Answer:
B) Ionic
Explanation:
A powder contains FeSO4⋅7H2O (molar mass=278.01 g/mol), among other components. A 2.810 g sample of the powder was dissolved in HNO3 and heated to convert all iron to Fe3+. The addition of NH3 precipitated Fe2O3⋅xH2O, which was subsequently ignited to produce 0.443 g Fe2O3.What was the mass of FeSO4⋅7H2O in the 2.810 g sample?
Answer:
the mass of FeSO4.7H2O in the 2.810 g sample was 1.5402 g
Explanation:
From the given information:
Two moles of FeSO4.7H2O = one mole of Fe2O3
Let recall that:
number of moles of Fe2O3 = mass of Fe2O3 / molar mass of Fe2O3
Given that :
mass of Fe2O3 = 0.443 g
number of moles of Fe2O3 = 0.443 g/ 159.69 g/mol
number of moles of Fe2O3 = 0.00277 mol
Thus;
number of moles of FeSO4.7H2O = 2 × Fe2O3
number of moles of FeSO4.7H2O = 2 × 0.00277 mol
number of moles of FeSO4.7H2O = 0.00554 mol
However from the usual stoichiometry formula; the mass of a substance = number of moles × molar mass
Now; the mass of FeSO4.7H2O = number of moles × molar mass
the mass of FeSO4.7H2O = 0.00554 mol × 278.01 g/mol
the mass of FeSO4.7H2O = 1.5402 g
Therefore; the mass of FeSO4.7H2O in the 2.810 g sample was 1.5402 g
What are the concentrations of [K+], [OH-], [CO32-] and [H+], in a 1.2 M solution of K2CO3 ? (Note: Question is asking for concentrations and not pH) g
Answer:
The concentrations are: [K⁺] = 1.2 M, [OH⁻] = 0.016 M, [CO₃²⁻] = 1.18 M and [H⁺] = 6.25x10⁻¹³ M.
Explanation:
The dissociation equation of K₂CO₃ in water is:
K₂CO₃(aq) ⇄ K⁺(aq) + CO₃²⁻(aq) (1)
Also, the CO₃²⁻ will react with water as follows:
CO₃²⁻(aq) + H₂O(l) ⇄ HCO₃⁻(aq) + OH⁻(aq) (2)
The constant of the reaction (2) is:
[tex] Kb = \frac{[OH^{-}][HCO_{3}^{-}]}{[CO_{3}^{-2}]} = 2.08 \cdot 10^{-4} [/tex]
The solution of K₂CO₃ is 1.2 M, and since the mole ratio of K₂CO₃ with K⁺ and CO₃²⁻ is 1:1, then we have:
[tex] [K_{2}CO_{3}] = [K^{+}] = [CO_{3}^{-2}] = 1.2 M [/tex]
Now, from equation (2) we have:
CO₃²⁻(aq) + H₂O(l) ⇄ HCO₃⁻(aq) + OH⁻(aq) (3)
1.2 - x x x
[tex] 2.08 \cdot 10^{-4} = \frac{[OH^{-}][HCO_{3}^{-}]}{[CO_{3}^{-2}]} [/tex]
[tex] 2.08 \cdot 10^{-4} = \frac{x^{2}}{1.2 - x} [/tex]
[tex] 2.08 \cdot 10^{-4}*(1.2 - x) - x^{2} = 0 [/tex] (4)
By solving equation (4) for x we have:
x = 0.016 M = [HCO₃⁻] = [OH⁻]
Hence, the CO₃²⁻ concentration is:
[CO₃²⁻] = 1.2 M - 0.016 M = 1.18 M
Finally, the concentration of [H⁺] is:
[tex] [H^{+}][OH^{-}] = 10^{-14} [/tex]
[tex][H^{+}] = \frac{10^{-14}}{[OH^{-}]} = \frac{10^{-14}}{0.016} = 6.25 \cdot 10^{-13} M[/tex]
Therefore, the concentrations are: [K⁺] = 1.2 M, [OH⁻] = 0.016 M, [CO₃²⁻] = 1.18 M and [H⁺] = 6.25x10⁻¹³ M.
I hope it helps you!
A four carbon chain; the second carbon is also single bonded to CH3. Spell out the full name of the compound
Answer:
This description shows a methyl group.
Explanation:
Nylon 88 is made from the monomers H2N(CH2)8NH2 and HOOC(CH2)6COOH. So, would you characterize nylon 88 as rather an addition or a condensation polymer? Please explain your answer.
Answer:
Combination of H2N(CH2)8NH2 and HOOC(CH2)6COOH leads to the loss of water molecules at each linkage position.
Explanation:
A condensation polymer is a polymer formed when two monomers combine with the elimination of a small molecule such as water. The removal of the small molecule occurs at the point where the two monomers are joined to each other.
Nylon is known to form condensation polymers. This is because it involves the linkage of an -OH group to an -NH2 group. Water is eliminated in the process.
In the case of H2N(CH2)8NH2 and HOOC(CH2)6COOH, linkage of the both monomers at the 8 position of each chain leads to the formation of nylon- 8,8 with loss of water molecules at each linkage position. This stepwise loss of water molecules at each linkage makes it a condensation polymer.
Using the periodic table provided, identify the atomic mass of sodium (Na) . Your answer should have 5 significant figures. Provide your answer below: __ amu
Answer:
Your answer will either be 22.9897 or 22.990 !!
Explanation:
Identify the solutions that will form a precipitate when mixed with aqueous barium chloride, BaCl2 (aq). Select all that apply.
The given question is incomplete, the complete question is:
Identify the solutions that will form a precipitate when mixed with aqueous barium chloride, BaCl2 (aq). Select all that apply. potassium carbonate, K2CO3 (aq) silver nitrate, AgNO3(aq) sulfuric acid, H2SO4 (aq) sodium hydroxide, NaOH(aq) sodium chloride, NaCl (aq) copper(II) nitrate, Cu(NO3)2 (aq)
Answer:
The aqueous solution of BaCl2 form precipitate with potassium carbonate, silver nitrate, sulfuric acid and sodium hydroxide.
Explanation:
The formation of an insoluble salt, which takes place when two solutions comprising soluble salts are reacted with each other, the reaction is termed as precipitation reaction. The insoluble salt produced in the given reaction is termed as the precipitate.
BaCl₂ reacts with K₂CO₃ to produce white precipitate in the form of BaCO₃,
BaCl₂ (aq) + K₂CO₃ (aq) ⇒ BaCO₃ (s) + 2KCl (aq)
BaCl₂ reacts with H₂SO₄ to produce white precipitate in the form of BaSO₄,
BaCl₂ (aq) + H₂SO₄ (aq) ⇒ BaSO₄ (s) + 2HCl (aq)
BaCl₂ (aq) reacts with NaOH to produce white precipitate in the form of Ba(OH)2,
BaCl₂ (aq) + 2NaOH (aq) ⇒ Ba(OH)2 (s) + 2NaCl (aq)
BaCl₂ reacts with AgNO₃ to produce white precipitate in the form of 2AgCl (s),
BaCl₂ (aq) + 2AgNO₃ (aq) ⇒ Ba(NO₃)₂ (aq) + 2AgCl (s)
Upon reaction of BaCl₂ with Cu(NO₃)₂ no formation of precipitate takes place, and BaCl₂ does not react with NaCl.
Osmosis is the process responsible for carrying nutrients and water from groundwater supplies to the upper parts of trees. The osmotic pressures required for this process can be as high as 19.1 atm . What would the molar concentration of the tree sap have to be to achieve this pressure on a day when the temperature is 32 ∘C ? Express your answer to three significant figures and include the appropriate units. View Available Hint(s)
Answer:
[tex]M=0.763\frac{mol}{L}=0.763M[/tex]
Explanation:
Hello,
In this case, as the osmotic pressure (π) is widely known as a colligative property, we can see that the solution in this case is formed by water and tree sap, that is mathematically defined by:
[tex]\pi =iMRT[/tex]
Thus, since tree sap is a covalent substance that is nonionizing, we can infer its van't Hoff factor to be 1, therefore, for the given osmotic pressure and temperature, we can compute the molar concentration (in molar units mol/L) as follows:
[tex]M=\frac{\pi }{RT} =\frac{19.1atm}{0.082\frac{atm*L}{mol*K}*(32+273.15)K} \\\\M=0.763\frac{mol}{L}=0.763M[/tex]
Best regards.
g Calculate the time (in min.) required to collect 0.0760 L of oxygen gas at 298 K and 1.00 atm if 2.60 A of current flows through water. (Hint: Ideal gas law)
Answer:
7.67 mins.
Explanation:
Data obtained from the question include the following:
Volume (V) = 0.0760 L
Temperature (T) = 298 K
Pressure (P) = 1 atm
Current (I) = 2.60 A
Time (t) =?
Next, we shall determine the number of mole (n) of O2 contained in 0.0760 L.
This can be obtained by using the ideal gas equation as follow:
Note:
Gas constant (R) = 0.0821 atm.L/Kmol
PV = nRT
1 x 0.0760 = n x 0.0821 x 298
Divide both side by 0.0821 x 298
n = 0.0760 / (0.0821 x 298)
n = 0.0031 mole
Next, we shall determine the quantity of electricity needed to liberate 0.0031 mole of O2.
This is illustrated below:
2O²¯ + 4e —> O2
Recall:
1 faraday = 1e = 96500 C
4e = 4 x 96500 C
4e = 386000 C
From the balanced equation above,
386000 C of electricity liberated 1 mole of O2.
Therefore, X C of electricity will liberate 0.0031 mole of O2 i.e
X C = 386000 X 0.0031
X C = 1196.6 C
Therefore, 1196.6 C of electricity is needed to liberate 0.0031 mole of O2
Next, we shall determine the time taken for the process. This can be obtained as follow:
Current (I) = 2.60 A
Quantity of electricity (Q) = 1196.6 C
Time (t) =?
Q = It
1196.6 = 2.6 x t
Divide both side by 2.6
t = 1196.6/2.6
t = 460.23 secs.
Finally, we shall convert 460.23 secs to minute. This can be achieved by doing the following:
60 secs = 1 min
Therefore,
460.23 secs = 460.23/60 = 7.67 mins
Therefore, the process took 7.67 mins.
Un globo lleno de helio tenia un volumen de 8.5 L en el suelo a 20°C y a una presión de 750 torr. Cuando se le soltó, el globo se elevo a una altitud donde la temperatura era de -20°C y la presión de 425 torr, ¿Cuál era el volumen del gas del globo en estas condiciones?
Answer:
El volumen del gas era 12.95 L
Explanation:
Se relaciona la presión y el volumen mediante la ley de Boyle, que dice:
“El volumen ocupado por una determinada masa gaseosa a temperatura constante, es inversamente proporcional a la presión”
La ley de Boyle se expresa matemáticamente como: P*V=k
Por otro lado, la Ley de Charles consiste en la relación que existe entre el volumen y la temperatura absoluta de una cierta cantidad de gas ideal, el cual se mantiene a una presión constante. Esta ley dice que cuando la cantidad de gas y de presión se mantienen constantes, el cociente que existe entre el volumen y la temperatura siempre tendrán el mismo valor:
[tex]\frac{V}{T}=k[/tex]
Por último, la Ley de Gay Lussac dice que la temperatura absoluta y la presión son directamente proporcionales. Es decir, cuando se mantiene todo lo demás constante, mientras suba la temperatura de un gas subirá también su presión. Y mientras la temperatura del gas baje, lo mismo ocurrirá con la presión:
[tex]\frac{P}{T}=k[/tex]
Combinado las mencionadas tres leyes se obtiene:
[tex]\frac{P*V}{T} =k[/tex]
Cuando se desean estudiar dos diferentes estados, uno inicial y una final de un gas, se puede aplicar:
[tex]\frac{P1*V1}{T1} =\frac{P2*V2}{T2}[/tex]
Recordando que la temperatura debe usarse en grados Kelvin, conoces los siguientes datos:
P1: 750 torrV1: 8.5 LT1: 20°C= 293°K (siendo 0°C=273°K)P2: 425 torrV2: ?T2: -20°C= 253 °KReemplazando:
[tex]\frac{750 torr*8.5 L}{293K} =\frac{425 torr*V2}{253 K}[/tex]
Resolviendo:
[tex]V2=\frac{750 torr*8.5 L}{293K} *\frac{253 K}{425 torr}[/tex]
V2= 12.95 L
El volumen del gas era 12.95 L
Phosphofructokinase catalyzes the phosphorylation of fructose 6‑phosphate to fructose 1,6‑bisphosphate in glycolysis. Fructose 1,6‑bisphosphatase catalyzes the hydrolysis of fructose 1,6‑bisphosphate to fructose 6‑phosphate in gluconeogenesis.
fructose 6- phosphate
phosphofructokinase fructose 1 ,6-bisphosphatase
fructose 1,6 - bisphosphate
How does fructose-2,6-bisphosphate (F26BP) affect the activity of the enzymes phosphofructokinase-1 (PFK) and fructose I ,6-bisphosphatase (FBPase)?
a. increases PFK activity, increases FBPase activity
b. decreases PFK activity, increases FBPase activity
c. decreases PFK activity, decreases FBPase activity
d. increases PFK activity, decreases FBPase activity
Answer:
d. increases PFK activity, decreases FBPase activity
Explanation:
Fructose-2,6-bisphophate is formed by the phosphorylation of fructose-6-phosphate catalyzed by phosphofructokinase-2, PFK-2.
Fructose-2,6-bisphophate functions as an allosteric effector of the enzymes phosphofructokinase-1, PFK-1 and fructose-1,6-bisphosphatase, FBPase.
Fructose-2,6-bisphophate has opposite effects on the enzymes, PFK-1 and FBPase. When it binds to the allosteric site of the enzyme, PFK-1, it increases the enzymes's activity by increasing its affinity for its substrate fructose-6-phosphate and reduces its affinity for its allosteric inhibitors ATP and citrate. However, when it binds to FBPase, it reduces its activity by reducing its affinity for glucose, its substrate
Calculate the pH of a buffer solution obtained by dissolving 25.025.0 g of KH2PO4(s)KH2PO4(s) and 38.038.0 g of Na2HPO4(s)Na2HPO4(s) in water and then diluting to 1.00 L.
Answer:
pH = 7.37
Explanation:
The buffer H₂PO₄⁻/HPO₄²⁻ has as pKa 7.21. To find the pH of a buffer you can use H-H equation:
pH = pKa + log₁₀ [HPO₄²⁻] / [H₂PO₄⁻]
Where [HPO₄²⁻] and [H₂PO₄⁻] are molar concentrations of each species. As volume is 1.00L, [HPO₄²⁻] and [H₂PO₄⁻] are MOLES.
Moles of 25.0g of KH₂PO₄ (Molar mass: 136.086g/mol):
25.0g KH₂PO₄ ₓ (1mol / 136.086g) = 0.1837 moles KH₂PO₄ = moles H₂PO₄⁻
Moles of 38.0g of Na₂HPO₄ (Molar mass: 141.96g/mol):
38.0g KH₂PO₄ ₓ (1mol / 141.96g) = 0.2677 moles Na₂HPO₄ = moles HPO₄²⁻
Replacing in H-H equation:
pH = pKa + log₁₀ [HPO₄²⁻] / [H₂PO₄⁻]
pH = 7.21 + log₁₀ [0.2677] / [0.1837]
pH = 7.37
If a balloon takes up 625L at 273K, what will the new volume be when the balloon is heated to 353K.
Answer:
The new volume will be 808 L
Explanation:
Charles's law is a law that says that the volume of gas at constant pressure is directly proportional to its absolute temperature (in degrees Kelvin), that is, when the amount of gas and pressure are kept constant, the quotient between volume and temperature will always have the same value:
[tex]\frac{V}{T}=k[/tex]
Having a certain volume of gas V1 that is at a temperature T1 at the beginning of the experiment, by varying the volume of gas to a new value V2, the temperature will change to T2 and the following will be fulfilled:
[tex]\frac{V1}{T1} =\frac{V2}{T2}[/tex]
In this case:
V1= 625 LT1= 273 KV2= ?T2= 353 KReplacing:
[tex]\frac{625 L}{273 K} =\frac{V2}{353 K}[/tex]
Solving:
[tex]V2=353 K*\frac{625 L}{273 K}[/tex]
V2= 808 L
The new volume will be 808 L
Explain why o-vanillin does not fully protonate p-toluidine. Reference appropriate pKa values and include a balanced chemical reaction and an appropriate reaction arrow in your answer.
Answer:
Here's what I get
Explanation:
pKₐ of o-vanillin = 7.81; pKₐ of p-toluidine = 4.44
The higher the pKₐ, the weaker the acid.
Thus, o-vanillin is the weaker acid and has a stronger conjugate base.
The conjugate acid of p-toluidine is the stronger and has the weaker conjugate base.
The equation for the equilibrium is
H-OC₆H₃(OCH₃)CHO + CH₃C₆H₄NH₂ ⇌ ⁻OC₆H₃(OCH₃)CHO + CH₃C₆H₄NH₃⁺
weaker acid weaker base stronger base stronger acid
The reaction between the stronger acid and the stronger base pushes the position of equilibrium to the left.
Thus, o-vanillin does not fully protonate p-toluidine.
O-vanillin is a weaker acid than p-toluidine and has a more stable conjugate base; hence, o-vanillin does not fully protonate p-toluidine.
The pKa is defined as the negative logarithm of Ka. The dissociation constant of an acid Ka shows the extent of dissociation of an acid in solution. The higher the pKa, the lower the Ka and the weaker the acid.
The pKₐ of o-vanillin is 7.81 while the pKₐ of p-toluidine is 4.44. This means that o-vanillin is a weaker acid than p-toluidine and has a more stable conjugate base. Hence, o-vanillin does not fully protonate p-toluidine.
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Nitrogen has different oxidation states in the following compounds: nitrite ion, nitrous oxide, nitrate ion, ammonia, and nitrogen gas. Arrange these species in order of increasing nitrogen oxidation state. Select the correct answer below: A. ammonia, nitrogen gas, nitrite, nitrous oxide, nitrate B. nitrogen gas, ammonia, nitrous oxide, nitrite, nitrate C. ammonia, nitrogen gas, nitrous oxide, nitrite, nitrate D. ammonia, nitrogen gas, nitrate, nitrite, nitrous oxide
Answer:
C. ammonia, nitrogen gas, nitrous oxide, nitrite, nitrate
Explanation:
To establish the oxidation number of nitrogen in each compound, we know that the sum of the oxidation numbers of the elements is equal to the charge of the species.
Nitrite ion (NO₂⁻)
1 × N + 2 × O = -1
1 × N + 2 × (-2) = -1
N = +3
Nitrous oxide (NO)
1 × N + 1 × O = 0
1 × N + 1 × (-2) = 0
N = +2
Nitrate ion (NO₃⁻)
1 × N + 3 × O = -1
1 × N + 3 × (-2) = -1
N = +5
Ammonia (NH₃)
1 × N + 3 × H = 0
1 × N + 3 × (+1) = 0
N = -3
Nitrogen gas (N₂)
2 × N = 0
N = 0
The order of increasing nitrogen oxidation state is:
C. ammonia, nitrogen gas, nitrous oxide, nitrite, nitrate
The cell potential for an electrochemical cell with a Zn, Zn2 half-cell and an Al, Al3 half-cell is _____ V. Enter your answer to the hundredths place and do not leave out a leading zero, if it is needed.
Answer:
0.900 V
Explanation:
Oxidation half cell;
2Al(s) -----> 2Al^3+(aq) + 6e
Reduction half equation;
3Zn^2+(aq) + 6e ----> 3Zn(s)
E°anode = -1.66V
E°cathode= -0.76 V
E°cell= E°cathode - E°anode
E°cell= -0.76-(-1.66)
E°cell= 0.900 V
Which of the following involves a decrease in entropy? Group of answer choices the dissolution of NaCl in water the evaporation of ethanol the sublimation of carbon dioxide the decomposition of N2O4(g) to NO2(g) the freezing of liquid water into ice
Answer:
the freezing of liquid water into ice
Explanation:
Entropy is the degree of disorderliness of a system, entropy is an extensive property of a thermodynamic system. An extensive property of a system is one whose value changes with the number of particles or the amount of matter present in the system.
Gases possess the greatest entropy among the States of matter followed by liquids and lastly solids. Solid particles do not translate because they are held by strong intermolecular forces.
Hence, a change from liquid to solid implies a decrease in entropy since the solid state possesses less entropy in comparison to the liquid state, hence the answer.
Which functional group does the molecule below have?
A. Ether
B. Ester
C. Hydroxyl
D. Amino
Answer:
Hydroxyl
Explanation:
A hydroxyl group is a functional group that attaches to some molecules containing an oxygen and hydrogen atom, bonded together. Also spelled hydroxy, this functional group provides important functions to both alcohols and carboxylic acids.
The functional groups are the part of the organic chemistry that confers the characteristic feature of a molecule. The molecule has a hydroxyl group in its structure. Thus, option C is correct.
What are hydroxyl functional groups?Hydroxyl functional groups are the atoms or molecules that provide a distinctive property to a compound. It has a chemical formula of -OH that has oxygen covalently bonded to the hydrogen atom.
The hydroxyl group is called the alcohol group that is seen in methanol, ethanol, propanol, etc. The presence of hydrogen allows the compound to form a water bond with other molecules and makes them soluble and polar.
Therefore, option C. the molecule has a hydroxyl or alcoholic functional group attached to its carbon atom.
Learn more about the hydroxyl functional group here:
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Eugenol is a molecule that contains the phenolic functional group. Which option properly identifies the phenol in eugenol
Answer:
Explanation:
Hello,
Among the options given on the attached document, since phenolic functional group is characterized by a benzene ring bonded with a hydroxyl group (C₆H₅OH) we can see that the first option correctly points out such description. Thus, answer is on the second attached picture. Other options are related with other sections found in eugenol that are not phenolic.
Best regards.
The first option identified the phenol in eugenol.
Phenolic functional groupAccording to the attached image, since the phenolic functional group should be characterized by a benzene ring bonded along with a hydroxyl group (C₆H₅OH) so here we can see that the first option correctly points out such description. However, other options are related to other sections found in eugenol that are not phenolic.
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g The combustion of ethylene proceeds by the following reaction: C2H4(g) + 3 O2(g) → 2 CO2(g) + 2 H2O(g) If the rate of O2 is -0.23 M/s, then what is the rate (in M/s) of disappearance of C2H4?
Answer:
Explanation:
C₂H₄(g) + 3O₂(g) → 2CO₂(g) + 2H₂O(g)
In this reaction we see that 3 moles of O₂ reacts with one mole of C₂H₄ .
Hence rate of disappearance of O₂ is 3 times faster .
- d [O₂] / dt = - 3 d [ C₂ H₄ ] / dt
Putting the given value
.23 = 3 d [ C₂ H₄ ] / dt
d [ C₂ H₄ ] / dt = .23 / 3
= .077 M / s
Hence the rate of disappearance of C₂ H₄ is .077 moles / s .
For dinner you make a salad with lettuce, tomatoes, cheese, carrots, and
croutons. Your salad would be classified as a(n)
O A. compound
OB. element
OC. homogeneous mixture
D. heterogeneous mixture
A heterogeneous mixture
identify the correct acid/conjugate base pair in this equation:
NaHCO3 + H20 = + H2CO3 + OH
+ Na
H20 is an acid and H2CO3 is its conjugate base.
HCO3 is an acid and OH is its conjugate base.
H20 is an acid and HCO3 is its conjugate base.
H20 is an acid and OH is its conjugate base.
Answer:
H20 is an acid and OH is its conjugate base.
Explanation:
Chemical reactions involving acids and bases occur. An acid is a substance that dissociates in water i.e. lose an hydrogen ion/proton. According to the Bronsted-Lowry acid-base theory, when an acid dissociates in water and loses its hydrogen ion, the resulting substance that forms is the CONJUGATE BASE. A conjugate base is the compound formed as a result of the removal of an H+ ion from an acid.
Based on the chemical reaction in the question, NaHCO3 + H20 = H2CO3 + OH- + Na+
The H20 loses its hydrogen ion (H+) to form an anion OH-. This anion formed is the conjugate base while H20 is its acid.
Diluting sulfuric acid with water is highly exothermic:
(Use data from the Appendix to find for diluting 1.00 mol of H2SO4(l) (d = 1.83 g/mL) to 1 L of 1.00 MH2SO4(aq) (d = 1.060 g/mL). )
Suppose you carry out the dilution in a calorimeter. The initial T is 25.2°C, and the specific heat capacity of the final solution is 3.458 J/gK. What is the final T in °C ?
Answer:
The correct answer is 51.2 degree C.
Explanation:
The standard enthalpy for H₂SO₄ (l) is -814 kJ/mole and the standard enthalpy for H₂SO₄ (aq) is -909.3 kJ/mole.
Now the dHreaction = dHf (product) - dHf (reactant)
= -909.3 - (-814)
dHreaction or q = -95.3 kJ of energy will be used for dissociating one mole of H₂SO₄.
The heat change in calorimetry can be determined by using the formula,
q = mass * specific heat capacity * change in temperature -----------(i)
Based on the given information, the density of H₂SO₄ is 1.060 g/ml
The volume of H₂SO₄ is 1 Liter
Therefore, the mass of H₂SO₄ will be, density/Volume = 1.060 g/ml / 1 × 10⁻³ ml = 1060 grams
The initial temperature given is 25.2 degrees C, or 273+25.2 = 298.2 K, let us consider the final temperature to be T₂.
ΔT = T₂ -T₁ = T₂ - 298.2 K
Now putting the values in equation (i) we get,
95.3 kJ = 1060 grams × 3.458 j/gK (T₂ - 298.2 K) (the specific heat capacity of the final solution is 3.458 J/gK)
(T₂ - 298.2 K) = 95300 J / 1060 × 3.458 = 26 K
T₂ = 298.2 K + 26 K
T₂ = 324.2 K or 324.2 - 273 = 51.2 degree C.
When you placed the chromatography paper in the Petri dish containing the salt-water solution solvent, what would have happened if the level of solvent was above the level of the dye spots on your paper
Answer:
It will not achieve the desired separation
Explanation:
Chromatography is a separation method that involves the use of a stationary phase and a mobile phase. The stationary phase is immobile, in the particular instance of this question, the stationary phase is paper. The mobile phase is the appropriate solvent, in this case, a salt-water solution.
If the level of solvent is above the dye spots, it will introduce error into the separation. The solvent (if volatile) may evaporate without drawing up and separating the solute. Secondly, the solvent may simply dissolve the spots without achieving any meaningful separation of the components in the system. This second reason is particularly why the salt solution must be below the dye spots in this chromatographic separation.
Mrs. Wilson leaves her freshly-baked blueberry pie on the windowsill to cool. The delicious fragrance diffuses through the air with a diffusion coefficient of D = 0.2 cm2/s. How long does it take for Dennis to smell the pie in his treehouse 10 meters away? Give your answer in days, without entering the unit.
Answer:
Poop Buttt.
Explanation:
A researcher places a reactant for decomposition in an expandable reaction chamber and purges the air from the vessel with nitrogen gas. The 500mL reaction vessel is sealed at a pressure of 1.00atm and 390K. If the decomposition reaction was triggered by an electrical shock, producing 3.1g of oxygen gas, what would the volume (L) of the reaction vessel be if the temperature and pressure were kept constant
Answer:
3.1 L
Explanation:
Step 1: Given data
Pressure (P): 1.00 atmTemperature (T): 390 KMass of oxygen (m): 3.1 gVolume (V): ?Step 2: Calculate the moles of oxygen
The molar mass of oxygen is 32.00 g/mol.
[tex]3.1g \times \frac{1mol}{32.00g} = 0.097mol[/tex]
Step 3: Calculate the volume of the container
We will use the ideal gas equation.
P × V = n × R × T
V = n × R × T / P
V = 0.097 mol × (0.0821 atm.L/mol.K) × 390 K / 1.00 atm
V = 3.1 L
What is the standard enthalpy of formation of liquid methylamine (CH3NH2) ?C(s)+O2(g) -> CO2(g); ?H=-393.5 kJ 2H2O(l) -> 2H2(g)+O2(g); ?H=571.6 kJ N2(g)+O2(g) -> NO2(g); ?H=33.10 kJ 4CH3NH2(l)+13O2(g) -> 4CO2(g)+4NO2(g)+10H2O(l); ?H=-4110.4 kJ The calculated answer is -47.3 kJ/mol. Show the work to confirm or deny the answer
Answer:
ΔH =-47.3kJ
Explanation:
You must know standard enthalpy is defined as change of enthalpy during the formation of 1 mole of the substance from its constituent elements
You can find the standard enthalpy of any reaction from the sum of another similar reactions (Hess's law) as follows:
For methylamine, CH₃NH₂ is:
C(s) + 5/2 H₂(g) + 1/2 N₂(g) → CH₃NH₂ (l)
1. C(s) + O₂(g) → CO₂(g); ΔH=-393.5 kJ
2. 2H₂O(l) → 2H₂(g) + O₂(g); ΔH=571.6 kJ
3. 1/2N₂(g) + O₂(g) → NO₂(g); ΔH=33.10kJ
4. 4CH₃NH₂(l) + 13O₂(g) → 4CO₂(g) + 4NO₂(g) + 10H₂O(l); ΔH=-4110.4 kJ
The sum of (1)+(3) produce:
C(s) + 2O₂(g) + 1/2N₂(g) → CO₂(g) + NO₂(g) ΔH=-393.5kJ + 33.10kJ = -360.4kJ
-5/4 (2):
C(s) + 13/4O₂(g) + 1/2N₂(g) + 5/2H₂(g) → CO₂(g) + NO₂(g) + 5/2 H₂O(l)
ΔH= -360.4kJ -5/4 (571.6kJ) = -1074.9kJ
And this reaction -1/4 (4):
C(s) + 5/2 H₂(g) + 1/2 N₂(g) → CH₃NH₂(l)
ΔH= -1074.9kJ -1/4(-4110.4kJ)
ΔH =-47.3kJNow, you can confirm the calculated answer!
Which accurately describes one impact of the atmosphere on Earth’s cycles?
Answer:
Produces Wind Currents
Explanation:
Answer:
produces wind currents
Explanation:
i just took the test and got it right :}
Some metal oxides, such as Sc2O3, do not react with pure water, but they do react when the solution becomes either acidic or basic. Do you expect Sc2O3 to react when the solution becomes acidic or when it becomes basic?
Write a balanced chemical equation to support your answer.
Answer:
[tex]Sc_2O_3[/tex] reacts with an acidic solution
Explanation:
Scandium Oxide [tex]Sc_2O_3[/tex] is a basic metal oxide which therefore reacts with acidic solution. An oxide is a compound that contains only two elements, one of which is oxygen .
The objective of this question is to Write a balanced chemical equation to support your answer.
The chemical equation to support the reaction of [tex]Sc_2O_3[/tex] with acidic solution is as follows:
Assuming the acidic solution to be HCl
[tex]\mathbf{Sc_2O_3_{(s)} + 6 HCl_{(aq)} ----> 2 ScCl_{3(aq)} + 3H_2O_{(l)}}[/tex]
The ionic equation :
[tex]\mathbf{Sc_2O_{3(s)} + 6H^+_{(aq)} ---> 2Sc^{3+}_{(aq)} + 3H_2O_{(l)}}[/tex]
Fermentation is a complex chemical process of making wine by converting glucose C6H12O6 into ethanol C2H5OH and carbon dioxide: C6H12O6(s) ---> 2 C2H5OH (l) + 2 CO2(g) Calculate the mass of ethanol produced if 500.0 grams of glucose reacts completely.
Answer:
[tex]255.71~g~C_2H_5OH[/tex]
Explanation:
First, we have to check if the reaction is balanced:
[tex]C_6H_1_2O_6_(_s_)~->~2C_2H_5OH_(_l_)~+~2CO_2_(_g_)[/tex]
We have 6 carbon atoms on both sides, 12 hydrogens, and 6 oxygens. So, we can continue with the problem. If we want to calculate the mass of ethanol ([tex]C_2H_5OH[/tex]) we need to know:
1) Molar mass of glucose ([tex]C_6H_12O_6[/tex])
In this case, we have to know the atomic mass of each atom:
-) C 12 g/mol
-) O 16 g/mol
-) H 1 g/mol
With the formula we can calculate the molar mass:
(12*6) + (16*6) + (1*12) = 180 g/mol
2) Molar ratio between glucose and ethanol
In the balanced equation we have 1 mol of [tex]C_6H_12O_6[/tex] and 2 moles of [tex]C_2H_5OH[/tex]. So the molar mass is 1:2
3) Molar mass of ethanol ([tex]C_2H_5OH[/tex])
With the formula, we can calculate the molar mass:
(12*2) + (6*1) + (16*1) = 46 g/mol
Finally, we can to the calculation:
[tex]500.0g~C_6H_12O_6\frac{1mol~C_6H_12O_6}{180g~C_6H_12O_6}\frac{2mol~C_2H_5OH}{1mol~C_6H_12O_6}\frac{46g~C_2H_5OH}{1mol~C_2H_5OH}=255.71g~C_2H_5OH[/tex]
I hope it helps!
The branch of science which deals with chemicals and bonds is called chemistry.
The correct answer for the question is 255g of ethanol.
The process of digesting the glucose in absence of oxygen is called fermentation.
Before solving the question, we must check the atoms of compound and balanced it.
The molar mass of the glucose is as follows:
[tex]C_6H_12O_6 = (12*6) + (16*6) + (1*12) = 180 g/mol[/tex].After balancing the equation, the mole ratio of the glucose vs ethanol is 1:2
The molar mass of the ethanol is as follows:-
[tex]C_2H_5OH[/tex] =[tex]12*2) + (6*1) + (16*1) = 46 g/mol[/tex]
After diving each molar mass of each compound with respect to its mole ratio. The mass of the ethanol produced is 255g.
Hence, the correct answer is 255g.
For more information, refer to the link:-
https://brainly.com/question/19524691
p32p32 is a radioactive isotope with a half-life of 14.3 days. if you currently have 63.163.1 g of p32p32 , how much p32p32 was present 8.008.00 days ago
Answer:
92.93 g
Explanation:
Number of half lives that have elapsed in eight days =8/14.3 = 0.559
Fraction of the radioactive nuclide that remains after 0.559 half lives is given by
N/No=(1/2)^0.559
Where N= mass of radioactive nuclides remaining after a time t
No= mass of radioactive nuclides originally present
N/No=(1/2)^0.559= 0.679
Mass of nuclides present eight days before= 63.1g/0.679
Mass of nuclides present eight days before=92.93 g