerably with water.

The reactivity of alkali metals with water increases down the group. Lithium reacts quietly with water, sodium and potassium react with water with increasing vigour while rubidium and cesium react with water with exceptional violence.

This little explanation, is the reason behind the option chosen as the answer.

What is the pressure in millimeters of mercury of 0.0150 mol of helium gas with a volume of 213 mL at 50. C? (Hint: You must convert each quantity into the correct units (L, atm, mol and K) before substituting into the ideal gas law.)

Answers

Explanation:

0.08206 L atm mol-1K-1

pv=nRT

Px213 x10^-³ = 0.0150 x 0.08206 x 323

px213 x10^-³ = 0.398

p = 0.398/213 x10^-³

p = 1.87 x 10^-6atm

p = 0.0014mmHg

please brainliest

30. A. An organic compound - A (C4H80) forms phenyl
hydrazone with phenyl hydrazine and reduces Fehling's
mpt any two questions:
solution. It has negative iodoform test. Identify the
organic compound A.​

Answers

Answer:

Methyl ethyl ketone

Explanation:

Compound 'A' forms phenyl hydrazone, so it must be a carbonyl compound.  Since it also gives a negative iodoform test, so it can't be an aldehyde.  

'A' on reduction gives propane. So, it must be butanone.  Ketone reacts with phenyl hydrazine to form phenyl hydrazone but gives a negative iodoform test.

Thus, the correct answer is - Methyl ethyl ketone

Hypochlorous acid is formed in situ by reaction of aq. sodium hypochlorite solution with acetic acid. Draw balanced chemical equations to show the formation of hypochlorous acid and protonated hypochlorous acid.

Answers

Answer:

NaClO + CH₃COOH ----> HClO + CH3CO- + Na

Explanation:

This reaction occurs between the combination of a salt and an acid, that is, an oxide-reduction reaction

Which correctly lists three characteristics of minerals?
solid, crystal structure, definite chemical composition
organic, crystal structure, definite chemical composition
human-made, solid, organic
crystal structure, definite chemical composition, human-made

Answers

Answer:a

Explanation:

The three characteristics of minerals are that they are solid, have definite crystal structure and definite chemical composition.

What are minerals?

Minerals are defined as a chemical compound  which has a well -defined composition and possesses a specific crystal structure.It occurs naturally in the pure form.

If a compound occurs naturally in different crystal structure then each structure is considered as a different mineral.The chemical composition of a mineral varies  depending on the presence of small impurities which are present in small quantities.

Some minerals  can have variable proportions of two or more chemical elements  which occupy equivalent position in the crystal structure.It may also have variable composition which is split into separate species.

Physical properties of minerals include color,streak, luster,specific gravity  and cleavage.

Learn more about minerals ,here:

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Identify the structure of S (molecular formula C7H14O2). Compound S the odor of rum, (1H NMR data (ppm) at 0.93 (doublet, 6 H), 1.15 (triplet, 3 H), 1.91 (multiplet, 1 H), 2.33 (quartet, 2 H), and 3.86 (doublet, 2 H) ppm.Compound S:_______.

Answers

Answer:

Following are the answer to this question:

Explanation:

The structure of the S molecular formula [tex]C_7H_{14}O_2[/tex] defined in the attachment file.

Please find the attachment file.

A(n) _____ reaction occurs when an acid and a base are present in the same solution.

Answers

Answer:

The answer is Neutralization reaction

It occurs when an acid and a base are present in the same solution and react to form salt and water only

Hope this helps you

Identify the correct structure of 5-bromo-4-isopropylheptanoic acid.

Answers

Answer:

See attached picture.

Explanation:

Hello,

In this case, given the IUPAC name, we can infer we have a seven-carbon carboxylic acid that has a bromine at the fifth carbon, an isopropyl at the fourth carbon and the carboxyl functional group (COOH) at the first carbon, thus, on the attached document, you will find the correct structure.

Best regards.

How much heat will be absorbed by a 63.1 g piece of aluminum (specific heat = 0.930 J/g・°C) as it changes temperature from 23.0°C to 67.0°C?

Answers

Answer:

[tex]Q=2582J=2.58kJ[/tex]

Explanation:

Hello,

In this case, for us to compute the absorbed heat, we apply the following equation:

[tex]Q=m_{Al}Cp_{Al}(T_2-T_1)[/tex]

Whereas we use the mass, specific heat and temperature change for the piece of aluminium, thus, we obtain:

[tex]Q=63.1g*0.930\frac{J}{g*\°C}*(67.0-23.0)\°C\\ \\Q=2582J=2.58kJ[/tex]

It is positive as the heat is entering, therefore the temperature raises.

Best regards.

The activation energy for the decomposition of HI is 183 kJ/mol. At 573 K, the rate constant was measured to be 2.91 x 10^{-6} M/s. At what temperature in Kelvin does the reaction have a rate constant of 0.0760 M/s

Answers

Answer:

[tex]T_2=453.05K[/tex]

Explanation:

Hello,

In this case, the temperature-variable Arrhenius equation is written as:

[tex]\frac{k(T_2)}{k(T_1)}=exp(\frac{Ea}{R}(\frac{1}{T_2}-\frac{1}{T_1} ))[/tex]

Now, for us to solve for the temperature by which the reaction rate constant is 0.0760M/s we proceed as shown below:

[tex]ln(\frac{k(T_2)}{k(T_1)})=\frac{Ea}{R}(\frac{1}{T_2}-\frac{1}{T_1} )\\ln(\frac{0.0760M/s}{0.00000291M/s} )=\frac{183000J/mol}{8.314J/(mol*K)} *(\frac{1}{T_2} -\frac{1}{573K} )\\\frac{1}{T_2} -\frac{1}{573K} =\frac{10.17}{22011.06K^{-1}} \\\\\frac{1}{T_2}=4.62x10^{-4}K^{-1}+\frac{1}{573K}\\\\\frac{1}{T_2}=2.21x10^{-3}K^{-1}\\\\T_2=453.05K[/tex]

Regards.

Determine the the mass of one molecule of hydrogen sulfide gas.

Answers

Answer:

the molecular mass of hydrogen sulphide, which contains two atoms of hydrogen and one atom of sulphur is = 2 — 1 + 1 — 32 = 34 a.m.u.

plllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllls help me

Answers

Answer:

Explanation:

equilibrium constant

Kc = [ C ]² / [ A ] [ B ]

= .5² / .2  x 3

= .4167

Let moles of A to be added be n

concentration of A unreacted becomes .2 + n M

increase of product C by .2 M will require use of A  and B be .1 M

So unreacted A = .2 + n - .1 = n + .1

Kc = [ C ]² / [ A ] [ B ]

.4167 = .7² / ( n + .1 ) ( 3 - .1 )

n + .1 = .4

n = . 3 moles .

So .3 moles of A to be added .

Perform the following
mathematical operation, and
report the answer to the correct
number of significant figures.
5.446 x 0.14156

Answers

0.7709 is the answer; there are 4 significant figures.

The activation energy of the uncatalyzed reaction is about 3.98 times that of the catalyzed reaction with activation energy of 4.6 kJ. The uncatalyzed reaction has such a large activation energy that its rate is extremely slow. What is the activation energy for the uncatalyzed reaction

Answers

Answer:

18.308 KJ

Explanation:

From the given above, we obtained the following:

Activation energy for the catalyzed reaction = 4.6 kJ.

Activation energy for the uncatalyzed reaction =..?

Now, a careful observation of the question revealed that the activation energy of the uncatalyzed reaction is about 3.98 times that of the catalyzed reaction.

With this vital information, we can thus, calculate the activation energy of the uncatalyzed reaction as follow:

Activation energy for the uncatalyzed reaction = 3.98 times that of the catalyzed reaction.

Activation energy for the uncatalyzed reaction = 3.98 x 4.6 kJ = 18.308 KJ

Therefore, the activation energy of the uncatalyzed reaction is 18.308 KJ.

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