PV=nRT
P = 85.0 x 10^3 Pa
T = 273 + 20 = 293 K
n = 2.0 moles
R = 8.314 m3. pa / mol .K
V = 2.0 x 293 x 8.314 / 85.0x10^3 = 0.195623529 m^3
What is the molarity of a solution prepared by dissolving 33.4 g of sodium chloride in some water and then diluting the solution to a volume of 300.00 mL?
1. Since you must need the value for moles to find the molarity you have to convert 33.4g into moles.
33.4gNaCl / 58.44gNaCl = .572molNaCl
2. Since you cannot use mL to find molarity, you must convert the value into liters by dividing the value by 1000.
300.00 / 1000 = .3L
3. Now plug in your values into the equation. ( M = mol/L) [ Just divide moles by the liters]
M = .572 / .300 = 1.91M OR 1.91mol/L (Units can be M, or mol/L)
How does an atom develop a positive charge
Answer choices?? Please
An atom that gains a negative electron, it becomes a negative ion. If it loses an electron it becomes a positive ion. ... It now has more positive protons than electrons so it has an overall positive charge. Therefore it is a positive ion.
Why does the sun appear to move from east to west across the sky? A. Due to Earth's revolution B. Due to Earth's rotation C. Due to the sun's revolution D. Due to the sun's rotation
B. Due to Earth's rotation
Which brønsted-lowry acid is not considered to be a strong acid in water?
HNO2 IS WEAK ACID IF U LOOK AT Ka VALUES U FIND ONE LISTED FOR HNO2
In the haber process, ammonia is synthesized from nitrogen and hydrogen: n2 (g) + 3h2 (g) → 2nh3 (g) δg° at 298 k for this reaction is -33.3 kj/mol. the value of δg at 298 k for a reaction mixture that consists of 1.9 atm n2, 1.6 atm h2, and 0.65 atm nh3 is __________. a -40.5 b -1.8 c -3.86 ´ 103 d -104.5 e -7.25 ´ 103
The change in free energy of the system under the given conditions is -40.5 KJ/mol.
N2 (g) + 3H2 (g) → 2NH3 (g)
We have to use the relation; ΔG = ΔG⁰ + RT ln Q
ΔG = Free energy change under the given conditions
ΔG⁰ = standard free energy change
R = gas constant
Q = reaction quotient
We can obtain the reaction quotient from;
Q = [NH3]^2/[N2] [H2]^3
Q = [0.65]^2/[1.9] [1.6]^3
Q = 0.4225/7.7824
Q= 0.0543
Substituting the values;
ΔG⁰ = -33.3 KJ/mol
R = 8.314 JK-1mol-1
T = 298 K
Q = 0.0543
ΔG = -33.3 KJ/mol + (8.314 JK-1mol-1 × 298 K) ln (0.0543)
ΔG = -33.3 KJ/mol + (-7.2 KJ/mol)
ΔG = -40.5 KJ/mol
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To find the value of δG for the reaction mixture, use the equation ΔG = ΔG° + RT ln(Q), where ΔG° is the standard Gibbs free energy change, R is the gas constant, T is the temperature in Kelvin, and Q is the reaction quotient.
Explanation:The value of δG at 298 K for a reaction mixture that consists of 1.9 atm N2, 1.6 atm H2, and 0.65 atm NH₃ can be calculated using the equation ΔG = ΔG° + RT ln(Q), where ΔG° is the standard Gibbs free energy change, R is the gas constant, T is the temperature in Kelvin, and Q is the reaction quotient.
First, calculate the reaction quotient using the given pressures: Q = (NH₃)² / (N₂)(H₂)³. Substitute the given pressures into the equation to find Q. Then, use the equation ΔG = ΔG° + RT ln(Q), where R is the gas constant (8.314 J/(mol·K)), T is the temperature in Kelvin (298 K), ΔG° is the standard Gibbs free energy change (-33.3 kJ/mol), and Q is the reaction quotient. Calculate the value of ΔG at 298 K for the given reaction mixture using the calculated Q value.
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Ultraviolet light from the Sun can A. damage nerve cells but not skin cells. B. damage molecules in skin cells, such as DNA. C. cause skin cells to release stinging chemicals. D. help skin cells to repair DNA faster.
B. It can damage molecules in skin cells such as DNA. Prolonged exposure can actually cause skin cancer.
Ultraviolet light from the Sun primarily damages molecules in skin cells, particularly the DNA, which can lead to skin aging, skin cancer, and wrinkle formation. However, it also aids in Vitamin D production in the skin that has several health benefits. Apart from this, UV light is also used effectively for disinfection purposes.
Explanation:Ultraviolet light from the Sun primarily damages molecules in skin cells, particularly the DNA. This happens through the formation of bonds between an adjacent pair of pyrimidine nucleotides, thymine, and cytosine, on the same strand of DNA. The extent of the damage is often believed to be proportional to the amount of ultraviolet radiation received, which is known as the linear hypothesis.
All types of UV radiation can damage collagen fibres, causing an acceleration of skin aging and wrinkle formation. Overexposure to the Sun when young has been linked to the development of skin cancer like melanoma in later life. However, UV-B radiation from sunlight does have some beneficial effects too, such as Vitamin D production in the skin, reducing risks of certain types of cancer and osteoporosis.
Lastly, UV light can sometimes be used effectively for disinfection. It forms thymine dimers in the DNA of microbes, leading to mutations that can kill the microorganisms. These properties of UV rays are used in water purification systems and germicidal lamps.
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Hydrogen gas can be prepared in the laboratory by a single-displacement reaction in which solid zinc reacts with hydrochloric acid.How much zinc in grams is required to make 14.4 g of hydrogen gas through this reaction?
Zn +2 HCl ---------> ZnCl2 + H2
17.0 g H2 * 1 mol H2 / (2 * 1.008 g H2) * (1 mol Zn / 1 mol H2) =
8.432 mol Zn required
8.432 mol Zn * (65.39 g Zn / mol Zn) = 551.4 g Zn
Answer:
Explanation:
the balanced chemical equation for the displacement reaction between zinc and hydrchloric acid is given as
Zn + 2HCl → ZnCl₂ + H₂
mole ratio; 1 : 2 1 : 1
atomic mass of Zinc = 65.38g
atomic mass of hydrogen =1g
molecular mass of hydrgen(H₂) = 1×2=2g
from the chemical reaction above, it can be deduced that;
2g of hydrgen gas is produced from 65.38g of solid zinc
∴ 14.4g of hydrogen will be needing xg of solid zinc
x=[tex]\frac{14.4 X 65.38}{2}[/tex]
x = 941.472 ÷ 2
x=470.7g
14.4g of hydrogen requires 470.7g of solid zinc
The pressure of a gas is reduced from 1200.0 mmHg to 850.0 mmHg as the temperature of its container is increased from 90.0 c to 785.75 c. What would be the final volume be if the original volume was 85.0 mL ?
Even tho I gotta get to be the first
How do you get the molar ratio and concentration of the acid?
Predict whether each of the following molecules is polar or non-polar:a) IF,b) CS2, c) SO3, d) PCl3, e) SF6, f) IF5.
Polar: IF, PCl3, IF5
Nonpolar: CS2, SO3, SF6
Explanation:Polar molecules form when two atoms do not share electrons equally in a covalent bond. A molecule is classified as a polar molecule when the arrangement of the atoms is such that one end of the molecule has a positive electrical charge and the other end has a negative charge. A non-polar molecule does not have electrical poles.The electrons are distributed more equally. Therefore, a non-polar molecule does not have a profusion of charges at the opposite ends. The majority of hydrocarbon liquids are non-polar molecules.which of the following causes the formation of ionic bonds
A. Two atoms sharing electrons
b. One atom transferring electrons to another atom
C. An element undergoing radioactive decay
D. Fission
Answer: b. One atom transferring electrons to another atom
Explanation: An ionic bond is formed when an element completely transfers its valence electron to another element. The element which donates the electron is known as electropositive element and the element which accepts the electrons is known as electronegative element. This bond is formed between a metal and an non-metal.
Covalent bonds are formed by sharing of electrons between non metals
For example, In calcium iodide the one electron from calcium metal gets transferred to iodine atom and thus form an ionic bond to give [tex]CaI_2[/tex]
Electronic configuration of calcium:
[tex][Ca]=1s^22s^22p^63s^23p^64s^2[/tex]
Calcium atom will lose two electron to gain noble gas configuration and form calcium cation with +2 charge.
[tex][Ca^{2+}]=1s^22s^22p^63s^23p^6[/tex]
Electronic configuration of iodine:
[tex][I]=1s^22s^22p^63s^23p^64s^23d^{10}4p^5[/tex]
Iodine atom will gain one electron to gain noble gas configuration and form iodide ion with -1 charge.
[tex][I^-]=1s^22s^22p^63s^23p^64s^23d^{10}4p^6[/tex]
If the equilibrium concentrations of products are much greater than those of reactants in this system, what would be the magnitude of the equilibrium constant, K,? K < 1 K = 1 K > 1
Answer:
K > 1.
Explanation:
∵ The equilibrium constant K = [products]/[reactants].
Since, [products] > [reactants].
∴ The equilibrium constant K > 1.
A balloon filled with helium occupies 20.0 l at 1.50 atm and 25.0◦c. How many moles of helium will there be in the balloon at stp?
Answer:
There will be 1.23 moles of helium in the balloon at STP
Explanation:
1) Initial conditions of the helium gas:
V = 20.0 literp = 1.50 atmT = 25.0 °C = 25.0 + 273.15 K = 298.15 K2) Ideal gas equation:
pV = n RTp, V, and T are given aboveR is the Universal constant = 0.0821 atm-liter / ( K - mol)n is the unknown number of moles3) Solve for n:
n = pV / (RT) = n = 1.50 atm × 20.0 liter / (0.0821 atm-liter /k -mol ×298.15K)n = 1.23 mol4) At STP:
STP stands for standard pressure and temperature.The amount (number of moles) of the gas will not change because the change of pressure and temperature, so the number of moles reamain the same: 1.23 mol.Convert the following values:
1. 1.35 atm to psi
2. 100.2 kPa to mm Hg
3. 10.83 psi to kPa
1. 1.35 atm to psi = 19.83953 psi.
2. 100.2 kPa to mm Hg = 751.5617 mm / 29.589043 Hg
3. 10.83 psi to kPa = 74.670221 kPa
Hope this helps,
Davinia.
There are several aromatic compounds with the formula c8h9cl. draw those that have a disubstituted ring where the chlorine is not attached to the ring.
The skeletal formulas of the desired ones are drawn.
The two aromatic compounds with the formula C₈H₉Cl are 1-Chloro-2-ethylbenzene and 1-Chloro-3-methylbenzene.
1.
1-Chloro-2-ethylbenzene: In this structure, the chlorine atom (Cl) is attached to the second carbon atom of the ethyl group, which is connected to the aromatic benzene ring.
2.
1-Chloro-3-methylbenzene: In 1-Chloro-3-methylbenzene, the chlorine atom (Cl) is attached to the third carbon atom of the methyl group, which is connected to the aromatic benzene ring. The structure consists of a benzene ring (six carbon atoms in a hexagonal arrangement) with a methyl group attached to the third carbon atom of the ring.
In both cases, the chlorine atom (Cl) is not directly attached to the aromatic ring but is instead attached to a carbon atom that is connected to the ring.
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Question 33 Saved
Volcanic rock has often been found early in the rock record but less frequently in later rock layers. What does this suggest to scientists about early earth?
Question 33 options:
1)
Volcanic activity is the same today as it was early in Earth's history.
2)
Volcanic rock is more easily preserved than other kinds of rock.
3)
Volcanic activity was more common in the past than it is today.
4)
Volcanic rock is easier to find.
Answer:
The third option
Explanation:
If people have found more volcanic rock layers in the past that would mean that volcanic activity was more common in the past.
what is the best form for writing the hypthesis
I think the best way is persuasive. plz forgive me if this is wrong or not the answer that was expected
What two ions are central to the arrhenius definitions of acids and bases?
Answer: hydrogen (H⁺) or hydronium (H₃O⁺) and hydroxide (OH⁻)
Explanation:
1) Arrhenius definitions
The chemist Svante Arrhenius, winner of the Noble Prize in chemistry by 1903, defined acids and bases based on the ability of such substances to give ions in water solutons.
2) Acids
Arrhenius defined an acid as a substance that dissociates in water to form hydrogen ions (H⁺), also called protons. This process is called protonation (formation of protons). Due to its small size and high activity, H⁺ does not exist in that form, but it forms H₃O⁺ ions in aqueous solution.
3) Bases
Arrhenius defined a base as a substance that dissociates in water to form hydroxide (OH⁻) ions.
4) Examples
That was the first definitions of acids and bases and is restricted to water solutions and does not include many compounds (like NH₃) which nowadays are identified as acids or bases.
Some examples of Arrhenius acids and bases are:
KOH (aq) → K⁺ + OH⁻ (base becaue it produces OH⁻)H₂SO₄ (aq) → H⁺ + HSO₄⁻ (acid because ir produced H⁺)According to the Arrhenius definition, acids release hydrogen ions (H⁺) and bases release hydroxide ions (OH⁻) in aqueous solutions. These two ions are central to identifying Arrhenius acids and bases.
According to the Arrhenius definition, acids and bases are classified based on the ions they release in an aqueous solution. An Arrhenius acid is a compound that produces hydrogen ions (H⁺), while an Arrhenius base is a compound that produces hydroxide ions (OH⁻). For example, HCl dissociates in water to form H⁺ ions and Cl⁻ ions, categorizing it as an Arrhenius acid. Conversely, NaOH dissociates to form Na⁺ ions and OH- ions, making it an Arrhenius base.
The central ions in these definitions are thus hydrogen ions (H⁺) for acids and hydroxide ions (OH⁻) for bases.
Question 1
Nuclear processes which involved the splitting of a heavy nucleus into 2 nuclei with smaller mass numbers is called _____________.
Fusion
Nuclear Chain Reaction
Spent rods
Fission
Question 2
When U-235 absorbs a neutron and becomes the unstable nucleus of U-236, this is the beginning of a(n) _________________.
a chain reaction
Nuclear Fusion
spent rods
Exothermic reaction
Question 3
Nuclear power plants use the principle of ________________.
Exothermic Reactions
Spent Rods
Nuclear Fusion
Controlled Nuclear Fission
Question 4
Two serious negatives about nuclear power plants:
1. a potential meltdown of the plant releasing huge amounts of radiation
2. _________________
Hard to find safe nuclear waste disposal of radioactive materials that are radioactive for thousands of years
unleashed chain reactions
Nuclear Fusion
Critical Mass
Question 5
Biological effects of exposure to radiation include:
1. the penetrating ability of the radiation into tissues
2. ___________________________________
uncontrolled chain reactions
nuclear fission
nuclear fusion
The ionizing ability of radiation to affect cell functions
Question 1: Fission
Question 2: Chain reaction
Question 3: Controlled nuclear fission
Question 4: Hard to find safe nuclear disposal of...
Question 5: The ionising ability of radiation to affect cell functions.
Strong bases are
a
also strong acids.
b
weak electrolytes.
c
nonelectrolytes.
d
strong electrolytes.
D strong electrolytes
Strong bases are strong electrolytes that completely dissociate in water to form hydroxide ions.
Explanation:Strong bases are strong electrolytes. They completely dissociate in water, forming hydroxide ions (OH-) that are able to conduct electricity. Examples of strong bases include sodium hydroxide (NaOH) and potassium hydroxide (KOH).
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What is temperature
Temperature is a degree of hotness or coldness the can be measured using a thermometer.Temperature is measured in degrees on the Fahrenheit, Celsius, and Kelvin scales. It's also a measure of how fast the atoms and molecules of a substance are moving.
(Only answer if you’re for certain) Which type of molecule is shown below?
A. Alkene
B. Aromatic
C. Alkyne
D. Alkane
i believe B is the answer
Answer: Alkene
Explanation: Apex
In an exothermic reaction, the water will _______ in a calorimeter. A. release heat B. drop in temperature C. drop in volume D. absorb heat
the correct answer is D, absorb heat
I think the answer is A because exo means release
please make brainliest if right
Which condition will release more energy when an electron changes levels? From n = 4 to n = 3 from n = 3 to n = 2 from n = 2 to n = 1
Answer:
n = 2 to n = 1
Explanation:
The electrons in the closest orbitals to the atomic nucleus experience the strongest attraction. Therefore, for an electron to jump up from the ground state to n = 2 a lot of energy is required to overcome the attractive force of the nucleus (also means this energy will be dispensed when the electron jumps down). The energy required to jump up to other energy levels decreases as these orbitals are shielded from the attractive force of the nucleus.
Answer:
From n = 2 to n = 1
Explanation:
There are 7 energy levels, numbered from 1 to 7, and in which electrons are distributed, logically in order according to their energy level. Electrons with less energy will be spinning at level 1. Level 1 is the innermost level or closest to the nucleus and is the one with the lowest energy level. Level 7 is the outermost or furthest from the core and is the level that has the highest energy level.
Then the electrons are spinning around the nucleus forming layers, as previously mentioned. In each of them, the energy that the electron possesses is different. In the layers very close to the nucleus, the force of attraction between it and the electrons is very strong, so they will be strongly bound (joined). The opposite occurs in remote layers, in which electrons are weakly bound, so it will be easier to make electronic exchanges in the last layers.
So, since electrons are more strongly attracted to the nucleus, for an electron to jump from the fundamental state at level n = 2 to level 1, a lot of energy is required to overcome the aforementioned force of attraction. At the other levels, the attraction of the nucleus is lower, so the energy required to jump to other energy levels will also be lower.
Hydrofluoric acid, hf, has a ka of 6.8 × 10−4. what are [h3o+], [f−], and [oh−] in 0.710 m hf?
Answer:
[H₃O⁺] = [F⁻] = 2.2 x 10⁻² M. & [OH⁻] = 4.55 x 10⁻¹³.
Explanation:
For a weak acid like HF, the dissociation of HF will be:HF + H₂O ⇄ H₃O⁺ + F⁻.
[H₃O⁺] = [F⁻].
∵ [H₃O⁺] = √Ka.C,
Ka = 6.8 x 10⁻⁴, C = 0.710 M.
∴ [H₃O⁺] = √Ka.C = √(6.8 x 10⁻⁴)(0.710) = 2.197 x 10⁻² M ≅ 2.2 x 10⁻² M.
∴ [H₃O⁺] = [F⁻] = 2.2 x 10⁻² M.
∵ [H₃O⁺][OH⁻] = 10⁻¹⁴.
∴ [OH⁻] = 10⁻¹⁴/[H₃O⁺] = 10⁻¹⁴/(2.2 x 10⁻²) = 4.55 x 10⁻¹³.
The value of concentration of [H₃O⁺] & [F⁻] is 2.2 x 10⁻² M and [OH⁻] is 4.55 x 10⁻¹³M in 0.710M HF.
How we calculate acid dissociation constant?
Acid dissociation constant for any HA acid will be calculated as:
Ka = [H⁺][A⁻] / [HA].
ICE table for given reaction is written as:
HF + H₂O ⇄ H₃O⁺ + F⁻
Initial: 0.710 0 0
Change: -x +x +x
Equilibrium: 0.710-x +x +x
Given value of Ka = 6.8 × 10⁻⁴
Putting all values on the above value of Ka, we get
6.8 × 10⁻⁴ = x.x / 0.710-x
We take the value of 0.710-x as 0.710, because value of x is negligible as compare to 0.710.
6.8 × 10⁻⁴ = x.x / 0.710
x² = 6.8 × 10⁻⁴ × 0.710
x = 2.197 x 10⁻² M ≅ 2.2 x 10⁻² M
So, [H₃O⁺] = [F⁻] = 2.2 x 10⁻² M
We know that pH + pOH = 14, we write this equation in base 10 form as:
[H₃O⁺][OH⁻] = 10⁻¹⁴
[OH⁻] = 10⁻¹⁴ / 2.2 x 10⁻² = 4.55 x 10⁻¹³
Hence, value of [H₃O⁺] & [F⁻] is 2.2 x 10⁻² M and [OH⁻] is 4.55 x 10⁻¹³M.
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A 25.0 ml sample of aqueous sodium hydroxide has been used to titrate to the second equivalence point 22.30 ml of 0.253 m sulfuric acid. What is the molarity of the sodium hydroxide solution?
The molarity of the sodium hydroxide is calculated by first calculating the number of moles of sulfuric acid in the 22.30 ml used.
This is done as follows since 0.253M is contained in 1000ml solution.
(0.253×22.3)/1000=0.00564 moles
The equation for the reaction is:
H₂SO₄₍ₐq)+2NaOH₍ₐq)⇒ Na₂SO₄₍aq) +2 H₂O₍l₎
Therefore the reacting ratios between sodium hydroxide and sulfuric acid is 2:1
therefore the number of moles that reacted with the sulfuric acid is calculated as follows:
(0.00564 moles ₓ 2)/1=0.01128moles
0.1128moles is in 25ml therefore, 1000ml has:
(1000ₓ0.01128moles)/25= 0.4512M NaOH
22.30 mL of 0.253 M H₂SO₄ titrates 25.0 mL of 0.448 M NaOH.
Let's consider the neutralization reaction between sulfuric acid and sodium hydroxide.
2 NaOH + H₂SO₄ ⇒ Na₂SO₄ + 2 H₂O
22.30 mL of 0.253 M H₂SO₄ react. The reacting moles of H₂SO₄ are:
[tex]0.02230 L \times \frac{0.253mol}{L} = 5.64 \times 10^{-3} mol[/tex]
The molar ratio of NaOH to H₂SO₄ is 2:1. The reacting moles of NaOH are:
[tex]5.64 \times 10^{-3} mol H_2SO_4 \times \frac{2molNaOH}{1 molH_2SO_4} = 0.0112 molNaOH[/tex]
0.0112 moles of NaOH are in 25.0 mL of solution. The molarity of NaOH is:
[tex]M = \frac{0.0112 mol}{0.0250L} = 0.448 M[/tex]
22.30 mL of 0.253 M H₂SO₄ titrates 25.0 mL of 0.448 M NaOH.
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Be sure to answer all parts. Predict the effect of decreasing the temperature on the amounts of reactants in the following reactions: (a) C2H2(g) + H2O(g) ⇌ CH3CHO(g) ΔH o rxn = −151 kJ The amounts of reactants stay the same. The amounts of reactants decrease. The amounts of reactants increase. (b) CH3CH2OH(l) + O2 (g) ⇌ CH3CO2H(l) + H2O(g) ΔH o rxn = −451 kJ The amounts of reactants stay the same. The amounts of reactants decrease. The amounts of reactants increase. (c) 2 C2H4(g) + O2(g) ⇌ 2 CH3CHO(g) (exothermic) The amounts of reactants stay the same. The amounts of reactants decrease. The amounts of reactants increase. (d) N2O4(g) ⇌ 2 NO2(g) (endothermic) The amounts of reactants decrease. The amounts of reactants stay the same. The amounts of reactants increase.
Answer:
(a) The amounts of reactants decrease.
(b) The amounts of reactants decrease.
(c) The amounts of reactants decrease.
(d) the amounts of reactants increase.
Explanation:
Le Châtelier's principle states that when there is an dynamic equilibrium, and this equilibrium is disturbed by an external factor, the equilibrium will be shifted in the direction that can cancel the effect of the external factor to reattain the equilibrium.(a) C₂H₂(g) + H₂O(g) ⇌ CH₃CHO(g) ΔH°rxn = − 151 kJ.
Since, ΔH is a negative value, the reaction is exothermic reaction.So, the reaction can be represented as:
C₂H₂(g) + H₂O(g) ⇌ CH₃CHO(g) + heat.
As the temperature is decreases, it is like that decreasing the concentration of products side, that shifts the reaction towards the right side (products side) to attain the equilibrium again. So, the amounts of reactants decrease.
So, the right choice is reaction is: The amounts of reactants decrease.
(b) CH₃CH₂OH(l) + O₂ (g) ⇌ CH₃CO₂H(l) + H₂O(g) ΔH°rxn = − 451 kJ.
Since, ΔH is a negative value, the reaction is exothermic reaction.So, the reaction can be represented as:
CH₃CH₂OH(l) + O₂ (g) ⇌ CH₃CO₂H(l) + H₂O(g) + heat.
As the temperature is decreases, it is like that decreasing the concentration of products side, that shifts the reaction towards the right side (products side) to attain the equilibrium again. So, the amount of reactants decrease.
So, the right choice is reaction is: The amounts of reactants decrease.
(c) 2C₂H₄(g) + O₂(g) ⇌ 2CH₃CHO(g) (exothermic).
Since, the reaction is exothermic reaction.So, the reaction can be represented as:
2C₂H₄(g) + O₂(g) ⇌ 2CH₃CHO(g) + heat.
As the temperature is decreases, it is like that decreasing the concentration of products side, that shifts the reaction towards the right side (products side) to attain the equilibrium again. So, the amounts of reactants decrease.
So, the right choice is reaction is: The amounts of reactants decrease.
(d) N₂O₄(g) ⇌ 2NO₂(g) (endothermic).
For endothermic reaction, it can be represented as:N₂O₄(g) + heat ⇌ 2NO₂(g)
As the temperature is decreased, it is like that decreasing the concentration of reactants side, that shifts the reaction towards the lift side (reactants side) to attain the equilibrium again. So, the amounts of reactants increase.
So, the right choice is reaction is: the amounts of reactants increase.
Can someone help me with #1 please?
Remark
HA is an acid that produces only 1 Hydrogen when it breaks down.
The NaOH is a base and a good strong one.
When you do the titration you can be guaranteed that ever mole of NaOH that is consumed represents 100% of what is needed.
So your first step is to find out how many moles of NaOH is needed.
Givens
C = 0.1 mol/L
V = 23.64 mL * [ 1 L / 1000 mL] = 0.02364 L
Formula
C = mol / L
Solution
What you are looking for is the number of mols of NaOH used.
0.1 mol/L = mol / 0.02364 L Multiply both sides by 0.02364
0.1 mol/L * 0.02364 L = mol
mol = 0.002364
Now the number of mols of HA is going to be exactly the same. That's because the titration formula is
HA + NaOH ==> NaA + HOH
==========================
mols = given mass / molar Mass
Molar Mass = given mass / mols
mols = 0.002364 moles
given mass = 0.5632
Molar Mass = ??
Molar Mass = 0.5632 grams / 0.002364
Molar Mass = 238.24 grams
Calculate the approximate volume of a 0.600 mol sample of gas at 15.0°C and a pressure of 1.10?12.9 L 24.6 L 22.4 L 129 L
A 12.9 L
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What is ΔH∘rxn for the following chemical reaction? CS2(g)+2H2O(l)→CO2(g)+2H2S(g) You can use the following table of standard heats of formation (ΔH∘f) to calculate the enthalpy of the given reaction. Element/ Compound Standard Heat of Formation (kJ/mol) Element/ Compound Standard Heat of Formation (kJ/mol) H(g) 218 N(g) 473 H2(g) 0 O2(g) 0 H2O(l) −285.8 O(g) 249 CS2(g) 116.7 H2S(g) −20.60kJ C(g) 71 CO2(g) −393.5kJ C(s) 0 HNO3(aq) −206.6 Express the standard enthalpy of reaction to three significant figures and include the appropriate units. View Available Hint(s)
Answer : The standard enthalpy of the reaction is 20.2 kJ
Explanation :
Enthalpy change : It is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as [tex]\Delta H^o[/tex]
The equation used to calculate enthalpy change is of a reaction is:
[tex]\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f(product)]-\sum [n\times \Delta H^o_f(reactant)][/tex]
The equilibrium reaction follows:
[tex]CS_2(g)+2H_2O(l)\rightleftharpoons CO_2(g)+2H_2S(g)[/tex]
The equation for the enthalpy change of the above reaction is:
[tex]\Delta H^o_{rxn}=[n_{(CO_2)}\times \Delta H^o_f_{(CO_2)}+n_{(H_2S)}\times \Delta H^o_f_{(H_2S)}]-[n_{(H_2O)}\times \Delta H^o_f_{(H_2O)}+n_{(CS_2)}\times \Delta H^o_f_{(CS_2)}][/tex]
We are given:
[tex]\Delta H^o_f_{(CS_2(g))}=116.7kJ/mol\\\Delta H^o_f_{(H_2O(l))}=-285.8kJ/mol\\\Delta H^o_f_{(CO_2(g))}=-393.5kJ/mol\\\Delta H^o_f_{(H_2S(g))}=-20.60kJ/mol[/tex]
Putting values in above equation, we get:
[tex]\Delta H^o_{rxn}=[(1mol\times -393.5kJ/mol)+(2mol\times -20.60kJ/mol)]-[(1mol\times 116.7kJ/mol)+(2mol\times -285.8kJ/mol)]=20.2kJ[/tex]
Therefore, the standard enthalpy of the reaction is 20.2 kJ
The standard enthalpy of reaction is 20.2 kJ/mol.
We can calculate the standard enthalpy of reaction from the standard heat of formation using the formula;
ΔHreaction = ∑ΔH∘f products - ΔH∘f reactants
The equation of the reaction is; CS2(g)+2H2O(l)→CO2(g)+2H2S(g)
ΔH∘f CS2 = 116.7 KJ/mol
ΔH∘f H2O(l) = −285.8 KJ/mol
ΔH∘f CO2 = −393.5 kJ/mol
ΔH∘f H2S= −20.60 kJ/mol
Substituting values, we have;
ΔHreaction = ∑[(−393.5) + 2 ×(−20.60) - (116.7) + 2 ×(−285.8)] kJ/mol
ΔHreaction = 20.2 kJ/mol
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