Answer
2MnO₄⁻(aq) + 10Cl⁻(aq) + 16H⁺ → 2Mn²⁺(aq) + 5Cl₂(g) + 8H₂O
Explanation
What is given:
The following reaction takes place in an acidic solution:
MnO₄⁻(aq) + Cl⁻(aq) → Mn²⁺(aq) + Cl₂(g) (unbalanced).
What to find:
To identify the half-reactions and balance the redox reaction.
Step-by-step solution:
Step 1: Identify the half-reactions.
The oxidation half-reaction is: Cl⁻(aq) → Cl₂(g)
The reduction half-reaction is: MnO₄⁻(aq) → Mn²⁺(aq)
Step 2: Balance the equations for atoms (except O and H).
Oxidation: 2Cl⁻(aq) → Cl₂(g)
Reduction: MnO₄⁻(aq) → Mn²⁺(aq)
Step 3: Balance the equations for atoms O and H using H₂O and H⁺.
Oxidation: 2Cl⁻(aq) → Cl₂(g)
Reduction: MnO₄⁻(aq) + 8H⁺ → Mn²⁺(aq) + 4H₂O
Step 4: Balance the charge in the half-reactions.
Oxidation: 2Cl⁻(aq) → Cl₂(g) + 2e⁻
Reduction: MnO₄⁻(aq) + 8H⁺ + 5e⁻ → Mn²⁺(aq) + 4H₂O
Step 5: Multiply each half-reaction by the proper number to balance charges in the reaction.
Oxidation: 2Cl⁻(aq) → Cl₂(g) + 2e⁻ x 5
Reduction: MnO₄⁻(aq) + 8H⁺ + 5e⁻ → Mn²⁺(aq) + 4H₂O x 2
↓
Oxidation: 10Cl⁻(aq) → 5Cl₂(g) + 10e⁻
Reduction: 2MnO₄⁻(aq) + 16H⁺ + 10e⁻ → 2Mn²⁺(aq) + 8H₂O
Step 6: Add the equations and simplify to get a balanced equation.
2MnO₄⁻(aq) + 10Cl⁻(aq) + 16H⁺ + 10e⁻ → 2Mn²⁺(aq) + 5Cl₂(g) + 8H₂O + 10e⁻
Simplifying the equation, we have:
2MnO₄⁻(aq) + 10Cl⁻(aq) + 16H⁺ → 2Mn²⁺(aq) + 5Cl₂(g) + 8H₂O
Which of the following can be mixed in solution with H₂CO3 to make a buffer?OA. Na₂CO3OB. HFOC. NH3OD. NaHCO3
A buffer system usually will need a weak acid, like H2CO3, which is carbonic acid, and a strong base, and the only compound that could figure a fairly strong base among the options given is Na2CO3. NH3 and NaHCO3 are weak bases and HF is a strong acid. Therefore the best answer will be letter A
What is the pH of 5.43 M solution of potassium trifluoroacetate KC2F302) at 25 C?
Potassium trifluoroacetate is the salt of a weak acid (trifluoroacetic acid) and a strong base (potassium hydroxide). In water it will dissociate into its ions.
KC₂F₃O₂ ------> K⁺ + C₂F₃O₂⁻
The trifluoroacetate ion is the conjugate base of a weak acid:
C₂F₃O₂⁻ + H₂O <-----> C₂F₃O₂H + OH⁻
The dissociation constant of this conjugate base will be:
Kb = [C₂F₃O₂H] * [OH⁻]/[C₂F₃O₂⁻]
I searched for the value of the Ka for the trifluoroacetatic acid. With this value we can find the Kb for the trifluoroacetate.
Ka = 1
Ka * Kb = 1 * 10^(-14)
Kb = 1 * 10^(-14)/1
Kb = 1 * 10^(-14)
So first we will have to find the concentration of OH⁻ and then we will be able to find the concentration of H+ and finally the pH.
We have to use the ICE table.
C₂F₃O₂⁻ + H₂O <-----> C₂F₃O₂H + OH⁻
I 5.43 0 0
C -x +x +x
E 5.43 - x x x
The initial concentration of the trifluoroacetate ion is 5.43, then some moles of it will convert into moles of C₂F₃O₂H and OH⁻. And finally we found the equilibrium concentrations that we can replace of the expression of the Kb.
Kb = [C₂F₃O₂H] * [OH⁻]/[C₂F₃O₂⁻]
Kb = x * x /(5.43 - x)
Kb = x²/(5.43 - x)
We can replace the Kb for the value that we found and solve this quadratic equation for x.
1 * 10^(-14) = x²/(5.43 - x)
1 * 10^(-14) * (5.43 - x) = x²
x² + 1 * 10^(-14) x - 5.43 * 10^(-14) = 0
This quadratic equation has two roots. We will use the positive one since it is a concentration.
x₁ = 2.33 * 10^(-7) x₂ = - 2.33 * 10^(-7)
The concentration of OH- in the equilibrium is x. We can determine the pOH.
[OH-]eq = x₁
[OH-]eq = 2.33 * 10^(-7)
pOH = - log [OH-]
pOH = - log (2.33 * 10^(-7))
pOH = 6.63
Finally we can find the pH using that value:
pH + pOH = 14
pH = 14 - pOH = 14 - 6.63
pH = 7.37
Answer: the pH of a 5.43 M solution of potassium trifluoroacetate is 7.37 supposing that ka of trifluoroacetic acid is 1.
Determine the overall and specific shapes around each of the specified atoms in the following molecule. Then, draw a 3D VSEPR structure on the right.
Explanation
A) Trigonal planar
B) Trigonal pyramid
C) Tetrahedral
D) Linear
E) Linear
F) Tetrahedral
G) Linear
A gas occupies 12.3 L at a pressure of 40 mmHg. What is the volume when the pressure is increases to 60 mmHg?
Step 1 - Understanding the relation between pressure and volume
There are three main variables that can modify the state of a gas: pressure (P), temperature (T) and volume (V). When one of them is kept constant, linear relationships arise between the other two variables, i.e., proportionality relations.
When the temperature is kept constant (isothermic transformation), the pressure and the volume become inversely proportional, i.e., increasing the pressure reduces the volume.
We can state it mathematically as:
[tex]P_1V_1=P_2V_2[/tex]Step 2 - Using the equation to solve the exercise
To calculate the new volume (V2), we can use the equation. The values we need are given in the exercise:
[tex]\begin{gathered} V_1=12.3L \\ P_1=40\operatorname{mm}Hg \\ P_2=60\operatorname{mm}Hg \\ V_2=\text{?} \end{gathered}[/tex]Substituting these values in the equation above, we have:
[tex]\begin{gathered} 12.3\times40=x\times60 \\ \\ x=\frac{12.3\times40}{60}=8.2L \end{gathered}[/tex]The final volume would be thus 8.2L.
What is the mass percent concentration of a solution containing 30.0 g NaCl and 170.0 mL of water? Assume a density of 1.00 g/mL for water.
The expression used to calculate the mass percent concentration is: % (m/m) = (mass of solute) / (mass of solution (solute + solvent)) x 100.
First let's transform mL into grams of water:
1 g of water ---- 1 mL of water
x g of water ---- 170 mL of water
x = 170 g of water
Now let's calculate the mass of solution:
mass of solute = 30 g of NaCl
mass of solvent = 170 g of water
solute + solvent = 30 + 170 = 200 g
% (m/m) = (mass of solute) / (mass of solution (solute + solvent)) x 100.
% (m/m) = (30g/200g) x 100.
% (m/m) = 15.0%
Answer: 15.0%
KH D m d c mEJOg3.65 g = [?] mg=Enter
In each 1 gram, we have 1000 miligrams, that is:
1 g = 1000 mg
So, we can use rule of three to find the answer:
x ---- 3.65 g
1000 mg ---- 1 g
Thus:
[tex]\begin{gathered} \frac{x}{1000mg}=\frac{3.65\cancel{g}}{1\cancel{g}} \\ x=1000mg\cdot3.65 \\ x=3650mg \end{gathered}[/tex]So, we have:
3.65 g = 3650 mg
What is the solution's freezing point: 15 g of CH4N2O (Molar mass = 60.055 g/mol) in 200. g of H2O? (Kf = 1.86 (°C·kg)/mol)
Ok to answer this question we firsst need to fin the number of mol of Urea (CH4N2O). to do this we simply :
1 mol of urea =15/60.055 = 0.25mol
therefore 200g of water contain 0.25mol
the next step is to determine the malality of our solution in 200g of water, to do this we say:
200 g = 1Kg/1000g = 0.2kg
therefor 0.25mol/0.2Kg = 1.25mol/kg
and from the equation:
we know that i = 1
we are given Kf
b is the molality that we just calculated
therefore;
the solutions freezing point is -2.325°C
The freezing point of the given solution was calculated to be -2.325 °C. The freezing point is the point at which a liquid changes from a liquid to a solid under atmospheric pressure.
The two phases, liquid and solid, exist in equilibrium at the freezing point. At the freezing point, both solid and liquid exist at the same time.
Given,
15 grams of CH₄N₂O is dissolved in 200 grams of water.
1 mol of urea = 60 grams
so 15 grams of urea will be found in 15/60 = 0.25 moles
So 0.25 moles of urea is present in 200 grams of water.
To determine the molality of the solution
200gm/1000gm = 0.2 kg water
0.25 moles/0.2 kg = 1.25 kg/mol
So to calculate the freezing point of the solution we use the formula:
ΔT = i× Kf × b
Given Kf = 1.86 (°C·kg)/mol), b is the molality of the solution that is = 1.25 kg/mol, and i = 1
ΔT = 1 × 1.86 × 1.25 = 2.325 °C
ΔT = 0 °C - 2.325 °C = -2.325 °C
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Which product (if any) will form a precipitate in the following reaction?3Li2CO3 + 2Al(NO3)3 --> 6LiNO3 + Al2(CO3)3a)LiNO3 b)No Precipitate Formed c)Not this choice d)Al2(CO3)3
For this reaction we have to look at which compound will be a solid at the end, this means that this specific compound is not soluble in water, therefore forming a precipitate
3 Li2CO3 (aq) + Al(NO3)3 (aq) -> 6 Li(NO3) (aq) + Al2(CO3)3 (s)
As we can see in the balanced reaction, we will have Al2(CO3)3 as a solid product from this reaction, therefore it will be forming a precipitate
Answer: D
If 8.54 g of CuNO3 is dissolved in water to make a 0.520 M solution, what is the volume of the solution in milliliters?
For solving this exercise we need the molar mass of CuNO3 which is 125.5 g/mol
Step 1) We calculate the moles of this compound:
8.54 g CuNO3 / 125.5 g/mol = 0.0680 moles CuNO3
Step 2) We calculate the volume of the solution in L:
(Remember M is equal to mol/L)
0.0680 moles CuNO3 / 0.520 mol/L = 0.131 L
Step 3) We change L to mL:
(Remember 1L=1000mL)
0.131 L x (1000mL/1L) = 131 mL
Answer: Volume = 131 mL
What is Le Chatelier's principle used for? OA. To determine what effect a change will have on a mixture at equilibrium. B. None of these OC. To determine the magnitude of a change in equilibrium. D. All of these
Le Chatelier's principle is used to determine what effect a change will have on a mixture at equilibrium.
It describes the effect that a mixture suffers when it is exposed to a change.
It means that the correct answer is A.
what volume will 2.0 moles of hydrogen occupy at STP?
In this question, at STP (standard temperature and pressure) it is conventional to assume that 1 mol of a substance is equal to 22.4 Liters of volume, therefore, if we have 2 moles of Hydrogen gas:
1 mol H2 = 22.4 L
2 moles H2 = x L
x = 44.8 Liters of volume
What is the molecular formula of a compound given the molar mass of the compound is 34.04 gram and the empirical formula isNH?(A)NHB) N4H4N2H6N2H2
1) List the known and unknown quantities.
Molar mass: 34.04 g.
Empirical formula: NH.
Molecular formula: unknown.
2) Find the molar mass of the empirical formula.
N: 14.007 g/mol.
H: 1.008 g/mol.
N + H = 14.007 + 1.008 = 15.015 g/mol.
3) Divide the molar mass by the molar mass of the empirical formula
[tex]X=\frac{34.04}{15.015}[/tex][tex]X=2.27[/tex]Multiply the subscripts of the empirical formula by 2.
[tex]N_2H_2[/tex].
According to the phase diagram for H₂O, what happens to the phases ofwater at 0.5 atm pressure as the temperature is decreased from 110°C to -10°C?A) water changes from a gas to a solid to a liquidB) Water changes from a solid to a liquid to a gasC) water changes from a liquid to a gas to a solidD) water changes from a gas to a liquid to a solid
Answer:
[tex]D[/tex]Explanation:
Here, we want to get what happens to the phases of water as the temperature is decreased at 5 atm
From what we have in the question, we would be moving from left to right
What would happen here is that we move from the gaseous state to the liquid and then to the ice state
This means we have a change from a gas to a liquid then a solid
How many atoms are in Hg3 PO 42
What is the percent of H inNH3?(H = 1.008 amu, N = 14.01 amu)=
In order to find the percentage of an element within a compound, we need to have the mass for this element and also the mass of the whole compound, and the masses are:
H = 1.008 g/mol or amu
and we have 3 H in the compound, therefore = 3.024 g/mol or amu
NH3 = 17.031 g/mol
Now we know that 17.031 is the total mass, so we will call it 100%
17.031 g/mol = 100%
3.024 g/mol = x %
x = 17.75% of H in NH3
Complete the second row of the table.Express the volume in liters to three significant figures.
In this case you need to use the volume and temperature equation from Charles law.
[tex]\frac{V_1}{T_1}=\frac{V_2}{T_2}[/tex]We are given T1 = 139 k
T2= 317 k
V2= 176 L
We want to find V1. Now lets re-arrange the equation.
[tex]V_1=\frac{V_2}{T_2}xT_1[/tex]V1 = (176 x 139)/317
V1= 77.2 L
what type of reaction is C6H12O6 + 6O2 => 6CO2 + 6H2O
Explanation
A combustion reaction is a type of chemical reaction in which a compound and an oxidant are reacted to produce heat and a new product. The general form of a combustion reaction can be represented as follows:
Hydrocarbond + O2 => CO2 + H2O
In addition to heat (also common)
Answer: a combustion reaction
Each part has its own answer & part 2/2 is a continuation of the first part 1/2 :)
Answer:
Explanations:
1) According to the first question, we are to find the molar mass of Na₂SO₄.
Find the molar mass of each element present on the compound
• Molar mass of Sodium (Na) = 23g/mol
,• Molar mass of sulfur (S) = 32g/mol
,• Molar mass of Oxygen (O) = 16g/mol
Get the molar mass of Na₂SO₄.
[tex]\begin{gathered} \text{Molar mass of }Na_{2}SO_{4}=2(23)+32+4(16) \\ \text{Molar mass of }Na_2SO_4=46+32+64 \\ \text{Molar mass of }Na_{2}SO_{4}=142g\text{/mol} \end{gathered}[/tex]Therefore the molar mass of Na₂SO₄ is 142g/mol
2) The formula for calculating the number of moles of the compound Na₂SO₄ is expressed as:
[tex]n=\text{Molar conc.}\times\text{volume}[/tex]Given the following parameters:
• Molar concentration = 0.202M = 0.202mol/L
,• Volume = 784mL= 0.784L
Substitute the given parameters into the formula:
[tex]\begin{gathered} n=0.202\frac{mol}{\cancel{L}}\times0.784\cancel{L} \\ n=0.1584\text{moles} \end{gathered}[/tex]Hence the number of Na₂SO₄ needed is 0.1584moles.
Write the balanced molecular chemical equation for the reaction of solid lithium with aqueous chromium(III) acetate. If no reaction occurs, simply write only NR. Be sure to include the proper phases for all species within the reaction.
The balanced molecular chemical equation for the reaction of solid lithium with aqueous chromium(III) acetate is given below:
3 Li (s) + Cr(CH₃CO₂)₃ (aq) ---> 3 LiCH₃CO₂ + Cr(s)
What are chemical reactions?Chemical reactions are reactions that involve the rearrangement of atoms of the substances reacting such that new substances are formed.
Chemical reactions are usually represented using chemical equations.
Chemical equations can be written as molecular equations or as ionic equations.
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Imagine you are a chef. You must make a lunch that does not involve any chemical changes in the preparation. Name three foods or drinks you will serve, and describe what physical changes you will use to prepare them.
So,
First of all, let's remember what physical and chemical changes are.
A physical change occurs when there's changes in the size or shape of matter. Changes of state, for example, from solid to liquid or from liquid to gas, are also physical changes. In this kind of change, the composition of the matter doesn't change.
A chemical change occurs when a substance combines with another to form a new substance, and the composition of the matter changes.
Now that we know what these changes are, we could give some examples of a physical change in cooking, for example:
1. Suppose we want to serve a freezed drink, like a popsicle. For this, we could freeze a fruit juice. This is a physical change because the composition of the juice changes from liquid to solid.
2. Suppose we want to serve pasta. We need to boil the pasta to make it soft, which is a physical change clearly.
3. Suppose we want to serve meat but this is freezed, so, we need to thaw it to start cooking, which is a physical change.
Empriical formula for Na=29.1 S=40.5. O=30.4
Answer:
Na2S2O3.
Explanation:
What is given?
If we suppose that the compound has a mass of 100 g:
Mass of Na = 29.1 g,
Mass of S = 40.5 g,
Mass of O = 30.4 g,
Molar mass of Na = 23 g/mol,
Molar mass of S = 32 g/mol,
Molar mass of O = 16 g/mol.
Step-by-step solution:
First, we have to convert all the given masses of each element to moles using their respective molar mass, like this:
[tex]\begin{gathered} 29.1\text{ g Na}\cdot\frac{1\text{ mol Na}}{23\text{ g Na}}=1.27\text{ mol Na,} \\ \\ 40.5\text{ g S}\cdot\frac{1\text{ mol S}}{32\text{ g }}=1.27\text{ mol S,} \\ \\ 30.4\text{ g O}\cdot\frac{1\text{ mol O}}{16\text{ g O}}=1.9\text{ mol O.} \end{gathered}[/tex]The next step is to divide each number of moles by the least number of moles obtained, which in this case is 1.27, as follows:
[tex]\begin{gathered} 1.27\text{ mol Na/1.27=1 mol Na,} \\ 1.27\text{ mol S/1.27=1 mol S,} \\ 1.9\text{ mol O/1.27=1.5 mol O.} \end{gathered}[/tex]Remember that we need to have the number of moles of each element as integers, so if we multiply each number of moles of each element by 2, we will obtain:
[tex]\begin{gathered} 1\text{ mol Na}\cdot2=2\text{ mol Na,} \\ 1\text{ mol S}\cdot2=2\text{ mol S,} \\ 1.5\text{ mol O}\cdot2=3\text{ mol O.} \end{gathered}[/tex]We would have 2 mol of Na, 2 mol of S, and 3 mol of O, so the empirical formula is Na2S2O3
ellusPerform the followingmathematical operation, andreport the answer to theappropriate number ofsignificant figures.5.1 - 2.3685 =-
1) Uncertainty in substraction. the result must have the same number of decimal places as the one with less.
5.1 - 2.3685 = 2.7315
The result must have one decimal place. Since the second decimal place (3) is smaller than five we round down.
The result is 2.7.
For the complete redox reaction given, write the half-reactions & identify the oxidizing & reducing agents.4Fe + 3O 2 -> 2Fe 2 O 3
4Fe + 3O2 ==> 2Fe2O3
Is the following statement about our solar system true or false?
Mercury orbits the Sun at more than four times the speed of Saturn.
The speed of Mercury around the sun is 47 Km/s while the speed of Saturn around the sun is 9.69 Km/s. Therefore, Mercury orbits the Sun at a speed more than four times the speed of Saturn. Therefore, the given statement is true.
What is planet Mercury?Mercury can be described as the smallest planet in the Solar System and also the closest to the Sun. Mercury orbits around the Sun and takes 88 Earth days. Mercury rotates in a unique way in the Solar System as it is tidally locked with the Sun in a 3 : 2 spin-orbit resonance.
Mercury can be described as a rocky body like Earth with an equatorial radius of 2,439.7 kilometers. Mercury contains approximately 70% metallic and 30% silicate material.
Saturn is the second-largest and the sixth planet from the sun in the Solar System, after Jupiter. Saturn is a gas giant with a radius of about nine and a half times that of Earth. The speed of Mercury orbits the Sun is more than four times the speed of Saturn.
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Three students (A, B, and C) are asked to determine the volume of a sample of water. Each student measures the volume three times with a graduated cylinder. The results in milliliters are: A (87.1, 88.2, 87.6); B (86.9, 87.1, 87.2); C (87.6, 87.8, 87.9). The true volume is 87.0 mL. Comment on the precision and the accuracy of each student’s results. aStudent A has results that are neither precise nor accurate. bStudent B has results that are both precise and accurate. cStudent C has results that are precise but not accurate. dStudent A has results that are precise and accurate. eStudent B has results that are neither precise and accurate. fStudent C has results that are accurate but not precise.
Answer:
a. Student A has results that are neither precise nor accurate.
b. Student B has results that are both precise and accurate.
c. Student C has results that are precise but not accurate.
Explanation:
First, let's remember the concepts of accuracy and precision:
Accuracy is a measure of how close a measurement is to the correct or accepted value of the quantity being measured.
Precision is a measure of how close a series of measurements are to one another.
You can see this better in the following picture:
Based on this logic, you can note that:
- The results of Student A are not accurate and are not precise.
- The results of Student B are very close to each other which is precise and in turn, they are close to the real value which is very accurate.
- The results of Student C are very close to each other which is precise but it isn't close to the real value, so it isn't accurate.
The answers would be:
a. Student A has results that are neither precise nor accurate.
b. Student B has results that are both precise and accurate.
c. Student C has results that are precise but not accurate.
35 ml of liquid has the mass of 41.6 g what is the density
INFORMATION:
We know that:
- 35 ml of liquid has the mass of 41.6 g
And we must find its density
STEP BY STEP EXPLANATION:
To find it, we must use the following formula
[tex]density=\frac{mass}{volume}[/tex]From given information:
- mass of the liquid: 41.6 g
- volume of the liquid: 35 ml
Now, replacing the values in the formula
[tex]\begin{gathered} density=\frac{41.6g}{35ml} \\ \text{ Simplifying,} \\ density=1.19\frac{g}{ml} \end{gathered}[/tex]Finally, the density of the liquid is 1.19 g/ml
ANSWER:
The density of the liquid is 1.19 g/ml
If 1219 Joules of heat raise the temperature of 250g of metal by 64degreesC. What is the specific heat of the metal in J/g degreesC
To answer this question, we have to use the following formula:
[tex]Q=Cp\cdot m\cdot\Delta T[/tex]Where Q is the heat, Cp is the specific heat, m is the mass and ΔT is the difference in temperature.
In this case, we know the heat, the mass and the difference in temperature and we need to find the specific heat. Solve the given equation for Cp:
[tex]Cp=\frac{Q}{m\cdot\Delta T}[/tex]Replace for the given values and solve:
[tex]\begin{gathered} Cp=\frac{1219J}{250g\cdot64\degree C} \\ Cp=0.076J/g\degree C \end{gathered}[/tex]It means that the specific heat of the metal is 0.076J/g°C.
why is boiling point of vegetable oil higher than boiling point of maple syrup?
Firstly, what you need to understand is that boiling points reflect the intermolecular attractive forces within the sustance.
The stronger the intermolecular forces, the lower the vapor pressure of the substance and the higher the boiling point.
In this case, vegetable oil has stronger intermolecular forces, these forces are difficult to break, which then leads to increased/higher temperatures needed to break them. That is why vegetable oil has a boiling point of approximately 300 degrees celsius.
Maple syrup has a lower boiling point because its intermolecular forces are less stronger compared to vegetable oil. Not a lot of energy is required to break the intermolecular bonds, thus lower boiling temperature.
Do structural isomers have the same number of Carbon atoms?
Structural isomers have the same formula, but they have different bonds, that's why they can have the same number of carbon atoms.
How does Boron achieve a noble gas electron configuration?
Boron gas can achieve noble gas configuration by losing or gaining electrons.
Noble gases were substances that have completed respective octets and have entirely filled respective valence shells. Helium (He), Neon(Ne), Argon(Ar), Krypton(Kr), Xenon (Xe), as well as Radon, are still examples of noble gases.
Atoms can adopt a noble gas configuration whether by acquiring new electrons or by giving and receiving electrons already present in even their own outermost shell.
The atomic number of Boron is 5 . the electronic configuration of B is 2,3. When it will gain 5 electrons with another atom or lose 3 electrons then it will attain a noble gas configuration.
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