Answer:
3.62×10³ L/mol
Explanation:
Beer-Lambert law relates the absorbance of a sample and its concentration. Its formula is:
A = ε×C×l
Where A is absorbance of the sample, ε is molar absorptivity (A constant f each sample), C its concentration and l is path length
Now, the formula obtained was:
y = (3.62×10³ L/mol) x
Where Y ia absorbance = A, x its concentration = C and 1cm is path length.
You can write:
A = (3.62×10³ L/mol)×C×l
That means, molar absorptivity of your sample under the meaured conditions is:
3.62×10³ L/mol
A multistep reaction can only occur as fast as its slowest step. Therefore, it is the rate law of the slow step that determines the rate law for the overall reaction. Consider the following multistep reaction:
A+B ----- AB(slow)
A+AB-----A2B(fast
....................................................
2A+B ----- A2B(overall)
Based on this mechanism, determine the rate law for the overall reaction.
a) rate = kA2BAB
b) rate = kAB
c) rate = kAAB
d) rate = kA2B
Answer:
b) rate = kAB.
Explanation:
Hello,
In this case, considering the given statement, we can notice that the rate law of the overall reaction will be determined for the slowest step, that is:
[tex]A+B \rightarrow AB\ \ (slow)[/tex]
In such a way, we can infer that the rate law will contain both the concentration of A and B to the first power both, since their stoichiometric coefficients in the chemical equation are both one:
[tex]rate=k[A][B][/tex]
Thereby, answer is b) rate = kAB, that should be better rate = k[A][B] by expressing the concentrations.
Best regards.
Stote 4 ways in which excesine alcohol conscuption is
harmful to humans
Answer:
An addiction could occur, maybe an overdose?, this could lead to death and maybe you would do unreasonable things which could get you fined or arrested.
Explanation:
Answer:
Excessive alcohol is harmful because you could get addicted.Alcohol can affect your nervous system.Your sugar levels will not be good.Parts of your body and organs will become inflamed.You can get a larger amount of muscle cramps.Also you will not be able to get enough vitamins in your body.Accidents that lead to deaths could occur.You would do crazy actions with things such as theft or breaking into a house which could get you fined or arrested.Too much alcohol can lead to high blood pressure, disease and even strokes.You can have birth defectsWith excessive alcohol you can get osteoporosis.You can also get your immune system weakened.Finally, alcohol can lead to cancer.Hope this helped,
Kavitha
A compound is found to contain 18.28 % phosphorus , 18.93 % sulfur , and 62.78 % chlorine by mass. To answer the question, enter the elements in the order presented above. QUESTION 1: The empirical formula for this compound is . QUESTION 2: The molar mass for this compound is 169.4 g/mol. The molecular formula for this compound is
Answer:
1. EF = PSCl₃; 2. MF = PSCl₃
Explanation:
1. Empirical formula
The empirical formula is the simplest whole-number ratio of atoms in a compound.
The ratio of atoms is the same as the ratio of moles.
So, our first job is to calculate the molar ratio of P:S:Cl.
Assume 100 g of the compound.
(a) Calculate the mass of each element.
Then we have 18.28 g P, 18.93 g S, and 67.28 g Cl.
(b) Calculate the moles of each element
[tex]\text{Moles of P} = \text{18.28 g C} \times \dfrac{\text{1 mol P}}{\text{30.97 g P}} = \text{0.5902 mol P}\\\\\text{Moles of S} = \text{18.93 g S} \times \dfrac{\text{1 mol S}}{\text{32.06 g S }} = \text{0.5905 mol S}\\\\\text{Moles of Cl} = \text{62.78 g Cl} \times \dfrac{\text{1 mol Cl}}{\text{35.45 g Cl }} = \text{1.771 mol Cl}[/tex]
(c) Calculate the molar ratio of the elements
Divide each number by the smallest number of moles
P:S:Cl = 0.5902:0.5905:1.898 = 1:1.000:3.000 ≈ 1:1:3
(d) Write the empirical formula
EF = PSCl₃
The empirical formula for this compound is PSCl₃.
2. Molecular formula
(a) Calculate the ratio of the molecular and empirical formula masses
n = (169.4 u)/(169.40 u) = 1.000 ≈ 1
(b) Calculate the molecular formula
MF = (EF)ₙ = (EF)₁ = PSCl₃
The molecular formula for this compound is PSCl₃.
what is the IUPAC name of KNO3
Which of the following reagents should be used to convert to Question 2 options: A) Na, NH3 B) H2, Pt C) H2, Lindlar's catalyst D) HgSO4, H2O
Answer:
A
Explanation:
If we intend to achieve the anti addition of Hex-3-yne to yield (E) Hex-3-ene, the we must use Na/NH3. The first step of the reaction involves the transfer of an electron from sodium to the alkene; this yields a radical anion. Strong electron replusion ensues between the single electron and the lone pair on the carbon. This now forces the both to be found at a trans position to each other and this is the basis of the stereochemistry of the product.
Secondly, the radical anion abstracts a proton from ammonia. Another sodium atom transfers an electron leading to the formation of a vinyl carbanion, the alkyl groups are now trans to each other.
This carbanion now abstracts a proton from ammonia and the final product is formed.
1. Rank the following compounds in order of decreasing acid strength using periodic trends. Rank the acids from strongest to weakest.
A. HCl
B. H2S
C. HBr
D. BH3
2. Without consulting the table of acid-dissociation constants, match the following acids to the given Ka1 values.
1. H2S
2. H2SO3
3. H2SO4
A. Kal = 1.7 x 10^-7
B. Kal = 1.7 x 10^-2
C. Kal = very large
Answer:
ESCALAS MAYORES (D, E, G, A, B) Porfavor necesito ayuda,te lo agradecería muchísimo!!
Es urgente
Sample gas has a volume of 3.40 L at 10°C what will be its volume at 100°C pressure remaining constant
Answer:
V2 = 4.48L
Explanation:
using charles law
V1/T1=V2/T2
3.4/283=V2/373
0.012=V2/373
V2= 0.012 x 373
V2 = 4.48L
The following balanced equation describes the reduction of iron(III) oxide to molten iron within a blast furnace: Fe2O3(s) + 3CO(g) ---> 2Fe(l) + 3CO2(g) Steve inserts 450. g of iron(III) oxide and 260. g of carbon monoxide into the blast furnace. After cooling the pure liquid iron, Steve determines that he has produced 288g of iron ingots. How much of the excess reactant (in grams) is left over after the theoretical yield of liquid iron is produced?
Answer:
Amount of excess Carbon (ii) oxide left over = 23.75 g
Explanation:
Equation of the reaction: Fe₂O₃ + 3CO ----> 2Fe + 3CO₂
Molar mass of Fe₂O₃ = 160 g/mol;
Molar mass of Carbon (ii) oxide = 28 g/mol
From the equation of reaction, 1 mole of Fe₂O₃ reacts with 3 moles of carbon (ii) oxide; i.e. 160 g of iron (iii) oxide reacts with 84 g (3 * 28 g) of carbon (ii) oxide
450 g of Fe₂O₃ will react with 450 * 84/180) g of carbon (ii) oxide = 236..25 g of carbon (ii) oxide
Therefore the excess reactant is carbon (ii) oxide.
Amount of excess Carbon (ii) oxide left over = 260 - 236.25
Amount of excess Carbon (ii) oxide left over = 23.75 g
When (R)-3-bromo-2,3-dimethylpentane is treated with sodium hydroxide, four different alkenes are formed. Draw all four products, and rank them in terms of stability.
Answer:
Most stable: 2,3-dimethylpent-2-ene > (E)-3,4-dimethylpent-2-ene > (Z)-3,4-dimethylpent-2-ene > 2-methyl-3-methylenepentane : Least stable
Explanation:
Treatment of NaOH with (R)-3-bromo-2,3-dimethylpentane results in the elimination of HBr. Each H atoms present on each [tex]\beta[/tex]-carbon atoms can be eliminated result in the formation of four possible products: (1) 2,3-dimethylpent-2-ene, (2) (E)-3,4-dimethylpent-2-ene, (3) (Z)-3,4-dimethylpent-2-ene and (4) 2-methyl-3-methylenepentane.
The stability of these alkenes depends on the number of hyperconjugative H atoms present with respect to the double bond. In accordance with this, 2,3-dimethylpent-2-ene is the most stable alkene (11-hyperconjugative H atoms). Then, 3,4-dimethylpent-2-ene is the second most stable alkene (7-hyperconjugative H atoms). Among (E)-3,4-dimethylpent-2-ene and (Z)-3,4-dimethylpent-2-ene, (E)-3,4-dimethylpent-2-ene is more stable due to it's less sterically hindered structure. 2-methyl-3-methylenepentane is the least stable alkene (3-hyperconjugative H atoms).
So, decreasing order of stability of alkenes from most stable to least stable:
2,3-dimethylpent-2-ene > (E)-3,4-dimethylpent-2-ene > (Z)-3,4-dimethylpent-2-ene > 2-methyl-3-methylenepentane
Three bromo As just a result of the nucleophilic substitution mechanism, 2,4 dimethylpentane generates a racemic mix containing both the r and s forms of the molecule.
In the first, sluggish phase, a two-degree cation with a positive charge is produced at carbon 3, and then it undergoes rearrangement to a 3-degree carbocation at carbon 2 either from the side. So because carbocation only has 6 electrons in its outermost shell, it is sp2 hybridized and thus planar in structure. In step two then occurs, as well as the attack of the -OH group, as the -OH (hydroxyl) group can strike from either side (top or bottom), leading to the formation of a racemic mixture of 2,4 dimethyl pentane-2-ol.This quantity of alpha-hydrogens in an alkyl group can be used to determine its stability. The greater the number of alpha-hydrogens inside an alkene, the overall larger the number of hyperconjugated structures, and thus the greater the stability. Due to the obvious symmetrical structure of a trans-isomer, whenever the amount of alpha-hydrogens is the same, the trans-alkene isomer is much more stable than that of the cis-alkene isomer.Please find the attached file.
Learn more about the 3-Bromo-2,3-dimethylpentane:
brainly.com/question/7139334
most vegetables substantially diminish in quality in as little as days
Answer:
As little as 2 days
Hope this is correct
HAVE A GOOD DAY!
For carbon: What is the effective nuclear charge? In which orbitals do the valence electrons reside? For silicon: What is the effective nuclear charge? In which orbitals do the valence electrons reside?
Answer:
For carbon, the effective nuclear charge is 3.25 and the valence electrons will reside in the orbitals 2s^2 and 2p^2
For silicon, the effective nuclear charge is 4.15 and its valance electrons will reside in the orbitals 3s^2 and 3p^2
Explanation:
Carbon
The effective nuclear charge of carbon is 3.25
To get the orbitals in which it’s valence electron reside, let’s write the electronic configuration
The atomic number of carbon is 6
So the configuration will be;
1s^2 2s^2 2p^2
So the valence electrons will reside in the orbitals 2s^2 and 2p^2
For silicon;
It’s effective nuclear charge is +4.15
The electronic configuration of silicon with atomic number 14 is;
1s^2 2s^2 2p^6 3s^2 3p^2
So the valence electrons will reside in the orbitals 3s^2 and 3p^2
When a 2.75g sample of liquid octane (C8H18) is burned in a bomb calorimeter, the temperature of the calorimeter rises from 22.0 °C to 41.5 °C. The heat capacity of the calorimeter, measured in a separate experiment, is 6.18 kJ/°C. Determine the ΔE for octane combustion in units of kJ/mol octane.
Answer:
THE HEAT OF COMBUSTION IS 4995.69 kJ/mol OF OCTANE.
Explanation:
Heat capacity = 6.18 kJ/C
Temperature change = 41.5 C - 22.0 C = 19.5 C
Heat required to raise the temperature by 19.5 °C is:
Heat = heat capacity * temperature change
Heat = 6.18 kJ/ C * 19.5 C
heat = 120.51 kJ of heat
120.51 kJ of heat is required to raise the temperature of 2.75 g sample of a liquid octane.
Molar mass of octane = ( 12* 8 + 1 * 18) = 114 g/mol
So therefore, the heat of the reaction per mole of octane will be:
120.51 kJ of heat is required for 2.75 g of octane
x J of heat will be required for 114 g of octane
x J = 120.51kJ * 114 / 2.75
x = 4995.69 kJ of heat per mole.
In conclusion, the heat of the combustion reaction in kJ / mole of octane is 4995.69 kJ/mol
Find the molecular formula of each compound CCl , 189.83 g/mol C3H2N , 156.23 g/mol
Answer:
Explanation:
The given formula is empirical formula
Let the molecular formula of first be
[tex]( CCl )_n[/tex]
molecular weight = n x ( 12 + 35.5 )
= 47.5 n
Given molecular weight = 189.83 so
47.5 n = 189.83
n = 3.99 or 4 approx
Molecular formula =
[tex]( CCl )_4[/tex]
= C₄ Cl₄
Let the molecular formula of second compound be
[tex]( C_3H_2N)_n[/tex]
molecular weight = n x ( 3 x 12 +2+14 )
= 52 n
Given molecular weight = 156.23 so
52 n = 156.23
n = 3.0044 or 3 approx
Molecular formula =
[tex]( C_3H_2N )_3[/tex]
= C₉H₆ N₃
An endothermic reaction proceeds in the forward direction. Which of the following statements will be true if the temperature at which the reaction occurs is decreased?
a. The equilibrium constant increases.
b. The reaction shifts toward the products.
c. The concentrations of the products decrease.
d. The equilibrium constant decreases
Answer:
The correct answers are:
c. The concentrations of the products decrease
d. The equilibrium constant decreases
Explanation:
Changes in temperature shift the equilibrium. In this problem, the reaction is endothermic, so it absorbs heat so heat is considered as a reactant:
Reactants + heat ⇒ Products
If the temperature is decreased, the heat is decreased, so reactants are removed from the reaction at equilibrium. According to Le Chaterlier's principle, the system will try to compensate the produced change. If reactants are removed, the systems will form reactants and the equilibrium will shift toward the left (formation of more reactants). In consequence, the amount of products will be decreased.
Thus, acorrect option is: c. The concentrations of the products decrease.
Since the equilibrium constant is given by the ratio of concentration of products over concentration of reactants, if the concentration of products decrease, the equilibrium constant also decreases. So, another correct option is: d. The equilibrium constant decreases.
For the reaction X + Y → Z, the reaction rate is found to depend only upon the concentration of X. A plot of 1/X verses time gives a straight line. What is the rate law for this reaction?
Answer:
r = k [X]²
Explanation:
X + Y → Z
Generally, the rate of reaction depends on the concentration of reactants. However, the question stated that the rate depends only on reactant X.
The plot of 1/X versus time giving a straight line signifies that this is a second order reaction.
For a second-order reaction, a plot of the inverse of the concentration of a reactant versus time is a straight line with a slope of k.
From this, our rate law is r = k [X]²
This substituent deactivates the benzene ring towards electrophilic substitution but directs the incoming group chiefly to the ortho and para positions.
A) -F
B) -OCH2CH3
C) -CF3
D) -NHCOCH3
E) -NO2
Answer:
F
Explanation:
Halogens may interact with the benzene ring via inductive or resonance effects. Halogens deactivate the benzene ring by inductive effect rather than by resonance effects.
The lone pairs of electrons present on the halogen atoms may be donated to the ring by resonance, but an opposite effect, the inductive pull (-I inductive effect) of the halogen atoms on electrons away from the benzene ring due to the high electro negativity of the halogens leads to a deactivation of the ring towards electrophilic substitution.
Hence inductive electron withdrawal by the halogen atom predominates over electron donation by resonance effect and the benzene ring g is deactivated towards electrophilic substitution at the ortho and para positions.
what bsic difference is between NMR and MS spectroscopic techniques?
Answer:
The Nuclear magnetic resonance is the process this technique does not use radiation.
The ms is an sensitive technology can be a massive number and small sample of the blood.
Explanation:
The Nuclear magnetic resonance we look at the both side of that coin.
The technique provides that fatty acid composition and various including amino acids.
These are contain the complementary these biomarkers, that are suitable for all kinds of studies. there are many types of research:-
(1) A powerful tool metabolic (2) A versatile tool research (3) Quick analysis (4) Low cost analysis.
The MS is an extremely sensitive technology using a very small number of the blood.
(1) Powerful techniques (2) Highly method (3) Large number of metabolites (4)Small sample volume
MS can be fine mapping metabolic pathways to sign analytical strategy.
How many milliliters of 0.0850 M NaOH are required to titrate 25.0 mL of 0.0720 M hydrobromic acid, HBr, to the equivalence point?
Answer:
21.2 mL
Explanation:
Step 1: Write the balanced equation.
NaOH + HBr ⇒ NaBr + H₂O
Step 2: Calculate the reacting moles of HBr
25.0 mL of 0.0720 M hydrobromic acid react.
[tex]0.0250 L \times \frac{0.0720mol}{L} = 1.80 \times 10^{-3} mol[/tex]
Step 3: Calculate the reacting moles of NaOH
The molar ratio of NaOH to HBr is 1:1. The reacting moles of NaOH are 1/1 × 1.80 × 10⁻³ mol = 1.80 × 10⁻³ mol.
Step 4: Calculate the required volume of NaOH
[tex]1.80 \times 10^{-3} mol \times\frac{1,000mL}{0.0850mol} = 21.2 mL[/tex]
BeH2 has no lone pairs of electrons. What's the structure of this molecule?
Answer:
linear
Explanation:
nothing bends off
A student is given an antacid tablet that weighs 5.8400 g. The tablet is crushed and 4.2800 g of the antacid is added to 200. mL of simulated stomach acid. It is allowed to react and then filtered. It is found that 25.00 mL of this partially neutralized stomach acid required 11.6 mL of a NaOH solution to titrate it to a methyl red end point. It takes 29.0 mL of this NaOH solution to neutralize 25.00 mL of the original stomach acid. How much of the original 200. mL of stomach acid (in mL) is neutralized by the 4.2800 g crushed sample of the tablet
Answer:
Explanation:
Given that:
mass of the antacid tablet = 5.8400 g
required mass of the antacid tablet = 4.2800 g was added to 200. mL of simulated stomach acid.
The amount of the original 200. mL of stomach acid (in mL) needed to neutralize the 4.2800 g crushed sample of the tablet can be calculated as:
= 11.6 mL of NaOH × 25.00 mL /29.0 mL NaOH
= 10.00 mL original stomach acid
Now; since it requires 11.6 mL of NaOH o neutralize 10.00 mL of original acid , then:
the antacid neutralized = 200 mL - 10.00 mL
the antacid neutralized = 190.00 mL
What was Ernest Rutherford experiment
what is radiologist
Radiologists are medical doctors that treat injuries using medical imaging (radiology)
Answer:
a person who uses X-rays or other high-energy radiation, especially a doctor specializing in radiology.
Explanation:
Ethane burns according to the following reaction: 2C2H6 + 7O2 → 4CO2 + 6H2O + 2.86 x 103 kJ How much heat can be generated when 185 grams of oxygen gas (MW = 32 g/mol) are consumed?
Answer:
question is not clear please send clear question
what mass of calcium nitrate can be prepared by the reaction of 18.9 grams of nitric acid with 7.4 grams of calcium hydroxide
Answer:
16.4 grams of calcium nitrate can be prepared by the reaction of 18.9 grams of nitric acid with 7.4 grams of calcium hydroxide
Explanation:
The balanced reaction is:
Ca(OH)₂ + 2 HNO₃ → Ca(NO₃)₂ + 2 H₂O
First, you must determine the limiting reagent. The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.
To determine the limiting reagent, it is possible to use the reaction stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction):
Ca(OH)₂: 1 moleHNO₃: 2 molesCa(NO₃)₂: 1 mole H₂O : 2 molesBeing:
Ca: 40 g/moleO: 16 g/moleH: 1 g/moleN: 14 g/moleThen, the molar mass of the compounds participating in the reaction is:
Ca(OH)₂: 40 g/mole + 2*(16 g/mole + 1 g/mole)= 74 g/moleHNO₃: 1 g/mole + 14 g/mole + 3*16 g/mole= 63 g/moleCa(NO₃)₂: 40 g/mole + 2*(14 g/mole + 3*16 g/mole)= 164 g/moleH₂O : 2*1 g/mole + 16 g/mole= 18 g/moleThen, by stoichiometry of the reaction the following amounts of reagents and products participate:
Ca(OH)₂: 1 mole* 74 g/mole= 74 gHNO₃: 2 moles* 63 g/mole= 126 gCa(NO₃)₂: 1 mole* 164 g/mole= 164 gH₂O : 2 moles* 18 g/mole= 36 gThen apply the following rule of three: if 126 grams of nitric acid reacts with 74 grams of calcium hydroxide, 18.9 grams of nitric acid with how much mass of calcium hydroxide does it react?
[tex]mass of calcium hydroxide=\frac{18.9 grams of nitric acid*74 grams of calcium hydroxide}{126 grams of nitric acid}[/tex]
mass of calcium hydroxide= 11.1 grams
But 11.1 grams of calcium hydroxide are not available, 7.4 grams are available. Since you have less mass than you need to react with 18.9 grams of nitric acid, calcium hydroxide will be the limiting reagent.
Then, it is possible to determine the amount of mass of calcium nitrate produced by another rule of three: if 164 grams of calcium nitrate are formed by stoichiometry from 74 grams of calcium hydroxide, how much mass of calcium nitrate will form from 7.4 grams of calcium hydroxide?
[tex]mass of calcium nitrate=\frac{7.4 grams of calcium hydroxide*164 grams of calcium nitrate}{74 grams of calcium hydroxide}[/tex]
mass of calcium nitrate= 16.4 grams
16.4 grams of calcium nitrate can be prepared by the reaction of 18.9 grams of nitric acid with 7.4 grams of calcium hydroxide
Write the condensed formula from left to right, starting with (CH3)x where x is a number.
Complete question:
Write the condensed formula from left to right, starting with (CH3)x where x is a number.
See attached image for the structure formula of the compound
Answer:
(CH₃)₂CHC(CH₃)₃ named as 2,2,3-Trimethylbutane
Explanation:
If we number the longest chain of the carbon starting from the left, we will observe that there are four carbons in the straight chain as shown in the image.
Starting from first carbon from the left of the carbon chain, at carbon number number 2, there two alkyl group, that is two methyl (CH3 is two). Also at carbon number 3, there are three alkyl group, that is three methyl (CH3 is three).
The condensed formula will be written as;
(CH₃)₂CHC(CH₃)₃
This compound is named as 2,2,3-Trimethylbutane, an isomer of Heptane
If 50 ml of 1.00 M of H2SO4 and 50 ml of 2.0 M KOH are mixed what is the concentration of the resulting solutes?
Answer:
0.5 M
Explanation:
First, let us look at the balanced equation of the reaction.
[tex]H_2SO_4 + 2KOH --> K_2SO_4 + 2H_2O[/tex]
The solute formed is [tex]K_2SO_4[/tex].
Recall that: mole = molarity x volume
Hence,
50 ml, 1.00 M H2SO4 = 0.05 x 1 = 0.05 mole
50 ml, 2.0 M KOH = 0.05 x 2 = 0.1 mole
From the equation
1 mole of H2SO4 reacts with 2 moles of KOH to give 1 mole of K2SO4.
Hence,
0.05 mole H2SO4 reacting with 0.1 mole KOH will give 0.05 mole [tex]K_2SO_4[/tex].
Also recall that: concentration = mole/volume
Total volume of resulting solution = 50 ml + 50 ml = 100 ml or 0.1 liter
Concentration of [tex]K_2SO_4[/tex] = mole of [tex]K_2SO_4[/tex]/volume of resulting solution
= 0.05/0.1 = 0.5 M
The concentration of the resulting solute = 0.5 M
Balance the following redox reaction in acidic solution: H+(aq)+Zn(s)→H2(g)+Zn2+(aq) Express your answer as a chemical equation. Identify all of the phases in your answer. nothing
Answer:
The balanced equation is: Zn(s) + 2H⁺(aq) → Zn²⁺(aq) + H₂(g)
Explanation:
Zn(s) is a simple substance (its oxidation number is zero) and it is oxidized to Zn²⁺. It loses two electrons, so the half reaction is the following:
Zn(s) → Zn²⁺(aq) + 2 e- (oxidation reaction)
Hydrogen ion (H⁺) is reduced to hydrogen gas (H₂). The oxidation number is decreased from +1 to 0 (because H₂ is a simple substance). H⁺ gains 1 electron per H atom, so the half reaction is the following:
2H⁺(aq) + 2 e- → H₂(g) (reduction reaction)
We obtain the overall reaction from the addition of the two half reactions. We write the reduction reaction first and then the oxidation reaction, as follows:
2H⁺(aq) + 2 e- → H₂(g)
+
Zn(s) → Zn²⁺(aq) + 2 e-
---------------------------------
Zn(s) + 2H⁺(aq) → Zn²⁺(aq) + H₂(g)
The two electrons at both sides of the equation (2 e-) are canceled. The overall reaction is in acidic solution due to the presence of H⁺ ions. The net charge at both sides is the same : +2, so the mass and the charge are balanced.
The vapor pressure of pure water at 250C is 23.77 torr. What is the vapor pressure of water above a solution that is 1.500 m glucose, C6H12O6?
Answer:
Vapor pressure of water = 23.14torr
Explanation:
When you made a solution, vapor pressure decreases following Raoult's law:
[tex]P_{solution} = X_{solvent} P_{solvent}[/tex]
Where P is vapor pressure and X mole fraction
As vapor pressure of water is 23.77torr we must find the mole fraction of water knowing the solution is 1.500m glucose (That is 1.500 moles of glucose per kg of water = 1000g of water).
1000g of H₂O are, in moles (Molar mass: 18.02g/mol):
1000g H₂O ₓ (1mole / 18.02g) = 55.5 moles of H₂O.
As we know now the solution contains 55.5 moles of water and 1.5 moles of glucose. Thus, mole fraction of water (Solvent) is:
[tex]X_{H_2O} = \frac{55.5molesH_2O}{55.5molesH_2O + 1.5 molesGlucose} = 0.9737[/tex]
Replacing in Raoult's law, pressure of water above the solution is:
[tex]P_{solution} = X_{solvent} P_{solvent}[/tex]
[tex]P_{solution} = 0.9737*23.77torr[/tex]
Vapor pressure of water = 23.14torrDraw the Lewis structure of ethyne (C₂H₂) and then choose the appropriate pair of molecular geometries of the two central atoms. Your answer choice is independent of the orientation of your drawn structure.
A) linear / linear
B) trigonal/pyramidal
C) pyramidal/trigonal
D) trigonal pyramidal/trigonal pyramidal
E) planar / linear
Answer:
A) linear / linear
Explanation:
In this case, we have a triple bond beetween the atoms (See figure 1). If we have this triple bond we will have an Sp hybridization (in both carbons). We have to remember the relationship between the geometry and the hybridization:
-) Sp3 = Tetrahedral
-) Sp2 = Trigonal
-) Sp = Linear
Due to the hybridization, we will have a linear structure between the atoms. The angle between the atoms is 180º (See figure 2).
So, if we have a hybridization Sp for both carbons, we will have a linear geometry in each carbon. Therefore, the answer is A.
Limiting Reagent
1.) A student chose the wrong result of the two calculations of BaSO4, namely, the higher value. What would you expect to happen to the value of the % yield? Explain.
2.) In the process of filtration, what do you think has happened to the excess reagent which has not reacted? Where does it go, and do you think you could recover it, if needed? Explain.
Answer:
a) the percentage yield will exceed 100%
b) the excess reactant is filtered along with the barium sulphate precipitate. It is possible to recover the excess reactant by carefully washing the precipitate with water.
Explanation:
In the precipitation of barium sulphate, the ions in the reactants exchange partners in the product leading to an insoluble product.
In every reaction, there is a limiting reactant whose amount determines the amount of product that can be obtained. The reactant in excess remains in the system even after the reaction is completed and may be recovered alongside the product which leads to a percentage yield above 100%.
If the excess reactant is soluble in water, it can be recovered from the precipitate if needed by washing the precipitate with water.