A compound D with the molecular formula C6H12 is optically inactive but can be resolved into enantiomers. On catalytic hydrogenation, D is converted to E (C6H14) and E is optically inactive. Propose structures for D and E. (Draw a three-dimensional formula for each using dashes and wedges around chiral centers.)
Answers
Answer:
D: CH2=CH-CH(CH3)-CH2-CH3 (R & S enantiomers)
E: CH3-CH2-(CH3)-CH2-CH3
(Please see the figures enclosed )
Explanation:
D is a racemic mixture (R & S) of 3-metyl-pent-1-ene, so it is optically inactive (although each of two enantiomers is optically active, the mixture is optically inactive. The reason is that two enantiomers are present in an equal amount).
E is optically inactive, so its structure has to be symmetric.
Related Questions
What is the atomic mass of AlNO2?
Answers
Answer:
I am not sure, but I think this is the answer 72.987 g/mol
Which option describes a similarity and a difference between isotopes of an element? A. same atomic number; different number of protons B. same number of protons; different atomic number C. same atomic number; different mass number D. same mass number; different atomic number E. same number of neutrons; same number of protons
Answers
Answer:
c
Explanation:
11. Caproic acid, which is responsible for the foul odor of dirty socks, is composed of C, H, and O atoms. Combustion of a 0.225-g sample of this compound produces 0.512 g CO2 and 0.209 g H2O. (a) What is the empirical formula of caproic acid
Answers
Answer:
C3H6O
Explanation:
Step 1:
Data obtained from the question include the following:
Mass of the compound = 0.225g
Mass of CO2 = 0.512g
Mass of H2O = 0.209g
Step 2:
Determination of the masses of carbon, hydrogen and oxygen present in the compound.
This is illustrated below:
For Carbon, C:
Molar mass of CO2 = 12 + (2x16) = 44g/mol
Mass of C in CO2 = 12/44 x 0.512 = 0.1396g
For Hydrogen, H:
Molar mass of H2O = (2x1) + 16 = 18g/mol
Mass of H in H2O = 2/18 x 0.209 = 0.0232g
For Oxygen, O:
Mass of O = 0.225 – (0.1396 + 0.0232)
Mass of O = 0.0622g
Step 3:
Determination of the empirical formula for caprioc acid.
This can be obtain as follow:
C = 0.1396g
H = 0.0232g
O = 0.0622g
Divide by their molar mass
C = 0.1396/12 = 0.0116
H = 0.0232/1 = 0.0232
O = 0.0622/16 = 0.0039
Divide by the smallest
C = 0.0116/0.0039 = 3
H = 0.0232/0.0039 = 6
O = 0.0039/0.0039 = 1
Therefore, the empirical formula for caprioc acid is C3H6O
For the aqueous reaction dihydroxyacetone phosphate↽−−⇀glyceraldehyde−3−phosphate dihydroxyacetone phosphate↽−−⇀glyceraldehyde−3−phosphate the standard change in Gibbs free energy is ΔG°′=7.53 kJ/molΔG°′=7.53 kJ/mol . Calculate ΔGΔG for this reaction at 298 K298 K when [dihydroxyacetone phosphate]=0.100 M[dihydroxyacetone phosphate]=0.100 M and [glyceraldehyde-3-phosphate]=0.00200 M[glyceraldehyde-3-phosphate]=0.00200 M .
Answers
Answer:
ΔG = -2.17 kJ/mol
Explanation:
ΔG of a reaction at any moment could be obtained thus:
ΔG = ΔG° + RT ln Q
Where ΔG° is standard change in free energy of a particular reaction (7.53kJ/mol for the reaction of the problem, R is gas constant (8.314×10⁻³kJ/molK), T is absolute temperature (298K) and Q is reaction quotient of the reaction.
For the reaction:
dihydroxyacetone phosphate ⇄ glyceraldehyde−3−phosphate
Q is defined as:
Q = [glyceraldehyde−3−phosphate] / [dihydroxyacetone phosphate]
Replacing values in ΔG formula:
ΔG = 7.53kJ/mol + 8.314×10⁻³kJ/molK × 298.15K ln [0.00200M] / [0.100M]
ΔG = -2.17 kJ/mol
Aqueous sulfuric acid H2SO4 will react with solid sodium hydroxide NaOH to produce aqueous sodium sulfate Na2SO4 and liquid water H2O. Suppose 62. g of sulfuric acid is mixed with 33.8 g of sodium hydroxide. Calculate the minimum mass of sulfuric acid that could be left over by the chemical reaction. Round your answer to 2 significant digits.
Answers
Answer:
Approximately [tex]21\; \rm g[/tex].
Explanation:
[tex]\rm H_2SO_4[/tex] (a diprotic acid) reacts with [tex]\rm NaOH[/tex] (a monoprotic base) at a one-to-two ratio:
[tex]\rm 2\; NaOH\, (s) + H_2SO_4\, (aq) \to Na_2SO_4\; (aq) + 2\; H_2O\, (l)[/tex].
In other words, if [tex]n(\mathrm{NaOH})[/tex] and [tex]n(\mathrm{H_2SO_4})[/tex] represent the number of moles of the two compounds reacted, then:
[tex]\displaystyle \frac{n(\mathrm{H_2SO_4})}{n(\mathrm{NaOH})} = \frac{1}{2}[/tex].
Look up the relative atomic mass data on a modern periodic table:
[tex]\rm H[/tex]: [tex]1.008[/tex].[tex]\rm S[/tex]: [tex]32.06[/tex].[tex]\rm O[/tex]: [tex]15.999[/tex].[tex]\rm Na[/tex]: [tex]22.990[/tex].Calculate the (molar) formula mass of [tex]\rm H_2SO_4[/tex] and [tex]\rm NaOH[/tex]:
[tex]M(\mathrm{H_2SO_4}) = 2 \times 1.008 + 32.06 + 4 \times 15.999 = 98.072\; \rm g \cdot mol^{-1}[/tex].
[tex]M(\mathrm{NaOH}) = 22.990 + 15.999 + 1.008 = 39.997\; \rm g \cdot mol^{-1}[/tex].
Calculate the number of moles of formula units in that [tex]33.8\; \rm g[/tex] of [tex]\rm NaOH[/tex]:
[tex]\begin{aligned}n(\mathrm{NaOH}) &= \frac{m(\mathrm{NaOH})}{M(\mathrm{NaOH})} \\ &= \frac{33.8\; \rm g}{39.997\; \rm g \cdot mol^{-1}} \approx 0.845\; \rm mol\end{aligned}[/tex].
Apply the ratio [tex]\displaystyle \frac{n(\mathrm{H_2SO_4})}{n(\mathrm{NaOH})} = \frac{1}{2}[/tex] to find the (maximum) number of moles of [tex]\rm H_2SO_4[/tex] that would react with the [tex]33.8\; \rm g[/tex] of [tex]\rm NaOH[/tex]:
[tex]\begin{aligned}n(\mathrm{H_2SO_4}) &= \frac{n(\mathrm{H_2SO_4})}{n(\mathrm{NaOH})} \cdot n(\mathrm{NaOH})\\ &= \frac{1}{2} \times 0.845 \approx 0.4225\; \rm mol\end{aligned}[/tex].
Calculate the mass of that [tex]0.4225\; \rm mol[/tex] of [tex]\rm H_2SO_4[/tex]:
[tex]\begin{aligned}m(\mathrm{H_2SO_4}) &= n(\mathrm{H_2SO_4}) \cdot M(\mathrm{H_2SO_4})\\ &= 0.4225 \; \rm mol \times 98.072\; \rm g \cdot mol^{-1} \approx 41.435\; \rm g \end{aligned}[/tex].
When the maximum amount of [tex]\rm H_2SO_4[/tex] is reacted, the minimum would be in excess. Hence, the minimum mass of
[tex]62\; \rm g - 41.435\; \rm g \approx 21\; \rm g[/tex] (rounded to two significant figures.)
URGENT!! This is timed, PLEASE HELP!
Nitrogen gas can be prepared by passing gaseous ammonia over solid copper (II) oxide at high temperatures, as described by the following balanced equation:
2 NH3(g) + 3 CuO(s) → 1N2(g) + 3 Cu(s) + 3 H2O(g)
How many grams of N2 are formed when 120.51 g of NH3 are reacted with excess CuO?
(Please explain using steps and show the whole process. Make sure the answer is in sig figs)
Answers
Answer:
99.24 gm of nitrogen .
Explanation:
molecular weight of ammonia = 17 , molecular weight of nitrogen = 28.
2 NH₃(g) + 3 CuO(s) → 1N₂(g) + 3 Cu(s) + 3 H₂O(g)
2 x 17 gm 28 gm
( 34 gm )
34 gm of ammonia forms 28 gms of nitrogen
1 gm of ammonia forms 28 / 34 gms of nitrogen
120.51 gn of ammonia forms 28 x 120.51 / 34 gms of nitrogen
28 x 120.51 / 34 gms
= 99.24 gms of nitrogen will be formed .
If 25.8 mL of an AgNO3 solution is needed to precipitate all Cl- ions in a 1570 mg of KCl (forming AgCl), what is the molarity of the AgNO3nsolution?
Answers
Answer:
M=0.816M
Explanation:
Hello,
In this case, we should consider the following reaction:
[tex]AgNO_3+KCl\rightarrow KNO_3+AgCl[/tex]
Thus, by knowing the 1:1 molar ratio of silver nitrate and potassium chloride, we can easily compute the moles of silver nitrate precipitating the 1570 mg of potassium chloride considering its molar mass of 74.5513 g/mol:
[tex]n_{AgNO_3}=1570mgKCl*\frac{1gKCl}{1000mgKCl} *\frac{1molKCl}{74.5513gKCl}*\frac{1molAgNO_3}{1molKCl} \\\\n_{AgNO_3}=0.021molAgNO_3[/tex]
Then, by using the volume of silver nitrate in liters (0.0258 L), we can directly compute the molarity:
[tex]M=\frac{0.021molAgNO_3}{0.0258L}\\ \\M=0.816M[/tex]
Regards.
When you look at an ant up close, using a convex lens, what do you see?
Answers
Answer:
You would be able to see the ants clearly with the unique body parts.
Explanation:
Convex lens is also known as Converging lens. It helps in converging rays of light to become a principal focus which is usually very clear and visible to the eyes. The converging lens when used to view the ant makes the ant appear very visible and the individual is able to see all the unique body parts of the ant. This type of lens is used in individuals who are longsighted.
At 25.0 °C the Henry's Law constant for sulfur hexafluoride (SP) gas in water is 2.4x 10 M/atm Calculate the mass in grams of SFo, gas that can be dissolved in S25. ml. of water at 25.0 C and a SF, partial pressure of 1.90 atm Be sure your answer has the correct number of significant digits.
Answers
Complete Question
The complete question is shown on the first uploaded image
Answer:
The mass is [tex]m = 0.0349 \ g[/tex]
Explanation:
From the question we are told that
The Henry's Law constant is [tex]k = 2.4 *10^{10} M/atm[/tex]
The volume of water is [tex]V = 525 \ ml = 0.525 \ L[/tex]
The partial pressure is [tex]P = 1.90 \ atm[/tex]
The temperature is [tex]T = 25 ^oC[/tex]
Henry's law is mathematically represented as
[tex]C = P * k[/tex]
Where C is the concentration of sulfur hexafluoride(SP)
substituting value
[tex]C = 1.90 * 2.4*10^{-4}[/tex]
[tex]C = 4.56*10^{-4} \ M[/tex]
The number of moles of SP is mathematically represented as
[tex]n = C * V[/tex]
substituting value
[tex]n = 0.525 * 4.56*10^{-4}[/tex]
[tex]n = 2.39 *10^{-4} \ moles[/tex]
The mass of SP that dissolved is
[tex]m = n * Z[/tex]
Where Z is the molar mass of SP which has a constant value of
[tex]Z = 146 g/mole[/tex]
So
[tex]m = 2.394*10^{-4} * 146[/tex]
[tex]m = 0.0349 \ g[/tex]
12.39 g sample of phosphorus (30.97 g/mol) reacts with 52.54 g of chlorine gas, Cl2
(70.91 g/mol) to form only phosphorus trichloride, PC13 (137.33 g/mol). Which is the
limiting reactant?
Answers
Answer:
P is the limiting reagent
Explanation:
P = phosphorus = 30.97g/mol
Cl2 = Chlorine = 70.91g/mol
PCl3 = Phosphorus Trichloride = 137.33g/mol
P + Cl2 = PCl3
Left Side
P = 1
Cl = 2
Right Side
P = 1
Cl = 3
So equation needs to be balanced first
2P + 3Cl = 2PCl3
Left Side
P = 2
Cl = 6
Right Side
P = 2
Cl = 6
That's better.
Ok so we have 12.39g of P so we have 0.4 moles of it
We then have 52.54g of Cl so we have 0.74 moles of it
For every P we need 1.5 Cl so we have an excess of Cl
Consider the following reaction. I– 2 H2O2 (l) 2 H2O (l) + O2 (g) A solution contains 15 mL 0.1 M KI, 15 mL of DI water and 5 mL of 3% H2O2. After the decomposition of H2O2 is complete, you titrate the solution with 0.1 M AgNO3. If the catalyst, I–, is not consumed in the reaction and is completely recovered, what volume of the 0.1 M AgNO3 is required to reach the end point?
Answers
Answer:
Explanation:
The given chemical reaction is:
[tex]2H_2O_{(l)} \to^{I^-}} 2H_2O_{(l)}+O_2_{(g)}[/tex]
From above equation [tex]I^-[/tex] serves as catalyst which is not consumed by the reaction and also it is completely recovered; as a result to that , the full volume of KI will definitely react with AgNO₃.
Given that :
the volume of potassium iodide [tex]V_{KI} = 15 \ ml[/tex]
the molarity of potassium [tex]M_{KI} = 0.1 \ M[/tex]
the volume of distilled water [tex]V_W = 15 \ mL[/tex]
The volume of 3% [tex]H_2O_2 \ \ V_{H_2O_2} = 5 \ mL[/tex]
Molarity of AgNO₃ [tex]M_{AgNO_3} = 0.1 \ M[/tex]
Let take an integral look with the reaction between KI and AgNO₃; we have
[tex]KI + AgNO_3 \to KNO_3 + AgI[/tex]
At the end point; the moles of KI will definitely be equal to the moles of AgNO₃
So;
[tex]M_{KI}V_{KI}= M_{AgNO_3}V_{AgNO_3} \\ \\ V_{AgNO_3} = \dfrac{M_{KI}V_{KI}}{M_{AgNO_3}} \\ \\ \\ V_{AgNO_3} = \dfrac{ 0.1*15}{0.1}[/tex]
[tex]V_{AgNO_3} = 15 \ ml[/tex]
Thus; the volume of 0.1 M AgNO₃ needed to reach the end point is 15 mL
An electrochemical cell is constructed with a zinc metal anode in contact with a 0.052 M solution of zinc nitrate and a silver cathode in contact with a 0.0042 M solution of silver(I) nitrate. What is the value of Q to use in the Nernst equation for this cell
Answers
Answer:
Q = 12.38
Explanation:
The Nernst equation is given as; Ecell = E°cell - (2.303RT/nF) log Q ;where Q is the reaction quotient.
The reaction quotient, Q in a reaction, is the product of the concentrations of the products divided by the product of the concentrations of the reactants.
In an electrochemical cell, Q is the ratio of the concentration of the electrolyte at the anode to that of the electrolyte at the cathode.
Q = [anode]/[cathode]
therefore , Q = 0.052/0.0042 = 12.38
1. Reaccionan 9.7 Kg de un mineral de níquel al 70% con 8L de una solución de ácido fosfórico al 60% y con una densidad de 1.36g/ml.
Answers
Answer:
The reaction produces 201.4 g of hydrogen gas and 12.2 kg of Nickel Phosphate.
Explanation:
English Translation
9.7 Kg of a 70% nickel mineral react with 8L of a 60% phosphoric acid solution and with a density of 1.36g / ml.
Solution
The problem doesn't seen to be complete as it doesn't ask a question in the end. But, we will just calculate the amount of each product expected to cover the grounds.
The balanced chemical reaction between Nickel and Phosphoric acid is given as
3Ni + 2H₃PO₄ → 3H₂ + Ni₃(PO₄)₂
We need to first obtain the limiting reagent, that is, the reagent that is used up during the reaction and is in short supply. This reagent determines the amount of products that will be formed.
Mass of nickel that is present at the start = 70% of 9.7 kg = 6.79 kg
Mass of Phosphoric acid present at the start of the reaction = 60% of (8000 mL × 1.36 g/mL) = 6528 g = 6.528 kg
Converting both of these to number of moles
Number of moles = (mass)/(Molar mass)
For nickel,
Mass = 6.79 kg = 6790 g
Molar mass = 58.6934 g/mol
Number of moles at the start = (6790/58.6934) = 115.7 moles
For Phosphoric acid
Mass = 6528 g
Molar mass = 97.994 g/mol
Number of moles = (6528/97.994) = 66.6 moles
3 moles of Ni reacts with 2 moles of H₃PO₄
From the number of moles present initially, shows that Phosphoric acid is in limited supply and is the limiting reagent.
From the stoichiometric balance of the reaction
2 moles of H₃PO₄ gives 3 moles of H₂
66.6 moles of H₃PO₄ will give (66.6×3/2) of H₂, that is, 99.9 moles of H₂.
Mass of H₂ liberated from the reaction = (Number of moles) × (molar mass) = 99.9 × 2.016 = 201.3984 g = 201.4 g
2 moles of H₃PO₄ gives 1 mole of Ni₃(PO₄)₂
66.6 moles of H₃PO₄ will give (66.6×1/2) of Ni₃(PO₄)₂, that is, 33.3 moles of Ni₃(PO₄)₂.
Mass of Ni₃(PO₄)₂ produced from the reaction = (Number of moles) × (molar mass) = 33.3 × 366.02 = 12,188.466 g = 12.2 kg
Hope this Helps!!!
Enter your answer in the provided box. To make use of an ionic hydrate for storing solar energy, you place 409.0 kg of sodium sulfate decahydrate on your house roof. Assuming complete reaction and 100% efficiency of heat transfer, how much heat (in kJ) is released to your house at night
Answers
Answer:
409.0 kg of sodium sulfate decahydrate will produce 4.49×10⁵ kJ
of heat energy.
Explanation:
CHECK THE COMPLETE QUESTION BELOW
To make use of an ionic hydrate for storing solar energy, you place 409.0 kg of sodium sulfate decahydrate on your house roof. Assuming complete reaction and 100% efficiency of heat transfer, how much heat (in kJ) is released to your house at night? Note that sodium sulfate decahydrate will transfer 354 kJ/mol
EXPLANATION
Here we were asked to calculate the amount of heat will be generated by 409.0 kg of sodium sulfate decahydrate at night assuming there Isa complete reaction and 100% efficiency of heat transfer in the process
The molecular weight of sodium sulfate decahydrate (H₂₀Na₂O₁₄S) is needed here, so it must be firstly calculated.
The molecular weight of sodium sulfate decahydrate (H₂₀Na₂O₁₄S)
( 1*20) + (22.98*2) + (16*14)+ (32*14)= 322.186 g/mol.
Thus 409.0 kg of H₂₀Na₂O₁₄S will have a value which is equivalent to = (409000g)/(322.186 g/mol.)
=1269.453mol of H₂₀Na₂O₁₄S.
But it was stated in the the question that per mole of H₂₀Na₂O₁₄S will transfer 354 kJ heat.
Therefore, 1269.453mol will transfer 1269.453× 354 kJ = 4.49×10⁵ kJ of heat.
Hence, 409.0 kg of sodium sulfate decahydrate will produce
4.49×10⁵ kJ of heat energy.
The aluminum in a package containing 75 ft2 of kitchen foil weighs approximately 12 ounces. Aluminum has a density of 2.70 g/cm3 . What is the approximate thickness of the aluminum foil in millimeters?(1 ounce = 28.4g)
Answers
Answer:
18130 mm
Explanation:
Now we have a lot of unit conversions to do in order to correctly answer this question. We shall do these conversions gradually.
First we convert the weight in ounce to grams.
If 1 ounce = 28.4g
12 ounces = 12×28.4 = 340.8 g
Next we convert the area of aluminum from ft2 to m2
1ft2= 0.0929 m2
75 ft2= 75 × 0.0929= 6.9675m2
Now density of aluminum= 2.70 gcm-3
Density= mass/volume
But volume= area× thickness
Density= mass/ area × thickness
Density × area × thickness= mass
Thickness= mass/ density × area
Thickness= 340.8g / 2.70gcm-3 × 6.9675m2
Thickness= 340.8/18.8
Thickness= 18.13 m
Since 1000 milimeters make 1 metre
Thickness= 18130 mm
In what unit do we usually measure the force of the earth gravity? Acceleration due to gravity is 9.8/s^2
Answers
Answer:
in short weight
Explanation:
weight is mass x gravitational pull on an object
Write the equilibrium constant: Pb3(PO4)2(s) = 3Pb2+ (aq) +
2PO2 (aq)
Answers
Answer:
Kc = [Pb²⁺]³.[PO₄³⁻]²
Explanation:
Let's consider the following reaction at equilibrium.
Pb₃(PO₄)₂(s) ⇄ 3 Pb²⁺(aq) + 2 PO₄³⁻(aq)
The concentration equilibrium constant is the product of the concentrations of the products raised to their stoichiometric coefficients divided by the product of the concentrations of the reactants raised to their stoichiometric coefficients. It only includes gases and aqueous species.
Kc = [Pb²⁺]³.[PO₄³⁻]²
This equilibrium constant is known as the solubility product of Pb₃(PO₄)₂.
• Briefly discuss the cause of errors in the measurements
Answers
Which of the following viewed the atom as having a nucleus made up of protons and neutrons,with electrons orbiting the nucleus in fixed, stable orbits, much like the planets orbit the sun?
Answers
The correct answer is C. Bohr's model
Explanation:
Bohr's model of the atom developed in 1913 proposed each atom contained a nucleus with protons and neutrons. Also, there were electrons that orbited the nucleus. About this, Niels Bohr proposed the orbits of electrons were similar to those of planets around the sun; however, these did not occur due to gravity but to attraction forces. This model integrated new accurate ideas about the atom. However, this model was still inaccurate because particles in an atom are electrically charged and electrons do not orbit in fixed stable orbits and cannot be compared to the movement of planets around a star.
Answer:
Bohr's model
Explanation:
A boy with pneumonia has lungs with a volume of 1.7 L that fill with 0.070 mol of air when he inhales. When he exhales, his lung volume decreases to 1.3 L. Enter the number of moles of gas that remain in his lungs after he exhales. Assume constant temperature and pressure.
Answers
Answer:
0.053moles
Explanation:
Hello,
To calculate the number of moles of gas remaining in his after he exhale, we'll have to use Avogadro's law which states that the volume of a given mass of gas is directly proportional to its number of moles provided that temperature and pressure are kept constant. Mathematically,
V = kN, k = V / N
V1 / N1 = V2 / N2= V3 / N3 = Vx / Nx
V1 = 1.7L
N1 = 0.070mol
V2 = 1.3L
N2 = ?
From the above equation,
V1 / N1 = V2 / N2
Make N2 the subject of formula
N2 = (N1 × V2) / V1
N2 = (0.07 × 1.3) / 1.7
N2 = 0.053mol
The number of moles of gas in his lungs when he exhale is 0.053 moles
Two scientists study data collected during an experiment and reach different conclusions. How would the scientific community address their disagreement?
Please
Answers
Answer: D. They would device an experiment that could test the two scientists conclusions.
Explanation:
The results of the scientific study must be verified by peer scientists or members of the scientific community to proof whether the research has been conducted produce a valid evidence.
In the given situation, the two scientists had developed different conclusion for the same experiment. This may mean either of the two may have put up an incorrect conclusion.
The scientific community may address this issue by performing the experiment. Every scientific conclusion is based upon the results of the experimental approach.
Answer:d
Explanation:
A temperature of 50°F is equal to °C.
Answers
Answer:
CONVERT IT:
50°F is equal to 10°C
Answer:
10 degrees Celsius
Explanation:
(50°F − 32) × 5/9 = 10°C
Given the information you now know, what is the effect of hyperventilation on blood pH?pH? During hyperventilation, the rapid in the blood CO2CO2 concentration shifts the equilibrium to the which the concentration of H+,H+, thereby the blood pH.
Answers
Answer:
When hypercapnia processes occur, where the concentration of carbon dioxide gas increases in the blood, the protonization of the blood increases, this means that the H + ions increase in concentration, thus generating metabolic acidosis.
This metabolic acidosis is regulated by various systems, but the respiratory system collaborates by generating hyperventilation, to increase blood oxygen pressures, decrease CO2 emissions, and indirectly decrease acidity.
Explanation:
This method of regulating the body is crucial, since the proteins in our body will not be altered if they do not happen.
The enzymes, the red globules, and many more fundamental things for life ARE PROTEINS, that in front of acidic media these modify their structure by denaturing themselves and ceasing to fulfill their functions. This is the reason why it seeks to neutralize the blood pH when it comes to an increase in CO2.
A 33.0−g sample of an alloy at 93.00°C is placed into 50.0 g of water at 22.00°C in an insulated coffee-cup calorimeter with a heat capacity of 9.20 J/K. If the final temperature of the system is 31.10°C, what is the specific heat capacity of the alloy? J g·°C
Answers
Answer:
THE SPECIFIC HEAT OF THE ALLOY IS 0.9765 J/g K
Explanation:
Mass of alloy = 33 g
Initial temperature of alloy = 93°C
Mass of water = 50 g
Initail temp. of water = 22 °C
Heat capacity of calorimeter = 9.20 J/K
Final temp. = 31.10 °C
specific heat of alloy = unknown
specific heat capacity of water = 4.2 J/g K
Heat = mass * specific heat * change in temperature = m c ΔT
Heat = heat capcity * chage in temperature = Δ H * ΔT
In calorimetry;
Heat lost by the alloy = Heat gained by water + Heat of the calorimeter
mc ΔT = mcΔT + Heat capacity * ΔT
33 * C * ( 93 - 31.10) = 50 * 4.2 * ( 31.10 -22) + 9.20 * ( 31.10 -22)
33 * C * 61.9 = 50 * 4.2 * 9.1 + 9.20 * 9.1
2042.7 C = 1911 + 83,72
C = 1911 + 83.72 / 2042.7
C = 1994.72 /2042.7
C =0.9765 J/g K
The specific heat of the alloy is 0.9765 J/ g K
g Reduction involves the A) loss of neutrons, gain of electrons, and an increase in oxidation state. B) loss of neutrons. C) increase in oxidation state. D) gain of electrons and an increase in oxidation state. E) gain of electrons.
Answers
Answer:
E. Gain of electrons
Explanation:
A reduction reaction is one part of the two concurrent reactions that take place in a redox (reduction-oxidation) reaction.
During reduction, an atom gains electrons from a donor atom, and it's oxidation number becomes smaller.
Option A is wrong because reduction does not increase oxidation state nor are neutrons involved
Option B is wrong because reduction is not a nuclear reaction (does not involve the nucleons)
Option C is wrong because reduction leads to reduction in oxidation state
Option D is wrong leads to a reduction in oxidation state when electrons are gained
Option E is correct because reduction involves gain of electrons
A sample of carbon dioxide gas at a pressure of 879 mm Hg and a temperature of 65°C, occupies a volume of 14.2 liters. Of the gas is cooled at constant pressure to a temperature of 23°C, the volume of the gas sample will be
Answers
Answer:
The correct answer is 12.43 Liters.
Explanation:
Based on the given question, the volume V₁ occupied by the sample of carbon dioxide gas is 14.2 liters at temperature (T₁) 65 degree C or 65+273 K = 338 K.
The gas is cooled at a temperature (T₂) 23 degree C or 273+23 K = 296 K
The volume of the gas (V₂) after cooling can be determined by using the formula,
V₁/T₁ = V₂/T₂
14.2/338 = V₂/296
0.0420 = V₂/296
V₂ = 0.0420 * 296
V₂ = 12.43 Liters.
Which of these tasks would a geologist be most likely to perform?
A. Determining the species of a recently collected specimen
O B. Hypothesizing how pieces of ancient pottery were used
O C. Creating a new kind of material using polymers
O D. Determining the best method to extract underground natural gas
SUBMIT
Answers
Answer:
Explanation:
O B. Hypothesizing how pieces of ancient pottery were used
The following data show the rate constant of a reaction measured at several different temperatures. Temperature (K) Rate Constant (1/s) 310 0.194 320 0.554 330 1.48 340 3.74 350 8.97 Part APart complete Use an Arrhenius plot to determine the activation barrier for the reaction. Express your answer using three significant figures. Ea
Answers
Complete Question
The complete question is shown on the first uploaded image
Answer:
Part A
activation barrier for the reaction [tex]E_a = 84 .0 \ KJ/mol[/tex]
Part B
The frequency plot is [tex]A = 2.4*10^{13} s^{-1}[/tex]
Explanation:
From the question we are told that
at [tex]T_1 = 300 \ K[/tex] [tex]k_1 = 5.70 *10^{-2}[/tex]
and at [tex]T_2 = 310 \ K[/tex] [tex]k_2 = 0.169[/tex]
The Arrhenius plot is mathematically represented as
[tex]ln [\frac{k_2}{k_1} ] = \frac{E_a}{R} [\frac{1}{T_1} - \frac{1}{T_2} ][/tex]
Where [tex]E_a[/tex] is the activation barrier for the reaction
R is the gas constant with a value of [tex]R = 8.314*10^{-3} KJ/mol \cdot K[/tex]
Substituting values
[tex]ln [\frac{0.169}{6*10^-2{}} ] = \frac{E_a}{8.314*10^{-3}} [\frac{1}{300} - \frac{1}{310} ][/tex]
=> [tex]E_a = 84 .0 \ KJ/mol[/tex]
The Arrhenius plot can also be mathematically represented as
[tex]k = A * e^{-\frac{E_a}{RT} }[/tex]
Here we can use any value of k from the data table with there corresponding temperature let take [tex]k_2 \ and \ T_2[/tex]
So substituting values
[tex]0.169 = A e ^{- \frac{84.0}{8.314*10^{-3} * 310} }[/tex]
=> [tex]A = 2.4*10^{13} s^{-1}[/tex]
Suppose an industrial quality-control chemist analyzes a sample from a copper processing plant in the following way. He adds powdered iron to a copper(II) sulfate sample from the plant until no more copper will precipitate. He then washes, dries, and weighs the precipitate, and finds that it has a mass of . Calculate the original concentration of copper(II) sulfate in the sample. Round your answer to significant digits.
Answers
Answer:
Concentration of Copper (II) Sulfate in the original sample in mol/L = 0.0035 M
Concentration of Copper (II) Sulfate in the original sample in g/L = 0.56 g/L
Explanation:
Complete Question
Fe(s) + CuSO₄(aq) → Cu(s) + FeSO₄(aq)
Suppose an industrial quality-control chemist analyzes a sample from a copper processing plant in the following way. He adds powdered iron to a 400.mL copper (II) sulfate sample from the plant until no more copper will precipitate. He then washes, dries, and weighs the precipitate, and finds that it has a mass of 89.mg. Calculate the original concentration of copper(II) sulfate in the sample. Round your answer to 2 significant figures.
Solution
Noting that the precipitate is Copper as it is the only solid by-product of this reaction.
89 mg of Copper is produced from this reaction.
We convert this into number of moles for further stoichiometric calculations
Mass of Copper = 89 mg = 0.089 g
Molar mass of Copper = 63.546 amu
Number of moles of Copper produced from the reaction = (0.089/63.546) = 0.0014005602 = 0.001401 mole
From the stoichiometric balance of the reaction,
1 mole of Copper is produced from 1 mole of Copper (II) Sulfate
0.001401 mole of Copper will be produced similarly from 0.001401 mole of Copper (II) Sulfate.
Number of moles of Copper (II) Sulfate in the original sample = 0.001401 mole
Concentration of Copper (II) Sulfate in the original sample in mol/L = (Number of moles) ÷ (Volume in L)
Number of moles = 0.001401 mole
Volume in L = (400/1000) = 0.4 L
Concentration of Copper (II) Sulfate in the original sample in mol/L = (0.001401/0.4) = 0.0035025 mol/L = 0.0035 mol/L to 2 s.f.
Concentration in g/L = (Concentration in mol/L) × (Molar Mass)
Concentration in mol/L = 0.0035025 M
Molar mass of Copper (II) Sulfate = 159.609 g/mol
Concentration of Copper (II) Sulfate in the original sample in g/L = 0.0035025 × 159.609 = 0.559 g/L = 0.56 g/L to 2 s.f
Hope this Helps!!!!
The concentration of the original copper solution is 0.035 M.
The equation of the reaction is;
Fe(s) + CuSO4(aq) -------> FeSO4(aq) + Cu(s)
Number of moles of copper obtained = 89 × 10^-3g/63.5 = 0.0014 moles
Since the reaction is 1:1, the number of moles of copper sulfate that reacted is c.
From the question, we are told that the volume of solution is 400.mL or 0.04L.
Hence, the concentration of the solution is; number of moles /volume
= 0.0014 moles/0.04L = 0.035 M
Learn more: https://brainly.com/question/9352088
Missing parts;
Suppose an industrial quality-control chemist analyzes a sample from a copper processing plant in the following way. He adds powdered iron to a 400.mL copper (II) sulfate sample from the plant until no more copper will precipitate. He then washes, dries, and weighs the precipitate, and finds that it has a mass of 89.mg. Calculate the original concentration of copper(II) sulfate in the sample. Round your answer to 2 significant figures.
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