The initial temperature of the copper metal was 27.38 degrees.
Explanation:
Data given:
mass of the copper metal sample = 215 gram
mass of water = 26.6 grams
Initial temperature of water = 22.22 Degrees
Final temperature of water = 24.44 degrees
Specific heat capacity of water = 0.385 J/g°C
initial temperature of copper material , Ti=?
specific heat capacity of water = 4.186 joule/gram °C
from the principle of:
heat lost = heat gained
heat gained by water is given by:
q water = mcΔT
Putting the values in the equation:
qwater = 26.6 x 4.186 x (2.22)
qwater = 247.19 J
qcopper = 215 x 0.385 x (Ti-24.4)
= 82.77Ti - 2019.71
Now heat lost by metal = heat gained by water
82.77Ti - 2019.71 = 247.19
Ti = 27.38 degrees
The human body can get energy by metabolizing proteins, carbohydrates or fatty acids, depending on the circumstances. Roughly speaking, the energy it gets comes mostly from allowing all the carbon atoms in the food molecules to become oxidized to carbon dioxide by reaction with oxygen from the atmosphere. Hence the energy content of food is roughly proportional to the carbon content. Let's consider stearic acid , a fatty acid from which fats are made, and fructose , one of the simplest carbohydrates. Using the idea above about energy content, calculate the ratio of the energy the body gets metabolizing each gram of stearic acid to the energy the body gets metabolizing each gram of fructose. Round your answer to the correct number of significant digits.
The ratio of energy obtained from metabolizing stearic acid to fructose is 3:1.
Explanation:To calculate the ratio of the energy the body gets metabolizing stearic acid to the energy the body gets metabolizing fructose, we need to compare their carbon content. The molecular formula of stearic acid is C18H36O2, which means it has 18 carbon atoms. The molecular formula of fructose is C6H12O6, which means it has 6 carbon atoms. Since the energy content of food is roughly proportional to the carbon content, we can calculate the ratio by dividing the number of carbon atoms in stearic acid (18) by the number of carbon atoms in fructose (6). This gives us a ratio of 3:1.
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The energy ratio of metabolizing one gram of stearic acid (a fatty acid) to one gram of fructose (a carbohydrate) is 2.25, with stearic acid providing 9 kcal/g and fructose providing 4 kcal/g of energy.
The question asks to calculate the ratio of the energy the body gets metabolizing each gram of stearic acid to the energy the body gets metabolizing each gram of fructose. Stearic acid is a fatty acid, and fructose is a simple carbohydrate. According to the data provided, each gram of carbohydrates yields approximately 4 kcal of energy, while each gram of fat yields about 9 kcal. Therefore, the ratio of energy from fats to carbohydrates is 9 kcal/g to 4 kcal/g.
To calculate the ratio of energy from stearic acid to fructose:
Identify the energy values: 9 kcal/g for stearic acid and 4 kcal/g for fructose.
Divide the energy value of stearic acid by the energy value of fructose: 9 kcal/g \/ 4 kcal/g = 2.25.
This means that metabolizing one gram of stearic acid yields 2.25 times the energy compared to metabolizing one gram of fructose.
Gasoline is a remarkably interesting soup of hydrocarbons of various sorts, with bits of this and that added, but the average chemistry is not too far from being carbon and hydrogen, with two hydrogen atoms for each carbon. Burning involves combining gasoline with oxygen to make water and carbon dioxide. (Other things that are made in small quantities, such as carbon monoxide, are not as nice.) The chemical formula for burning gasoline can then be written something like: CH2+1.5 O2 --> CO2+H2O (If you don’t like having one-and-a-half oxygen molecules, you can think of two hydrocarbons plus three oxygens making two carbon dioxides and two waters; it is the same thing, really.) In burning, each carbon atom, C, in gasoline eliminates two hydrogens and replaces them with two oxygens each carbon atom weighs 12 atomic mass units each hydrogen weighs 1 each oxygen weighs 16; So, CH2 starts out weighing 14 (12 from carbon and 2 from hydrogen), and CO2 ends up weighing 44 (12 from carbon and 32 from oxygen)—the weight has more than tripled. Rounding that off a little, the total weight of CO2 put out by a typical U.S. driver is three times larger than the weight of gasoline burned. To get the number of pounds of CO2 per year from a typical car, then, multiply your answer from the previous question by 3.
Answer:GASOLINE IS NOT FUYKING SOUP IF YOU EAT IT YOULL DIE
Explanation:
Just a fair warning
This answer explains that during the combustion of gasoline, each carbon atom forms carbon dioxide (CO₂) and each hydrogen atom forms water (H₂O). This process increases the total mass due to the addition of oxygen atoms, resulting in CO₂ emissions being roughly three times the weight of the gasoline burned.
In this question, we explore the combustion of gasoline, a complex mixture of hydrocarbons. The main reaction for burning gasoline can be approximated by the equation:
CH₂ + 1.5 O₂ → CO₂ + H₂O
This indicates that each carbon atom in the gasoline molecule combines with oxygen to form carbon dioxide (CO₂), while each hydrogen atom combines with oxygen to form water (H₂O). The average chemical formula for gasoline is close to CH₂; thus, if you burn hydrocarbons, the products will primarily be CO₂ and H₂O.
For example, burning gasoline with the empirical formula CH₂ involves replacing two hydrogen atoms with oxygen, resulting in carbon dioxide and water. Here's a more balanced version of the reaction:
2 CH₂ + 3 O₂ → 2 CO₂ + 2 H₂O
This reaction results in a mass increase as the products (CO₂ and H₂O) weigh more than the original reactants (CH₂ and O₂). Specifically, for each carbon atom (12 atomic mass units) and two hydrogen atoms (2 atomic mass units) in the gasoline, the resulting CO₂ and H₂O (carbon dioxide: 44 AMU; water: 18 AMU) demonstrate an overall tripling of mass due to the addition of oxygen atoms.
By understanding this principle, one can estimate that the weight of CO₂ released by burning gasoline is roughly three times the weight of the gasoline itself. This is why a typical car emits a substantially larger mass of CO₂ relative to the fuel it consumes.
Can convection occur in both liquids and
gases? Suggest a reason for your answer
using the particle theory.
Answer:
Yes, Convection can occur in both liquids and gases
Explanation:
The Particle Theory suggests that Particles are always moving. Convection occurs in the breeze you feel, when the warm particles quickly move upwards resulting in cold air quickly sinking and creating a breeze. Convection also occurs when you boil water, the molecules of warm water quickly move to the top and cold water quickly moves to the bottom resulting in a circle the constantly keeps the water circulating, making it boil.
Which of the following statements is TRUE? Question 1 options: There is a "heat tax" for every energy transaction. A spontaneous reaction is always a fast reaction. The entropy of a system always decreases for a spontaneous process. Perpetual motion machines are a possibility in the near future. None of these are true.
Question:
Which of the following statements is TRUE?
A. Perpetual motion machines are a possibility in the near future.
B. The entropy of a system always decreases for a spontaneous process.
C. A spontaneous reaction is always a fast reaction.
D. There is a "heat tax" for every energy transaction.
E. None of the above are true.
Answer:
The correct answer is D)
There is a "heat tax" for every energy transaction.
Explanation:
Heat and work are two different ways in which energy is moved from one device to another. In the field of thermodynamics the distinction between Heat and Work is significant. The transfer of thermal energy between systems is heat. This is what is referred to as "heat tax".
No other statement in the question above is correct.
Cheers!
A 25.888 g sample of aqueous waste leaving a fertilizer manufacturer contains ammonia. The sample is diluted with 73.464 g of water. A 10.762 g aliquot of this solution is then titrated with 0.1039 M HCl . It required 31.89 mL of the HCl solution to reach the methyl red endpoint. Calculate the weight percent NH3 in the aqueous waste.
Answer:
Weight % of NH₃ in the aqueous waste = 2.001 %
Explanation:
The chemical equation for the reaction
[tex]\\\\NH_3} + HCl -----> NH_4Cl[/tex]
Moles of HCl = Molarity × Volume
= 0.1039 × 31.89 mL × [tex]\frac{1 \ L}{1000 \ mL}[/tex]
= 0.0033 mole
Total mass of original sample = 25.888 g + 73.464 g
= 99.352 g
Total HCl taken for assay = [tex]\frac{10.762 \ g}{99.352 \ g}[/tex]
= 0.1083 g
Moles of NH₃ = [tex]\frac{0.0033 \ mol}{0.1083}[/tex]
= 0.03047 moles
Mass of NH₃ = number of moles × molar mass
Mass of NH₃ = 0.03047 moles × 17 g
Mass of NH₃ = 0.51799
Weight % of NH₃ = [tex]\frac{0.51799 \ g}{25.888 \ g} * 100%[/tex]%
Weight % of NH₃ in the aqueous waste = 2.001 %
A 40.0-mL sample of 0.100 M HNO2 (Ka = 4.6 x 10-4 .) is titrated with 0.200 M KOH. Calculate: a. the pH when no base is added b. the volume of KOH required to reach the equivalence point. c. the pH after adding 5.00 mL of KOH d .the pH at one-half the equivalence point e. the pH at the equivalence point f. the pH after 30 mL of the base is added
The initial pH of 0.100 M HNO2 is approximately 2.17. It takes 20.0 mL of 0.200 M KOH to reach the equivalence point. The subsequent pH values at various points in the titration reflect the changing concentrations of HNO2 and OH-.
A 40.0-mL sample of 0.100 M HNO2 (Ka = 4.6 x 10-4) is titrated with 0.200 M KOH.
a. The pH when no base is added
To find the initial pH, we first need to calculate the concentration of H3O+ using the equilibrium expression for the weak acid:
HNO2 ⇌ H+ + NO2-
Using Ka, we get:
Ka = [H+][NO2-] / [HNO2]
4.6 x 10-4 = x2 / 0.100
Solving for x, we get:
x = √(4.6 x 10-4 * 0.100) = 0.00678 M
pH = -log(0.00678) ≈ 2.17
b. The volume of KOH required to reach the equivalence point
The moles of HNO2 are:
0.040 L * 0.100 M = 0.004 mol
Since KOH and HNO2 react in a 1:1 molar ratio, the volume of 0.200 M KOH required is:
0.004 mol / 0.200 M = 0.020 L = 20.0 mL
c. The pH after adding 5.00 mL of KOH
Moles of KOH added:
0.005 L * 0.200 M = 0.001 mol
Moles of HNO2 remaining:
0.004 mol - 0.001 mol = 0.003 mol
Concentration of HNO2 remaining = 0.003 mol/0.045 L = 0.0667 M
Concentration of NO2- formed = 0.001 mol/0.045 L = 0.0222 M
Using the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
pH = -log(4.6 x 10-4) + log(0.0222/0.0667) ≈ 3.35
d. The pH at one-half the equivalence point
At one-half the equivalence point, the concentration of [A-] = [HA], so:
pH = pKa = -log(4.6 x 10-4) ≈ 3.34
e. The pH at the equivalence point
At the equivalence point, all HNO2 has been converted to NO2-:
NO2- will hydrolyze to produce OH-:
NO2- + H2O ⇌ HNO2 + OH-
Using Kb for NO2-:
Kb = Kw/Ka = 1.0 x 10-14 / 4.6 x 10-4 = 2.17 x 10-11
Setting up the equation:
Kb = [OH-][HNO2] / [NO2-]
2.17 x 10-11 = x2 / 0.100
Solving for x:
x = √(2.17 x 10-11 * 0.100) = 1.47 x 10-6
pOH = -log(1.47 x 10-6) ≈ 5.83
pH = 14 - 5.83 = 8.17
f. The pH after 30 mL of the base is added
Moles of KOH added:
0.030 L * 0.200 M = 0.006 mol
Excess moles of OH-:
0.006 mol - 0.004 mol = 0.002 mol
Concentration of OH- in the total volume:
0.002 mol / 0.070 L = 0.02857 M
pOH = -log(0.02857) ≈ 1.54
pH = 14 - 1.54 = 12.46
The correct answers are: (a). [H⁺] = 2.17; (b). [tex]\text{Volume} = 20.0\ mL[/tex]; (c). [tex][NO_2^-] = \text{0.0222 M}[/tex]; (d). pH = 3.34; (e). [tex]\text{pH} = 8.02[/tex]; (f). pH = 12.46.
Let's solve the different parts of the titration problem step-by-step:
a. pH when no base is added:
First, we need to find the pH of a 0.100 M HNO₂ solution. HNO₂ is a weak acid and it partially ionizes in water:
HNO₂ ⇌ H⁺ + NO₂⁻The expression for the acid dissociation constant Ka is:
[tex]K_a = 4.6 \times 10^{-4} = \frac{[H^+][NO_2^-]}{[HNO_2]}[/tex]Assuming that the initial concentration of HNO₂ is C0 = 0.100 M and the change in concentration is x:
[tex]4.6 \times 10^{-4} = \frac{(x \times x)}{(0.100 - x)}[/tex]Assuming x is small relative to 0.100 M:
[tex]4.6 \times 10^{-4} \approx \frac{x^2} {0.100}[/tex]x² = 4.6 x 10⁻⁵x = 6.78 x 10⁻³ M[H⁺] = 6.78 x 10⁻³ M, pH = -log[H⁺] = -log(6.78 x 10⁻³) = 2.17b. Volume of KOH required to reach the equivalence point:
At the equivalence point, moles of HNO₂ = moles of KOH.Moles of HNO₂ = 0.100 M × 0.040 L = 0.00400 molFor KOH: 0.00400 mol = volume × 0.200 M[tex]\text{Volume} = \frac{\text{0.00400 mol}} {\text{0.200 M}} = 0.0200\ L = 20.0\ mL[/tex]c. pH after adding 5.00 mL of KOH:
Moles of KOH added = 0.200 M × 0.00500 L = 0.00100 molRemaining moles of HNO₂ = 0.00400 mol - 0.00100 mol = 0.00300 molTotal volume = 40.0 mL + 5.0 mL = 45.0 mL = 0.0450 L[tex][HNO_2] = \frac{\text{0.00300 mol}} {\text{0.0450 L}} = \text{0.0667 M}[/tex][tex][NO_2^-] = \frac{\text{0.00100 mol}} {\text{0.0450 L}} = \text{0.0222 M}[/tex]Using the Henderson-Hasselbalch equation:
[tex]pH = pKa + \log(\frac{[\text{NO}_2^-]}{[\text{HNO}_2]}) \\[/tex][tex]pK_a = -\log(4.6 \times 10^{-4}) = 3.34 \\[/tex][tex]pH = 3.34 + \log(\frac{0.0222}{0.0667}) = 3.34 - 0.477 = 2.86[/tex]d. pH at one-half the equivalence point:
At one-half the equivalence point, [HNO₂] = [NO₂⁻], so pH = pKa.
pH = pKa = 3.34e. pH at the equivalence point:
At the equivalence point, all HNO₂ has reacted to form NO₂⁻. The solution contains NO₂⁻ ions, which hydrolyze:
NO₂⁻ + H₂O ⇌ HNO₂ + OH⁻[tex]Kb = \frac{K_w}{K_a} = \frac{1.0 \times 10^{-14}}{4.6 \times 10^{-4}} = 2.17 \times 10^{-11}[/tex]Let x be the concentration of OH⁻:
[tex]K_b = \frac{x^2}{0.0500\text{ M}} \\[/tex][tex]2.17 \times 10^{-11} = \frac{x^2}{0.0500} \\[/tex][tex]x^2 = 1.085 \times 10^{-12} \quad \Rightarrow \quad x = \sqrt{1.085 \times 10^{-12}} = 1.04 \times 10^{-6}\text{ M} \\[/tex][tex][\text{OH}^-] = 1.04 \times 10^{-6}\text{ M} \\[/tex][tex]\text{pOH} = -\log[\text{OH}^-] = -\log(1.04 \times 10^{-6}) = 5.98 \\[/tex][tex]\text{pH} = 14 - \text{pOH} = 14 - 5.98 = 8.02[/tex]f. pH after 30 mL of the base is added:
Moles of KOH added = 0.200 M × 0.030 L = 0.00600 molExcess moles of KOH = 0.00600 mol - 0.00400 mol = 0.00200 molTotal volume = 40 mL + 30 mL = 70 mL = 0.070 L[tex][OH^-] = \frac{\text{0.00200 mol}}{\text{0.070 L}} = \text{0.0286 M}[/tex]pOH = -log(0.0286) = 1.54pH = 14 - pOH = 14 - 1.54 = 12.46onsider two aqueous solutions of nitrous acid (HNO2). Solution A has a concentration of [HNO2]= 0.55 M and solution B has a concentration of [HNO2]= 1.55 M . You may want to reference (Page 743) Section 16.6 while completing this problem. Part A Which statement about the two solutions is true? Which statement about the two solutions is true? Solution A has the higher percent ionization and solution B has the higher pH. Solution B has the higher percent ionization and the higher pH. Solution B has the higher percent ionization and solution A has the higher pH. Solution A has the higher percent ionization and the higher pH.
Answer:
Solution A has the higher percent ionization and the higher pH.
Explanation:
Percent ionization depends on the concentration of acid in a solution. If the solution having more concentration of acid so the percent ionization will be lower while if the solution have low amount of acid i. e. dilute solution so the percent ionization will be higher. In solution A, the concentration of HNO2 is lower which is an acid so the percent ionization is higher and the pH of the solution is also higher as compared to solution B.
Answer:
Solution A has the higher percent ionization and the higher pH
Explanation:
Addition of AgNO3 to aqueous solutions of the complex results in a cloudy white precipitate, presumably AgCl. You dissolve 0.1000 g of the complex in H2O and perform a precipitation titration with 0.0500 M AgNO3 as the titrant. Using an electrode that is sensitive to [Ag ], you reach the endpoint after 9.00 mL of titrant is added. How many grams of chloride ion were present in the 0.1000-g sample
Answer:
0.016 grams of chloride ion were present in the 0.1000 grams of sample.
Explanation:
According to question, 9.00 mL of titrant was added to solution with 0.1000 grams of complex to reach the end point.
Molarity of the silver nitrate solution = 0.0500 M
Volume of the silver nitrate solution = V = 9.00 mL = 0.009 L
1 mL = 1000 L
Moles of silver nitrate = n
[tex]Molarity=\frac{Moles}{Volume (L)}[/tex]
[tex]0.0500 M=\frac{n}{0.009 L}[/tex]
n = 0.00045 mol
[tex]Cl^-+AgNO_3\rightarrow AgCl+NO_3^{-}[/tex]
According to 1 mole of silver nitrate reacts with 1 mol of chloride ion, then 0.00045 moles of silver nitrate will :
[tex]\frac{1}{1}\times 0.00045 mol=0.00045 mol[/tex] of chloride ions
Mass of chloride ions :
0.00045 mol × 35.5 g/mol = 0.016 g
0.016 grams of chloride ion were present in the 0.1000 grams of sample.
The half-equivalence point of a titration occurs halfway to the equivalence point, where half of the analyte has reacted to form its conjugate, and the other half still remains unreacted. If 0.300 moles of a monoprotic weak acid ( Ka=3.6×10−5 M) is titrated with NaOH , what is the pH of the solution at the half-equivalence point?
Answer:
pH=pKa
pH=4.44
Explanation:
Since the titration occur between a weak acid and a strong base.
then at half -equivalence point, the pH of the solution is equals to the pKa of the weak acid.
Therefore, pH=pKa
Ka of weak acid=3.6×10^−5
To calculate the pKa of the weak acid using the express below;
pKa =- log(Ka)
pKa=−log(3.6×10−5)=4.44
From the question, the pKa of the solution is at half -equivalence point
Then,
pH=pKa
pH=4.44
The question says that the titration occurred between a weak acid and a
strong base at half-equivalence point. Then we can deduce that the pH of
the solution is equal to the pKa of the weak acid.
pH=pKa
Ka of monoprotic weak acid=3.6×10⁻⁵
The pKa of the monoprotic weak acid will be calculated by :
pKa = - log(Ka)
pKa = −log(3.6×10⁻⁵) = 4.44
Since the pKa of the solution is at half -equivalence point
pH=pKa
pH=4.44
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Propose a mechanism for the formation of the monobrominated product. Draw all missing reactants and/or products in the appropriate boxes by placing atoms on the canvas and connecting them with bonds. Add charges where needed. Electron flow arrows should start on the electron(s) of an
Answer:
The mechanism is SN2
Explanation:
See mechanism of monobromination of alkane attached
Answer:
Explanation:
find the solution below
You have two beakers. One beaker contains 100 mL of NaOH (a strong base); the other contains 100 mL of aqueous Na3PO4 (a weak base). You test the pH of each solution. Which of the following statements is true ?
Answer:
D. If the pH of NaOH is 12, then that of Na₃PO₄ solution has to be lesser than 12.
Explanation:
In this problem, we are comparing the pH of a strong base to a weak base. A strong base is one that ionizes completely in aqueous solutions where as a weak base ionizes slightly.
The pH scale is good tool for measuring the acidity and alkalinity of various substances. It ranges from 1 - 14;
1 7 14
← →
increasing acidity increasing alkalinity
neutrality
Strong bases have their pH value close to 14 and weak bases are close to 7.
Since Na₃PO₄ is a weak base, it will have lesser pH value compared to a strong base such as NaOH
The question lack options, that are as follows"
a. The Na3PO4 has a higher pH because it has more sodium ions than NaOH. NaOH(aq- Na3PO4(aq)b. It is possible for the solutions in each beaker to have the same pHC. If the pH of the NaOH solution is 12.00, the pH of the Na3PO4 solution has to be greater than 12.00.d. If the pH of the NaOH solution is 12.00, the pH of the Na3PO4 solution has to be less than 12.00.The following statements are true in the given question:
d. If the pH of the NaOH solution is 12.00, the pH of the Na3PO4 solution has to be less than 12.00.We know that:
NaOH is a strong base and Na3PO4 is a weak base. As the strength of OH- in 100ml NaOH and that of OH- in 100ml Na3PO4 is different. They have different pH values. As we know that pH is inversely proportional to H+ ion concentration. This means the higher the pH value lower is H+ ion concentration.Again,
PH +POH =14
From the above equation,
the higher the PH value of a solution less will be its value of POH vice versa. Hence greater is the OH- ion concentration.So,
the pH of a solution is directly proportional to OH-ion concentration.Thus, If pH NaOH = 12.00, NaOH is a strong base and Na3PO4 is a weak base. For the solution, the weak base OH-ion concentration is less. Clearly, the pH of Na3PO4 is less than 12.00.
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Which reaction is an example of heterogeneous catalysis?
Answer:
Explanation:
Industrial examples
Process Reactants, Product(s)
Ammonia synthesis (Haber–Bosch process) N2 + H2, NH3
Nitric acid synthesis (Ostwald process) NH3 + O2, HNO3
Hydrogen production by Steam reforming CH4 + H2O, H2 + CO2
Ethylene oxide synthesis C2H4 + O2, C2H4O
Answer:
The answer is A - Ethene gas reacts with hydrogen gas by using a nickel catalyst.
Explanation:
just did on Edge
A beaker holds 962 g of a brine solution that is 6.20 percent salt. If 123g of water are evaporated from the beaker, how much salt must be added to have an 8.60 percent brine solution? How many grams of the 8.6% brine solution will be produced?
To increase the salt concentration to 8.6%, you need to add 8794g of salt to the solution. The final mass of the 8.6% brine solution will be 961g.
Explanation:To have an 8.60 percent brine solution, you would need to add salt to compensate for the loss of water due to evaporation. First, calculate the mass of water after evaporation by subtracting 123g from the initial mass of the brine solution (962g - 123g = 839g).
Then, find the mass of the salt needed by multiplying the final mass of the solution by the desired percent of salt (839g / 0.0860 = 9756g). Subtract the initial mass of the brine solution to determine the amount of salt that must be added (9756g - 962g = 8794g).
To find the mass of the 8.6% brine solution that will be produced, subtract the mass of the added salt from the final mass of the solution (9756g - 8794g = 961g).
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What will be the pH of 1.0 mol dm-3 of NH4OH, which is 1% dissociated
The pH of this solution is 12.
We can solve this question knowing that the ammonium hydroxide, NH₄OH, dissociates in water as follows:
NH₄OH(aq) ⇄ NH₄⁺(aq) + OH⁻(aq)
Based on the reaction, 1 mole of NH₄OH produces 1 mole of OH⁻
With this molarity and the 1% dissociated we can find the molarity of OH⁻. With molarity of OH⁻ we can find pOH (pOH = -log[OH⁻]) and pH (pH = 14-pOH) as follows:
Molarity OH⁻:
A solution 1.0mol dm⁻³ = 1M of NH₄OH produce 1% of OH⁻ ions because only 1% is dissociate, that is:
[tex]1M NH_4OH*(\frac{1MOH^-}{100MNH_4OH}) = 0.01M OH^-[/tex]
Now, we can find pOH as follows:
pOH:
pOH = -log [OH⁻] = 2
And pH:
pH:
pH = 14 - pOH
pH = 12
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Pure platinum is too soft to be used in jewelry because it scratches easily. To increase the hardness so that it can be used in jewelry, platinum is combined with other metals to form an alloy. To determine the amount of platinum in an alloy, a 8.528 g sample of an alloy containing platinum and cobalt is reacted with excess nitric acid to form 2.49 g of cobalt(II) nitrate. Calculate the mass percent of platinum in the alloy.
Answer:
percentage mass of platinum in the alloy ≈ 90.60 %
Explanation:
The alloy is 8.528 g sample of an alloy containing platinum and cobalt . The alloy react with excess nitric acid to form cobalt(ii) nitrate . Platinum is resistant to acid so it will definitely not react with the acid only the cobalt metal in the alloy will react with the acid.
The chemical reaction can be represented as follows:
Co (s) + HNO₃ (aq) → Co(NO₃)₂ (aq) + NO₂ (l) + H₂O (l)
The balanced equation
Co (s) + 4HNO₃ (aq) → Co(NO₃)₂(aq) + 2NO₂ (l) + 2H₂O (l)
Cobalt is the limiting reactant
atomic mass of cobalt = 58.933 g/mol
Molar mass of Co(NO₃)₂ = 58.933 + 14 × 2 + 16 × 6 = 58.933 + 28 + 96 = 182.933 g
58.933 g of cobalt produce 182.933 g of Co(NO₃)₂
? gram of cobalt will produce 2.49 g of Co(NO₃)₂
cross multiply
grams of cobalt that will react = (58.933 × 2.49)/182.933
grams of cobalt that will react = 146.74317000/182.933
grams of cobalt that will react= 0.8021689362 g
grams of cobalt that will react = 0.802 g
mass of platinum in the alloy = 8.528 g - 0.802 g = 7.726 g
percentage mass of platinum in the alloy = 7.726/8.528 × 100 = 772.600/8.528 = 90 .595 %
percentage mass of platinum in the alloy ≈ 90.60 %
How many moles of helium gas are contained in 4.0: flask at STP
To calculate the number of moles of helium in a 4.0-liter flask at STP, divide the volume of the gas by the molar volume of a gas at STP (22.4 L/mol). This yields approximately 0.17857 moles of helium gas.
First, we need to know the volume of the flask, but since the student has not specified the volume correctly, let's assume it is '4.0 liters'. At STP, one mole of any gas occupies 22.4 liters. To find the number of moles in the flask, we use the molar volume of a gas at STP.
Here's the calculation:
Divide the volume of the gas in the flask by the molar volume of a gas at STP.[tex]\frac{4.0\ L}{ 22.4\ L/mol} = number\ of\ moles\ of\ helium.[/tex]Calculate: [tex]\frac{4.0\ L}{22.4\ L/mol} = 0.17857\ mol.[/tex]Therefore, the flask contains approximately 0.17857 moles of helium gas.
At a certain temperature, 0.660 mol SO 3 is placed in a 4.00 L container. 2 SO 3 ( g ) − ⇀ ↽ − 2 SO 2 ( g ) + O 2 ( g ) At equilibrium, 0.110 mol O 2 is present. Calculate K c .
Answer:
[tex]Kc=6.875x10^{-3}[/tex]
Explanation:
Hello,
In this case, for the given chemical reaction at equilibrium:
[tex]2 SO_3 ( g ) \rightleftharpoons 2 SO_ 2 ( g ) + O_ 2 ( g )[/tex]
The initial concentration of sulfur trioxide is:
[tex][SO_3]_0=\frac{0.660mol}{4.00L}=0.165M[/tex]
Hence, the law of mass action to compute Kc results:
[tex]Kc=\frac{[SO_2]^2[O_2]}{[SO_3]^2}[/tex]
In such a way, in terms of the change [tex]x[/tex] due to the reaction extent, by using the ICE method, it is modified as:
[tex]Kc=\frac{(2x)^2*x}{(0.165-2x)^2}[/tex]
In that case, as at equilibrium 0.11 moles of oxygen are present, [tex]x[/tex] equals:
[tex]x=[O_2]=\frac{0.110mol}{4.00L}=0.0275M[/tex]
Therefore, the equilibrium constant finally turns out:
[tex]Kc=\frac{(2*0.0275)^2*0.0275}{(0.165-2*0.0275)^2} \\\\Kc=6.875x10^{-3}[/tex]
Best regards.
1) Analysis subquestions (7 points): (a) Draw the mechanism of the reaction - remember, there are two main parts to the aldol condensation, the addition step, followed by the elimination. (b) Explain why your reaction forms the enone product, rather than the hydroxyketone intermediate. 2) Critical analysis (7 points): a) You have been given a 1H NMR spectrum of your product. Fully assign this spectrum (i.e. determine which peaks in the 1H NMR correspond to which hydrogens in the product). The peaks have been labeled 1-8 on the spectrum, and the relevant hydrogens Ha-Hh below. b) Calculate the coupling constant between He and Hf. Explain how can this can help determine the stereochemistry (i.e. cis vs. trans) of the double bond. (7) Acetone is a symmetrical molecule, so there are two positions that can react. Draw the product you would expect to obtain if you used two molar equivalents of vanillin rather than one. c) Acetone is a symmetrical molecule, so there are two positions that can react. Draw the product you would expect to obtain if you used two molar equivalents of vanillin rather than one.
Find the attachments
In the equation KClO3 -> KCl + O2, how many moles of oxygen are produced when 3.0 mol of KClO3 decomposes completely?
Answer:
4.5
Hope this helps!
In the given chemical equation according to stoichiometry, 4.5 moles of oxygen are produced when 3.0 mole of KClO₃ decomposes completely.
What is stoichiometry?Stoichiometry is the determination of proportions of elements or compounds in a chemical reaction. The related relations are based on law of conservation of mass and law of combining weights and volumes.
Stoichiometry is used in quantitative analysis for measuring concentrations of substances present in the sample. It is important while making solutions and balancing chemical equations.In the given balanced chemical equation, 2 moles of KClO₃ gives 3 moles of oxygen , thus 3 mole of KClO₃ will give 3×3/2=4.5 moles.
Thus, 4.5 moles of oxygen are produced when 3.0 mole of KClO₃ decomposes completely.
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7. If you fill a balloon with 5.2 moles of gas and it creates a balloon with a volume of 23.5 liters, how many moles are in a balloon at the same temperature and pressure that has a volume of 14.9 liters
To solve this problem, we can use the ideal gas law equation PV=nRT. We can find the number of moles in the first balloon using the given information, and then use that value to find the volume of the second balloon.
Explanation:To solve this problem, we can use the ideal gas law, which states that PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.
First, we can use the given information to find the number of moles in the first balloon. Rearranging the ideal gas law equation, we have n = PV / RT. Plugging in the values, we get n = (5.2 mol)(23.5 L) / (0.0821 atm L/mol K)(T in Kelvin).
Once we have the number of moles for the first balloon, we can use this value to find the volume of the second balloon. Rearranging the ideal gas law equation, we have V = nRT / P. Plugging in the values and solving for V, we get V = (5.2 mol)(0.0821 atm L/mol K)(T in Kelvin) / (P)
You are given 25.00 mL of an acetic acid solution of unknown concentration. You find it requires 35.75 mL of a 0.1950 M NaOH solution to exactly neutralize this sample (phenolphthalein was used as an indicator). What is the molarity of the acetic acid solution? What is the percentage of acetic acid in the solution?
Answer:
0.2788 M
1.674 %(m/V)
Explanation:
Step 1: Write the balanced equation
NaOH + CH₃COOH → CH₃COONa + H₂O
Step 2: Calculate the reacting moles of NaOH
[tex]0.03575 L \times \frac{0.1950mol}{L} = 6.971 \times 10^{-3} mol[/tex]
Step 3: Calculate the reacting moles of CH₃COOH
The molar ratio of NaOH to CH₃COOH is 1:1.
[tex]6.971 \times 10^{-3} molNaOH \times \frac{1molCH_3COOH}{1molNaOH} = 6.971 \times 10^{-3} molCH_3COOH[/tex]
Step 4: Calculate the molarity of the acetic acid solution
[tex]M = \frac{6.971 \times 10^{-3} mol}{0.02500L} =0.2788 M[/tex]
Step 5: Calculate the mass of acetic acid
The molar mass of acetic acid is 60.05 g/mol.
[tex]6.971 \times 10^{-3} mol \times \frac{60.05g}{mol} =0.4186 g[/tex]
Step 6: Calculate the percentage of acetic acid in the solution
[tex]\frac{0.4186g}{25.00mL} \times 100\% = 1.674 \%(m/V)[/tex]
Answer:
Concentration acetic acid = 0.27885 M
% acetic acid = 0.69%
Explanation:
You are given 25.00 mL of an acetic acid solution of unknown concentration. You find it requires 35.75 mL of a 0.1950 M NaOH solution to exactly neutralize this sample (phenolphthalein was used as an indicator). What is the molarity of the acetic acid solution?
what is the percentage of acetic acid in the solution? Assume the density of the solution is 1 g/ml.
Step 1: Data given
Volume of acetic acid = 25.00 mL = 0.025 L
Volume of NaOH = 35.75 mL = 0.03575 L
Molarity of NaOH = 0.1950 M
Step 2: The balanced equation
CH3COOH + NaOH → CH3COONa + H2O
Step 3: Calculate moles
Moles = molarity * volume
Moles NaOH = 0.1950 M * 0.03575 L
Moles NaOH = 0.00697125 moles
Step 4: Calculate concentration of acetic acid
We need 0.00697125 moles of acetic acid to neutralize NaOH
Concentration = moles / volume
Concentration = 0.00697125 moles / 0.025 L
Concentration = 0.27885 M
Step 5: Calculate mass of acetic acid
Mass acetic acid = moles * molar mass
Mass acetic acid = 0.00697125 moles * 60.05g/mol
Mass acetic acid = 0.4186 grams
Step 6: Calculate mass of sample
Total volume = 60.75 mL = 0.06075 L
Mass of sample 60.75 mL * 1g/mL = 60.75 grams
Step 7: Calculate the percentage of acetic acid in the solution
% acetic acid = (0.4186 grams / 60.75 grams ) * 100 %
% acetic acid = 0.69%
A chemist fills a reaction vessel with 9.20 atm nitrogen monoxide (NO) gas, 9.15 atm chlorine (CI) gas, and 7.70 atm nitrosyl chloride (NOCI) gas at a temperature of 25.0°C. Under these conditions, calculate the reaction free energy AG for the following chemical reaction: 2NO(g) +CI (8) - 2NOCI (8) Use the thermodynamic information in the ALEKS Data tab. Round your answer to the nearest kilojoule
Answer:
The reactions free energy [tex]\Delta G = -49.36 kJ[/tex]
Explanation:
From the question we are told that
The pressure of (NO) is [tex]P_{NO} = 9.20 \ atm[/tex]
The pressure of (Cl) gas is [tex]P_{Cl} = 9.15 \ atm[/tex]
The pressure of nitrosly chloride (NOCl) is [tex]P_{(NOCl)} = 7.70 \ atm[/tex]
The reaction is
[tex]2NO_{(g)} + Cl_2 (g)[/tex] ⇆ [tex]2 NOCl_{(g)}[/tex]
From the reaction we can mathematically evaluate the [tex]\Delta G^o[/tex] (Standard state free energy ) as
[tex]\Delta G^o = 2 \Delta G^o _{NOCl} - \Delta G^o _{Cl_2} - 2 \Delta G^o _{NO}[/tex]
The Standard state free energy for NO is constant with a value
[tex]\Delta G^o _{NO} = 86.55 kJ/mol[/tex]
The Standard state free energy for [tex]Cl_2[/tex] is constant with a value
[tex]\Delta G^o _{Cl_2} = 0kJ/mol[/tex]
The Standard state free energy for [tex]NOCl[/tex] is constant with a value
[tex]\Delta G^o _{NOCl} =66.1kJ/mol[/tex]
Now substituting this into the equation
[tex]\Delta G^o = 2 * 66.1 - 0 - 2 * 87.6[/tex]
[tex]= -43 kJ/mol[/tex]
The pressure constant is evaluated as
[tex]Q = \frac{Pressure \ of \ product }{ Pressure \ of \ reactant }[/tex]
Substituting values
[tex]Q = \frac{(7.7)^2 }{(9.2)^2 (9.15) } = \frac{59.29}{774.456}[/tex]
[tex]= 0.0765[/tex]
The free energy for this reaction is evaluated as
[tex]\Delta G = \Delta G^o + RT ln Q[/tex]
Where R is gas constant with a value of [tex]R = 8.314 J / K \cdot mol[/tex]
T is temperature in K with a given value of [tex]T = 25+273 = 298 K[/tex]
Substituting value
[tex]\Delta G = -43 *10^{3} + 8.314 *298 * ln [0.0765][/tex]
[tex]= -43-6.36[/tex]
[tex]\Delta G = -49.36 kJ[/tex]
A sample of NaOH (sodium hydroxide) contains a small amount of Na2CO3 (sodium carbonate). For titration to the phenolphthalein endpoint, 0.200 g of this sample requires 23.98 mL of 0.100 M HCl. An additional 0.700 mL of 0.100 M HCl is required to reach the methyl orange endpoint. What is the percentage of Na2CO3 by mass in the sample?
Answer:
3.71%
Explanation:
The phenolphthalein endpoint refers to the reactions:
OH⁻ + H⁺ → H₂O
CO₃⁻² + H⁺ → HCO₃⁻
While the methyl orange endpoint to:
HCO₃⁻ + H⁺ → H₂CO₃
So the additional volume required for the second endpoint tells us the amount of HCO₃⁻ species, which in turn is the total amount of Na₂CO₃ in the sample:
0.700 mL * 0.100 M * [tex]\frac{1mmolHCO_{3}^{-}}{1mmolHCl}[/tex] = 0.07 mmol HCO₃⁻
Now we calculate the mass of Na₂CO₃, using its molecular weight:
0.07 mmol HCO₃⁻ = 0.07 mmol Na₂CO₃
0.07 mmol Na₂CO₃ * 106 mg/mmol = 7.42 mg Na₂CO₃
No calculations using the volume of the first equivalence point are required because the problem already tells us the mass of the sample is 0.200 g.
0.200 g ⇒ 0.200 * 1000 = 200 mg
%Na₂CO₃ = 7.42 mg/200 mg * 100 = 3.71%
Problem 19.24a Using the reagents below, list in order (by letter, no period) those necessary to convert the starting material into the given product: Note: More than 1 mole of the starting material may be used. Not all spaces provided may be needed. Type "na" in any space where you have no reagent. a. heat, -CO2 b. NaOEt c. (CH3CH2)2CuLi d. CH2Cl2, PCC e. C3H7C(O)CH(C2H5)C(O)C2H5 f. CH3CH2Li g. CH3C(O)Cl, AlCl3 h. NBS, ROOR i. H2NC(O)NH2 j. HN(CH3)2 k. OH-, H2O, heat then H3O l. H3O
Answer:
Step 1: The answer is option (b) NaOEt
Step 2: The answer is option (k) OH-, H2O, heat then H3O
Step 3: The answer is option (a) heat, -CO2
Step 4: na (no reagent)
Explanation:
See the attached file for the explanation.
3. Calculate the theoretical value for the number of moles of CO2 that should have been produced in each balloon assuming that 1.45 g of NaHCO3 is present in an antacid tablet. Use stoichiometry (a mole ratio conversion must be present) to find your answers (there should be three: one answer for each balloon). (6 pts)
Answer:
the theoretical value for the number of moles of [tex]CO_{2(aq)}[/tex] is 0.0173 moles
Explanation:
The balanced chemical equation for the reaction is represented by:
[tex]H_3C_6H_5O_{7(aq)} + 3NaHCO_{3(aq)} ------>3CO_{2(g)}+3H_2O+Na_3C_6H_5O_{7(aq)}[/tex]
From above equation; we would realize that 3 moles of [tex]NaHCO_{3(aq)}[/tex] reacts with [tex]H_3C_6H_5O_{7(aq)}[/tex] to produce 3 moles of [tex]CO_{2(aq)}[/tex]
However ; the molar mass of [tex]NaHCO_{3(aq)}[/tex] = 84 g/mol
mass given for [tex]NaHCO_{3(aq)}[/tex] = 1.45 g
therefore , we can calculate the number of moles of [tex]NaHCO_{3(aq)}[/tex] by using the expression :
number of moles of [tex]NaHCO_{3(aq)}[/tex] = [tex]\frac{mass \ given}{ molar \ mass}[/tex]
number of moles of [tex]NaHCO_{3(aq)}[/tex] = [tex]\frac{1.45}{84}[/tex]
number of moles of [tex]NaHCO_{3(aq)}[/tex] = 0.0173 mole
Since the ratio of [tex]NaHCO_{3(aq)}[/tex] to [tex]CO_{2(aq)}[/tex] is 1:1; that implies that number of moles of [tex]NaHCO_{3(aq)}[/tex] is equal to number of moles of [tex]CO_{2(aq)}[/tex] produced.
number of moles of [tex]CO_{2(aq)}[/tex] = [tex]\frac{mass \ given}{ molar \ mass}[/tex]
0.0173 = [tex]\frac{mass \ given}{ 44 \ g/mol}[/tex]
mass of [tex]CO_{2(aq)}[/tex] = 0.0173 × 44
mass of [tex]CO_{2(aq)}[/tex] = 0.7612 g
Thus; the theoretical value for the number of moles of [tex]CO_{2(aq)}[/tex] is 0.0173 moles
Heat capacity is the amount of heat needed to raise the temperature of a substance 1 ∘ ∘ C or 1 K. Open Odyssey. In the Molecular Explorer, choose Measuring Specific Heat (16). Follow the directions for water only. What variable is plotted on the y - y- axis? total energy What variable is plotted on the x - x- axis? temperature What is the molar heat capacity? molar heat capacity = J ⋅ K − 1 ⋅ mol − 1 J⋅K−1 ⋅mol−1 What is the specific heat capacity?
Answer:J.K^-1. kg^-1
Explanation:
Heat capacity= H
Mass=m
Specific heat capacity= c
H=mc
J.K^-1 = c ×kg
c= J.K^-1/kg
=J.K^-1.kg^-1
Which of the following statements concerning gas pressure is/are correct? (1) Gas pressure arises from gas molecules sticking to the wall of the container holding the gas. (2) The force exerted on the inside walls of a gas-filled container is inversely proportional to the number of gas molecules within the container. (3) As the temperature of a gas increases, gas molecules exert more force on the walls of their container.
Answer:
As the temperature of a gas increases, gas molecules exert more force on the walls of their container.
Explanation:
Pressure is the force exerted by a substance per unit area on another substance. The pressure of a gas is the force that the gas exerts on the walls of its container.
Gases collide frequently with each other and the walls of the container. This pressure of the gas increases with increase in temperature since increase in temperature increases the kinetic energy of gas molecules. They now collide more frequently with the walls of the container hence the answer.
The gas pressure is defined as the force exerted by the gas particles when they collide with the walls of the container. It is the pressure exerted per unit area.
The correct option is:
Option C. As the temperature of a gas increases, gas molecules exert more force on the walls of their container.
The correct explanation can be given as:
The gas molecules are in random motions, and continuously exert pressure on the walls of the container. As the temperature rises, the kinetic energy of the particles is also increased, which causes a faster collision. Thus, the gas pressure is increased as the temperature is increased.
Therefore, option C is correct.
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what is the molecule OH- called?
Answer:
it hydroxide ion
Hope it will help you
The molecule OH- is termed as a hydroxide ion. It comprises one hydrogen ion and one oxygen ion and holds a negative charge. Hydroxide ions are observed in water-based solutions and play a crucial role in chemical reactions such as hydrolysis; they also impact the pH levels in solutions.
Explanation:The molecule OH- is called a hydroxide ion. This comes from the fact that it consists of one hydrogen ion (H) and one oxygen ion (O). An ion is an atom or molecule that carries a charge, in this case, the charge of the hydroxide ion is -1 because it has gained one electron. Hydroxide ions are found in solutions resulting from the ionization of water.
The molecule OH- plays an important role in various chemical reactions. For example, in hydrolysis, a molecule of water disrupts a compound; the water splits into H and OH. One part of the divided compound bonds with the hydrogen atom, and the other part bonds with the hydroxide group. The presence of hydroxide ions also has a significant effect in determining the pH of a solution; the higher the concentration of hydroxide ions, the more basic or alkaline the solution is.
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Conceptual Checkpoint 18.13 When 1,3-dinitrobenzene is treated with nitric acid and sulfuric acid at elevated temperature, the product is 1,3,5-trinitrobenzene. Draw the sigma complex for each possible pathway to compare the relative stability of each sigma complex. For the mechanism, draw the curved arrows as needed. Include lone pairs and charges in your answer. Do not draw out any hydrogen explicitly in your products. Do not use abbreviations such as Me or Ph.
Using sulfuric acid in the nitration of benzene generates a more reactive nitronium ion, which then reacts with benzene. The detailed mechanisms of nitration and sulfonation involve sigma complexes with resonance stabilization. An energy diagram for nitration shows intermediates and activation energy.
The importance of using sulfuric acid in the nitration of benzene by nitric acid is to activate the HNO₃, creating a more reactive electrophile, the nitronium ion (NO₂+). This ion then reacts with benzene to form nitrobenzene. In the mechanism of the sulfonation of benzene, sulfur trioxide (SO₃) acts as the electrophile, which is generated from sulfuric acid. The detailed mechanisms for both reactions involve the formation of a sigma complex with resonance forms and the restoration of aromaticity through deprotonation.
For the nitration of benzene, an energy diagram shows the activation energy and intermediates formed. The sigma complex for 1,3-dinitrobenzene undergoing further nitration to form 1,3,5-trinitrobenzene illustrates how the meta position stabilizes the complex by avoiding the creation of high-energy intermediates with adjacent positive charges, as seen in ortho and para substitution.
The equations representing the nitration and sulfonation of benzene are as follows:
Nitration: C₆H₆ + HNO₃ -> C₆H₅NO₂ + H₂OSulfonation: C₆H₆ + SO₃/H₂SO₄ -> C₆H₅SO₃H + H₂Oif you have 3.0 moles of argon gas at STP, how much volume will the argon take up?