Answer:
[tex]\Delta H=-11897J[/tex]
Explanation:
Hello,
In this case, it is widely known that for isochoric processes, the change in the enthalpy is computed by:
[tex]\Delta H=\Delta U+V\Delta P[/tex]
Whereas the change in the internal energy is computed by:
[tex]\Delta U=nCv\Delta T[/tex]
So we compute the initial and final temperatures for one mole of the ideal gas:
[tex]T_1= \frac{P_1V}{nR}=\frac{10.90atm*4.86L}{0.082*n}=\frac{646.02K }{n} \\\\T_2= \frac{P_2V}{nR}=\frac{1.24atm*4.86L}{0.082*n}=\frac{73.49K }{n}[/tex]
Next, the change in the internal energy, since the volume-constant specific heat could be assumed as ³/₂R:
[tex]\Delta U=1mol*\frac{3}{2} (8.314\frac{J}{mol*K} )*(73.49K-646.02K )=-7140J[/tex]
Then, the volume-pressure product in Joules:
[tex]V\Delta P=4.86L*\frac{1m^3}{1000L} *(1.24atm-10.90atm)*\frac{101325Pa}{1atm} \\\\V\Delta P=-4756.96J[/tex]
Finally, the change in the enthalpy for the process:
[tex]\Delta H=-7140J-4757J\\\\\Delta H=-11897J[/tex]
Best regards.
The change in enthalpy is 70.42J
Data;
Volume of the gas = 4.86LInitial Pressure = 10.90 atmFinal Pressure = 1.24 atmChange in Enthalpy = ?Change in EnthalpyThe change of enthalpy is calculated as
[tex]\delta H = \delta V + \delta nRT\\\delta n = 0\\\delta H = \delta U \\[/tex]
The volume change is negligible
The change in enthalpy here is equal to change in internal energy over ΔE
[tex]\delta H = \delta U = nCv\delta T\\\delta H = \frac{3}{2}(nR\delta T)\\\delta H = \frac{3}{2}\{\delta PV)\\ \delta H = \frac{3}{2}[(10.90-1.24)*4.86] \\\delta H = 70.42J[/tex]
The change in enthalpy is 70.42J
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The transfer of surface water into the ground to become groundwater is known as
and it can replenish an aquifer.
Answer: Recharge
Explanation:
To solve this we must be knowing each and every concept related to groundwater recharge. Therefore, the transfer of surface water into the ground to become groundwater is known as groundwater recharge.
What is groundwater recharge?The water that is added to the aquifer and through unsaturated zone after percolation (or infiltration) following any storm rainfall event is known as groundwater recharge.
In the natural world, rivers, lakes, streams, rain, and snowmelt all contribute to groundwater recharge. Other surface water trickles and through soil, eventually connecting with a source of water underneath the surface, while other surface water has evaporated or enters another watershed.
Therefore, the transfer of surface water into the ground to become groundwater is known as groundwater recharge.
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ch3-ch2-ch-ch(cl)-ch=o IUPAC name
Answer:
2-chloropentanal
Answer:
2-chloropentanal
Explanation:
ch3-ch2-ch-ch(cl)-ch=o IUPAC name
H H H H
H - C - C - C - C - C = O
H H H Cl
So as can be seen 2 as the Chlorine is on the second carbon.
Chloro because of the chlorine.
Pent because there's 5 carbon
al because there's an aldehydes
Aldehyde = −CHO
2-chloropentanal
which best describes a mixture.
A it has a single composition and it has a set of characteristics
B it can have different compositions but it has a set of charactaristics that does not change
C it has a single composition but it has a set of characteristics that does change
D it can have different compositions and it has a set of characteristics that does change
Answer:
B) It can have different compositions, but it has a set of characteristics that does not change.
Explanation:
On e d g e n u i t y
I believe the answer is d lmk if im wrong or right
The equilibrium constant, Kc, for the following reaction is 4.76×10-4 at 431 K. PCl5(g) PCl3(g) + Cl2(g) When a sufficiently large sample of PCl5(g) is introduced into an evacuated vessel at 431 K, the equilibrium concentration of Cl2(g) is found to be 0.233 M. Calculate the concentration of PCl5 in the equilibrium mixture. M
Answer:
Explanation:
Step 1: Data given
The equilibrium constant, Kc, for the following reaction is 4.76 * 10^-4 at 431 K
The equilibrium concentration of Cl2(g) is 0.233 M
Step 2: The balanced equation
PCl5(g) ⇄ PCl3(g) + Cl2(g)
Step 3: The initial concentration
[PCl5]= Y M
[PCl] = 0M
[Cl2] = 0M
Step 4: Calculate the concentration at equilibrium
[PCl5] = Y + X M = Y - 0.233 M
[PCl]= XM = 0.233 M
[Cl2]= XM = 0.233 M
Step 5: Define Kc
Kc = [Cl2]* [PCl3] / [PCl5]
4.76 * 10^-4 = 0.233² / (Y -0.233)
0.000476 = 0.05429 / (Y - 0.233)
Y - 0.233 = 0.05429 / 0.000476
Y - 0.233 = 114.05 M
Y = 114.283 M = the initial concentration
The concentration of PCl5 at the equilibrium is 114.05 M
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 .
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
how do you create flu vaccine,
Answer:
Explanation:
The fluid containing virus is harvested from the eggs. For inactivated influenza vaccines (i.e., flu shots), the vaccine viruses are then inactivated (killed), and the virus antigen is purified. The manufacturing process continues with quality testing, filling and distribution.
Select the correct answer.
What effect does an increase in products have on the reaction rate of a mixture at equilibrium?
A.
The forward reaction rate increases.
B.
Both the forward and the reverse reaction rates decrease.
Both the forward and the reverse reaction rates increase.
D.
The reverse reaction rate increases.
Reset
Next
Answer:
At equilibrium the rate of the forward reaction is equal to the rate of the backward reaction.
When the product of a reaction at equilibrium is increased the equilibrium will shift left or to the reactant side. As a result the excess product will get converted to reactant. This is in accordance to Le Chatelier's principle.
Le Chatelier's principle states that when a system is subjected to stress the equilibrium will shift in a direction to minimize effect of the stress.
Thus the products added to the system at equilibrium will make the equilibrium shift to the reactant side, the rate of the reverse or backward reaction will increase.
Explanation:
Hope This Helps Amigo!
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
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
Calculate the osmotic pressure of a solution prepared by dissolving 65.0 g of Na2SO4 in enough water to make 500 mL of solution at 20°C. (Assume no ion pairing – in other words, assume that the electrolyte completely dissociates into its constituent ions.)
Answer:
66.0 atm
Explanation:
We can calculate the osmotic pressure (π) using the following expression.
[tex]\pi = i \times M \times R \times T[/tex]
where,
i: van 't Hoff indexM: molarityR: ideal gas constantT: absolute temperatureStep 1: Calculate i
Sodium sulfate completely dissociates according to the following equation.
Na₂SO₄ ⇒ 2 Na⁺ + SO₄²⁻
Since it produces 3 ions, i = 3.
Step 2: Calculate M
We can calculate the molarity of Na₂SO₄ using the following expression.
[tex]M = \frac{mass\ of\ solute }{molar\ mass\ of\ solute\ \times liters\ of\ solution} = \frac{65.0g}{142.04g/mol \times 0.500L} =0.915M[/tex]
Step 3: Calculate T
We will use the following expression.
K = °C + 273.15
K = 20°C + 273.15 = 293 K
Step 4: Calculate π
[tex]\pi = 3 \times 0.915M \times \frac{0.08206atm.L}{mol.K} \times 293K =66.0 atm[/tex]
Wine goes bad soon after opening because the ethanol dissolved in it reacts with oxygen gas to form water and aqueous acetic acid , the main ingredient in vinegar. Calculate the moles of water produced by the reaction of of oxygen. Be sure your answer has a unit symbol, if necessary, and round it to significant digits.
Answer:
1.7 moles of ethanol would be needed.
Explanation:
* Calculate the moles of ethanol needed to produce 1.70mol of water. Be sure your answer has a unit symbol, if necessary, and round it to the correct number of significant digits.
First off, we have to state the equation for the reaction.
So we know that;
ethanol + oxygen → acetic acid + water
This leads us to;
C2H5OH + O2 → CH3COOH + H2O
1 1 1 1
To obtain the moles of ethanol needed to produce 1.70mol of water, we look at the stoichiometry of the reaction above.
1 mol of ethanol produces 1 mole of water
x mol of thanol would produce 1.7 mol of water
Thus we have;
1 = 1
x = 1.7
x = 1.7 moles of ethanol would be needed.
Fractionation of Crude Oil Select the correct ranking of the following alkanes according to the height reached in a fractionating column, highest first: butane, heptadecane, dodecane, ethane, decane Select the correct ranking of the following alkanes according to the height reached in a fractionating column, highest first:
butane, heptadecane, dodecane, ethane, decane
A. ethane > butane > decane > dodecane > heptadecane
B. heptadecane > > dodecane > decane butane > ethane
C. ethane > butane > decane> heptadecane >
D. dodecane butane > ethane > decane > dodecane > heptadecane
Answer:
A. ethane > butane > decane > dodecane > heptadecane
Explanation:
In fractionating column, crude oil is separated by means of fractional distillation due to the wide range of boiling point of the crude products such as ethane, propane, butane pentane etc.
The product with the least weight rises to top height while the product with highest weight will move down.
For the given hydrocarbon products, the ranking according to their molecular weight, starting with the lighter product to heavier product is
ethane (C2), butane (C4), decane(C10), dodecane (C12), heptadecane(C17).
Thus, the correct ranking, starting with the product that will rise highest is ethane > butane > decane > dodecane > heptadecane
In this reaction: Mg (s) + I₂ (s) → MgI₂ (s)
If 2.34 moles of Mg react with 3.56 moles of I₂, and 1.76 moles of MgI₂ form, what is the percent yield?
Answer:
98.9%
Explanation:
2 moles of I₂ are required for each mole of Mg, so the reaction is limited by the available I₂. The 3.56 moles of I₂ should react with 1.78 moles of Mg to produce 1.78 moles of MgI₂. Instead, we get 1.76 moles of MgI₂.
The yield is 1.76/1.78 × 100% ≈ 98.876%
The yield is 98.9% of the quantity expected based on available reactants.
A thermometer is placed in water in order to measure the water’s temperature. What would cause the liquid in the thermometer to rise? The molecules in the water move closer together. The molecules in the thermometer’s liquid spread apart. The kinetic energy of the water molecules decreases. The kinetic energy of the thermometer’s liquid molecules decreases.
Answer: The molecules in the thermometer's liquid spread apart.
Explanation:
Mercury is the only metal that remains liquid at room temperature. It has a high coefficient of expansion therefore the its level rises when exposed to a temperature range. It can detect a slight change in temperature. It has a high boiling point.
When the thermometer is placed in the water to measure the temperature, the molecules of thermometer liquid that is mercury only will spread due to high coefficient of expansion. This can be seen as rise in temperature.
Answer:
B
Explanation:
Just did the test
A Carbon-10 nucleus has 6 protons and 4 neutrons. Through radioactive beta decay, it turns into a Boron-10 nucleus, with 5 protons and 5 neutrons, plus another particle. What kind of additional particle, if any, is produced during this decay
Answer:
No additional particle was produced during the decay.
Explanation:
The equation of decay is given as;
¹⁰₆C + ⁰₋₁ e → ¹⁰₅B + x
To identify x, we have to calculate its atomic and mass number.
In the reactants side;
Atomic Number = 6 + (-1) = 5
Mass number = 10 + 0 = 10
In the products side;
Atomic Number = 5 + x
Mass Number = 10 + x
Generally, reactant = product
Atomic Number;
5 = 5 + x
x = 5 - 5 = 0
Mass Number;
10 = 10 + x
x = 10 - 10 = 0
This means no additional particle was produced during the decay.
Consider a solution containing 0.100 M fluoride ions and 0.126 M hydrogen fluoride. The concentration of fluoride ions after the addition of 5.00 mL of 0.0100 M HCl to 25.0 mL of this solution is __________ M.
a. 0.0980
b. 0.0817
c. 0.0167
d. 0.0850
e. 0.00253
Answer:
The answer is "Option b"
Explanation:
In this question first we calculates the moles in F-, HF, and in HCL, which can be defined as follows:
Formula:
[tex]\ Number \ of \ moles\ = \ Molarity \times \ Volume \ in \ litter[/tex]
[tex]\ moles \ in\ F- = 0.100 \ M \times 0.0250 L\\\\[/tex]
[tex]=\ 0.0025 \ moles[/tex]
[tex]\ moles \ in \ HF \ = 0.126M \times 0.0250 L[/tex]
[tex]= 0.00315 \ moles[/tex]
[tex]\ moles \ in \ HCl = 0.0100M \times 0.00500 L[/tex]
[tex]= 0.00005 \ moles[/tex]
[tex]\ Reaction: \\\\F - + H+ \rightarrow HF[/tex]
[tex]\Rightarrow \ moles \ in \ F- = 0.0025 \\\\\Rightarrow \ moles \ in \ H+ = 0.00005 \\\\ \Rightarrow \ moles \ in \ HF = 0.00315\\\\ \ total \ moles = 0.00250 -0.0000500 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.00315 + 0.00005\\\\\ total \ moles =0.00245 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.00245[/tex]
[tex]\ total \ volume \ in \ the \ solution = \ V = \ 0.0300 L\\\\ after \ addition \ of \ HCl \ the \ concentration \ of \ F- \ = 0.00245\ moles \div V[/tex]
[tex]=\frac{ 0.00245 \ moles }{0.0300L}\\\\= \frac{245 \times 10^4}{300 \times 10^5} \\\\= \frac{245}{3000} \\\\ = 0.0817 M[/tex]
Which of the following provides evidence to support Thomson's hypothesis about electrons??
A.) Gold foil experiment
B.) Cathode ray experiments
C.)Spectrum of colors emitted by gas
D.) Radiation produced when beryllium is bombarded with alpha particles
Cathode ray experiments of the following provides evidence to support Thomson's hypothesis about electrons.
What is the hypothesis of Thomson's atomic model?Rutherford's gold leaf experiment demonstrated that the atom is essentially empty space with such a tiny, compact, positively-charged nucleus. Thomson had proposed the plum pie model of the atom, which featured negatively-charged electrons buried within a favorably "soup." Since most of the alpha particles flow through an atom directly without being deflected, contrary to what Thomson's model predicted, the majority of a space inside of an atom is empty. As a result, the Thomson model of a molecule was disproved.
Who disproved Thomson's theory?According to Thomson's model, every atom is made up of negative charges "plums" surrounded in positively charged "pudding," or electrons with a soup of positive ion to balance their negative charges. Hans Geiger and Arthur Marsden's 1909 gold foil test refuted the 1904 Thomson model.
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what is the correct ionic equation, including all coefficients, charges, and phases for the following sets of reactants? Assume that the contribution of protons from H2SO4 is near 100%.
Ba(OH)2(aq)+H2SO4(aq) —>
help, I have no clue
Answer:
Ba(OH)2(aq)+H2SO4(aq) gives us 2BaH+H2O
Explanation:
The calculated yield for the production of carbon dioxide was 73.4g. When the
experiment was performed in the lab, a yield of 72.3g was produced. What is the
percent yield of carbon dioxide?
Answer:10 grams of CO2
Explanation:
Yeild= exp. yeild÷ thoretical yeild × 100
Yeild= 73.3÷73.4 × 100
Yeild= 0.1 ×100
Yeild= 10
Quantum number of boron
Answer:
The answer is 5.
Write the equilibrium constant: Pb3(PO4)2(s) = 3Pb2+ (aq) +
2PO2 (aq)
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.
how many moles of helium gas occupy 22.4 L at 0 degreeC at 1 atm pressure
Answer:
1 mole of the gas occupies 22. 4L at O0C at 1atm pressure. Hence, the correct option is C.
Explanation:
ions always have the same electronic structure as elements in which group of the periodic table?
Answer:
In 0 group of the periodic table
Explanation:
So they will not react with other atoms because they have a full outer shell of electrons and an overall charge of 0.
Hope it helps.
Benzene can be converted to 1,3,5-tribromobenzene in five reaction steps and four intermediate compounds. Select the appropriate reagent from the followings.
Br2, R2O2
CH3Cl, AlCl3
CH3COCl, AlCl3
NaNO2, HCl
HNO3, H2SO4
H3PO2
H3PO4
KMnO4
Answer:
The appropriate reagent is: H3PO2.
Explanation:
H3PO2 is in charge of eliminating the amino group by diazotization, remember that the amino group had previously achieved bromination at positions m; that is to say that it achieved in the beginning that the three bromine atoms of 1,2,4 tribromobenzene were introduced in the meta positions among themselves, which finally corresponds as part of the last reaction to the 1,3,5-tribromobenzene position.
Hi I have a lab for Chemistry I am struggling with. I have to do calculations given the following information
1. Mass of evaporating dish plus sample 26.57 g
2. Mass of evaporating dish 24.29 g
3. Mass of evaporating plus NaCl 68.66 g
4. Mass of evaporating dish 67.84 g
5. Mass of filter paper plus sand 37.69 g
6. Mass of filter paper 36.34 g
CALCULATIONS AND CONCLUSIONS
1. Calculate the mass of unknown mixture g
2. Calculate the mass of NaCl recovered g
3. Calculate the mass of sand recovered g
4. Calculate the percentage of NaCl in your unknown mixture %
5. Calculate the percent sand in your unknown mixture %
6. Calculate the total mass of sand and salt recovered g
7. Calculate the percent recovery of the components %
Answer:
1. 2.28 g
2. 0.82 g
3. 1.35 g
4. 36 %
5. 59 %
6. 2.17 g
7. 95 %
Explanation:
Hello,
1. In this case, the mass of the unknown mixture is obtained by subtracting the mass of the dish plus sample and the mass of the dish:
m = 26.57 g- 24.29 g = 2.28 g
2. In this case, the mass of the NaCl recovered is obtained by subtracting the mass of the dish plus NaCl and the mass of the dish:
m = 68.66 g- 67.84 g = 0.82 g
3. In this case, the mass of the sand recovered is obtained by subtracting the mass of the filter paper plus sand and the mass of the filter paper:
m = 37.69 g- 36.34 g = 1.35 g
4. The percentage of NaCl is computed by considering its mass and the mass of the unknown mixture:
% NaCl = 0.82 g / 2.28 g * 100 % = 36 %
5. The percentage of sand is computed by considering its mass and the mass of the unknown mixture:
% sand = 1.35 g / 2.28 g * 100 % = 59 %
6. Here,we have to add the mass of NaCl and sand:
m = 0.82 g + 1.35 g = 2.17 g
7. Finally, the percent recovery is obtained by diving the total recovered mass by the total obtained mass of the mixture:
% recovery = 2.17 g / 2.28 g * 100 % = 95 %
Best regards.
a binary ionic compound is made of two components name one of them
Answer:
CATION
Explanation:
It's one is the action and the mother is a cation.
Classify the following unbalanced chemical reaction Fe(OH)2(s) + HCl(aq) = FeCl2(aq) + H2O(l)
1. Acid-Base Reaction
2. Precipitation Reaction
3. Oxidation-Reduction Reaction
4. Combustion Reaction
Answer:
1. Acid-Base Reaction
Explanation:
Fe(OH)2(s) + HCl(aq) = FeCl2(aq) + H2O(l)
base acid
This a reaction between base and acid.
Ferrous hydroxide is an inorganic alkaline compound whereas hydrochloric acid is an acid. The reaction between Fe(OH)₂and HCl is an acid-base reaction. Thus, option 1 is correct.
What is an acid-base reaction?An acid-base reaction is a chemical change that occurs and takes place when the reactant constitutes an acid and a base. They are characterized by the exchange of protons that results in the formation of conjugate bases and acids or salt.
The acid-base chemical reaction is shown as,
Fe(OH)₂(s) + HCl(aq) ⇒ FeCl₂(aq) + H₂O(l)
Here, ferrous hydroxide is a base with hydroxide ions and hydrochloric acid is an acid with hydrogen ions. HCl donates its proton to form water molecules with hydroxide ions of ferrous hydroxide.
Therefore, in option 1. the reaction is an acid-base reaction.
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what would happen if you place two positive charges next to each other and let go. would they attract, stay still, or they would repel
Answer:
they would repel
Explanation:
unlike charges attract while like ones repel.
Ethanol, , boils at 78.29 °C. How much energy, in joules, is required to raise the temperature of 2.00 kg of ethanol from 26.0 °C to the boiling point and then to change the liquid to vapor at that temperature? (The specific heat capacity of liquid ethanol is 2.44 J/g ∙ K, and its enthalpy of vaporization is 855 J/g.)
Answer:
THE HEAT REQUIRED TO CHANGE 2 KG OF ETHANOL FROM 26 °C TO THE BOILING POINT AND TO VAPOR AT THAT TEMPERATURE IS 1965.175 KJ.
Explanation:
Boiling point of ethanol = 78.29 °C = 78.29 + 273 K = 351.29 K
Mass = 2 kg = 2000 g
Final temp. = 26.0 °C = 26 + 273 K= 299 K
Change in temperature = (78.29 - 26) °C = 52.29 °C
1. Heat required to raise the temperature from 26 °C to the boiling point?
Heat = mass * specific heat * change in temperature
Heat = 2000 * 2.44 * 52.29
Heat = 255 175.2 J
2. Heat required to change the liquid to vapor at that temperature?
Heat = mass * enthalphy of vaporization
Heat = 2000 * 855
Heat =1 710000 J
The total heat required to raise the temperature of 2 kg of ethanol from 26 °C to the boiling point and then to change the liquid to vapor at that temperature will be:
Heat = mcT + m Lv
Heat = 255 175.2 J + 1710000 J
Heat = 1965175.2 J
Heat = 1965.175 kJ of heat.
The core of the pyruvate dehydrogenase complex is made up of eight catalytic________that make up the_______component.
a. monomers; E1b. dimers; E2c. dimers; E3d. trimers; E2
Answer:
(D.)
The core of the pyruvate dehydrogenase complex is made up of eight catalytic trimers that make up the E2 component.
Explanation:
Eight trimers assemble as a hollow truncated cube, which forms the core of the multi-enzyme complex, known as the E2 complex in human pyruvate dehydrogenase complex.
A growing concern in agricultural and food chemistry is the presence of residues in food. We use many forms of organic chemicals in agriculture and food chemistry and there is growing concern as to how safe these materials are. Choose an organic chemical used in agricultural of food chemistry and report on the functional groups contained in your compound, the uses of the compound, and the safety of that compound.
Answer:
Monosodium Glutamate (MSG) is a chemical which is used in agricultural of food chemistry.
Explanation:
Monosodium Glutamate (MSG) is a chemical that is used in types of different food as food additives. The functional group that is present in Glutamate are carboxylic acid and amine. This chemical is used in different types of foods which is responsible for enhancing the taste of the food. Monosodium Glutamate is safe if it is used in moderate dose but adversely affected when it is used in large amount.
