Answer:
Water isn't combustible. There is nothing you can add to it other than gasoline that will make it even remotely combustible. Now, through electrolysis, it can be broken down into hydrogen and oxygen, which could be burned in an internal combustion engine.
because the water in itself does not produce energy
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
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.
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.
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
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
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:
The constant pressure molar heat capacity of argon, C_{p,m}C
p,m
, is
20.79\text{ J K}^{-1}\text{ mol}^{-1}20.79 J K
−1
mol
−1
at 298\text{ K}298 K. What
will be the value of the constant volume molar heat capacity of argon,
C_{V,m}C
V,m
, at this temperature?
Answer:
Constant-volume molar heat capacity of argon is 12.47 J K ⁻¹mol⁻¹
Explanation:
Argon is a monoatomic gas that behaves as an ideal gas at 298K.
Using the first law of thermodinamics you can obtain:
Work, Q, for constant pressure molar heat capacity,CP:
CP = (5/2)R
For constant-volume molar heat capacity,CV:
CV = (3/2)R
That means:
2CP/5 = 2CV/3
3/5 = CV / CP
As CP of Argon is 20.79 J K ⁻¹mol⁻¹, CV will be:
3/5 = CV / CP
3/5 = CV / 20.79 J K ⁻¹mol⁻¹
12.47 J K ⁻¹mol⁻¹ = CV
Constant-volume molar heat capacity of argon is 12.47 J K ⁻¹mol⁻¹Do you think you could go a week without causing any chemical reactions?
yes yes yes yes
yes
yes
yes
yes
yes
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?
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 requireIdentify 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:_______.
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.
Identify the correct structure of 5-bromo-4-isopropylheptanoic acid.
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.
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
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.
A(n) _____ reaction occurs when an acid and a base are present in the same solution.
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
What is the standard cell potential for the spontaneous voltaic cell formed from the given half-reactions
Answer:
because it is
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
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
2) Which type movement do pivot joints allow?
What is the main side reaction that competes with elimination when a primary alkyl halide is treated with alcoholic potassium hydroxide, and why does this reaction compete with elimination of a primary alkyl halide but not a tertiary alkyl halide
Answer:
The main competing reaction when a primary alkyl halide is treated with alcoholic potassium hydroxide is SN2 substitution.
Explanation:
The relative percentage of products of the reaction between an alkyl halide and alcoholic potassium hydroxide generally depends on the structure of the primary alkylhalide. The attacking nucleophile/base in this reaction is the alkoxide ion. Substitution by SN2 mechanism is a major competing reaction in the elimination reaction intended.
A more branched alkyl halide will yield an alkene product due to steric hindrance, similarly, a good nucleophile such as the alkoxide ion may favour SN2 substitution over the intended elimination (E2) reaction.
Both SN2 and E2 are concerted reaction mechanisms. They do not depend on the formation of a carbocation intermediate. Primary alkyl halides generally experience less steric hindrance in the transition state and do not form stable carbocations hence they cannot undergo E1 or SN1 reactions.
SN2 substitution cannot occur in a tertiary alkyl halides because the stability of tertiary carbocations favours the formation of a carbocation intermediate. The formation of this carbocation intermediate will lead to an SN1 or E1 mechanism. SN2 reactions is never observed for a tertiary alkyl halide due to steric crowding of the transition state. Also, with strong bases such as the alkoxide ion, elimination becomes the main reaction of tertiary alkyl halides.
Select the oxidation reduction reactions??
Answer:
Explanation:
1 ) Cl₂ + ZnBr₂ = ZnCl₂ + Br₂
In this reaction , oxidation number of Cl decreases from 0 to -1 so it is reduced and oxidation number of Br increases from -1 to 0 so it is oxidised . Hence this reaction is oxidation - reduction reaction .
2 )
Pb( ClO₄)₂ + 2KI = PbI₂ + 2KClO₄
In this reaction oxidation number of none is changing so it is not an oxidation - reduction reaction.
3 )
CaCO₃ = CaO + CO₂
In this reaction also oxidation number of none is changing so it is not an oxidation - reduction reaction.
So only first reaction is oxidation - reduction reaction.
2nd option is correct.
Draw the structure 2 butylbutane
Answer:
please look at the picture below.
Explanation:
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
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
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?
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.
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.
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
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.
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
plllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllls help me
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 .
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.
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
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
Answer:
Hey there!
CS2) Carbon Disulfide.
PBr3) Phosphorus Tribromide
NO) Nitric Oxide
CF4) Carbon Tetrafluoride
P2O5) Phosphorus Pentoxide
Let me know if this helps :)
2.
Name the following compounds:
a. Rb20
Answer:
Rubidium oxide
Explanation:
Determine the the mass of one molecule of hydrogen sulfide gas.
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.
Which substance is a base? HCOOH RbOH H2CO3 NaNO3
Answer:
RbOH
Explanation:
For this question, we have to remember what is the definition of a base. A base is a compound that has the ability to produce hydroxyl ions [tex]OH^-[/tex], so:
[tex]AOH~->~A^+~+~OH^-[/tex]
With this in mind we can write the reaction for each substance:
[tex]HCOOH~->~HCOO^-~+~H^+[/tex]
[tex]RbOH~->~Rb^+~+~OH^-[/tex]
[tex]H_2CO_3~->~CO_3^-^2~+~2H^+[/tex]
[tex]NaNO_3~->~Na^+~+~NO_3^-[/tex]
The only compound that fits with the definition is [tex]RbOH[/tex], so this is our base.
I hope it helps!
RbOH or rubidium hydroxide is a base.
• RbOH or rubidium hydroxide is the inorganic compound.
• It comprises hydroxide anions and an equal number of rubidium cations.
• Rubidium hydroxide, like other strong bases is highly corrosive.
• The formation of rubidium hydroxide takes place when the metal rubidium reacts with water.
• The bases refers to the substance, which gets dissociate in an aqueous solution to produce OH- or hydroxide ions.
• Rubidium hydroxide is a base, when it is dissolved in water it give rise to an alkali, when reacts with acid it generates a rubidium salt.
Thus, RbOH is a base.
To know more about:
https://brainly.com/question/18616956
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?
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
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)
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
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
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 moleBeing:
Fe: 55.85 g/moleCu: 63.54 g/moleS: 32 g/moleO: 16 g/molethe 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/moleThen, 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 gThen 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.
Suppose that you add 26.7 g of an unknown molecular compound to 0.250 kg of benzene, which has a K f of 5.12 oC/m. With the added solute, you find that there is a freezing point depression of 2.74 oC compared to pure benzene. What is the molar mass of the unknown compound
Answer: The molar mass of the unknown compound is 200 g/mol
Explanation:
Depression in freezing point is given by:
[tex]\Delta T_f=i\times K_f\times m[/tex]
[tex]\Delta T_f=2.74^0C[/tex] = Depression in freezing point
i= vant hoff factor = 1 (for molecular compound)
[tex]K_f[/tex] = freezing point constant = [tex]5.12^0C/m[/tex]
m= molality
[tex]\Delta T_f=i\times K_f\times \frac{\text{mass of solute}}{\text{molar mass of solute}\times \text{weight of solvent in kg}}[/tex]
Weight of solvent (benzene)= 0.250 kg
Molar mass of solute = M g/mol
Mass of solute = 26.7 g
[tex]2.74^0C=1\times 5.12\times \frac{26.7g}{Mg/mol\times 0.250kg}[/tex]
[tex]M=200g/mol[/tex]
Thus the molar mass of the unknown compound is 200 g/mol
The molar mass of an unknown solute compound in the solution has been 199.626 g/mol.
With the addition of the solute to the solution, there has been a depression in the freezing point. The depression in the freezing point can be expressed as:
Depression in freezing point = Van't Hoff factor × Freezing point constant × molality
The molality can be defined as the moles of solute per kg solvent
Molaity = [tex]\rm \dfrac{Mass\;of\;solute\;(g)}{Molecular\;mass\;of\;solute}\;\times\;\dfrac{1}{Mass\;of\;solvent\;(kg)}[/tex]
The depression in freezing point can be given as:
Depression in freezing point = Van't Hoff factor × Freezing point constant × [tex]\rm \dfrac{Mass\;of\;solute\;(g)}{Molecular\;mass\;of\;solute}\;\times\;\dfrac{1}{Mass\;of\;solvent\;(kg)}[/tex] ......(i)
Given, the depression in freezing point = 2.74 [tex]\rm ^\circ C[/tex]
Van't Hoff factor = 1 (Molecular compound)
Freezing point constant (Kf) = 5.12 [tex]\rm ^\circ C[/tex]/m
Mass of solute = 26.7 g
Mass of solvent = 0.250 kg
Substituting the values in equation (i):
2.74 [tex]\rm ^\circ C[/tex] = 1 × 5.12
[tex]\rm \dfrac{2.74}{5.12}[/tex] = [tex]\rm \dfrac{1}{Molecular\;mass\;of\;solute}\;\times\;\dfrac{26.7}{0.250\;kg}[/tex]
0.535 = [tex]\rm \dfrac{1}{Molecular\;mass\;of\;solute}\;\times\;106.8[/tex]
Molecular mass of solute = [tex]\rm \dfrac{106.8}{0.535}[/tex] g/mol
Molecular mass of solute = 199.626 g/mol
The molar mass of an unknown solute compound in the solution has been 199.626 g/mol.
For more information about the molality of the compound, refer to the link:
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