Answer:
3.97 L
Explanation:
Data obtained from the question include the following:
Initial volume (V1) = 3.5 L
Initial temperature (T1) = 19 °C
Final temperature (T2) = 58 °C
Final volume (V2) =..?
Next, we shall convert celsius temperature to Kelvin temperature. This can be done as shown below:
Temperature (K) = temperature (°C) + 273
T (K) = T (°C) + 273
Initial temperature (T1) = 19 °C
Initial temperature (T1) = 19 °C + 273 = 292 K
Final temperature (T2) = 58 °C
Final temperature (T2) = 58 °C + 273 = 331 K
Finally, we shall determine the new volume of the gas by using Charles' law equation as shown below:
Initial volume (V1) = 3.5 L
Initial temperature (T1) = 292 K
Final temperature (T2) = 331 K
Final volume (V2) =..?
V1 /T1 = V2 /T2
3.5 /292 = V2 /331
Cross multiply
292 x V2 = 3.5 x 331
Divide both side by 292
V2 = (3.5 x 331) / 292
V2 = 3.97 L
Therefore, the new volume of the gas is 3.97 L.
Why are antiparallel beta sheets more stable than parallel beta sheets?
Answer:
The side chains of the amino acids alternate above and below the sheet
Explanation:
Hydrogen bonds are formed between the amine and carbonyl groups across strands. ... Antiparallel ß sheets are slightly more stable than parallel ß sheets because the hydrogen bonding pattern is more optimal.
If D+2 would react with E-1, what do you predict to be the formula?
Answer:
DE2
Explanation: for every one D+2 you need two E-1 because +2=-2
If the H+ concentration is 0.00001 M, what is the OH- concentration?
Answer:
1.00x10^-9
Explanation:
The reaction, 2 SO3(g) <--> 2 SO2(g) + O2(g) is endothermic. Predict what will happen if the temperature is increased.
Explanation:
This reaction is in equilibrium and would hence obey lechatelier's principle. This principle states that whenever a system at equilibrium undergoes a change, it would react in way so as to annul that change.
Since it is an endothermic reaction, increasing the temperature would cause the reaction to shift towards the right.
This means that it favours product formation and more of the product would be formed.
For each of the processes, determine whether the entropy of the system is increasing or decreasing. The system is underlined.
1. a snowman melts on a spring day
2. a document goes through a paper shredder
3. a water bottle cools down in a refrigerator
4. silver tarnishes
5. dissolved sigar precipitates out of water to form rock candy
A. Entropy is increasing
B. Entropy is decreasing
Entropy is INCREASING when a snowman melts, a document goes through paper shredder, silver tarnishes, while it is DECREASING when dissolved sugar precipitates, water vapor forms droplets and water cools down.
Entropy can be defined as the degree of randomness or disorder of a particular system.
Entropy is equal to zero (0) for a perfectly ordered system.
Heat increases the entropy of the system because more energy excites the molecules and it increases the amount of random activity.
Moreover, the cooling decreases the entropy of the system because molecules are more ordered and it decreases the amount of random activity.
In conclusion, entropy is INCREASING when a snowman melts, a document goes through paper shredder, silver tarnishes, while it is DECREASING when dissolved sugar precipitates, water vapor forms water droplets and the water cools down.
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Which of the following is an alkaline earth metal?
A. Silicon (Si)
B. Magnesium (Mg)
C. Carbon (C)
D. Aluminum (AI)
Answer:
B
Explanation:
The alkaline earth metals are the elements located in Group 2. The only element out of our choices that is in Group 2 is Magnesium.
Balance the following redox reaction in acidic solution: H+(aq)+Zn(s)→H2(g)+Zn2+(aq) Express your answer as a chemical equation. Identify all of the phases in your answer. nothing
Answer:
The balanced equation is: Zn(s) + 2H⁺(aq) → Zn²⁺(aq) + H₂(g)
Explanation:
Zn(s) is a simple substance (its oxidation number is zero) and it is oxidized to Zn²⁺. It loses two electrons, so the half reaction is the following:
Zn(s) → Zn²⁺(aq) + 2 e- (oxidation reaction)
Hydrogen ion (H⁺) is reduced to hydrogen gas (H₂). The oxidation number is decreased from +1 to 0 (because H₂ is a simple substance). H⁺ gains 1 electron per H atom, so the half reaction is the following:
2H⁺(aq) + 2 e- → H₂(g) (reduction reaction)
We obtain the overall reaction from the addition of the two half reactions. We write the reduction reaction first and then the oxidation reaction, as follows:
2H⁺(aq) + 2 e- → H₂(g)
+
Zn(s) → Zn²⁺(aq) + 2 e-
---------------------------------
Zn(s) + 2H⁺(aq) → Zn²⁺(aq) + H₂(g)
The two electrons at both sides of the equation (2 e-) are canceled. The overall reaction is in acidic solution due to the presence of H⁺ ions. The net charge at both sides is the same : +2, so the mass and the charge are balanced.
The second-order decomposition of HI has a rate constant of 1.80 · 10-3 M-1s-1. How much HI remains after 27.3 s if the initial concentration of HI is 4.78 M?
Answer: 3.87M of HI remains after 27.3 s
Explanation:
Using the Second order decomposition equation of
1/[H]t =K x t +1/[A]o
Given initial concentration ,[A]o = 4.78M
time, t = 27.3 s
rate of constant , k= 1.80 x 10^-3 M-1s-1
1/[H] t= 1/[A] t= concentration after time, t=?
SOLUTION
1/[A] t =kt +1/[A]o
1/[A] t =(1.80 x 10^-3 (27.3)+1/4.78
0.04914+0.2092=0.2583
1/[A] t =0.2583
[A] t =1/0.2583= 3.87M
If the equilibrium constant of the reaction is 0.85, then which statement is true if the mass of A is 10.5 grams; the density of B is 0.82 g/ml; the concentration of C is 0.64 M; and the concentration of D is 0.38 M.
A(s) + 3 B(l) _____ 2(aq) + D(aq)
Pick the correct statement about this system.
A. Q < K and reaction shifts left
B. Q > K and reaction shifts left
C. Q > K and reaction shifts right
D. Q = K and reaction does not shift
E. Q < K and reaction shifts right
Answer:
E. Q < K and reaction shifts right
Explanation:
Step 1: Write the balanced equation
A(s) + 3 B(l) ⇄ 2(aq) + D(aq)
Step 2: Calculate the reaction quotient (Q)
The reaction quotient, as the equilibrium constant (K), only includes aqueous and gaseous species.
Q = [C]² × [D]
Q = 0.64² × 0.38
Q = 0.15
Step 3: Compare Q with K and determine in which direction will shift the reaction
Since Q < K, the reaction will shift to the right to attain the equilibrium.
Which is most likely to happen during a precipitation reaction?
A. A solid substance will break down into two new substances that
are gases.
B. An insoluble solid will form when ions in dissolved compounds
switch places.
C. A substance will react with oxygen to form water and carbon
dioxide.
D. A gas will form when positive ions switch places to form new
compounds.
Answer:
I think its B
Explanation:
Precipitation reactions leave a solid behind. The solid is called a precipitate.
Answer:
B
Explanation:
An insoluble solid will form when ions in dissolved compounds switch places.
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.
A chemist adds of a mercury(I) chloride solution to a reaction flask. Calculate the micromoles of mercury(I) chloride the chemist has added to the flask.
Answer:
3.383x10⁻³ micromoles of HgCl
Explanation:
The chemist adds 170mL of a 1.99x10⁻⁵mmol/L Mercury (I) chloride, HgCl.
The solution contains 1.99x10⁻⁵milimoles of HgCl in 1L. That means in 170mL = 0.170L there are:
0.170L × (1.99x10⁻⁵milimoles HgCl / L) = 3.383x10⁻⁶ milimoles of HgCl.
Now, in 1milimole you have 1000 micromoles. That means in 3.383x10⁻⁶ milimoles of HgCl you have:
3.383x10⁻⁶ milimoles of HgCl ₓ (1000micromoles / 1milimole) =
3.383x10⁻³ micromoles of HgClSulfuric acid is commonly used as an electrolyte in car batteries. Suppose you spill some on your garage floor. Before cleaning it up, you wisely decide to neutralize it with sodium bicarbonate (baking soda) from your kitchen. The reaction of sodium bicarbonate and sulfuric acid is
Answer:
The mass of NaHCO3 required is 235.22 g
Explanation:
*******
Continuation of Question:
2NaHCO3(s) + H2SO4(aq) → Na2SO4(aq) + 2CO2(g) + 2H2O(l)
You estimate that your acid spill contains about 1.4 mol H2SO4. What mass of NaHCO3 do you need to neutralize the acid?
********\
The question requires us to calculate the mass of NaHCO3 to neutralize the acid.
From the balanced chemical equation;
1 mol of H2SO4 requires 2 mol of NaHCO3
1.4 would require x?
Upon solving for x we have;
x = 1.4 * 2 = 2.8 mol of NaHCO3
The relationship between mass and number of moles is given as;
Mass = Number of moles * Molar mass
Mass = 2.8 mol * 84.007 g/mol
Mass = 235.22 g
How many moles of aqueous magnesium ions and chloride ions are formed when 0.250 mol of magnesium chloride dissolves in water
Answer:
0.250 mol Mg²⁺
0.500 mol Cl⁻
Explanation:
Magnesium chloride (MgCl₂) dissociates into ions according to the following equilibrium:
MgCl₂ ⇒ Mg²⁺ + 2 Cl⁻
1 mol 1 mol 2 mol
1 mol of Mg²⁺ and 2 moles of Cl⁻ are formed per mole of MgCl₂. If we have 0.250 mol of MgCl₂, the following amounts of ions will be formed:
0.250 mol MgCl₂ x 1 mol Mg²⁺/mol MgCl₂= 0.250 mol Mg²⁺
0.250 mol MgCl₂ x 2 mol Cl⁻/mol MgCl₂= 0.500 mol Cl⁻
Answer:
HEY THE ANSWER ABOVE ME IS RIGHT!! i defientely misclicked my rating :/
5/5 all the way.
Explanation:
Review the reversible reactions given, along with the associated equilibrium constant Kat room temperature. In each case, determine whether the forward or reverse reaction is favored.
CH3COOH → CH3C00^- + H^+
Ka=1.8 x 10^-5
AgCl → Ag^+ + Cl^-
Ksp=1.6 x 10^-10
Al(OH)3 → Al^3+ + 3OH^-
Ksp=3.7 x 10^-15
A+B → C
K=4.9 x 10^3
Answer:
The answers to your questions are given below
Explanation:
The following data were obtained from the question:
CH3COOH → CH3C00^- + H^+
Equilibrium constant, Ka = 1.8 x 10^-5
AgCl → Ag^+ + Cl^-
Equilibrium constant, Ksp = 1.6 x 10^-10
Al(OH)3 → Al^3+ + 3OH^-
Equilibrium constant, Ksp = 3.7 x 10^-15
A+B → C
Equilibrium constant, K = 4.9 x 10^3
When the value of the equilibrium constant is grater than 1, it shows that the concentration of product is higher than that of the reactant and it implies that the forward reaction is favored.
When the value of the equilibrium constant is 1, it shows that the the concentration of the product and reactant are the same. Therefore neither the forward nor the reverse reaction is favored.
When the value of the equilibrium constant is lesser than 1, it shows that the concentration of the reactant is higher than the concentration of the product. Therefore, the reversed reaction is favored.
Now, we shall the question given above as follow:
A. CH3COOH → CH3C00^- + H^+
Equilibrium constant, Ka = 1.8 x 10^-5
Since the value of the equilibrium constant is lesser than 1, it means that the reverse reaction is favored.
B. AgCl → Ag^+ + Cl^-
Equilibrium constant, Ksp = 1.6 x 10^-10
Since the value of the equilibrium constant is lesser than 1, it means that the reverse reaction is favored.
C. Al(OH)3 → Al^3+ + 3OH^-
Equilibrium constant, Ksp = 3.7 x 10^-15
Since the value of the equilibrium constant is lesser than 1, it means that the reverse reaction is favored.
D. A+B → C
Equilibrium constant, K = 4.9 x 10^3
Since the value of the equilibrium constant is greater than 1, it means that the forward reaction is favored.
The reaction conditions are:
A. The reverse reaction is favored.
B. The reverse reaction is favored.
C. The reverse reaction is favored.
D. The forward reaction is favored.
Chemical reaction:
A. [tex]CH_3COOH[/tex] → [tex]CH_3COO^- + H^+[/tex]
Equilibrium constant, Ka = [tex]1.8 * 10^{-5}[/tex]
B. [tex]AgCl[/tex] → [tex]Ag^+ + Cl^-[/tex]
Equilibrium constant, Ksp = [tex]1.6 * 10^{-10}[/tex]
C. [tex]Al(OH)_3[/tex] → [tex]Al^{3+} + 3OH^-[/tex]
Equilibrium constant, Ksp = [tex]3.7 * 10^{-15}[/tex]
D. A+B → C
Equilibrium constant, K = [tex]4.9 * 10^3[/tex]
Conditions for Equilibrium constant:When the value of the equilibrium constant is greater than 1, it shows that the concentration of product is higher than that of the reactant and it implies that the forward reaction is favored.
When the value of the equilibrium constant is 1, it shows that the the concentration of the product and reactant are the same. Therefore neither the forward nor the reverse reaction is favored.
When the value of the equilibrium constant is lesser than 1, it shows that the concentration of the reactant is higher than the concentration of the product. Therefore, the reversed reaction is favored.
Thus, the reactions will be:
A. The reverse reaction is favored.
B. The reverse reaction is favored.
C. The reverse reaction is favored.
D. The forward reaction is favored.
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Solid iron(II) oxide reacts with oxygen gas to produce solid iron(III) oxide. Balance the equation for this reaction (in lowest multiple integers). Write the unbalanced equation for this reaction.
Answer
Hello
I think the reaction is like this FeO+OFe²O³
And the balance reaction is 2Fe+OFe²O³
Explanation:
At first we should find sth that has more atoms than the other then for example we realized that we have two atoms of Fe in Fe²O³ then put 2 before FeO and now we have 2 atoms of Fe in right side and 2 atoms of Fe in left then Oxygen in FeO change to 2 atoms of Oxygen and we have an other one in right side that they become 3 atoms of Oxygen and now we have 3 atoms of Oxygen in both right and left side.
Finally our reaction balanced.
Good luck
Iron(II) is available to bond with chloride ion. How many of each type of ion will bond to form an ionic compound?
A) 3 iron(II), 1 chloride
B) 2 iron(II), 3 chloride
C) 2 iron(II), 1 chloride
D) 1 iron(II), 2 chloride
Answer:
D) 1 iron(II), 2 chloride
Explanation:
Iron II chloride is the compound; FeCl2. It is formed as follows, ionically;
Fe^2+(aq) + 2Cl^-(aq) -----> FeCl2
The formation of one mole of FeCl2 involves the reaction one mole of iron and two moles of chloride ions. This means that in FeCl2, the ratio of iron to chlorine is 1:2 as seen above.
Therefore there is one iron II ion and two chloride ions in each mole of iron II chloride, hence the answer.
Identify the elements that have the following abbreviated electron configurations.
A) [Ne] 3s23p5.
B) [Ar] 4s23d7.
C) [Xe] 6s1.
Answer:
A) Chlorine (Cl)
B) Cobalt (Co)
C) Caesium (Cs)
Hope this helps.
The abbreviated electron configurations that was given in the question belongs to
Chlorine (Cl)
Cobalt (Co)
Caesium (Cs) respectively.
Electronic configurations can be regarded as the electronic structure, which is the way an electrons is arranged in energy levels towards an atomic nucleus.The electron configurations is very useful when describing the orbitals of an atom in its ground state.To calculate an electron configuration, we can put the periodic table into sections, and this section will represent the atomic orbitals which is the regions that house the electrons. Groups one of the period table and two belongs to s-block, group 3 through 12 belongs to the d-block, while 13 to 18 can be attributed to p-block ,The rows that is found at bottom are the f-blockTherefore, electron configurations explain orbitals of an atom when it is in it's ground state.
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Given a fixed amount of gas help at a constant pressure, calculate the temperature to which the gas would have to be changed if a 1.75 L sample at 23.0*C were to have a final volume of 3.50 L.
A. 46.0*C
B. 89.5*C
C. 169*C
D. 319*C
E. 592*C
Answer:
592 K or 319° C
Explanation:
From the statement of Charles law we know that the volume of a given mass of gas is directly proportional to its absolute temperature at constant pressure. Thus;
V1/T1= V2/T2
Initial volume V1 = 1.75 L
Initial temperature T1= 23.0 +273 = 296 K
Final volume V2= 3.50 L
Final temperature T2 = the unknown
T2= V2T1/V1= 3.50 × 296 / 1.75
T2 = 592 K or 319° C
2
22. A sodium chloride solution is 15.0% m/m%. Calculate mass of sodium chloride in 219 g solution.
14.2g
80.38
11.2 g
32.9 g
Answer: The mass of sodium chloride in 219 g solution is 32.9 g
Explanation:
To calculate the mass percent of element in a given compound, we use the formula:
[tex]\text{Mass percent of A}=\frac{\text{Mass of A}}{\text{mass of A +mass of B}}\times 100[/tex]
To find mass of sodium chloride in solution:
[tex]\text{Mass percent of sodium chloride}=\frac{\text{Mass of sodium chloride}}{\text{mass of solution}}\times 100[/tex]
Mass percent of sodium chloride= 15.0 %
Mass of solution = 219g
[tex]15=\frac{\text{Mass of sodium chloride}}{219}\times 100[/tex]
[tex]{\text{Mass of sodium chloride}=32.9g[/tex]
Thus mass of sodium chloride in 219 g solution is 32.9 g
What is a major product of the reaction in the box?
Answer:
Molecule C
Explanation:
In this case, on the first reaction, we will have the production of a Grignard reagent. This molecule will react with [tex]D_2O[/tex] and a deuterium atom will be transferrred to the benzene ring. Then at the top of the molecule, we will have an acetal structure. This acetal can be broken by the action of the acid [tex]DCl[/tex], In the mechanism at the end, we will obtain a carbonyl group bonded to a hydrogen atom. Therefore we will have in the final product the aldehyde group. See figure 1 to further explanations.
I hope it helps!
The accepted value of the number of Liters of gas in a mole is 22.4. List two possible reasons on why our experiment yielded a different value for the number of Liters in a mole of a gas.
Hint: Our experiment was conducted in July, in St. Paul, Minnesota.
Answer:
- Pressure in St. Paul, Minnesota
- Temperature in St. Paul, Minnesota
Explanation:
22.4 L or dm³ is the volume for a gas under Standard pressure and temperature conditions.
It is logically to say, that tempereature value at the day of the experiment was not 273.15 K, which is 32°F
We can say, that the pressure was not 1 atm. St Paul Minnesota has a minimum, but a little height, so the pressure differs by few figures from the standard pressure values.
We also have to mention, that 22.4 L is the value for the Ideal gases at standards conditions. Ideal gases does not exisist on practice, we always talk about real gases. Don't forget the Ideal Gases Law equation:
P . V = n . R . T
Pressure . Volume = number of moles . 0.082 L.atm /mol. K . 273.15K
Number of moles must be 1 at STP, to determine a volume of 22.4L
Draw a Lewis structure for one important resonance form of HBrO4 (HOBrO3). Include all lone pair electrons in your structure. Do not include formal charges in your structure.
Answer:
The Lewis structure is attached with the answer -
Explanation:
Lewis structure or Lewis dot diagram are diagrams or representation of showing the bonding between different or same atoms of a molecule in any and also shows lone pairs of electrons that may exist in the molecule as dots.
HBrO₄ is bromine oxoacid which is also known as perbromic acid. It is a unstable inorganic compound.
The Lewis structure is attached in form of image with representation of lone pairs of electrons.
How many atoms of hydrogens are found in 3.21 mol of
C3H8?
Answer:
1.55 × 10²⁵ atoms of H
Explanation:
3.21mol C₃H₈ × 8mol H × (6.022×10²³)
Find the [OH−] of a 0.32 M methylamine (CH3NH2) solution. (The value of Kb for methylamine (CH3NH2) is 4.4×10−4.) Express your answer to two significant figures and include the appropriate units.
Answer:
[tex][OH^-]=0.01165M[/tex]
Explanation:
Hello,
In this case, for the dissociation of methylamine:
[tex]CH_3NH_2(aq)+H_2O(l)\rightleftharpoons CH_3NH_3^+(aq)+OH^-(aq)[/tex]
We can write the basic dissociation constant as:
[tex]Kb=\frac{[CH_3NH_3^+][OH^-]}{[CH_3NH_2]}[/tex]
That in terms of the reaction extent [tex]x[/tex], turns out:
[tex]Kb=\frac{x*x}{[CH_3NH_2]_0-x}[/tex]
[tex]4.4x10^{-4}=\frac{x^2}{0.32M-x}[/tex]
That has the following solution for [tex]x[/tex]:
[tex]x_1=-0.01209M\\x_2=0.01165M[/tex]
Yer 0.01165M is valid only as no negative concentrations are eligible. It means that it is the concentration of hydroxyl ions in the solution:
[tex][OH^-]=0.01165M[/tex]
Best regards.
Determine the volume occupied by 10 mol of helium at
27 ° C and 82 atm
Answer:
3.00 L
Explanation:
PV = nRT
(82 atm × 101325 Pa/atm) V = (10 mol) (8.314 J/mol/K) (27 + 273) K
V = 0.00300 m³
V = 3.00 L
Using the Bohr model, determine the lowest possible energy, in joules, for the electron in the Li2+ ion.
Answer: E = - 19.611×[tex]10^{-18}[/tex] J
Explanation: The lowest possible energy can be calculated using the formula:
[tex]E_{n} = - Z^{2}.\frac{k}{n^{2}}[/tex]
where:
Z is atomic number of the atom;
k is a constant which contains other constants and is 2.179×[tex]10^{-18}[/tex] J
n is a layer;
For the lowest possible, n=1.
Atom of Lithium has atomic number of Z=3
Substituing:
[tex]E_{1} = - 3^{2}.\frac{2.179.10^{-18}}{1}[/tex]
[tex]E_{1} =[/tex] [tex]-19.611.10^{-18}[/tex] J
The energy for the electron in the [tex]Li^{+2}[/tex] ion is - 19.611 joules
The lowest possible energy, in Joules, for the electron in the [tex]Li^{2+}[/tex] ion is equal to [tex]1.96\times 10^{-17}\; Joules[/tex]
To determine the lowest possible energy, in Joules, for the electron in the [tex]Li^{2+}[/tex] ion, we would use the Bohr model:
Mathematically, Bohr's model is given by the equation:
[tex]Energy = -Z^2 \frac{k}{n^2}[/tex]
Where:
Z is the atomic number of an atom.n is the number of energy level.k is Rydberg constant.We know that the atomic number of lithium (Li) is equal to 3.
Also, at the lowest possible energy, n = 1.
Rydberg constant = [tex]2.179 \times 10^{-18}[/tex]
Substituting the parameters into the equation, we have;
[tex]E_1 = -3^2 \times \frac{2.179 \times 10^{-18}}{1^2} \\\\E_1 =9 \times 2.179 \times 10^{-18}\\\\E_1 =1.96\times 10^{-17}\; Joules[/tex]
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Balance the following
Na+02-→ Na20
Al+O2 ->Al2O3
H2+12+ ->HI
Mg+H2O → Mg(OH)2+H2
Ca+O2 -> Cao
Answer:
1. Na + O2 → Na2O (Balanced)
2. 4Al + 3O2 → 2(Al2O3) (Balanced)
3. H2 + i2 → 2HI (Balanced)
4. Mg + 2H2O → Mg(OH)2+ H2 (Balanced)
5. 2Ca +O2 → 2CaO (Balanced)
A saturated solution was formed when 5.16×10−2 L of argon, at a pressure of 1.0 atm and temperature of 25 ∘C, was dissolved in 1.0 L of water.
Calculate the Henry's law constant for argon. it must be im M/atm
Answer:
The Henry's law constant for argon is [tex]k=2.11*10^{-3}\frac{ M}{atm}[/tex]
Explanation:
Henry's Law indicates that the solubility of a gas in a liquid at a certain temperature is proportional to the partial pressure of the gas on the liquid.
C = k*P
where C is the solubility, P the partial pressure and k is the Henry constant.
So, being the concentration [tex]C=\frac{ngas}{V}[/tex]
where ngas is the number of moles of gas and V is the volume of the solution, you must calculate the number of moles ngas. This is determined by the Ideal Gas Law: P*V=n*R*T where P is the gas pressure, V is the volume that occupies, T is its temperature, R is the ideal gas constant, and n is the number of moles of the gas. So [tex]n=\frac{P*V}{R*T}[/tex]
In this case:
P=PAr= 1 atmV=VAr= 5.16*10⁻² LR=0.082 [tex]\frac{atm*L}{mol*K}[/tex]T=25 °C=298 °KThen:
[tex]n=\frac{1 atm*5.16*10^{-2} L}{0.082 \frac{atm*L}{mol*K} *298K}[/tex]
Solving:
n= 2.11 *10⁻³ moles
So: [tex]C=\frac{ngas}{V}=\frac{2.11*10^{-3} moles}{1 L} =2.11*10^{-3} \frac{moles}{L}= 2.11*10^{-3} M[/tex]
Using Henry's Law and being C=CAr and P =PAr:
2.11*10⁻³ M= k* 1 atm
Solving:
[tex]k=\frac{2.11*10^{-3} M}{1 atm}[/tex]
You get:
[tex]k=2.11*10^{-3}\frac{ M}{atm}[/tex]
The Henry's law constant for argon is [tex]k=2.11*10^{-3}\frac{ M}{atm}[/tex]
The Henry's law constant for argon gas in 1 litre of water is 2.1 × 10⁻³M/atm.
What is Henry's law?Henry's law of gas states that solubility of a gas in any liquid at particular temperature is directly proportional to the partial pressure of the gas.
C∝P
C = kP, where
k = Henry's constant
P = partial pressure of gas
C is the solubility and it is present in the form of concentration and will be calculated as:
C = n/V
n = no. of moles
V = volume
And moles of the gas will be calculated by using the ideal gas equation as:
PV = nRT
n = (1)(5.16×10⁻²) / (0.082)(298) = 2.1 × 10⁻³ moles
And Concentration in liquid will be:
C = 2.1 × 10⁻³mol / 1L = 2.1 × 10⁻³ M
Now we put all these values in the first equation to calculate the value of k as:
k = (2.1 × 10⁻³M) / (1atm) = 2.1 × 10⁻³M/atm
Hence required value of k is 2.1 × 10⁻³M/atm.
To know more about Henry's law, viist the below link:
https://brainly.com/question/12823901
Given a fixed amount of gas in a rigid container (no change in volume), what pressure will the gas exert if the pressure is initially 1.50 atm at 22.0oC, and the temperature is changed to 11.0oC?
A. 301 atm
B. 1.56 atm
C. 0.750 atm
D. 1.44 atm
E. 3.00 atm
Answer:
The pressure the gas will have if the pressure is initially 1.50 atm at 22.0 ° C and the temperature changes at 11.0 ° C is 1.44 atm (option D)
Explanation:
Gay Lussac's law indicates that, as long as the volume of the container containing the gas is constant, as the temperature increases, the gas molecules move more rapidly. Then the number of collisions against the walls increases, that is, the pressure increases. That is, the gas pressure is directly proportional to its temperature.
Gay-Lussac's law can be expressed mathematically as follows:
[tex]\frac{P}{T}=k[/tex]
Where P = pressure, T = temperature, K = Constant
You have a gas that is at a pressure P1 and at a temperature T1. When the temperature varies to a new T2 value, then the pressure will change to P2, and then:
[tex]\frac{P1}{T1}=\frac{P2}{T2}[/tex]
In this case:
P1= 1.50 atmT1= 22 °C= 295 °K (being 0°C= 273 °K)P2= ?T2= 11 °C= 284 KReplacing:
[tex]\frac{1.5 atm}{295 K}=\frac{P2}{284 K}[/tex]
Solving:
[tex]P2= 284 K*\frac{1.5 atm}{295 K}[/tex]
P2=1.44 atm
The pressure the gas will have if the pressure is initially 1.50 atm at 22.0 ° C and the temperature changes at 11.0 ° C is 1.44 atm (option D)