Answer:
[tex]\large \boxed{5.1}[/tex]
Explanation:
1. Initial concentrations of reactants
[SO₂] = (3.8 mol)/(75 L) = 0.0507 mol·L⁻¹
[O₂] = (7.0 mol)/(75 L) = 0.0933 mol·L⁻¹
2. Equilibrium concentration of SO₃
[SO₃] = (1.5 mol)/(75 L) = 0.0200 mol·L⁻¹
3. Set up an ICE table
2SO₂ + O₂ ⇌ 2SO₃
I/mol·L⁻¹: 0.0507 0.0933 0
C/mol·L⁻¹: -2x -x +2x
E/mol·L⁻¹: 0.0507-2x 0.0933-x 2x
4. Calculate x
We know the final concentration of SO₃ is 0.0200 mol·L⁻¹, so
2x = 0.0200
x = 0.0100
5. Find the final concentrations of the reactants
Insert the numbers into the ICE table.
2SO₂ + O₂ ⇌ 2SO₃
I/mol·L⁻¹: 0.0507 0.0933 0
C/mol·L⁻¹: -0.0200 -0.0100 +0.0200
E/mol·L⁻¹: 0.0307 0.0833 0.0200
6. Calculate K
[tex]K_{\text{eq}} = \dfrac{\text{[SO$_{3}$]}^{2}}{\text{[SO}_{2}]^{2}\text{[O$_{2}$]}} = \dfrac{0.0200^{2}}{0.0307^{2}\times0.0833} =\mathbf{5.1}\\\\\text{The value of the equilibrium constant is $\large \boxed{\mathbf{5.1}}$}[/tex]
How long should you hold the iron on the hair to heat the strand and set the base ?
A) 5 seconds
B) 15 seconds
C) 30 seconds
D) 1 minute
A 8.22-g sample of solid calcium reacted in excess fluorine gas to give a 16-g sample of pure solid CaF2. The heat given off in this reaction was 251 kJ at constant pressure. Given this information, what is the enthalpy of formation of CaF2(s)
Answer:
The enthalpy of formation of CaF₂ is -1224.4 kJ.
Explanation:
The enthalpy of formation of CaF₂ can be calculated as follows:
[tex] \Delta H_{f} = \frac{q}{n_{CaF_{2}}} [/tex]
Where:
q: is the heat liberated in the reaction = -251 kJ
The number of moles of CaF₂ is:
[tex] n_{CaF_{2}} = \frac{m}{M} [/tex]
Where:
m: is the mass of CaF₂ = 16 g
M: is the molar mass of CaF₂ = 78.07 g/mol
[tex] n_{CaF_{2}} = \frac{m}{M} = \frac{16 g}{78.07 g/mol} = 0.205 moles [/tex]
Now, the enthalpy of formation of CaF₂ is:
[tex]\Delta H_{f} = \frac{q}{n_{CaF_{2}}} = \frac{-251 \cdot 10^{3} J}{0.205 moles} = -1224.4 kJ/mol[/tex]
Therefore, the enthalpy of formation of CaF₂ is -1224.4 kJ.
I hope it helps you!
Resonance Structures are ways to represent the bonding in a molecule or ion when a single Lewis structure fails to describe accurately the actual electronic structure. Equivalent resonance structures occur when there are identical patterns of bonding within the molecule or ion. The actual structure is a composite, or resonance hybrid, of the equivalent contributing structures. Draw Lewis structures for thecarbonate ion and for phosphine in which the central atom obeys the octet rule. ... How many equivalent Lewis structures are necessary to describe the bonding in CO32-
Answer:
See explanation
Explanation:
A Lewis structure is also called a dot electron structure. A Lewis structure represents all the valence electrons on atoms in a molecule as dots. Lewis structures can be used to represent molecules in which the central atom obeys the octet rule as well as molecules whose central atom does not obey the octet rule.
Sometimes, one Lewis structure does not suffice in explaining the observed properties of a given chemical specie. In this case, we evoke the idea that the actual structure of the chemical specie lies somewhere between a limited number of bonding extremes called resonance or canonical structures.
The canonical structure of the carbonate ion as well as the lewis structure of phosphine is shown in the image attached to this answer.
The following balanced equation describes the reduction of iron(III) oxide to molten iron within a blast furnace: Fe2O3(s) + 3CO(g) ---> 2Fe(l) + 3CO2(g) Steve inserts 450. g of iron(III) oxide and 260. g of carbon monoxide into the blast furnace. After cooling the pure liquid iron, Steve determines that he has produced 288g of iron ingots. What is the theoretical yield of liquid iron, in grams? Just enter a numerical value. Do not enter units.
Answer: 313.6
Explanation:
To calculate the moles :
[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]
[tex]\text{Moles of} Fe_2O_3=\frac{450g}{160g/mol}=2.8moles[/tex]
[tex]\text{Moles of} CO=\frac{260g}{28g/mol}=9.3moles[/tex]
[tex]Fe_2O_3(s)+3CO(g)\rightarrow 2Fe(l)+3CO_2(g)[/tex]
According to stoichiometry :
1 mole of [tex]Fe_2O_3[/tex] require 3 moles of [tex]CO[/tex]
Thus 2.8 moles of [tex]Fe_2O_3[/tex] will require=[tex]\frac{3}{1}\times 2.8=8.4moles[/tex] of [tex]CO[/tex]
Thus [tex]Fe_2O_3[/tex] is the limiting reagent as it limits the formation of product and [tex]CO[/tex] is the excess reagent.
As 1 mole of [tex]Fe_2O_3[/tex] give = 2 moles of [tex]Fe[/tex]
Thus 2.8 moles of [tex]Fe_2O_3[/tex] give =[tex]\frac{2}{1}\times 2.8=5.6moles[/tex] of [tex]Fe[/tex]
Mass of [tex]Fe=moles\times {\text {Molar mass}}=2.6moles\times 56g/mol=313.6g[/tex]
Theoretical yield of liquid iron is 313.6 g
How many molecules of CaCl2 are equivalent to 75.9g CaCl2 (Ca=40.08g/mol, CL=35.45g/mol)
Answer:
[tex]\large \boxed{4.12 \times 10^{23}\text{ formula unis of CaCl}_{2}}$}[/tex]
Explanation:
You must calculate the moles of CaCl₂, then convert to formula units of CaCl₂.
1. Molar mass of CaCl₂
CaCl₂ = 40.08 + 2×35.45 = 40.08 + 70.90 = 110.98 g/mol
2. Moles of CaCl₂ [tex]\text{Moles of CaCl}_{2} = \text{75.9 g CaCl}_{2} \times \dfrac{\text{1 mol CaCl}_{2}}{\text{110.98 g CaCl}_{2}} = \text{0.6839 mol CaCl}_{2}[/tex]
3. Formula units of CaCl₂
[tex]\text{No. of formula units} = \text{0.6839 mol CaCl}_{2} \times \dfrac{6.022 \times 10^{23}\text{ molecules CaCl}_{2}}{\text{1 mol P$_{2}$O}_{5}}\\\\= \mathbf{4.12 \times 10^{23}}\textbf{ formula units CaCl}_{2}\\\text{There are $\large \boxed{\mathbf{4.12 \times 10^{23}}\textbf{ formula units of CaCl}_{2}}$}[/tex]
What would be the voltage (Ecell) of a voltaic cell comprised of Cr (s)/Cr3+(aq) and Fe (s)/Fe2+(aq) if the concentrations of the ions in solution were [Cr3+] = 0.75 M and [Fe2+] = 0.25 M at 298K?
Answer:
0.35 V
Explanation:
(a) Standard reduction potentials
E°/V
Fe²⁺ + 2e- ⇌ Fe; -0.41
Cr³⁺ + 3e⁻ ⇌ Cr; -0.74
(b) Standard cell potential
E°/V
2Cr³⁺ + 6e⁻ ⇌ 2Cr; +0.74
3Fe ⇌ 3Fe²⁺ + 6e-; -0.41
2Cr³⁺ + 3Fe ⇌ 2Cr + 3Fe²⁺; +0.33
3. Cell potential
2Cr³⁺(0.75 mol·L⁻¹) + 6e⁻ ⇌ 2Cr
3Fe ⇌ 3Fe²⁺(0.25 mol·L⁻¹) + 6e-
2Cr³⁺(0.75 mol·L⁻¹) + 3Fe ⇌ 2Cr + 3Fe²⁺(0.25 mol·L⁻¹)
The concentrations are not 1 mol·L⁻¹, so we must use the Nernst equation
[tex]E = E^{\circ} - \dfrac{RT}{zF}\ln Q[/tex]
(a) Data
E° = 0.33 V
R = 8.314 J·K⁻¹mol⁻¹
T = 298 K
z = 6
F = 96 485 C/mol
(b) Calculations:
[tex]Q = \dfrac{\text{[Fe}^{2+}]^{3}}{ \text{[Cr}^{3+}]^{2}} = \dfrac{0.25^{3}}{ 0.75^{2}} =\dfrac{0.0156}{0.562} = 0.0278\\\\E = 0.33 - \left (\dfrac{8.314 \times 298}{6 \times 96485}\right ) \ln(0.0278)\\\\=0.33 -0.00428 \times (-3.58) = 0.33 + 0.0153 = \textbf{0.35 V}\\\text{The cell potential is }\large\boxed{\textbf{0.35 V}}[/tex]
The final overall chemical equation is Upper Ca upper O (s) plus upper C upper O subscript 2 (g) right arrow upper C a upper C upper O subscript 3 (s).. When the enthalpy of this overall chemical equation is calculated, the enthalpy of the second intermediate equation is halved and has its sign changed. is halved. has its sign changed. is unchanged.
Answer:
the enthalpy of the second intermediate equation is halved and has its sign changed.
Explanation:
Let us take a look at the first and second intermediate reactions as well as the overall reaction equation for the process under review;
First reaction;
Ca (s) + CO₂ (g) + ½O₂ (g) → CaCO₃ (s) ΔH₁ = -812.8 kJ
Second reaction;
2Ca (s) + O₂ (g) → 2CaO (s) ΔH₂ = -1269 kJ
Hence the overall equation is now;
CaO (s) + CO₂ (g) → CaCO₃ (s) ΔH = ?
According to the Hess law of constant heat summation, the enthalpy of the overall reaction is supposed to be obtained as a sum of the enthalpy of both reactions but this will not give the enthalpy of the overall reaction in this case. The enthalpy of the overall reaction is rather obtained by halving the enthalpy of the second intermediate reaction and reversing its sign before taking the sum as shown below;
Enthalpy of Intermediate reaction 1 + ½(- Enthalpy of Intermediate reaction 2) = Enthalpy of Overall reaction
Answer:
A.
Explanation:
Did it on Edge.
3. There are many different primary standards that could be used in a standardization titration. What are the criteria for a primary standard
Answer:
High purity.
Stability (low reactivity)
Low hygroscopicity (to minimize weight changes due to humidity)
Explanation:
There are different primary standards that could be used in a standardization titration in order to achieve the best and accurate result possible. These standards include high purity,stability and low hygoscropicity .
A high purity means the reactants lack impurities which could affect the result. Stability also ensures that there is non reactivity with elements/compounds in the atmosphere while low hygroscopicity ensures weight changes are minimized due to humidity.
Choose the compound that exhibits hydrogen bonding as its strongest intermolecular force.
A. C2H6
B. CH3OH
C. CH2Br2
D. SBr2
E. None of the above compounds exhibit hydrogen bonding.
Answer:
B
Explanation:
To form hydrogen bondings between the molecules, the compound needs a highly electronegative atom (usually N, O, or F) bonded with a hydrogen atom;
and that the highly electronegative atom has lone pair outermost shell electrons.
In the 5 options, only B (CH3OH) has an N, O, or F atom that has lone pair outermost shell electrons (2 lone pairs on each O atom), so it can form hydrogen bonds within its molecules.
Hydrogen bonds are stronger than the van der Waals' forces between its molecules (that exist regardless of whether there are hydrogen bonds).
The compound that exhibits hydrogen bonding as its strongest intermolecular force is CH₃OH as electronegative oxygen atom is bonded to hydrogen atom.
What is compound?Compound is defined as a chemical substance made up of identical molecules containing atoms from more than one type of chemical element.
Molecule consisting atoms of only one element is not called compound.It is transformed into new substances during chemical reactions. There are four major types of compounds depending on chemical bonding present in them.They are:
1)Molecular compounds where in atoms are joined by covalent bonds.
2) ionic compounds where atoms are joined by ionic bond.
3)Inter-metallic compounds where atoms are held by metallic bonds
4) co-ordination complexes where atoms are held by co-ordinate bonds.
They have a unique chemical structure held together by chemical bonds Compounds have different properties as those of elements because when a compound is formed the properties of the substance are totally altered.
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write the IUPAC name OF THE FOLLOWING COMPOUNDS
Answer:
Explanation:
a) 2 chloro butane
b) 2-3 dimethyl butane
c) 2 bromo 3 nitro pentane
d) 2-3 trimethyl pentane
e) 2-bromo,3-methyl,4-nitro hexane
f) 2-methyl cyclo butane
Use your periodic table and calculator as needed for the following question.
How much stock solution is needed to make 250 mL of a 6.0M solution. The molarity of the stock solution is 18M.
Selections may be rounded so choose the best answer.
56 mL
83 mL
2.3 mL
4.7 ml
For a single substance at atmospheric pressure, classify the following as describing a spontaneous process, a nonspontaneous process, or an equilibrium system.
A) Solid melting below its melting point
B) Gas condensing below its condensation point
C) Liquid vaporizing above its boiling point
D) Liquid freezing below its freezing point
E) Liquid freezing above its freezing point
F) Solid melting above its melting point
G) Liquid and gas together at boiling point with no net condensation or vaporization
H) Gas condensing above its condensation point
I) Solid and liquid together at the melting point with no net freezing or melting
Answer:
Spontaneous process- This is the process that occurs on its own without the application of any external energy or other factor. They include
B) Gas condensing below its condensation point
C) Liquid vaporizing above its boiling point
D) Liquid freezing below its freezing point
F) Solid melting above its melting point
Non spontaneous - This is the process that doesn’t occurs on its own and requires the application of any external energy or factor. They include
A) Solid melting below its melting point
E) Liquid freezing above its freezing point
H) Gas condensing above its condensation point
Equilibrium system
G) Liquid and gas together at boiling point with no net condensation or vaporization
I) Solid and liquid together at the melting point with no net freezing or melting
A) Solid melting below its melting point - nonspontaneous process
B) Gas condensing below its condensation point - spontaneous process
C) Liquid vaporizing above its boiling point - spontaneous process
D) Liquid freezing below its freezing point - spontaneous process
E) Liquid freezing above its freezing point - nonspontaneous process
F) Solid melting above its melting point - spontaneous process
G) Liquid and gas together at boiling point with no net condensation or vaporization - Equilibrium system
H) Gas condensing above its condensation point - nonspontaneous process
I) Solid and liquid together at the melting point with no net freezing or melting - Equilibrium system
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Amanda is doing a report for her Earth Science class about the four seasons. Which of the following would be an effective scientific model to incorporate in her project? a. A calendar indicating the first days of autumn, winter, spring, and summer b. Four earth models, each with a different tilt to represent the earth’s position relative to the sun, lined up in order of season c. A poster board with pictures of weather characteristic of the four seasons d. A bar graph indicating average high and low temperatures for Amanda’s area in the autumn, winter, spring, and summer Please select the best answer from the choices provided
Answer:
(B)
Explanation:
edg 2020
The four seasons in earth is originating from the earth's revolution around the sun. Therefore, the most suitable model for Amanda is Four earth models, each with a different tilt to represent the earth’s position relative to the sun, lined up in order of season.
What are seasons?Seasons in earth is originating from the difference in the distance from the sun over each time period. Hence, revolution of earth around sun make these seasons.
The time period at which earth comes closer to sun more hot will be earth's surface and we experience summer season. When we far from sun winter season occurs.
Therefore, different season are coming based on the distance of earth from sun at each revolution point. This is also affected by the tilt of earth's in its own axis.
Hence, different earth models with different distance from sun is most suitable model here for Amanda. Thus option B is correct.
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g Air contains nitrogen, oxygen, argon, and trace gases. If the partial pressure of nitrogen is 592 mm Hg, oxygen is 160 mm Hg, argon is 7 mm Hg, and trace gas is 1 mm Hg, what is the atmospheric pressure
Answer:
Explanation:
Atmospheric pressure = partial pressure of nitrogen + partial pressure of oxygen + partial pressure of argon + partial pressure of trace element
putting the given values
Atmospheric pressure = 592 + 160 + 7 + 1
= 760 mm of Hg .
Which relationship can be used to aid in the determination of the heat absorbed by bomb calorimeter? 
Answer:
ΔH = [tex]q_{p}[/tex]
Explanation:
In a calorimeter, when there is a complete combustion within the calorimeter, the heat given off in the combustion is used to raise the thermal energy of the water and the calorimeter.
The heat transfer is represented by
[tex]q_{com}[/tex] = [tex]q_{p}[/tex]
where
[tex]q_{p}[/tex] = the internal heat gained by the whole calorimeter mass system, which is the water, as well as the calorimeter itself.
[tex]q_{com}[/tex] = the heat of combustion
Also, we know that the total heat change of the any system is
ΔH = ΔQ + ΔW
where
ΔH = the total heat absorbed by the system
ΔQ = the internal heat absorbed by the system which in this case is [tex]q_{p}[/tex]
ΔW = work done on the system due to a change in volume. Since the volume of the calorimeter system does not change, then ΔW = 0
substituting into the heat change equation
ΔH = [tex]q_{p}[/tex] + 0
==> ΔH = [tex]q_{p}[/tex]
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.
Charcoal from the dwelling level of the Lascaux Cave in France gives an average count of 0.97 disintegrations of ^14 C per minute per gram of sample. Living wood gives 6.68 disintegrations per minute per gram. Estimate the date of occupation and hence the probable date of the wall painting in the Lascaux Cave. Hint: Disintegrations per minute per gram" has the same units as the time-derivative of concentration for a radioactive decay model. (You may use the fact that the half-life of ^14C is 5568 years.)
Answer:
Explanation:
count given by old sample = .97 disintegrations per minute per gram
count given by fresh sample = 6.68 disintegrations per minute per gram
Half life of radioactive carbon = 5568 years
rate of disintegration
dN / dt = λ N
In other words rate of disintegration is proportional to no of radioactive atoms present . As number reduces rate also reduces .
Let initial no of radioactive be N₀ and after time t , number reduces to N
N₀ / N = 6.68 / .97
Now
[tex]N=N_0e^{-\lambda t}[/tex]
[tex]\frac{N}{N_0} =e^{-\lambda t}[/tex]
[tex]\frac{6.68}{.97} = e^{\lambda t}[/tex]
λ is disintegration constant
λ = .693 / half life
= .693 / 5568
= .00012446 year⁻¹
Putting the values in the equation above
[tex]\frac{6.68}{.97} = e^{.00012446\times t}[/tex]
[tex]6.8866 = e^{.00012446\times t}[/tex]
1.929577 = .00012446 t
t = 15503.6 years .
BEARINGS
Question 12 (SSCE 1994 May/June)
(a) A village P is 10km from a lorry station, Q, on
a bearing 065º. Another village R, is 8km from
Q on a bearing 155º. Calculate
(i) the distance of R from P to the nearest
kilometer
(ii) the bearing of R from P to the nearest degree
(b) M is a village on PR such that QM is
perpendicular to PR. Find the distance of M
from P to the nearest kilometer.
Answer:
a. (i) the distance of R from P is 13 Km to the nearest kilometer
(ii) the bearing of R from P = 360 - (38 + 25) = 297° to the nearest degree
b. the distance of m from P is 11 Km to the nearest kilometre
Explanation:
a) A triangle PQR is formed. Q = 90°, p = 8 km; r = 10 km; distance of R from P is q is to be found.
(i) Using the cosine rule: q² = p² + r² - 2prCosQ
q² = 8² + 10² - 2 * 8 * 10 * Cos90
q² = 64 + 100 + 0
q² = 164
q = 13 Km to the nearest kilometre
Therefore, the distance of R from P is 13 Km to the nearest kilometer
(ii) the bearing of R from P
The angle at P is found using the formula Cos P = (q² + r² - p²)/2qr
Cos P = 13² + 10² - 8²/2 * 13 *10
Cos P = 0.7884
P = Cos⁻¹ 0.7884
P = 38°
Therefore, the bearing of R from P = 360 - (38 + 25) = 297° to the nearest degree
Note : 25° is alternate (Northwest) to 65°at P
b) A right-angled triangle QMP is formed
using the trigonometrical ratios; cos Θ = adjacent/hypotenuse
where the hypotenuse side = 10 km, adjacent side = distance of M from P, x
cos P = x/10
x = cos 38 * 10
x = 11 Km to the nearest kilometre
Therefore, the distance of m from P is 11 Km to the nearest kilometre
An atom with 19 protons and 18 neutrons is a(n)
A. Isotope of potassium(K)
B. Standard atom of argon(Ar)
C. Standard atom of (K)
D. Isotope of argon (Ar)
Answer:
A
Explanation:
The number of protons indicates the element so we know it's potassium. To get the number of neutrons you subtract the number of protons (19) from the mass number which for potassium is 39.
39-19=20 neutrons
Because you have 18 neutrons then yours would be an isotope.
Answer: A. Isotope of potassium(K)
Explanation: Founders Educere answer.
Determine whether each of the following salts will form a solution that is acidic, basic, or pH-neutral. Drag the appropriate items to their respective bins.
Al(NO3)3
C2H5NH3NO3
NaClO
RbI
CH3NH3CN
Answer:
Al(NO₃)₃: Acidic.
C₂H₅NH₃NO₃: Acidic.
NaClO: Basic
RbI: pH-neutral
CH₃NH₃CN: Solution basic
Explanation:
The general rules to determine if a solution is acidic, basic or neutral are:
If it is a salt of a strong acid and base, the solution will be pH-neutral. If it is a salt of a strong acid and a weak base, the solution will be acidic due to the hydrolysis of the weak base component (cation). If it is a salt of a strong base and a weak acid, the solution will be basic due to the hydrolysis of the weak acid component (anion).For the salts:
Al(NO₃)₃. The repective acid is HNO₃ (Strong acid) and the base is Al(OH)₃ (Weak base). As the salt comes from strong acid and weak base. SOLUTION ACIDIC
C₂H₅NH₃NO₃. The acid is HNO₃ (Strong acid) and the base C₂H₅NH₃OH (Weak base). SOLUTION ACIDIC.
NaClO. Tha acid is HClO (weak acid), and the base NaOH (Strong base). SOLUTION BASIC.
RbI: The acid is HI (Strong acid) and the base RbOH (Strong base). pH-NEUTRAL
CH₃NH₃CN. The acid is HCN (weak acid; pKb = 4.79) and the base CH₃NH₃OH (weak base; pKa = 10.64). Both weak acid and base will produce each hydrolisis. The lower pK will predominate. That is the weak acid. SOLUTION BASIC
Solution of Al(NO₃)₃ and C₂H₅NH₃NO₃ salts is acidic, NaClO is basic and of RbI & CH₃NH₃cyanide is neutral in nature.
What is pH?pH of any solution tells about the acidity or basicity of the solution, pH of any solution ranges from 0 to 14 and from acidity to basicity.
Al(NO₃)₃ is a salt which is formed by the mixing of strong acid HNO₃ (Nitric acid) and weak base Al(OH)₃, so the resultant solution of the salt is acidic in nature.C₂H₅NH₃NO₃ salt is formed by the mixing of strong acid HNO₃ (Nitric acid) and weak base C₂H₅NH₃OH, so the resultant solution of the salt is acidic in nature.NaClO is a salt of weak acid is HClO and strong base NaOH, so the resultant solution of the salt is basic in nature.RbI salt is formed by the combination of strong acid HI and strong base RbOH, so the resultant solution of the salt is neutral in nature.CH₃NH₃Cyanide is a salt of weak acid hydrogen cyanide and weak base CH₃NH₃OH, so the resultant solution of the salt is neutral in nature.Hence, appropriate differentiation was done above.
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what is the pH of a solution that has a hydronium ion concentration 100 times greater than a solution with a pH of 4
Answer:
2
Explanation:
First, find the hydronium ion concentration of the solution with a pH of 4.
[H₃O⁺] = 10^-pH
[H₃O⁺] = 10⁻⁴
[H₃O⁺] = 1 × 10⁻⁴
Next, multiple the hydronium ion concentration by 100 to find the hydronium ion concentration of the new solution.
[H₃O⁺] = 1.0 × 10⁻⁴ × 100 = 0.01
Lastly, find the pH.
pH = -log [H₃O⁺]
pH = -log (0.01)
pH = 2
The pH of a solution that has a hydronium ion concentration 100 times greater than a solution with a pH of 4 is 2.
Hope this helps.
According to the collision theory, all collisions do not lead to reaction. Which choice gives both reasons why not all collisions between reactant molecules lead to reaction? 1. The total energy of two colliding molecules is less than some minimum amount of energy. 2. Molecules cannot react with each other unless a catalyst is present. 3. Molecules that are improperly oriented during collision will not react. 4. Solids cannot react with gases.
Answer: 3. Molecules that are improperly oriented during collision will not react.
Explanation:
According to the collision theory , the number of collisions that take place per unit volume of the reaction mixture is called collision frequency. The effective collisions are ones which result into the formation of products.
It depends on two factors:-
1. Energy factor: For collision to be effective, the colliding molecules must have energy more than a particular value called as threshold energy.
2. Orientation factor: Also the colliding molecules must have proper orientation at the time of collision to result into formation of products.
Thus not all collisions between reactant molecules lead to reaction because molecules that are improperly oriented during collision will not react.
Generally, systems move spontaneously in the direction of increasing entropy. TRUE FALSE
Answer:
true
Explanation:
The volume of ammonia gas at 1.14 atm of pressure is gradually decreased from 90.3 mL to 43.4 mL. What is the final pressure of ammonia if there is no change in temperature
Answer:
THE FINAL PRESSURE OF AMMONIA IF THERE IS NO CHANGE IN TEMPERATURE AND A DECREASE IN VOLUME FROM 90.3 mL TO 43.4 mL IS 2.91 atm.
Explanation:
At constant temperature, the pressure of a given mass of gas is inversely proportional to the volume. This question follows Boyle's law of gas laws.
Mathematically written as:
P1V1 = P2V2
Re-arranging the formula by making P2 the subject of the formula;
P2 = P1V1 / T2
P1 = 1.4 atm
V1 = 90.3 mL
V2 = 43.4 mL
P2 = unknown
So therefore, we have:
P2 = 1.4 * 90.3 / 43.4
P2 = 2.91 atm
The final pressure of ammonia is therefore 2.91 atm.
Answer:
2.37 atm
Explanation:
Step 1: Given data
Initial pressure of ammonia (P₁): 1.14 atmInitial volume of ammonia (V₁): 90.3 mLFinal pressure of ammonia (P₂): ?Final volume of ammonia (V₂): 43.4 mLConstant temperatureStep 2: Calculate the final pressure of ammonia
Since the temperature is kept constant, we can calculate the final pressure of ammonia using Boyle's law.
[tex]P_1 \times V_1 = P_2 \times V_2\\P_2 = \frac{P_1 \times V_1}{V_2} = \frac{1.14atm \times 90.3mL}{43.4mL} = 2.37 atm[/tex]
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
A chemistry student weighs out of chloroacetic acid into a volumetric flask and dilutes to the mark with distilled water. He plans to titrate the acid with solution. Calculate the volume of solution the student will need to add to reach the equivalence point. Round your answer to significant digits.
Answer:
11.6mL of the 0.1400M NaOH solution
Explanation:
0.154g of chloroacetic acid diluted to 250mL. Titrated wit 0.1400M NaOH solution.
The reaction of chloroacetic acid, ClCH₂COOH (Molar mass: 94.5g/mol) with NaOH is:
ClCH₂COOH + NaOH → ClCH₂COO⁻ + Na⁺ + H₂O
Where 1 mole of the acid reacts per mole of the base.
That means the student will reach equivalence point when moles of chloroacetic acid = moles NaOH.
You will titrate the 0.154g of ClCH₂COOH. In moles (Using molar mass) are:
0.154g ₓ (1mol / 94.5g) = 1.63x10⁻³ moles of ClCH₂COOH
To reach equivalence point, the student must add 1.63x10⁻³ moles of NaOH. These moles comes from:
1.63x10⁻³ moles of NaOH ₓ (1L / 0.1400moles NaOH) = 0.0116L of the 0.1400M NaOH =
11.6mL of the 0.1400M NaOH solutionAn amphoteric salt is one that contains an anion that can act as either an acid or a base in water. Baking soda, NaHCO3, is an example. By combining the ionization and hydrolysis reactions of the anion, you get the principle reaction that occurs when this salt is dissolved in water.
2HCO3-(aq) CO32-(aq) + H2CO3(aq)
The pH for such, a solution is given by
pH = pKa1 + pKa2/2
where Ka1 and Ka2 are the ionization constants of the acid (in this case, H2CO3). Note that the pH of the solution is independent of the salt concentration. Calculate the pH of a NaHCO3 solution.
Answer:
pH = 8.34
Explanation:
The equilbriums of the amphoteric HCO₃⁻ (Ion of NaHCO₃) are:
H₂CO₃ ⇄ HCO₃⁻ + H⁺ Ka1 -Here, HCO₃⁻ is acting as a base-
HCO₃⁻⇄ CO₃²⁻ + H⁺ Ka2 -Here, is acting as an acid-
Where Ka1 = 4.3x10⁻⁷ and Ka2 = 4.8x10⁻¹¹. As pKa = -log Ka:
pKa1 = 6.37; pKa2 = 10.32
As the pH of amphoteric salts is:
pH = (pKa1 + pKa2) / 2
pH = 8.34bonding is similar to ionic bonding, except there are no high-electronegativity atoms present to accept any electrons that the present atoms are willing to donate. Group of answer choices
The question is incomplete, the complete question is;
["covalent", "Van der Waals", "ionic", "hydrogen", "metallic"] bonding is similar to ionic bonding, except their are no high-electronegativity atoms present to accept any electrons that the present atoms are willing to donate.
Answer:
metallic
Explanation:
The metallic bond bears a striking similarity to the ionic bonds only that there are no electronegative elements present to accept electrons in a metallic bond as in an ionic bond.
Most metals usually have a few valence electrons which are loosely bound to the outermost shell of the metal atom. Metallic bonds are usually comprised of metal ions bonded together by a sea of mobile electrons
These mobile electrons exert an attractive force on the positive ions and hold them together in the metallic crystal lattice. This force of attraction that holds the metal atoms together in the metallic crystal lattice is known as the metallic bond.
A chemist studies the reaction below. 2NO(g) + Cl2(g) 2NOCl(g) He performs three experiments using different concentrations and measures the initial reaction rates (The data from the three experiments is in the table). 1. Write the rate law 2. Solve for k.
Answer:
1. [tex]Rate =k [NO]^{2}[Cl_{2}][/tex]
2. [tex]k= 0.42 \frac{L^{2}}{mol^{2}*s}[/tex]
Explanation:
[tex]Rate =k [NO]^{m}[Cl_{2}]^{n}[/tex]
[tex]Rate1 = k[0.4]^{m}[0.3]^{n}=0.02\\Rate 2=k [0.8]^{m}[0.3}]^{n}=0.08\\\\\frac{Rate1}{Rate2}=\frac{0.02}{0.08} =\frac{k[0.4]^{m}[0.3]^{n}}{k[0.8]^{m}[0.3]^n}} \\\\\frac{1}{4} =(\frac{1}{2} )^{m},\\m=2[/tex]
[tex]Rate3 =k [0.8]^{m}[0.6]^{n}=0.16\\Rate 2= k[0.8]^{m}[0.3}]^{n}=0.08\\\\\frac{Rate3}{Rate2}=\frac{0.16}{0.08} =\frac{k[0.8]^{m}[0.6]^{n}}{k[0.8]^{m}[0.3]^n}} \\\\\frac{2}{1} =(\frac{2}{1} )^{n},\\n=1[/tex]
[tex]Rate =k [NO]^{2}[Cl_{2}]^{1}[/tex]
[tex]Rate =k [NO]^{2}[Cl_{2}]^{1}\\Rate 1=k [0.4]^{2}[0.3]^{1} =0.02\\k*0.16*0.3=0.02\\k=\frac{0.02}{0.16*0.3}=\frac{1}{8*(\frac{3}{10} )}=\frac{5}{12} = 0.42 \frac{L^{2}}{mol^{2}*s}[/tex]
Which of the following sets of quantum numbers is NOT allowed? a. n = 5, l= 4, ml= –2, ms = +1/2 b. n = 2, l = 1, ml= 0, ms = +1/2 c. n = 4, l= 4, ml= 0, ms = –1/2 d. n = 3, l= 2, ml= –1, ms = +1/2 e. n = 2, l= 0, ml= 0, ms = -1/2
Answer:
C
Explanation:
the n value must always be greater than the l value
Out of the following set of quantum numbers ,set C is not allowed as the azimuthal and principal quantum numbers are same.
What are quantum numbers?Quantum numbers are the numbers which describe the values of conserved quantities with respect to the dynamics of a quantum system.They correspond to the Eigen values of operators which commute with the Hamiltonian quantities.
The Hamiltonian quantities can be known with precision simultaneously with the system's energy.Quantum numbers can take values of discrete sets of integers or even half-integers even though they can approach infinity in some cases.
They can specifically describe energy levels of electrons, and can also explain angular momentum,spin,etc.These are used to describe the path of an electron in an atom ,when the quantum numbers of all atoms are combined they must comply with the Schrodinger equation.
Learn more about quantum numbers,here:
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