Answer:
C. The enthalpy of the products is higher than that of the reactants
Explanation:
An endothermic process is any process with an increase in the enthalpy of the system
The enthalpy of the products is higher than that of the reactants describes an endothermic process. Therefore, option (C) is correct.
What is the endothermic process?An endothermic process can be described as any thermodynamic process with an increase in the enthalpy H of the system. In such a process, a closed system commonly absorbs thermal energy from its surroundings.
An endothermic reaction leads to an increase in the temperature of the system and a decrease temperature of the surroundings. A physical process, such as the melting of ice cubes is an endothermic process.
Whether a process can take place spontaneously depends not only on the enthalpy change but also on the entropy change and absolute temperature.
An endothermic process commonly needs a favorable entropy increase (∆S > 0) in the system. In an endothermic process, the energy of the products is higher than that of the reactants.
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Suppose you are studying the Ksp of CaCl2, which has a molar mass of 110.98 g/mol, at multiple temperatures. You dissolve 4.99 g of CaCl2 in 10.0 mL of water at 100 oC and cool the solution. At 90 oC, a solid begins to appear. What is the Ksp of CaCl2 at 90 oC
Answer:
Hence the Solubility product,
Ksp = [Ca2+] [Cl-]2
or, Ksp = (4.5) (9)2
or, Ksp = 364.5
Explanation:
Mass of CaCl2 = 4.99 g
Molar mass of CaCl2 = 110.98 g/mol
Moles of CaCl2
= given mass/ molar mass
= 4.99/ 110.98
= 0.045
Volume = 10.0 mL = 0.01 L
CaCl2 dissociates into its ion as:
CaCl2 (s) \rightleftharpoons Ca2+ (aq) + 2 Cl- (aq)
At 90°C, the solution is saturated with Ca2+ and Cl- ions.
Moles of Ca2+ = Moles of CaCl2 dissolved = 0.045
Moles of Cl- = 2 x ( Moles of CaCl2 dissolved) = 2 x 0.045 = 0.09
[Ca2+] = Moles/ Volume = 0.045/ 0.01 = 4.5 M
[Cl-] = 0.09/ 0.01 = 9 M
Solubility product,
Ksp = [Ca2+] [Cl-]2
or, Ksp = (4.5) (9)2
or, Ksp = 364.5
A sample of a compound is found to consist of 0.44g H and 6.92g O what’s its formula
Answer:
Explanation:
H = 0.44/1.01 = 0.4356
O = 6.92/16 = 0.4319
This gives a 1:1 ratio. So the closest thing you could say is the formula is 0H
Going to your chemical storage room, you could justify that it is H2O2 or hydrogen peroxide. The question needs one more fact to make the answer certainty.
A reaction was performed, and the dichloromethane solvent was dried by adding magnesium sulfate drying agent. When the reaction flask was shaken, it was observed that the magnesium sulfate clumped together at the bottom of the flask. What does this observation indicate
The clumping of magnesium sulfate means that the wrong kind of drying agent have been used for the sample.
What is a drying agent?A drying agent is also referred to as a desiccant. It is a substance that is used to remove moisture from a sample. We must recall that the drying agent to be used must not react with the sample.
Since the magnesium sulfate was found to clump together at the bottom of the flask, it means that the wrong kind of drying agent have been used for the sample.
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GM 2 all ,What is an atom define it .Good Day
Answer:
An atom is the smallest particle of an element that can take part in chemical reaction.
Explanation:
hope it will help u Amri
Calculate the mass percent of each component in the following solution.
159 g NiCl2 in 500 g water
% Nicla
% water
Answer:
% NiCl2 = 24.13%
% water = 78.57%
Explanation:
Mass percentage = mass of solute/mass of solution × 100
According to this question, a solution contains 159 g of NiCl2 in 500 g of water. Hence, mass of the solution is calculated as follows:
Mass of solution = 159g + 500g
Mass of solution = 659g
Therefore;
A) % Mass of NiCl2 in solution = mass of NiCl2/mass of solution × 100
% Mass of NiCl2 in solution = 159/659 × 100
% Mass of NiCl2 in solution = 0.2413 × 100
= 24.13%
B) % Mass of water in solution = mass of water/mass of solution × 100
% Mass of water in solution = 500/659 × 100
% Mass of water in solution = 0.7587 × 100
% Mass of water in solution = 75.87%
Which of the following ligands is not capable of exhibiting linkage isomerism?
a. NCO-
b. -OH
c. -CN
d. -SCN
Answer:
a
...
........
...........
The conversion of CICH=CHCI to CI2C=CC12 can be carried out with:
C12 / H20
C12
C12/ hv
C12 / aq. NaOH
Answer:
I think third C12/hv is right answer
If a buffer is composed of 23.34 mL of 0.147 M acetic acid and 33.66 mL of 0.185 M sodium acetate, how many mL of 0.100 M NaOH can be added before the buffer capacity is reached
Answer:
25.5mL of 0.100M NaOH are needed to reach buffer capacity.
Explanation:
The buffer capacity is reached when the ratio between moles of conjugate base (Sodium acetate) and moles of weak acid (Acetic acid) is 10:
Moles sodium acetate / Moles Acetic acid = 10
The reaction of acetic acid, HA, with NaOH, to produce sodium acetate, NaA is:
HA + NaOH → H2O + NaA
That means the moles of NaOH added = Moles of HA that are being subtracted and moles of NaA that are been produced.
The initial moles of each species is:
Acetic acid:
23.34mL = 0.02334L * (0.147mol / L) = 0.00343 moles Acetic Acid
Sodium Acetate:
33.66mL = 0.03366L * (0.185mol / L) = 0.00623 moles Sodium Acetate
We can write the moles of each species when NaOH is added as:
Moles sodium acetate / Moles Acetic acid = 10
0.00623 moles + X / 0.00343 moles - X = 10
Where X are moles of NaOH added
Solving for X:
0.00623 moles + X = 0.0343 moles - 10X
11X = 0.0281
X = 0.00255 moles of NaOH are needed
In Liters:
0.0255mol NaOH * (1L / 0.100mol) = 0.0255L of 0.100M NaOH are needed =
25.5mL of 0.100M NaOH are needed to reach buffer capacity
Preganglionic axons are myelinated; postganglionic axons are unmyelinated.
a. True
b. False
Answer: True no need for an explanation..
Preganglionic axons are myelinated; postganglionic axons are unmyelinated True
In the reoxidation of QH2 by purified ubiquinone-cytochrome c reductase (Complex III) from heart muscle, the overall stoichiometry of the reaction requires 2 mol of cytochrome c per mole of QH2 because:
Answer: Options related to your question is missing below are the missing options
a. cytochrome c is a one-electron acceptor, whereas QH2 is a two-electron donor.
b. cytochrome c is a two-electron acceptor, whereas QH2 is a one-electron donor.
c. cytochrome c is water soluble and operates between the inner and outer mitochondrial membranes
d. heart muscle has a high rate of oxidative metabolism, and therefore requires twice as much cytochrome c as QH2 for electron transfer to proceed normally.
e. two molecules of cytochrome c must first combine physically before they are catalytically active.
answer:
cytochrome c is a one-electron acceptor, whereas QH2 is a two-electron donor. ( A )
Explanation:
The overall stoichiometry of the reaction requires 2 mol of cytochrome per mole of QH2 because a cytochrome is simply a one-electron acceptor while QH2 is not a one-electron donor ( i.e. it is a two-electron donor )
An electron donor in a reaction is considered a reducing agent because it donates its electrons to another compound thereby self oxidizing itself in the process.
Identify “A” in the following reaction: CH3¬COOH + Na2CO3 → A + CO2 + H2O
During the reaction of 2-methyl-2-butanol with the nucleophile-solvent mixture two layers are formed after shaking the reaction for 5 minutes. After removing the aqueous layer with a Pasteur pipette the organic layer is diluted with 1 mL dichloromethane. The organic phase is washed with 1 mL water. Two layers are obtained.
a. Top layer is Aqueous (H20/ H2SO4/NH4CI)
b. Top layer is Organic (CH2Cl2 and product)
c. Bottom layer is Organic (CH2Cl and product)
d. Top layer is Aqueous (H20)
Answer:
Top layer is Organic (CH2Cl2 and product)
Explanation:
In a solvent mixture, there are usually two phases, the organic phase and the aqueous phase.
It is usual that the organic phase is almost always less dense than the aqueous phase hence the organic phase tend to remain on top of the aqueous phase.
Hence, the top layer is expected to be the organic CH2Cl2 and product.
The peptide alanylglutamylalanylglycylalanylleucine has:
No free carboxyl groups
Five peptide bonds
A disulphide bridge
Two free amino groups
Peptide bonds are formed between carboxyl groups of one amino acid and amino group of another dehydration synthesis.
The correct answer is Five Peptide bonds
The peptide of "alanylglutamylalanylglycylalanylleucine" will have four or more peptide bonds.
It will also have free carboxyl groups.
The chain of amino acids is polypeptide.
Usually C terminal amino acid is the residual at the end of peptide bonds formation.
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Nitric acid can be formed in two steps from the atmospheric gases nitrogen and oxygen, plus hydrogen prepared by reforming natural gas. In the first step, nitrogen and hydrogen react to form ammonia: N2 (g) + 3H2 (g) â 2NH3 (g) =ÎHâ92.kJ In the second step, ammonia and oxygen react to form nitric acid and water:
NH3 (g) + 2O2 (g) â HNO3 (g) + H2O (g) =ÎHâ330.kJ
Required:
Calculate the net change in enthalpy for the formation of one mole of nitric acid from nitrogen, hydrogen and oxygen from these reactions.
Answer:
-376 kJ
Explanation:
The first step equation:
[tex]\mathsf{N_{2(g)} + 3H_2{(g)} \to 2NH_3{(g)} \ \ \ \Delta H = -92\ kJ}[/tex] ---- (1)
The second step equation:
[tex]\mathsf{NH_{3(g)} + 2O_2{(g)} \to HNO_3{(g)} +H_2O_{(g)} \ \ \ \Delta H = -330\ kJ}[/tex] ---- (2)
To determine the enthalpy of formation for 1 mole of HNO₃ (nitric acid), we have the following.
From the above equations; let multiply equation (1) by 1 and equation (2) by 2.
[tex]\mathsf{N_{2(g)} + 3H_2{(g)} \to 2NH_3{(g)} \ \ \ \Delta H = -92\ kJ}[/tex] ---- (3)
[tex]\mathsf{2NH_{3(g)} + 4O_2{(g)} \to 2HNO_3{(g)} +2H_2O_{(g)} \ \ \ \Delta H = 2(-330)\ kJ}[/tex] ----- (4)
adding the above two equations, we have:
[tex]\mathsf{N_{2(g)} + 3H_2{(g)}+ 2NH_{3(g)} + 4O_{2(g)} \to 2HNO_{3(g)} + 2NH_3{(g)} +2H_2O_{(g)} \ \ \ \Delta H = (-660 \ kJ -92\ kJ)}[/tex][tex]\mathsf{N_{2(g)} + 3H_2{(g)} + 4O_{2(g)} \to 2HNO_{3(g)} +2H_2O_{(g)} \ \ \ \Delta H = (-752 \ kJ)}[/tex]
Now, from the recent equation, we have:
2 moles of nitric acid = -752 kJ
∴
1 mole of nitric acid will be: = (1 mole × (-752 kJ)) ÷ 2 moles
1 mole of nitric acid will be: = -376 kJ
What process occurs during the corrosion of iron?
Answers
A.
Iron is oxidized.
B.
Iron is reduced.
C.
Iron (III) is oxidized.
D.
Iron (III) is reduced.
Answer:
A
Explanation:
The iron corrodes so it oxidized
Na Sa Bant HCL -> 50g Hao pt Soy
North America and south africa
What conversion factor would we need to convert moles of helium to atoms of helium?
Explanation:
The conversion factor required to convert moles of helium to atoms of helium is equal to 6.023*10²³ (NA). Helium is a monoatomic gas. Each molecule of helium gas contains one atom of helium. Therefore, one mole of helium gas will contain one mole atoms of helium.
Lead of mass 0.75kg is heated from 21°c to its melting point and continues to be heated unit it has all melted. Calculate how much energy is supplied to the lead. [Melting point of lead 372.5°c specific latent heat of fusion of lead = 23000 Jkg 'k ']
Answer:
65.5J
Explanation:
ML=Q
ML=MC(change in temperature)
0.75 X 23000 =0.75 X 351 X C
C= 65.5J
The energy supplied to the lead to melt from 21°c to its melting point is 51521 Joules.
What is the specific heat capacity?Specific heat is the amount of heat energy supplied to change the temperature of one unit mass of a substance by 1 °C. The SI unit of the specific heat capacity of a substance is J/Kg.
The mathematical expression for the specific heat capacity can be written as:
Q = mCΔT Where C is the specific heat of the substance.
The specific heat capacity depends upon the starting temperature and is an intensive characteristic of the material.
Given, the melting point of the lead T₂ = 327.5° C
The initial temperature of the lead, T₁ = 21° C
The latent heat of the lead given, L = 230000 J/Kg K
The specific heat of the lead, C = 130 J/Kg K
The heat required to melt the lead from 12°C to 327.5 °C is :
Q = m× [C × (T₂ - T₁) + L ]
Q = 0.75 × [0.130 (327.5 - 21) + 23000]
Q = 51521 J
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công thức phân tử của glucozo
C₆H₁₂O₆ is the molecular formula of gulcozo.
using the balanced equation below how many grams of lead(||) sulfate would be produced from the complete reaction of 23.6 g lead (|V) oxide
Answer:
59.8 g of PbSO₄.
Explanation:
The balanced equation for the reaction is given below:
Pb + PbO₂ + 2H₂SO₄ —> 2PbSO₄ + 2H₂O
Next, we shall determine the mass of PbO₂ that reacted and the mass of PbSO₄ produced from the balanced equation. This can be obtained as follow:
Molar mass of PbO₂ = 207 + (16×2)
= 207 + 32
= 239 g/mol
Mass of PbO₂ from the balanced equation = 1 × 239 = 239 g
Molar mass of PbSO₄ = 207 + 32 + (16×4)
= 207 + 32 + 64
= 303 g/mol
Mass of PbSO₄ from the balanced equation = 2 × 303 = 606 g
SUMMARY:
From the balanced equation above,
239 g of PbO₂ reacted to produce 606 g of PbSO₄.
Finally, we shall determine the mass of PbSO₄ that will be produced by the reaction of 23.6 g of PbO₂. This can be obtained as follow:
From the balanced equation above,
239 g of PbO₂ reacted to produce 606 g of PbSO₄.
Therefore, 23.6 g of PbO₂ will react to produce = (23.6 × 606) / 239 = 59.8 g of PbSO₄.
Thus, 59.8 g of PbSO₄ were obtained from the reaction.
If the balance were not tared prior to weighing out the KHP.... how would you expect this to affect the molarity of NaOH calculated? What type of error is this?
Answer:
Following are the response to the given question:
Explanation:
In the given scenario, When the balance has never been tainted before the KHP is weighted, which can affect the molar concentration of NaOH because its molarity is directly proportional to the weight including its substance. In this question it is the mistake is systemic because it may be corrected by modifying balancing parameters.
12 g of powdered magnesium oxide reacts with nitric acid to
form magnesium nitrate and water.
Calculate the mass of magnesium nitrate formed.
[Relative atomic mass of Mg = 24, N = 14,0 = 16] *
Answer:
80.8 g
Explanation:
First, let's write a balanced equation of this reaction
MgO + 2HNO₃ → Mg(NO₃)₂ + H₂O
Now let's convert grams to moles
We gotta find the weight of MgO
24 + 16 = 40 g/mol
12/40 = 0.3 moles of MgO
We can use this to find out how much Magnesium Nitrate will be formed
0.3 x 1 MgO / 1 Mg(NO₃)₂ = 0.3 moles of Magnesium Nitrate formed
Convert moles to grams
Find the weight of Mg(NO₃)₂ but don't forget that 2 subscript acts as a multiplier of whatever is inside that parenthesis.
24 + 14 x 2 + 16 x 3 x 2 = 148 g/mol
148 x 0.3 = 80.8 g
Which of the following ions is the less likely to be formed?
A) Li+3
B) Na+
C) I-
D) Sr2+
Ε) Η+
Answer:
Li^3+
Explanation:
The electronic configuration of lithium is ; 1s2 2s1. This means that lithium has one electron in its outermost shell and two core electrons.
We know that it is difficult to remove these core electrons during ionization. Lithium belongs to group 1 hence Li^+ is formed more easily.
It is very difficult to form Li^3+ because it involves loss of core electrons which requires a lot of energy.
What would be the name of this compound?
Answer:
2,3 Dimethyl hexane
Explanation:
First, start the count from which side is given the CH3 smallest number
first; the longest carbon chain in this compound is 6
and you don't have any double and triple bonds or functional groups so it is Hexane
you start to count from the right side to give the branch molecules the smallest number ..
CH3 = methyl
and you have 2 methyl in this compound ..
and 2 mean you must write ( Di )
you write the name in this way
2,3 Dimethyl hexane
hope this helps you.
stay safe ...
To prepare a standard (calibration) curve for a spectroscopy experiment, start by preparing ___________ with ______________ Then, measure the ______________ of each solution at _____________ and create a plot of ____________ for the measured values. Finally, find the best-fit line of the data set.
Answer: See explanation
Explanation:
The calibration curve is the method used for the determination of the concentration of a substance such that the unknown sample will be compared to some standard samples of the known concentration.
To prepare a standard (calibration) curve for a spectroscopy experiment, start by preparing (multiple solutions) with (different known concentrations). Then, measure the (absorbance) of each solution at (thesame wavelength) and create a plot of (absorbance vs. concentration) for the measured values. Finally, find the best-fit line of the data set.
A couple coffee changes from a temperature of 75ﾟC to 12ﾟC what is the change in temperature in ﾟC
Explanation:
Answer
Open in answr app
The initial temperature of the coffee T0=100∘C.
The temperature of the room 15∘C.
Let T be the temperature at time t in minutes.
By Newton's law of cooling,
dtdTα(T−15)
∴dtdT=k(T−15)
⇒T−15dT=kdt
⇒∫T−15dT=∫kdt
⇒log(T−15)=kt+logc
⇒log(cT−15)=kt
⇒T−15=cekt
At t=0,T=100
We have 100−15=ce0⇒85=c
Thus we get T−15=85ekt.
When t=5minutes,T=60
⇒60−15=85e5k.
⇒e
Which of the following is classified as paramagnetic?
1.NO(+)
2.Be2
3.B2(+)
4.+N2(+)
calculate the hydrogen ion concentration of a solution who's pH is 2.4
Answer:
I don't know sorry yyyyyyy6yyyyyyyyyyyyyyyyyyyyyyyyyyy
Write down the molecular formula and molecular weight of carbon dioxide .
Answer:
CO2
Molar mass: 44.01 g/mol
Explanation:
CO2
In what areas of the periodic table do you find the most highly reactive elements?
Answer:
The elements toward the bottom left corner of the periodic table are the metals that are the most active in the sense of being the most reactive.
The most highly reactive elements are typically found at the far left (Group 1) and far right (Group 17) of the periodic table.
Highly reactive elements in the periodic tableGroup 1 elements, also known as alkali metals, are located on the far left of the periodic table. They have one electron in their outermost energy level and are highly reactive due to their tendency to lose that electron to achieve a stable electron configuration. This makes them very reactive with water and other substances.
Group 17 elements, known as halogens, are located on the far right of the periodic table. They have seven electrons in their outermost energy level and are highly reactive due to their strong tendency to gain one electron to achieve a stable electron configuration. This makes them reactive with metals and other elements.
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