Answer:
When the solution (with phenolphthalein) changes to colorless
Explanation:
When titrating with HCl is common to add phenolphthalein as an acid-base indicator.
Phenolphthalein is pink or fucsia when added into a basic solution. On the other hand when it is in acid solutions, is colorless.
So, when titrating, the NaOH solution will be initialy pink due to the phenolphthalein and when reaching the equivalence point, that color will fade out into colorless. This is how you know you hace reached the equivalent point.
In the industrial "chlor-alkali" process, pure chlorine and sodium hydroxide are produced by electrolyzing brine, essentially an aqueous solution of sodium chloride.
Suppose a current of 18.0 A is passed through an aqueous solution of NaCl for 67.0 seconds.
Calculate the mass of pure chlorine produced.
Be sure your answer has a unit symbol and the correct number of significant digits.
Answer:
0.443 g
Explanation:
In the electrolysis of an aqueous solution of NaCl, the following half-reactions take place:
Reduction: Na⁺(aq) + 1 e⁻ ⇒ Na(s)
Oxidation: 2 Cl⁻(aq) ⇒ Cl₂(g) + 2 e⁻
Let's consider the following relations:
1 A = 1 c/s1 mole of e⁻ has a charge of 96468 c (Faraday's constant)1 mole of Cl₂(g) is released when 2 moles of e⁻ circulateThe molar mass of Cl₂ is 70.9 g/molSuppose a current of 18.0 A is passed through an aqueous solution of NaCl for 67.0 seconds. The mass of chlorine produced is:
[tex]67.0s.\frac{18.0c}{s} .\frac{1mole^{-} }{96468c} .\frac{1molCl_{2}}{2mole^{-} } .\frac{70.9gCl_{2}}{1molCl_{2}} =0.443gCl_{2}[/tex]
Using Faraday's law of electrolysis, we can calculate the mass of chlorine produced by an 18.0 A current over 67.0 seconds in the chlor-alkali process. The result is approximately 0.44 grams.
Explanation:To determine the mass of chlorine produced in the chlor-alkali process using a current of 18.0 A for a duration of 67.0 seconds, we need to use Faraday's law of electrolysis. This states that the amount of a substance produced at an electrode during electrolysis is directly proportional to the quantity of electricity that passes through the solution.
The amount of a substance produced can be calculated using the formula: It = zF, where I is the current in amperes, t is the time in seconds, z is the ionic charge(1 for Cl-), and F is Faraday’s constant (96500 coulombs per mole of electrons).
First, let’s find the quantity of electric charge (Q) that has passed through the solution using the formula Q = It. In this case, I = 18 A and t = 67.0 seconds, so Q = 18 * 67.0 = 1206 Coulombs.
Now we can use this Q in the formula of m = QM/zF, where M is the molar mass of chlorine, which is approximately 35.5 g/mol. After plugging the numbers in, we find the mass (m) of the chlorine produced is 0.44 g.
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A cylinder with a moving piston expands from an initial volume of 0.250 L against an external pressure of 2.00 atm. The expansion does 288 J of work on the surroundings. What is the final volume of the cylinder? (1 LLaTeX: \cdot⋅atm = 101.3 J) Group of answer choices
Answer:
The final volume of the cylinder is 1.67 L
Explanation:
Step 1: Data given
Initial volume = 0.250 L
external pressure = 2.00 atm
Expansion does 288 J of work on the surroundings
Step 2: Definition of reversible work:
Wrev = -P(V2-V1) = -288 J
The gas did work, so V2>V1 (volume expands) and the work has a negative sign.(Wrev<0)
V2 = (-Wrev/P) + V1
⇒ with Wrev = reverse work (in J)
⇒ with P = the external pressure (in atm)
⇒ with V1 = the initial volume
We can see that your pressure is in atm and energy in J
To convert from J to L * atm we should use a convenient conversion unit using the universal gas constants :
R = 8.314472 J/mol *K and R= 0.08206 L*atm/K*mol
V2 =- (-288 J * (0.08206 L*atm/K*mol /8.314 J/mol *K))/2.00 atm + 0.250L
V2 = 1.67 L
The final volume of the cylinder is 1.67 L
The final volume of the cylinder after the expansion is 1.673 liters. This was found by calculating the change in volume using the work done on the surroundings and the external pressure, and adding it to the initial volume.
Explanation:To calculate the final volume of the cylinder after expansion, we need to use the work done on the surroundings and the external pressure. Work (W) is related to pressure (P) and volume change (ΔV) by the equation W = -PΔV, where the pressure is constant and work done on the surroundings is negative.
In this case, we know that the work done on the surroundings is 288 J and the external pressure is 2.00 atm. Since 1 L·atm is equivalent to 101.3 J, we convert the external pressure to joules by multiplying by the volume change ΔV in liters:
288 J = -(2.00 atm) · ΔV · 101.3 J/L·atm
Therefore, to find ΔV we divide 288 J by the product of 2.00 atm and 101.3 J/L·atm:
ΔV = -288 J / (2.00 atm · 101.3 J/L·atm)
ΔV = -1.423 L (taking the absolute value, since volume change is positive upon expansion)
The initial volume was 0.250 L, thus the final volume is:
V_final = V_initial + ΔV
V_final = 0.250 L + 1.423 L
V_final = 1.673 L
Are the bonds in each of the following substances ionic, nonpolar covalent, or polar covalent? (a)KCl (b)P4 (c)BF3 (d)SO2 (e)Br2 (f)NO2 For those substances with polar covalent bonds, which has the least polar bond? For those substances with polar covalent bonds, which has the most polar bond?
Answer:
(a) Ionic
(b) Nonpolar covalent
(c) Polar covalent
(d) Polar covalent
(e) Nonpolar covalent
(f) Polar covalent
For those substances with polar covalent bonds, which has the least polar bond? NO₂
For those substances with polar covalent bonds, which has the most polar bond? BF₃
Explanation:
Are the bonds in each of the following substances ionic, nonpolar covalent, or polar covalent?
The nature of a bond depends on the modulus of the difference of electronegativity (|ΔEN|) between the atoms that form it.
If |ΔEN| = 0, the bond is nonpolar covalent.If 0 < |ΔEN| ≤ 2, the bond is polar covalent.If |ΔEN| > 2, the bond is ionic.(a) KCl |ΔEN| = |EN(K) - EN(Cl)| = |0.8 - 3.0| = 2.2. The bond is ionic.
(b) P₄ |ΔEN| = |EN(P) - EN(P)| = |2.1 - 2.1| = 0.0. The bond is nonpolar covalent.
(c) BF₃ |ΔEN| = |EN(B) - EN(F)| = |2.0 - 4.0| = 2.0. The bond is polar covalent.
(d) SO₂ |ΔEN| = |EN(S) - EN(O)| = |2.5 - 3.5| = 1.0. The bond is polar covalent.
(e) Br₂ |ΔEN| = |EN(Br) - EN(Br)| = |2.8 - 2.8| = 0.0. The bond is nonpolar covalent.
(f) NO₂ |ΔEN| = |EN(N) - EN(O)| = |3.0 - 3.5| = 0.5. The bond is polar covalent.
The nature of bonding between atoms depends on the electro negativity between the atoms in the bond.
An ionic bond is formed by transfer of electrons from one atom to another. It commonly occurs between a metal and a nonmetal. Covalent bonds are formed when electrons are shared between bonding atoms. If the electronegativity difference between the two bonding atoms is small, electrons are equally shared in the molecule and the bonds are nonpolar. However, when the difference in electro negativity is significant (about 0.5) a significant magnitude of polarity of the bond is observed and the bond us polar because electrons lie closer to the more electronegative atom.
The classification of the compounds according to nature of bonding between atoms is done as follows;
KCl - ionic bond
P4 - nonpolar covalent bond
BF3 - polar covalent bond
SO2 - polar covalent bond
Br2 - nonpolar covalent
NO2 - polar covalent bond.
The most polar bond occurs between bromine and bromine because of a large electro negativity difference between the both atoms.
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3 Ni2+(aq) + 2 Cr(OH)3(s) + 10 OH− (aq) → 3 Ni(s) + 2 CrO42−(aq) + 8 H2O(l) ΔG∘ = +87 kJ/molGiven the standard reduction potential of the half-reaction Ni2+(aq) + 2 e− → Ni(s) E∘red = -0.28 V, calculate the standard reduction potential of the half-reactionCrO42−(aq) + 4 H2O(l) + 3 e− → Cr(OH)3(s) + 5 OH−(aq)A. -0.43 VB. -0.28 VC. 0.02 VD. -0.13 VE. -0.15 V
Answer:
The standard reduction potential E°cell (Cr6+/Cr3+) is -0.13V
Explanation:
Step 1: Data given
3 Ni^2+(aq) + 2 Cr(OH)3(s) + 10 OH− (aq) → 3 Ni(s) + 2 CrO4^2−(aq) + 8 H2O(l) ΔG∘ = +87000 J/mol
Ni2+(aq) + 2 e− → Ni(s) E∘red = -0.28 V
Step 2: The half reactions:
Cathode: Ni2+(aq) + 2 e− → Ni(s) E° = -0.28 V
Anode: CrO4^2-(aq) + 4H2O(l) +3e- → Cr(OH)3(s) + 5OH- (aq) E°= unknown
Step 3: Calculate E°cell
ΔG° = -n*F*E°cell
⇒ with ΔG° = the gibbs free energy
⇒ n = the number of electrons in the net reaction = 6
⇒ F = the Faraday constant = 96485 C
⇒ E°cell= the standard cell potential
Step 4: Calculate E°(Cr6+/Cr3+
E°cell= ΔG°/(-n*F)
E°cell = 87000 /(-6*96485)
E°cell = -0.150 V
E°cell = E°(Ni2+/Ni) - E°(Cr6+/Cr3+)
E°(Cr6+/Cr3+) = -0.13V
The standard reduction potential E°cell (Cr6+/Cr3+) is -0.13V
As the temperature of a reaction is increased, the rate of the reaction increases because the______________. a.reactant molecules collide less frequently and with greater energy per collision b. reactant molecules collide more frequently and with greater energy per collision c. reactant molecules collide less frequently d. activation energy is lowered e. reactant molecules collide more frequently with less energy per collision
Answer:
b. reactant molecules collide more frequently and with greater energy per collision
Explanation:
As the temperature of a reaction is increased, the rate of the reaction increases because the reactant molecules collide more frequently and with greater energy per collision. When temperature is increased there is an increase in the kinetic energy of the molecules. The more the molecules move the more they collide. According to the collision theory there should be enough energy to allow bonds to be formed during a chemical reaction hence the need for greater energy per collision. Also the rate of reaction is directly proportional to the number of collisions that occur.
A reaction at 17.0 degree C evolves 788.mmol of dinitrogen difluoride gas. Calculate the volume of dinitrogen difluoride gas that is collected. You can assume the pressure in the room is exactly 1atm. Be sure your answer has the correct number of significant digits.
Answer:
V = 18.8 L
Explanation:
This question is solved by using the ideal gas law :
PV = nRT ∴ V= nRT/P
where n= 0.788 mmol = 0.788 mol (1 mmol is a thousandths of a mol)
R = R constant for ideal gases = 0.08206 Latm/Kmol
T = 17.0 ºC = (17.0 + 273) K = 290 K
and V = volume in liters our unknown
Plugging our values and solving for V,
V = 0.788 mol x 0.08206 Latm/Kmol x 290 K / 1atm = 18.8 L
Applying the Ideal Gas Law with the given number of moles, temperature and pressure shows that the volume of dinitrogen difluoride gas collected is approximately 19.0 L.
Explanation:The volume of a gas can be calculated using the Ideal Gas Law if the number of moles, temperature and pressure are known. In your question, we're given that 788mmol of dinitrogen difluoride gas is evolved at a temperature of 17.0 degree C and a pressure of 1 atm. Firstly, convert the temperature to Kelvins by adding 273.15 to the Celsius temperature, giving us 290.15K.
The Ideal Gas Law is expressed as PV=nRT, where P is pressure, V is volume, n is the number of moles, R is the gas constant (0.0821 L·atm/ K·mol for this situation) and T is temperature. We can rearrange the equation to solve for volume: V=nRT/P. Therefore, substituting in the known values: V = 0.788 mol * 0.0821 L·atm/ K·mol * 290.15K / 1 atm, we find that the volume of dinitrogen difluoride gas collected is approximately 19.0 L, given correct to the number of significant digits.
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At constant pressure, which of these systems do work on the surroundings?
a. 2A(g) + 3B (g) ------> 4C (g)
b. 2A(g) + 2B (g) -------> 5C(g)
c. 2A (g) + B (g) -----> C(g)
d. A(s) + B(g) -------> 2C (g)
Answer:
The system does work on the surroundings.
Explanation:
The work (w) exerted in a chemical reaction depends on the change in the number of gaseous moles (Δn(g)), where
Δn(g) = n(gas, products) - n(gas, reactants)
These variables are linked through the following expression.
w = -R.T.Δn(g)
where,
R is the ideal gas constant
T is the absolute temperature
If Δn(g) > 0, w < 0 and the system does work on the surroundings.If Δn(g) < 0, w > 0 and the surroundings do work on the system.If Δn(g) = 0, w = 0 and no work is done.At constant pressure, which of these systems do work on the surroundings?
a. 2A(g) + 3B (g) ------> 4C (g)
Δn(g) = 4 -5 = -1. The surroundings do work on the system.
b. 2A(g) + 2B (g) -------> 5C(g)
Δn(g) = 5 -4 = 1. The system does work on the surroundings.
c. 2A (g) + B (g) -----> C(g)
Δn(g) = 1 - 3 = -2. The surroundings do work on the system.
d. A(s) + B(g) -------> 2C (g)
Δn(g) = 2 -2 = 0. No work is done.
QUICK WILL MARK BRAINLIEST!!!!
to calculate molar mass add the ______ mass for each atom in the chemical _____
from the periodic table.
(fill in the blanks)
Answer:
To calculate the molar mass add the atomic mass of each atom in the chemical formula.
Explanation:
For example Molar mass of NaHCO₃ calculated as
Molar mass is the sum of masses of all atom present in formula.
Atomic mass of sodium = 23 g/mol
Atomic mass of hydrogen = 1.008 g/mol
Atomic mass of carbon = 12 g/mol
Atomic mas of Oxygen = 16 g/mol
Molar mass of NaHCO₃ = 23 + 1.008 + 12 + 16× 3
Molar mass of NaHCO₃ = 23 + 1.008 + 12 + 48
Molar mass of NaHCO₃ = 84.008 g/mol
Given thatH2(g) + F2(g) -> 2HF(g) => ∆H = -546.6 kJ . mol-12H2(g) + O2(g) -> 2H20(l) => ∆H = -571.6 kJ. mol-1
Calculate the value of ∆H for:2F2(g) + 2H20(l) -> 4HF(g) + O2(g)
Answer: The [tex]\Delta H^o_{rxn}[/tex] for the reaction is -521.6 kJ.
Explanation:
Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.
According to this law, the chemical equation is treated as ordinary algebraic expressions and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the intermediate reactions.
The chemical equation for the reaction of fluorine and water follows:
[tex]2F_2(g)+2H_2O(l)\rightarrow 4HF(g)+O_2(g)[/tex] [tex]\Delta H^o_{rxn}=?[/tex]
The intermediate balanced chemical reaction are:
(1) [tex]H_2(g)+F_2(g)\rightarrow 2HF(g)[/tex] [tex]\Delta H_1=-546.6kJ[/tex] ( × 2)
(2) [tex]H_2(g)+O_2(g)\rightarrow 2H_2O(g)[/tex] [tex]\Delta H_2=-571.6kJ[/tex]
The expression for enthalpy of reaction follows:
[tex]\Delta H^o_{rxn}=[2\times \Delta H_1]+[1\times (-\Delta H_2)][/tex]
Putting values in above equation, we get:
[tex]\Delta H^o_{rxn}=[(2\times (-546.6))+(1\times (571.6))]=-521.6kJ[/tex]
Hence, the [tex]\Delta H^o_{rxn}[/tex] for the reaction is -521.6 kJ.
Arsine, AsH3, is a highly toxic compound used in the electronics industry for the production of semiconductors. Its vapor pressure is 35 Torr at 111.95 C and 253 Torr at 83.6 C. Using these data, calculate (a) the standard enthalpy of vaporization;
Answer:
-79.8 × 10⁴ J/mol
Explanation:
Arsine, AsH₃, is a highly toxic compound used in the electronics industry for the production of semiconductors. Its vapor pressure is 35 Torr at 111.95 °C and 253 Torr at 83.6 °C.
Then,
P₁ = 35 torr
T₁ = 111.95 + 273.15 = 385.10 K
P₂ = 253 torr
T₂ = 83.6 + 273.15 = 356.8 K
We can calculate the standard enthalpy of vaporization (ΔH°vap) using the two-point Clausius-Clapeyron equation.
[tex]ln(\frac{P_{2}}{P_{1}} )=\frac{-\Delta H\°_{vap}}{R} .(\frac{1}{T_{2}} -\frac{1}{T_{1}} )[/tex]
where,
R is the ideal gas constant
[tex]ln(\frac{253torr}{35torr})=\frac{-\Delta H\°_{vap}}{8.314J/K.mol} .(\frac{1}{356.8K}-\frac{1}{385.10K})\\ \Delta H\°_{vap}=-79.8 \times 10^{4} J/mol[/tex]
Consider the following equilibrium.2 NOBr(g)<=> 2 NO(g) + Br2(g)If nitrosyl bromide, NOBr, is 34 percent dissociated at 25°C and the total pressure is 0.25 atm, calculate Kp and Kc for the dissociation at this temperature.
Answer:
[tex]K_{p}[/tex] of the reaction is [tex]9.521 \times 10^{-3}[/tex].
[tex] K_{c}[/tex] of the reaction is [tex]3.89 \times 10^{-4}[/tex].
Explanation:
Initial pressure of NOBr is "x" atm.
34% = 0.34
The equilibrium chemical reaction is a follows.
[tex]2NOBr(g)\leftrightarrow 2NO(g)+Br_{2}(g)..............(1)[/tex]
Initial x 0 0
Change -0.34x +0.34x +0.17x
Final (x-0.34x) +0.34x +0.17x
Total pressure = sum of the pressure at equilibrium.
[tex]0.25atm=x-0.34x+0.34x+0.17x[/tex]
[tex]0.25atm= 1.17x[/tex]
[tex]x= \frac{0.25}{1.17}=0.213[/tex]
Partial pressure of NOBr= x-0.34x =x (1-0.34)
= 0.213(0.66) = 0.14 atm.
Hence, partial pressure of [tex][]P_{NOBr}][/tex] is 0.14 atm.
[tex]P_{NO}=0.34x = 0.34 \times 0.213 = 0.072[/tex]
[tex]P_{Br_{2}}=0.17x = 0.17 \times 0.213 = 0.036\,atm[/tex]
From equation (1)
[tex]K_{p}= \frac {P^{2}_{NO} P_{Br_{2}}}{P^{2}_{NOBr}}[/tex]
[tex]= \frac{(0.072)^{2} (0.036)}{(0.14)^{2}}= 9.521 \times 10^{-3}[/tex]
Therefore, [tex]K_{p}[/tex] of the reaction is [tex]9.521 \times 10^{-3}[/tex].
[tex]K_{p}= K_{c}(RT)^{\Delta n}[/tex]
Rearrange the equation is as follows.
[tex]K_{c}= \frac{K_{p}}{(RT)^{\Delta n}}[/tex]
[tex]\Delta n[/tex] = number of moles of reactants - Number of moles of products
= 3-2 = 1
[tex]= \frac{9.521 \times 10^{-3}}{(0.0821 \times 298)^{1}}= 0.389 \times 10^{-3} = 3.89\times 10^{-4}[/tex]
Therefore,[tex] K_{c}[/tex] of the reaction is [tex]3.89 \times 10^{-4}[/tex].
The [tex]K_p[/tex] and [tex]K_c[/tex] is " [tex]0.009649615 \ atm[/tex] and [tex]0.00039\ mol\ L^{-1}[/tex]"for the dissociation at this temperature.
Equation:[tex]2NOBr\rightleftharpoons 2NO+Br_2[/tex]
When the initial pressure of [tex]NOBr[/tex] is p, then at equilibrium, therefore the dissociation percentage of [tex]NOBr \ \ is\ \ 34\%[/tex]
[tex]\to P(NOBr)=0.66\ p\\\\\to P(NO)=0.34\ p\\\\\to P(Br_2)= \frac{0.34p}{2}=0.17\ p\\\\[/tex]
Calculating the total pressure:
[tex]\to 0.66\ p+0.34\ p+0.17\ p=1.17\ p=0.25\ atm\\\\[/tex]
[tex]\to p=0.214 \ atm\\\\[/tex]
The equilibrium partial pressures:
[tex]\to P(NOBr)=0.66\ p=0.141\ atm\\\\\to P(NO)=0.34\ p=0.073\ atm\\\\\to P(Br_2)=0.17\ p=0.036\ atm\\\\[/tex]
Calculating the [tex]K_p[/tex]:
[tex]\to K_p =\frac{(P_{NO})^2 (P_{Br_2})}{(P_{NOBr})^2}\\\\[/tex]
[tex]=\frac{(0.073)^2 \times (0.036)}{(0.141)^2}\\\\=\frac{(0.005329) \times (0.036)}{(0.019881)}\\\\=\frac{(0.000191844)}{(0.019881)}\\\\=0.009649615 \ atm[/tex]
Calculating the [tex]K_c[/tex]:
Using formula:
[tex]\to \Delta ng = \text{Number of moles of product - Number of moles of reactant}[/tex]
[tex]= (2+1) - 2\\\\= 3 - 2\\\\=1 \\[/tex]
The relationship between [tex]K_p[/tex] and [tex]K_c[/tex] is:
[tex]\to K_p = K_C(RT)^{\Delta ng}[/tex]
Hence, by putting the values of [tex]K_p[/tex], gas constant [tex]R, T,[/tex] and [tex]\Delta ng[/tex], we can calculate the value of [tex]K_c[/tex].
[tex]\to 0.0096\ atm = - K_c(0.0821 \ atm \ mol L^{-1}K^{-1} \times 298 K)^1[/tex]
[tex]\to K_c = \frac{0.0096 }{0.0821\times 298}[/tex]
[tex]= \frac{0.0096}{ 24.46}\\\\ = 0.00039\ mol\ L^{-1}[/tex]
Therefore, the [tex]K_p[/tex] and [tex]K_c[/tex] is " [tex]0.009649615 \ atm[/tex] and [tex]0.00039\ mol\ L^{-1}[/tex]"for the dissociation at this temperature.
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A chemical reaction occurring in a cylinder equipped with a moveable piston produces 0.621 mol of a gaseous product.
If the cylinder contained 0.120 mol of gas before the reaction and had an initial volume of 2.18 L, what was its volume after the reaction?
(Assume constant pressure and temperature and that the initial amount of gas completely reacts.)
Answer:
11.3 L
Explanation:
Initially, the cylinder had n₁ = 0.120 mol of gas in an initial volume V₁ = 2.18 L. At the end, it had n₂ = 0.621 mol in an unknown volume V₂. According to the Avogadro's law, in the same conditions of pressure and temperature, the volume is directly proportional to the number of moles.
[tex]\frac{V_{1}}{n_{1}} =\frac{V_{2}}{n_{2}} \\V_{2}=\frac{V_{1}\times n_{2} }{n_{1}} =\frac{2.18L \times 0.621mol}{0.120mol} =11.3L[/tex]
1) A certain amount of chlorine gas was placed inside a cylinder with a movable piston at one end. The initial volume was 3.00 L \rm L and the initial pressure of chlorine was 1.45atm The piston was pushed down to change the volume to 1.00 LCalculate the final pressure of the gas if the temperature and number of moles of chlorine remain constant..
2) In an air-conditioned room at 19.0C a spherical balloon had the diameter of 50.0 cm When taken outside on a hot summer day, the balloon expanded to 51.0cm cm in diameter. What was the temperature outside? Assume that the balloon is a perfect sphere and that the pressure and number of moles of air molecules remains the same.
Answer:
1) 4.35 atm
2) 36.88 °C
Explanation:
1) Because the temperature and number of moles remained constant, we can use the formula P₁V₁=P₂V₂
3.00 L * 1.45 atm = P₂ * 1.00 L
P₂ = 4.35 atm
2) First we use a geometrical formula to calculate the volume of the spherical balloon when it has a diameter of 50.0 and of 51.0 cm.
V₁ = 4/3 * π*(50/2)³ = 65449.85cm³
V₂ = 4/3 * π*(51/2)³ = 69455.90cm³
Then we use T₁V₂=T₂V₁, keeping in mind using Kelvin as the unit for temperatures:
292.16 K * 69455.90cm³ = T₂ * 65449.85cm³
T₂ = 310.04 K = 36.88 °C
The final pressure of the chlorine gas in the cylinder is 4.35 atm as per Boyle's Law. The temp on a hot summer day outside can be found using Charles's Law after a series of steps starting with converting the diameter to radius then to volume, adjusting for Kelvin, and then solving for T2 with Charles's Law.
Explanation:The subjects of these problems pertain to gas laws, specifically, Boyle's Law and Charles's Law.
1) According to Boyle's Law, the pressure and volume of a gas have an inverse relationship when the temperature and the number of moles remain constant. Thus, if the volume is decreased, the pressure should increase. We can calculate the final pressure with the formula P1*V1 = P2*V2. Substituting the given figures: 1.45 atm * 3.00 L = P2 * 1.00 L. Solving for P2, the final pressure of the gas is 4.35 atm.
2) With Charles's Law, the volume and temperature of a gas have a direct relationship when pressure and the number of moles remain constant. Converting the diameters to radii (25.0 cm to 26.0 cm) and using the volume formula of a sphere, you find the volume before and after. The formula for Charles's Law is V1/T1 = V2/T2. However, all temperatures need to be in Kelvin, so 19.0 C converts to 292.15 K. Substituting the calculated volumes and temperatures, solve for T2. This will give you the temperature outside in Kelvin. Convert to Celsius for a meaningful interpretation.
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When Fe2O3(s) reacts with H2(g) according to the following reaction, 2.00 kJ of energy are evolved for each mole of Fe2O3(s) that reacts. Complete the following thermochemical equation. 3Fe2O3(s) + H2(g)2Fe3O4(s) + H2O(g)
Answer:
3 Fe₂O₃(s) + H₂(g) ⇒ 2 Fe₃O₄(s) + H₂O(g) ΔH° = -6.00 kJ
Explanation:
Let's consider the following balanced equation.
3 Fe₂O₃(s) + H₂(g) ⇒ 2 Fe₃O₄(s) + H₂O(g)
When 1 mole of Fe₂O₃(s) reacts, 2.00 kJ of energy are evolved. Energy is an extensive property. In the balanced equation there are 3 moles of Fe₂O₃(s), so the evolved energy is:
[tex]3molFe_{2}O_{3}.\frac{2.00kJ}{1molFe_{2}O_{3}} =6.00kJ[/tex]
By convention, when energy is evolved it takes the negative sign. At constant pressure, the thermochemical equation is:
3 Fe₂O₃(s) + H₂(g) ⇒ 2 Fe₃O₄(s) + H₂O(g) ΔH° = -6.00 kJ
where
ΔH° is the standard enthalpy of reaction (heat released at constant pressure)
Do sample problem 13.8 in the 8th ed Silberberg book. You add 1.4 kg of ethylene glycol (C2H6O2) antifreeze to 4,192 g of water in your car's radiator. What is the boiling point of the solution? The Kb for water is 0.512 °C/m. Enter to 2 decimal places.
Answer:
The solution boils at 102.76 °C
Explanation:
Δt = m* Kb x i
i = 1 with 1 particle in solution
m = molality
Kb = 0.512 °C/molal
Step 2: Calculate moles C2H6O2
molar mass of C2H6O2 = 62.068 g/mol
Calculate number of moles:
Moles = Mass / molar mass
Moles = 1400 grams / 62.068 g/mol
Moles C2H6O2 = 22.56 moles
Step 3: Calculate molality
these are in 4192 g of water:
22.56 moles / 4.192 kg water
⇒ moles / kg of water = 5.38 moles / Kg = m olal
Step 4: Calculate ΔT
ΔT= Kb * molal = 5.38 molal* 0.512 °C/m
ΔT = 2.76 °C
Boiling point = 100°C + 2.75 °C = 102.76 °C
the solution boils at 102.76 °C
The boiling point of the ethylene glycol solution is calculated by first finding the molality of the solution and then using this to find the boiling point elevation. The normal boiling point of water is thus raised by this amount, giving a final boiling point of the solution as 102.75 degrees Celsius.
Explanation:To solve this problem, we need to calculate the molality of the ethylene glycol solution and then use this to determine the boiling point elevation of the solution.
Firstly, we need to convert the mass of ethylene glycol (C2H6O2) and water into moles. The molar mass of ethylene glycol is approximately 62.07 g/mol, so 1.4 kg (or 1400 g) of ethylene glycol is equivalent to approximately 22.54 moles. The mass of water is 4192 g, or roughly 4.192 kg.
The molality (m), is thus given by the formula m = moles of solute / kg of solvent, therefore the molality of the ethylene glycol solution is 22.54 moles / 4.192 kg = 5.38 m.
Next, we use the formula for boiling point elevation: ΔTb = Kb * m, where Kb is the ebullioscopic constant for water (given as 0.512 °C/m), and m is the molality calculated earlier. Therefore, the boiling point of the solution rises by ΔTb = 0.512 × 5.38 = 2.75 °C.
The normal boiling point of water is 100 °C, thus, the boiling point of the ethylene glycol solution is 100 °C + 2.75 °C = 102.75 °C.
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The carbon atoms of acetic acid (CH3COOH) exhibit what type of hybridization?A.) spB.) sp^2C.) sp^3D.) sp^2 and sp^3
Answer:
D.) sp² and sp³
Explanation:
Acetic acid CH₃COOH
H3C-----sp³ C(O)OH----sp²
One of the many reactions which occur when iron ore is smelted in a blast furnace is given below. For this reaction, the equilibrium constant in terms of partial pressures, Kp, is 0.900 at 873 K and 0.396 at 1273 K. Assume the reaction takes place in a vessel containing only product and reactant molecules and that AH and AS do not change with temperature. FeO(s) + CO(g) = Fe(s) + CO2(g) a) Calculate AH, AG and AS for the reaction at 873 K b) What is the mole fraction of CO2(g) at 873 K? Include only the gas-phase species in this mole fraction.
Answer:
Explanation: AH, AG and AS for the reaction at 873 K
delta G =-RT ln Kp
G= -8.314 x 873 ln 0.9
= 764.7J/mol
AG = AH-TAS
AH = 764.7 - 8314(0.9)
Kp =pCO/pCo2
Substitute the values at different temp. Solve simultaneously
Since mole fraction =kp/pCo
Mf =1.5
Answer:
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Explain why cis−1−chloro−2−methylcyclohexane undergoes E2 elimination much faster than its trans isomer.
The reacting conformation of the cis isomer has ___.
group(s) axial, making it ___ stable and present in a ____.
concentration than the reacting conformation of the trans isomer.
cis−1−chloro−2−methylcyclohexane undergoes E2 elimination more quickly than its trans isomer due to its axial configuration allowing for an anti-periplanar orientation, which is favorable for E2 elimination.
Explanation:The E2 elimination reaction of cis−1−chloro−2−methylcyclohexane happens faster than its trans isomer due to the orientation of its reacting groups. In the cis isomer, the leaving group (the chlorine) and the beta-proton are both axial. This axial positioning on the same side of the cyclohexane ring allows for an anti-periplanar orientation, which is favorable for an E2 elimination.
However, for the trans isomer, the beta-proton is equatorial while the leaving group (the chlorine) is axial, preventing them from assuming the anti-periplanar configuration. Hence, it becomes less stable and present in a lower concentration. Because the cis isomer can more readily achieve this favored geometry, it reacts more quickly in E2 reactions.
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A substance is found to be nonconductive, to have a relatively low melting point, and to be insoluble in water. This is most likely
A.) A metallic solid
B.) An ionic solid
C.) A molecular solid
D.) A network covalent solid
Answer:
C.) A molecular solid
Explanation:
Molecular solids have low melting points, are non-conductive and are insoluble in water. The interactions between the molecules or atoms can be hydrogen bonds, dipole dipole or London dispersion forces.Because they have relatively weak bonds there are easily vaporized and therefore have a low melting point. Although a network covalent solid is non-conductive it has a high melting point due to the strong covalent bonds. The metallic and the ionic solids are both conductive, with the metallic solids having a high melting point.
A substance that is nonconductive, with a low melting point, and insoluble in water is most likely a molecular solid, due to its weak intermolecular forces and neutral molecular composition.
Explanation:The substance described as nonconductive, having a relatively low melting point, and insoluble in water is most likely C.) A molecular solid. Molecular solids are typically composed of molecules held together by relatively weak intermolecular forces such as London dispersion forces, dipole-dipole interactions, or hydrogen bonds. These characteristics result in their low melting points and the inability to conduct electricity. Since they are made up of neutral molecules, they usually do not dissolve in water which is a polar solvent. On the other hand, metallic solids conduct electricity and are malleable, ionic solids conduct electricity when molten and are usually water soluble, and network covalent solids are typically hard and have high melting points.
When 4.51 g of CaCl2 dissolved in 50.00 mL of water in a coffee cup calorimeter, the temperature of the solution rose from 22.6°C to 25.8°C.
Specific heat of the solution is equal to the specific heat of water = 4.18 J/gºC.
Density of the solution is equal to the density of water = 1.00 g/mL.
What is qsolution?
What is qreaction ?
What is ÎHrxn in kJ/mol of CaCl2 ?
Explanation:
The heat gained by the solution = q
[tex]q=mc\times (T_{final}-T_{initial})[/tex]
where,
q = heat gained = ?
c = specific heat of solution= [tex]4.18 J/^oC[/tex]
Mass of the solution(m) = mass of water + mass of calcium chloride
Mass of water = ?
Volume of water = 50.00 mL
Density of water = 1.00 g/mL
Mass = Density × Volume
m = 1.00 g/mL × 50.00 mL = 50.00 g
Mass of the solution (m) = 50.00 g + 4.51 g =54.51 g
[tex]T_{final}[/tex] = final temperature = [tex]25.8 ^oC[/tex]
[tex]T_{initial}[/tex] = initial temperature = [tex]22.6 ^oC[/tex]
Now put all the given values in the above formula, we get:
[tex]q=54.51 g\times 4.18 J/g^oC\times (25.8-22.6)^oC[/tex]
[tex]q=729.126 J[/tex]
The heat gained by the solution is 729.126 J.
Heat energy released during the reaction = q'
q' = -q ( law of conservation of energy)
q' = -729.126 J
The heat energy released during the reaction is -729.126 J.
Moles of calcium chloride, n = [tex]\frac{4.51 g}{111 g/mol}=0.04063 mol[/tex]
[tex]\Delta H_{rxn}=\frac{q'}{n}=\frac{-729.126 J}{0.04063 mol}=-17,945.23 J/mol= -17.945 kJ/mol[/tex]
The ΔH of the reaction is -17.945 kJ/mol.
Mutations in genes encoding cell signaling proteins contribute to many cancers. For example, a chromosomal translocation fuses the Bcr gene to the Abl gene, leading to a constitutively active Bcr–Abl kinase and chronic myeloid leukemia (CML). CML is successfully treated with the drug imatinib (trade name Gleevec), which mimics an Abl kinase substrate and thus inhibits kinase activity. What normal cell molecule, and kinase substrate, does Gleevec mimic?
Imatinib is a small molecule kinase inhibitor. The BCR-ABL kinase can phosphorylate a series of downstream substrates, leading to proliferation of mature granulocytes. Bcr-Abl kinase substrate is the tyrosine. The Protein Tyrosine Kinase activity is an important requirement for malignant transformation, and that it cannot be complemented by any downstream effector, though not all interactions of BCR-ABL with other proteins are phosphotyrosine dependent.
if a chlorine atom were to attract an electron from sodium, the sodium atom would become blank charged.
When a chlorine atom attracts an electron from sodium, the sodium atom becomes positively charged, forming a sodium cation with a +1 charge, while chlorine becomes a chloride ion with a -1 charge, resulting in the formation of NaCl.
Explanation:If a chlorine atom were to attract an electron from sodium, the sodium atom would become positively charged. This occurs because chlorine has a high affinity for electrons due to its seven valence electrons, and it is more energy-efficient for chlorine to gain one electron than to lose seven. When chlorine gains an extra electron, it becomes a chloride ion with a net negative charge. Conversely, when sodium loses its single valence electron, it becomes a sodium ion with a +1 charge, also known as a cation.
The transfer of an electron from sodium to chlorine results in the formation of two oppositely charged ions that are held together by an ionic bond, creating the ionic compound NaCl. This electron transfer satisfies the octet rule for both ions, resulting in complete outermost shells with stable electron configurations.
Three solutions contain a certain acid. The first contains 10% acid, the second 30%, and the third 50%. A chemist wishes to use all three solutions to obtain a 50-liter mixture containing 32% acid. If the chemist wants to use twice as much of the 50% solution as of the 30% solution, how many liters of each solution should be used?
Answer:
To prepare 50L of 32% solution you need: 11L of 30% solution, 22L of 50% solution and 17L of 10% solution.
Explanation:
A 32% solution of acid means 32L of acid per 100L of solution. As the chemist wants to make a solution using twice as much of the 50% solution as of the 30% solution it is possible to write:
2x*50% + x*30% + y*10% = 50L*32%
130x + 10y = 1600 (1)
-Where x are volume of 30% solution, 2x volume of 50% solution and y volume of 10% solution-
Also, it is possible to write a formula using the total volume (50L), thus:
2x + x +y = 50L
3x + y = 50L (2)
If you replace (2) in (1):
130x + 10(50-3x) = 1600
100x + 500 = 1600
100x = 1100
x = 11L -Volume of 30% solution-
2x = 22L -Volume of 50% solution-
50L - 22L - 11L = 17 L -Volume of 10% solution-
I hope it helps!
Final answer:
To create a 50-liter mixture with 32% acid concentration, the chemist should use 20 liters of the 10% acid solution, 10 liters of the 30% acid solution, and 20 liters of the 50% acid solution.
Explanation:
The question involves solving a system of linear equations to determine the volume of each acid solution required to create a 50-liter mixture with a 32% acid concentration. Let's denote the amount of the 10% solution as x liters, the 30% solution as y liters, and the 50% solution as 2y liters (since it's twice the amount of the 30% solution).
The total amount of acid should be 32% of the 50-liter mixture: 0.10x + 0.30y + 0.50(2y) = 0.32 × 50.
We've also established the relationship between the 30% and 50% solutions: 2y is twice the amount of y.
x + 3y = 50
0.10x + 0.30y + 1.00y = 16
Combining and rearranging gives us x = 20 liters, y = 10 liters, and thus 2y = 20 liters. Therefore, to obtain the desired mixture, the chemist should use 20 liters of the 10% solution, 10 liters of the 30% solution, and 20 liters of the 50% solution.
Consider the neutralization reaction 2HNO3(aq) + Ba(OH)2 ( aq ) ⟶ 2H2O ( l ) + Ba ( NO3)2 ( aq ). A 0.125 L sample of an unknown HNO 3 solution required 32.3 mL of 0.200 M Ba ( OH ) 2 for complete neutralization. What is the concentration of the HNO 3 solution?
Answer:
The concentration of the HNO3 solution is 0.103 M
Explanation:
Step 1: Data given
Volume of the unknow HNO3 sample = 0.125 L
Volume of 0.200 M Ba(OH)2 = 32.3 mL = 0.0323 L
Step 2: The balanced equation
2HNO3(aq) + Ba(OH)2 ( aq ) ⟶ 2H2O ( l ) + Ba( NO3)2 (aq)
Step 3:
n2*C1*V1 = n1*C2*V2
⇒ n2 = the number of moles of Ba(OH)2 = 1
⇒ C1 = the concentration of HNO3 = TO BE DETERMINED
⇒ V1 = the volume of the HNO3 solution = 0.125 L
⇒ n1 = the number of moles of HNO3 = 2
⇒ C2 = the concentration of Ba(OH)2 = 0.200 M
⇒ V2 = the volume of Ba(OH)2 = 0.0323 L
1*C1 * 0.125 L = 2*0.200M * 0.0323 L
C1 = (2*0.200*0.0323)/0.125
C1 = 0.103 M
The concentration of the HNO3 solution is 0.103 M
Predict the product of the following reaction: CH3CH=CHCH3+H2OH3PO4⟶product. Enter the IUPAC name of the product
The product of the given reaction, an acid-catalyzed hydration of an alkene, would be 2-butanol, as per Markovnikov's Rule. This rule predicts the placement of the hydrogen and halide groups in a reaction.
Explanation:The reaction given is an example of an acid-catalyzed hydration of an alkene. In such reactions, an alkene reacts with water in the presence of an acid (in this case, H3PO4) to form an alcohol. The prediction of the product involves recognizing the reaction type and the reagents involved.
For the given reaction: CH3CH=CHCH3 + H2OH3PO4⟶, the product would be 2-butanol.
But why does this process form 2-butanol? It all comes down to Markovnikov's rule, which predicts that in the addition of a protic acid HX to an alkene, the acid hydrogen (H) becomes attached to the carbon with fewer alkyl substituents, and the halide (X) group becomes attached to the carbon with more alkyl substituents.
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A mixture of argon and krypton gases, in a 5.38 L flask at 67 °C, contains 6.18 grams of argon and 9.66 grams of krypton. The partial pressure of krypton in the flask is________atm and the total pressure in the flask is______atm?
Answer:
The partial pressure of krypton in the flask is 0.59 atm and the total pressure in the flask is 1.39 atm
Explanation:
This must be solved with the Ideal Gas Law equation.
First of all we need the moles or Ar and Kr in the mixture
Moles = Mass / Molar mass
Molar mass Ar 39.95g/m
Moles Ar = 6.18 g/39.95 g/m → 0.154 moles
Molar mass Kr 83.8 g/m
Moles Kr = 9.66 g/ 83.8g/m → 0.115 moles
Total moles in the mixture: 0.154 moles + 0.115 moles = 0.269moles
Now, we have the total moles, we can calculate the total pressure.
P . V = n . R . T
(T° in K = T° in C + 273)
P. 5.38L = 0.269mol . 0.082 L.atm/mol.K . 340K
P = (0.269mol . 0.082 L.atm/mol.K . 340K) / 5.38 L
P = 1.39 atm
Now we have the total pressure, we can apply molar fraction so we can know the partial pressure of Kr.
Kr pressure / Total Pressure = Kr moles / Total moles
Kr pressure / 1.39 atm = 0.115 moles / 0.269 moles
Kr pressure = (0.115 moles / 0.269 moles) / 1.39atm
Kr pressure = 0.59 atm
Which of the following represents a rule for balancing a nuclear equation?
I. The total number of protons plus neutrons in the products and reactants must be the same
II. The total number of each type of element in the products and reactants must be the same
III. The total number of nuclear charges in the products and reactants must be the same
IV. The total number of each type of elementary particle in the products and reactants must be the same.
Answer:
The correct answer is 1 the total number of proton plus neutrons in the products and reactants must be same.
Explanation:
During a radioactive reaction the nucleus of the radioactive atom undergo disintegration or breakdown to form new nucleus.
Radioactive disintegration deals with the emission of α particle or β particle from a radioactive nucleus resulting in the formation of new nucleus.Although new nucleus is formed the total number of proton plus neutrons in product and reactant must be same.
A rule for balancing a nuclear equation is: I. total number of protons + neutrons in the products and reactants must be the same.
What is a Nuclear Equation?A nuclear equation for a radioactive reaction, shows the reactants and product, where the proton number and atomic mass are conserved, unlike in chemical equations.
In a radioactive decay, there is emission of α particle or β particle from a radioactive nucleus to form a new nucleus.
Therefore, a rule for balancing a nuclear equation is: I. total number of protons + neutrons in the products and reactants must be the same.
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A sample consisting of n mol of an ideal gas undergoes a reversible isobaric expansion from volume Vi to volume 3Vi. Find the change in entropy of the gas by calculating, ∫dQ / T, where dQ = nCPdT. (Use the following as necessary: Cp and n.)
Answer:
The change in entropy of gas is [tex]\Delta S= nC_{P}ln3[/tex]
Explanation:
n= Number of moles of gas
Change in entropy of gas = [tex]ds= \int \frac{dQ}{T}[/tex]
[tex]dQ= nC_{p}dT[/tex]
From the given,
[tex]V_{i}=V[/tex]
[tex]V_{f}=3V[/tex]
Let "T" be the initial temperature.
[tex]\frac {V_{i}}{T_{i}}=\frac {V_{f}}{T_{f}}[/tex]
[tex]\frac {V}{T}=\frac {3V}{T_{f}}[/tex]
[tex]{T_{f}} = 3T[/tex]
[tex]\int ds = \int ^{T_{f}}_{T_{i}} \frac{nC_{P}dT}{T}[/tex]
[tex]\Delta S = nC_{p}ln(\frac{T_{f}}{T_{i}})[/tex]
[tex]\Delta S = nC_{p}ln3[/tex]
Therefore, The change in entropy of gas is [tex]\Delta S= nC_{P}ln3[/tex]
Calculate the hydronium ion, [ H30+], and hydroxide ion, [OH-], concentrations for a
0.0117 M HCl Solution.
Answer:
[H3O+] = 0.0117 M
[OH-] = 8.5 * 10^-13 M
Explanation:
Step 1: Data given
Concentration of HCl = 0.0117 M
Step 2:
HCl is a strong acid
pH of a strong acid = -log[H+] = - log[H3O+]
[H3O+] = 0.0117 M
pH = -log(0.0117)
pH = 1.93
pOH =14 - 1.93 = 12.07
pOH = -log[OH+] = 12.07
[OH-] = 10^-12.07 = 8.5 * 10^-13
Or
Kw / [H3O+] = [ OH-]
10^-14 / 0.0117 = 8.5*10^-13
Final answer:
The hydronium ion concentration [H3O+] in a 0.0117 M HCl solution is 0.0117 M, and the hydroxide ion concentration [OH-] is calculated to be 8.55 × 10^-13 M using the water dissociation constant.
Explanation:
To calculate the hydronium ion concentration, [H3O+], of a 0.0117 M HCl solution, we first need to understand that HCl is a strong acid which dissociates completely in water. This means that for every mole of HCl dissolved, there will be one mole of H3O+ ions in solution. Hence, the hydronium ion concentration in a 0.0117 M HCl solution is also 0.0117 M.
As for the hydroxide ion concentration, [OH-], we use the water dissociation constant (Kw), which is 1.0 × 10-14 at 25 °C. The product of the concentrations of the hydronium and hydroxide ions in any aqueous solution is equal to Kw. We can find [OH-] by rearranging the expression Kw = [H3O+][OH-] to solve for [OH-], giving us [OH-] = Kw / [H3O+]. Substituting in the values we get [OH-] = 1.0 × 10-14 M / 0.0117 M = 8.55 × 10-13 M.
Assuming 1 mol of Fe3+ and 2 mol of SCN- were allowed to react and reach equilibrium. 0.5 mol of product was formed. The total volume at equilibrium was 1 L. How much Fe3+ remained at equilibrium? (in mol) (1 point) Your Answer: How much SCN- remained at equilibrium? (in mol) (1 point) Your Answer: What’s the equilibrium constant, Kc? (1 point)
Answer:
a. 0.5 mol
b. 1.5 mol
c. 0.67
Explanation:
Fe3+ + SCN- -----> [FeSCN]2+
a. The ratio of the product to Fe3+ is 1:1. Meaning that if 0.5 mol of product was produced up then 0.5 mol of Fe3+ was used. Leaving 0.5 mol remaining at equilibrium
b. The ratio of the product to SCN= is 1:1. Meaning that if 0.5 mol of product was produced up then 0.5 mol of SCN- was used. Leaving 1.5 mol remaining at equilibrium
c. KC = 0.5/(0.5*1.5) = 0.67