For each of the following systems at equilibrium, predict whether the reaction will shift to the right, left, or not be affected by a decrease in the reaction container volume.
A) PCl3(g) + Cl2(g) ⇌ PCl5(g) 1. no shift
2. right shift
3. left shift B) 2 NO(g) ⇌ N2(g) + O2(g)
1. no shift 2. right shift3. left shift C) 2 NO2(g) ⇌ N2O4(g)
1. no shift 2. right shift3. left shift
Answer:
A - right shift
B - no shift
C - right shift
Explanation:
According to Le Chatelier's principle, when a reaction is in equilibrium and one of the factors affecting the rate of reaction is introduced, the equilibrium will shift so as to annul the effects of the constraint.
In this case, decreasing the volume of the reaction's container is equivalent to increasing the pressure of the reaction. When pressure is increased, the reaction will shift towards the side with a lower number of moles.
In A, the total number of moles on the left-hand side of the reaction is two while it is one on the right-hand side. An increase in pressure will, therefore, see the equilibrium shifting to the right-hand side.
In B, the total number of moles on both the right and the left-hand side is two each. An increase in the pressure will have no effect on the equilibrium.
In C, the total number of moles on the left-hand side is two while it is one on the right-hand side. Hence, an increase in the pressure of the reaction will cause a shift in the equilibrium to the right.
What is the concentration of A after 50.7 minutes for the second order reaction A → Products when the initial concentration of A is 0.250 M? (k = 0.117 M⁻¹min⁻¹)
Answer:
0.101 M
Explanation:
Step 1: Given data
Initial concentration of A ([A]₀): 0.250 MFinal concentration of A ([A]): ?Time (t): 50.7 minRate constant (k): 0.117 M⁻¹.min⁻¹Step 2: Calculate [A]
For a second-order reaction, we can calculate [A] using the following expression.
1/[A] = 1/[A]₀ + k × t
1/[A] = 1/0.250 M + 0.117 M⁻¹.min⁻¹ × 50.7 min
[A] = 0.101 M
A 100.0-mL sample of 1.00 M NaOH is mixed with 50.0 mL of 1.00 M H2SO4 in a large Styrofoam coffee cup calorimeter fitted with a lid through which a thermometer passes. The acid-base reaction is as follows:
2 NaOH(aq) + H2SO4(aq) → Na2SO4(aq) + 2 H2O(l)
The temperature of each solution before mixing is 22.3 °C. After mixing, the temperature of the solution mixture reaches a maximum temperature of 31.4 °C. Assume the density of the solution mixture is 1.00 g/mL, its specific heat is 4.18 J/g.°C, and no heat is lost to the surroundings. Calculate the enthalpy change, in kj, per mole of H2SO4 in the reaction.
a. +85.6 kJ/mol.
b. -85.6 kJ/mol.
c. +5.71 kJ/mol.
d. -5.71 kJ/mol.
e. -114 kJ/mol.
Answer:
THE ENTHALPY CHANGE IN KJ/MOLE IS +114 KJ/MOLE.
Explanation:
Heat = mass * specific heat capacity * temperature rise
Total volume = 100 + 50 = 150 mL
Total mass = density * volume
Total mass = 1 * 150 mL = 150 g
So therefore, the heat evolved during the reaction is:
Heat = 150 * 4.18 * ( 31.4 - 22.3)
Heat = 150 * 4.18 * 9.1
Heat = 5705.7 J
Equation for the reaction:
2 NaOH(aq) + H2SO4(aq) → Na2SO4(aq) + 2 H2O(l)
From the equation, 2 moles of NaOH reacts with 1 mole of H2SO4 to produce 1 mole of Na2SO4 and 2 moles of water
50 mL of 1 M of H2SO4 contains
50 * 1 / 1000 mole of acid
= 0.05 mole of acid
The production of 1 mole of water evolved 5705.7 J of heat and hence the enthalpy changein kJ per mole will be:
0.05 mole of H2SO4 produces 5705.7 J of heat
1 mole of H2SO4 will produce 5705.7 / 0.05 J
= 114,114 J / mole
In kj/mole = 114 kJ/mole.
Hence, the enthalpy change of the reaction in kJ /mole is +114 kJ/mole.
Why do prices increase when demand for a product is high? Companies know they can make more money by selling fewer products at higher prices. Companies know that people will be willing to spend more to get an in-demand product. Companies take advantage of the demand to make people spend more money on excess products. Companies know they can stop production and still make money on sales.
Answer:
Companies know that people will be willing to spend more to get an in-demand product.
Explanation:
When a product is really in demand, many customers are willing to part with more money order to purchase the product, as a result of that, many companies may take advantage of the increasing demand for the product to hike it's price.
Hence, the increase in price may not really have a negative impact on the quantity demanded because the demand for the product is high and customers are willing to spend more money in order to purchase an in-demand product, hence the answer above.
Prices increase when demand is high because companies know that people will be willing to spend more to get in-demand products.
Prices generally increase with higher demand for goods because the higher demand creates pressure for the supply to meet up.
Manufacturing companies can either increase their production to meet up with demand at the same price or capitalize on the situation to make more money by increasing the price without increasing the supply.
Since there is a buying pressure on the product in the market already, people would still be open to buying even at higher prices.
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The following reactions all have K < 1. 1) a. C6H5COO- (aq) + C6H5OH (aq) → C6H5COOH (aq) + C6H5O- (aq) b. F- (aq) + C6H5OH (aq) → C6H5O- (aq) + HF (aq) c. C6H5COOH (aq) + F- (aq) → HF (aq) + C6H5COO- (aq) Arrange the substances based on their relative acid strength.
Answer:
the acid strength is the order of [tex]\mathsf{HF _{(aq)} }[/tex] > [tex]\mathsf{C_6H_5COOH _{(aq)} }[/tex] > [tex]\mathsf{C_6H_5OH _{(aq)} }[/tex]
Explanation:
Given that :
a . [tex]\mathsf{C_6H_5COO^- _{(aq)} + C_6H_5OH _{(aq)} \to C_6H_5COOH _{(aq)} + C_6H_5O^- _{(aq)}}[/tex]
b. [tex]\mathsf{ F^- _{(aq)} + C_6H_5OH _{(aq)} \to C_6H_5O^- _{(aq)} + HF _{(aq)} }[/tex]
c. [tex]\mathsf{C_6H_5COOH _{(aq)} + F^- _{(aq)} \to HF _{(aq)} + C_6H_5COO^- _{(aq)} }[/tex]
Acid strength is the ability of an acid to dissociate into a proton and an anion. Take for instance.
HA ↔ H⁺ + A⁻
The acid strength of the following compounds above are:
[tex]\mathsf{C_6H_5OH _{(aq)} }[/tex] = 1.00 × 10⁻¹⁰
[tex]\mathsf{HF _{(aq)} }[/tex] = 6.6 × 10⁻⁴
[tex]\mathsf{C_6H_5COOH _{(aq)} }[/tex] = 6.3 × 10⁻⁵
As the acid dissociation constant increases the relative acid strength also increases.
From above, the acid strength is the order of [tex]\mathsf{HF _{(aq)} }[/tex] > [tex]\mathsf{C_6H_5COOH _{(aq)} }[/tex] > [tex]\mathsf{C_6H_5OH _{(aq)} }[/tex]
[tex]\mathsf{C_6H_5COO^- }[/tex], [tex]\mathsf{C_6H_5O^- _{(aq)}}[/tex] and F⁻ are Bronsted- Lowry acid
Bronsted- Lowry acid are molecule or ion that have the ability to donate a proton.
The second-order decomposition of NO2 has a rate constant of 0.255 M-1s-1. How much NO2 decomposes in 8.00 s if the initial concentration of NO2 (1.00 L volume) is 1.33 M
Answer:
0.9718M
Explanation:
Rate constant, k = 0.255 M-1s-1
time, t = 8.00 s
Initial concentration, [A]o = 1.33 M
Final concentration, [A] = ?
These quantities are represented by the equation;
1 / [A] = 1 / [A]o + kt
1 / [A] = 1 /1.33 + (0.255 * 8)
1 / [A] = 0.7519 + 2.04
[A] = 1 / 2.7919 = 0.3582 M
How much of NO2 decomposed is obtained from the change in concentration;
Change in concentration = Initial - Final
Change = 1.33 - 0.3582 = 0.9718M
1. The manufacturer of the vinegar used in the experiment stated that the vinegar contained 5.0% acetic acid. What is the percent error between your result and the manufacturer statement
Answer:
72.8 % (But verify explanation).
Explanation:
Hello,
In this case, with the following obtained results, the percent error is computed as follows:
Volume of vinegar= 7.0 mL
Volume of NaOH= (7+6.6+6.4)/3= 6.7 mL
Used concentration of NaOH= 1.5M
Concentration of acetic acid= (concentration of NaOH*volume of NaOH)/volume of vinegar= (6.7mL*1.5M)/7.0M= 1.44M
Assuming we have 100 mL (0.100L) of vinegar, moles of acetic acid in vinegar = 1.44M x 0.100 L= 0.144 mol
Mass of acetic acid in 100g of vinegar = 0.144 mol x 60.0g/mol= 8.64 g
% of acetic acid in vinegar=8.64 %
% error in percentage of acetic acid = [(8.64% - 5.0%)/5.0] x 100=72.8 %
Clearly, this result depend on your own measurements, anyway, you can change any value wherever you need it.
Regards.
A student mixed 50 ml of 1.0 M HCl and 50 ml of 1.0 M NaOH in a coffee cup calorimeter and calculate the molar enthalpy change of the acid-base neutralization reaction to be –54 kJ/mol. He next tried the same experiment with 100 ml of 1.0 M HCl and 100 ml of 1.0 M NaOH. The calculated molar enthalpy change of reaction for his second trial was:
Answer: The calculated molar enthalpy change of reaction for his second trial was -108 kJ.
Explanation:-
Molarity of a solution is defined as the number of moles of solute dissolved per Liter of the solution.
[tex]\text{no of moles}={\text{Molarity}\times {\text{Volume in L}}[/tex]
Thus [tex]\text{no of moles}of HCl={1.0M}\times {0.05L}=0.05moles[/tex]
Thus [tex]\text{no of moles}of NaOH={1.0M}\times {0.05L}=0.05moles[/tex]
[tex]HCl(aq)+NaOH(aq)\rightarrow NaCl(aq)+H_2O(l)[/tex]
Given for second trial:
[tex]\text{no of moles}of HCl={1.0M}\times {0.1L}=0.1moles[/tex]
[tex]\text{no of moles}of NaOH={1.0M}\times {0.1L}=0.1moles[/tex]
0.05 moles of [tex]HCl[/tex] reacts with 0.05 moles of [tex]NaOH[/tex] to release heat = 54 kJ
0.1 moles of [tex]HCl[/tex] reacts with 0.05 moles of [tex]NaOH[/tex] to release heat =[tex]\frac{54}{0.05}\times 0.1=108kJ[/tex]
Thus calculated molar enthalpy change of reaction for his second trial was -108 kJ.
16. The concentration of a solution of potassium hydroxide is determined by titration with nitric
acid. A 30.0 mL sample of KOH is neutralized by 42.7 mL of 0.498 M HNO3. What is the
concentration of the potassium hydroxide solution?
Answer:
[tex]M_{base}=0.709M[/tex]
Explanation:
Hello,
In this case, since the reaction between potassium hydroxide and nitric acid is:
[tex]KOH+HNO_3\rightarrow KNO_3+H_2O[/tex]
We can see a 1:1 mole ratio between the acid and base, therefore, for the titration analysis, we find the following equality at the equivalence point:
[tex]n_{acid}=n_{base}[/tex]
That in terms of molarities and volumes is:
[tex]M_{acid}V_{acid}=M_{base}V_{base}[/tex]
Thus, solving the molarity of the base (KOH), we obtain:
[tex]M_{base}=\frac{M_{acid}V_{acid}}{V_{base}} =\frac{0.498M*42.7mL}{30.0mL}\\ \\M_{base}=0.709M[/tex]
Regards.
Explain why only the lone pairs on the central atom are taken into consideration when predicting molecular shape
Answer:
Lone pairs cause more repulsion than bond pairs
Explanation:
A lone pair takes up more space around the central atom than bond pairs of electrons. This is because, a lone pair is attracted to only one nucleus while bond pairs are attracted to two nuclei.
Hence the repulsion between lone pairs is far greater than the repulsion between bond pairs or repulsion between a lone pair and a bond pair. The presence of a lone pair therefore distorts a molecule away from the ideal shape predicted on the basis of the valence shell electron pair repulsion theory.
Lone pairs are found to decrease the observed bond angles in a molecule.
Determine the molar solubility of AgBr in a solution containing 0.150 M NaBr. Ksp (AgBr) = 7.7 × 10-13.
Answer:
Molar solubility of AgBr = 51.33 × 10⁻¹³
Explanation:
Given:
Amount of NaBr = 0.150 M
Ksp (AgBr) = 7.7 × 10⁻¹³
Find:
Molar solubility of AgBr
Computation:
Molar solubility of AgBr = Ksp (AgBr) / Amount of NaBr
Molar solubility of AgBr = 7.7 × 10⁻¹³ / 0.150
Molar solubility of AgBr = 51.33 × 10⁻¹³
When The Molar solubility of AgBr is = 51.33 × 10⁻¹³
Calculation of Solubility of AgBr
Given as per question:
The Amount of NaBr is = 0.150 M
Then Ksp (AgBr) is = 7.7 × 10⁻¹³
Now we Find:
The Molar solubility of AgBr
The we Computation is:
The Molar solubility of AgBr is = Ksp (AgBr) / Amount of NaBr
After that Molar solubility of AgBr is = 7.7 × 10⁻¹³ / 0.150
Therefore, Molar solubility of AgBr is = 51.33 × 10⁻¹³
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A. Rank the following substances in order of decreasing standard molar entropy (S∘).
Rank the gases from largest to smallest standard molar entropy. To rank items as equivalent, overlap them.
Br(g)
Cl2(g)
I2(g)
F2(g)
B. Rank the following substances in order of decreasing standard molar entropy (S∘).
Rank the gases from largest to smallest standard molar entropy. To rank items as equivalent, overlap them.
H2S(g)
H2O(g)
H2O2(g)
C. Rank the following substances in order of decreasing standard molar entropy (S∘).
Rank the gases from largest to smallest standard molar entropy. To rank items as equivalent, overlap them.
C(s, amorphous)
C(s, diamond)
C(s, graphite)
Answer:
A. Rank the following substances in order of decreasing standard molar entropy (S∘).
Rank the gases from largest to smallest standard molar entropy
I2(g)>Br2(g)>Cl2(g)>F2(g)
B. Rank the gases from largest to smallest standard molar entropy. To rank items as equivalent, overlap them.
H2O2(g)>H2S(g) >H2O(g)
C. Rank the gases from largest to smallest standard molar entropy. To rank items as equivalent, overlap them.
C(s, amorphous) >C(s, graphite)>C(s, diamond)
Explanation:
Hello,
In this case, we can apply the following principles to explain the order:
- The greater the molar mass, the larger the standard molar entropy.
- The greater the molar mass and the structural complexity, the larger the standard molar entropy.
- The greater the structural complexity, the larger the standard molar entropy.
A. Rank the following substances in order of decreasing standard molar entropy (S∘).
Rank the gases from largest to smallest standard molar entropy
I2(g)>Br2(g)>Cl2(g)>F2(g)
This is due to the fact that the greater the molar mass, the larger the standard molar entropy.
B. Rank the gases from largest to smallest standard molar entropy. To rank items as equivalent, overlap them.
H2O2(g)>H2S(g) >H2O(g)
This is due to the fact that the greater the molar mass and the structural complexity, the larger the standard molar entropy as the hydrogen peroxide has four bonds and weights 34 g/mol as well as hydrogen sulfide that has two bonds only.
C. Rank the gases from largest to smallest standard molar entropy. To rank items as equivalent, overlap them.
C(s, amorphous) >C(s, graphite)>C(s, diamond)
Since the molecular complexity is greater in the amorphous carbon (messy arrangement), mid in the graphite and lower in the diamond (well organized).
Regards.
what volume in liters of carbon monoxide will be required to produce 18.9 L of nitrogen in the reaction below
2co(g) + 2no(g) -> n2(g) + 2co2(g)
Answer:
37.8 L OF CARBON MONOXIDE IS REQUIRED TO PRODUCE 18.9 L OF NITROGEN.
Explanation:
Equation for the reaction:
2 CO + 2 NO ------> N2 + 2 CO2
2 moles of carbon monoxide reacts with 2 moles of NO to form 1 mole of nitrogen
At standard temperature and pressure, 1 mole of a gas contains 22.4 dm3 volume.
So therefore, we can say:
2 * 22.4 L of CO produces 22.4 L of N2
44.8 L of CO produces 22.4 L of N2
Since, 18.9 L of Nitrogen is produced, the volume of CO needed is:
44.8 L of CO = 22.4 L of N
x L = 18.9 L
x L = 18.9 * 44.8 / 22.4
x L = 18.9 * 2
x = 37.8 L
The volume of Carbon monoxide required to produce 18.9 L of N2 is 37.8 L
Answer:
37.8
Explanation:
How many moles of gold are equivalent to 1.204 × 1024 atoms? 0.2 0.5 2 5
Answer:
c
Explanation:
How many moles of gold are equivalent to 1.204 × 1024 atoms?
0.2
0.5
2
5
C) 2 Is the correct answer, I took the test and it was correct.
According to the concept of Avogadro's number, there are 2 moles of gold which are equivalent to 1.204×10²⁴ atoms.
What is Avogadro's number?
Avogadro's number is defined as a proportionality factor which relates number of constituent particles with the amount of substance which is present in the sample.
It has a SI unit of reciprocal mole whose numeric value is expressed in reciprocal mole which is a dimensionless number and is called as Avogadro's constant.It relates the volume of a substance with it's average volume occupied by one of it's particles .
According to the definitions, Avogadro's number depend on determined value of mass of one atom of those elements.It bridges the gap between macroscopic and microscopic world by relating amount of substance with number of particles.
Number of atoms can be calculated using Avogadro's number as follows: mass/molar mass×Avogadro's number.Number of moles=number of atoms/Avogadro's number=1.204×10²⁴ /6.023×10²³=1.999≅2
Thus, there are 2 moles of gold which are equivalent to 1.204×10²⁴ atoms.
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At 25 °C, only 0.0990 mol of the generic salt AB3 is soluble in 1.00 L of water. What is the Ksp of the salt at 25 °C? AB3(s)↽−−⇀A3+(aq)+3B−(aq)
Answer:
[tex]Ksp=2.59x10^{-3}[/tex]
Explanation:
Hello,
In this case, given the 0.0990 moles of the salt are soluble in 1.00 L of water only, we can infer that the molar solubility is 0.099 M. Next, since the dissociation of the salt is:
[tex]AB_3\rightleftharpoons A^{3+}+3B^-[/tex]
The concentrations of the A and B ions in the solution are:
[tex][A]=0.099 \frac{molAB_3}{L}*\frac{1molA}{1molAB_3} =0.0099M[/tex]
[tex][B]=0.099 \frac{molAB_3}{L}*\frac{3molB}{1molAB_3} =0.000.297M[/tex]
Then, as the solubility product is defined as:
[tex]Ksp=[A][B]^3[/tex]
Due to the given dissociation, it turns out:
[tex]Ksp=[0.099M][0.297M]^3\\\\Ksp=2.59x10^{-3}[/tex]
Regards.
Consider the reaction: C(s) + O2(g)CO2(g) Write the equilibrium constant for this reaction in terms of the equilibrium constants, Ka and Kb, for reactions a and b below: a.) C(s) + 1/2 O2(g) CO(g) Ka b.) CO(g) + 1/2 O2(g) CO2(g) Kb
Answer:
A. Ka = [CO2] / [C] [O2]^1/2
B. Kb = [CO2] / [CO] [O2]^1/2
Explanation:
Equilibrium constant is simply defined as the ratio of the concentration of the products raised to their coefficient to the concentration of the reactants raised to their coefficient.
Now, we shall obtain the expression for the equilibrium constant for the reaction as follow:
A. Determination of the expression for equilibrium constant Ka.
This is illustrated below:
C(s) + 1/2 O2(g) <==> CO(g)
Ka = [CO2] / [C] [O2]^1/2
B. Determination of the expression for equilibrium constant Kb.
This is illustrated below:
CO(g) + 1/2 O2(g) <==> CO2(g)
Kb = [CO2] / [CO] [O2]^1/2
Most of the costs associated with using renewable resources are due to
а. overuse of resources
b. atmospheric pollution
C.lack of availability
d.global warming
Answer:
The answer is a.
Explanation:
Most of the costs associated with using renewable resources are due to overuse of the resources.
Most of the costs associated with using renewable resources are due to overuse of resources.
What are renewable resources?Renewable resources are those resources which will be generated naturally and continously from the nature and these are also inexhaustible means non ended.
As from the definition it is clear that we can reuse or will use again and again these types of resources, that's why cost associated with these renewable resources are high.
Atmospheric pollution and global warming causes hazardous effect on the environment, so it will not be the reason with the associated cost.Lack of availability makes its important, not costly and in our daily life we used many kinds of renewable resorces so it is not possible to use costly resources daily.Hence, overuse of resouces is one of the reason.
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Fill in the blanks with the words given below- [Atoms, homogeneous, metals, true, saturated, homogeneous, colloidal, compounds, lustrous] 1.An element which are sonorous are called................ 2.An element is made up of only one kind of .................... 3.Alloys are ............................. mixtures. 4.Elements chemically combines in fixed proportion to form ........................ 5. Metals are................................... and can be polished. 6. a solution in which no more solute can be dissolved is called a .................... solution. 7. Milk is a .............. solution but vinegar is a .................. solution. 8. A solution is a ................... mixture. pls help, could not get these answers
Answer:
1. metals
2. atom
3. homogeneous
4. compounds
5. lustrous
6. saturated
7. colloidal
8. homogeneous
Explanation:
How many atoms are in 65.0g of zinc?
from
1moles=iatom
Mole=mass÷avogardos
Where
Avogadro's= 6.02×10²³
So moles = 65.0÷6.02×10²³
Atoms of zinc = 391.6 ×10²³
The number of atoms present in the given mass of Zinc that is 65.0gm is [tex]5.99\times10^{ 23}[/tex].
Atoms are the basic building blocks of matter. They are the smallest units of an element that retain the chemical properties of that element.
Now, to determine the number of atoms in a given number of moles, we can use Avogadro's number, which is approximately [tex]6.022 \times10^{23}[/tex]atoms per mole.
First, we calculate the number of moles of zinc in 65.0g by dividing the given mass by the molar mass of zinc. The molar mass of zinc (Zn) is 65.38 g/mol.
Number of moles = Mass / Molar mass
Number of moles = 65.0g / 65.38 g/mol ≈ 0.9942 mol
Next, multiply the number of moles by Avogadro's number to find the number of atoms.
Number of atoms =[tex]Number of moles \times Avogadro's number[/tex]
Number of atoms = [tex]0.9942[/tex]mol × [tex]6.022 \times10^{23}[/tex] atoms/mol
Therefore, approximately [tex]5.99\times10^{ 23}[/tex] atoms are present in 65.0g of zinc.
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?
Which statement about energy transfer in a wave is ture
If 75.4 J of energy is absorbed by 0.25 mol of CCl4 at constant pressure, what is the change in temperature? The specific heat of CCl4 is 0.861 J/g·°C.
Answer:
ΔT = 2.28°C
Explanation:
Heat, H = 75.4J
Number of moles = 0.25 mol
Specific heat capacity, c = 0.861 J/g·°C
Change in temperature, ΔT = ?
These quantities are related by the following equation;
H = mc ΔT
Mass, m = Number of moles * Molar mass
m = 0.25mol * 153.82 g/mol
m = 38.455g
S back to the equation;
H = mc ΔT
Substituting the values;
75.4 = 38.455 * 0.861 * ΔT
ΔT = 75.4 / 33.11
ΔT = 2.28°C
The change in temperature is 2.28 °C
First, we will determine the mass of CCl₄ absorbed
From the given information,
Number of moles of CCl₄ absorbed = 0.25 mol
Using the formula
Mass = Number of moles × Molar mass
Molar mass of CCl₄ = 153.82 g/mol
∴ Mass of CCl₄ absorbed = 0.25 × 153.82
Mass of CCl₄ absorbed = 38.455 g
Now, using the formula
Q = mcΔT
Where Q is the quantity of heat
m is the mass
c is the specific heat of substance
and ΔT is the change in temperature
From the given information
Q = 75.4 J
c = 0.861 J/g.°C
Putting the parameters into the formula, we get
75.4 = 38.455 × 0.861 ×ΔT
75.4 = 33.109755 × ΔT
∴ ΔT = 75.4 ÷ 33.109755
ΔT = 2.28 °C
Hence, the change in temperature is 2.28 °C
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An electrolysis cell has two electrodes. Which statement is correct? A. Reduction takes place at the anode, which is positively charged. B. Reduction takes place at the cathode, which is positively charged. C. Reduction takes place at the dynode, which is uncharged. D. Reduction takes place at the cathode, which is negatively charged. E. Reduction takes place at the anode, which is negatively charged.
Answer:
D. Reduction takes place at the cathode, which is negatively charged.
Explanation:
In an electrolytic cell there are two electrodes; the cathode and the anode. The anode is the positive electrode while the cathode is the negative electrode. Oxidation occurs at the anode while reduction occurs at the cathode.
At the anode, species give up electrons and become positively charged ions while at the cathode species accept electrons and become reduced.
If the dissolution of borax in water is spontaneous, is the change in enthalpy positive or negative - or are both signs possible?
Answer:
"Both signs are possible" is the correct choice.
Explanation:
The indication including its change in enthalpy depends on the concentration.As when the enthalpy seems to be negative besides purposeful unexpected temperatures lower, as well as successful for necessary response at extremely high temperatures.As such that when both signals could have been the enthalpy change.
If the dissolution of borax in water is spontaneous, so the change in enthalpy should be negative.
What is enthalpy?Enthalpy of the reaction tells about the total amount of energy released or absorbed during any chemical reaction.
And for the spontaneous reaction, value of standard Gibb's free energy change of the reaction should be negative and formula will be represented as:
ΔG = ΔH - TΔS, where
ΔH = change in enthalpy
For the spontaneous reaction, value of enthalpy should be negative so that we get the negative value of ΔG.
Hence change in enthalpy should be negative.
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Which of the following pairs of chemical reactions are inverses of each other? Answer options: a. Hydrogenation and alkylation b.Halogenation and hydrolysis c. Ammoniation and alkylation d. Oxidation and reduction
Answer:
d. Oxidation and reduction
Explanation:
For this question we have to remember the definition of each type of reaction:
-) Hydrogenation
In this reaction, we have the addition of hydrogen to a molecule. Usually, an alkene or alkyne. In the example, molecular hydrogen is added to a double bond to produce an alkane.
-) Alkylation
In this reaction, we have the addition of a chain of carbon to another molecule. In the example, an ethyl group is added to a benzene ring.
-) Hydrolysis
In this reaction, we have the breaking of a bond by the action of water. In the example, a water molecule can break the C-O bond in the ester molecule.
-) Halogenation
In this reaction, we have the addition of a halogen (atoms on the VIIIA group). In the example, "Cl" is added to the butene.
-) Ammoniation
In this reaction, we have the addition of the ammonium ion ([tex]NH_4^+[/tex]). In the example, the ammonium ion is added to an acid.
-) Oxidation and reduction
In this reaction, we have opposite reactions. The oxidation is the loss of electrons and the reduction is the gain of electrons. For example:
[tex]Ag^+~+~e^-~->~Ag[/tex] Reduction
[tex]Al~->~Al^+^3~+~3e^-[/tex] Oxidation
A solution contains 0.0150 M Pb2+(aq) and 0.0150 M Sr2+(aq) . If you add SO2−4(aq) , what will be the concentration of Pb2+(aq) when SrSO4(s) begins to precipitate?
Answer:
[tex]\large \boxed{1.10 \times 10^{-3}\text{ mol/L}}[/tex]
Explanation:
1. Concentration of SO₄²⁻
SrSO₄(s) ⇌ Sr²⁺(aq) +SO₄²⁻(aq); Ksp = 3.44 × 10⁻⁷
0.0150 x
[tex]K_{sp} =\text{[Sr$^{2+}$][SO$_{4}^{2-}$]} = 0.0150x = 3.44 \times 10^{-7}\\x = \dfrac{3.44 \times 10^{-7}}{0.0150} = \mathbf{2.293 \times 10^{-5}} \textbf{ mol/L}[/tex]
2. Concentration of Pb²⁺
PbSO₄(s) ⇌ Pb²⁺(aq) + SO₄²⁻(aq); Ksp = 2.53 × 10⁻⁸
x 2.293 × 10⁻⁵
[tex]K_{sp} =\text{[Pb$^{2+}$][SO$_{4}^{2-}$]} = x \times 2.293 \times 10^{-5} = 2.53 \times 10^{-8}\\\\x = \dfrac{2.53 \times 10^{-8}}{2.293 \times 10^{-5}} = \mathbf{1.10 \times 10^{-3}} \textbf{ mol/L}\\\\\text{The concentration of Pb$^{2+}$ is $\large \boxed{\mathbf{1.10 \times 10^{-3}}\textbf{ mol/L}}$}[/tex]
How many moles of oxygen are in 49.2 grams of carbon dioxide?
Answer:
Number of mole in Oxygen = 2.24 mol
Explanation:
Given:
Amount of carbon dioxide (CO₂) = 49.2 gram
Find:
Number of moles in Oxygen
Computation:
Molecular weight of CO₂ = 44 gram (Approx)
Number of mole in CO₂ = 49.2 / 44
Number of mole in CO₂ = 1.11818182 mol
CO₂ has 2 mole of Oxygen,
So,
Number of mole in Oxygen = 2 × 1.11818182
Number of mole in Oxygen = 2.23636364
Number of mole in Oxygen = 2.24 mol
If the lead concentration in water is 1 ppm, then we should be able to recover 1 mg of lead from _____ L of water.
Answer:
1 L
Explanation:
ppm means parts per million. Generally the relationship between mass and litre is given as;
1 ppm = 1 mg/L
This means that 1 ppm is equivalent to 1 mg of a substance dissolved in 1 L of water.
2) What is the concentration (M) of CH3OH in a solution prepared by dissolving 11.7 g of CH3OH in sufficient water to give exactly 230 mL of solution?
Write a balanced nuclear equation for the following: The nuclide thorium-234 undergoes beta decay to form protactinium-234 .
Answer:
²³⁴₉₀Th --> ²³⁴₉₁Pa + ⁰₋₁e
Explanation:
thorium-234 = ²³⁴₉₀Th
beta decay = ⁰₋₁e
protactinium-234 = ²³⁴₉₁Pa
The balanced nuclear equation is given as;
²³⁴₉₀Th --> ²³⁴₉₁Pa + ⁰₋₁e
What are periodic trends if ionic radii
Answer:
Explan ionization energy, atomic radius, and electron affinityation:
This question most likely has answer choices. The possible answer choices are as followed:
Ionic radii tend to increase down a group.Ionic radii tend to decrease across a period.Anionic radii tend to increase across a period.Cationic radii tend to decrease across a period.Ionic radii increase when switching from cations to anions in a period.The answers are Ionic radii tend to increase down a group, Cationic radii tend to decrease across a period, and Ionic radii increase when switching from cations to anions in a period (1st, 4th, and 5th options).
