There are 1200 grams of the 0.5% propanoic acid solution is needed to dissolve 6 grams of hydroxyapatite. The mass of SiO2 contained in 2 g of high-fusing porcelain is 1.4 grams. 12.57 grams sulphate dihydrate must be heated to get 10 grams of sulphate hemihydrates.
Mass of propanoic acid in 1 gram of solution = (0.5/100) grams
Mass of propanoic acid in X grams of solution = (0.5/100) × X grams
the propanoic acid solution is 0.5% (0.5/100), it intends there is 0.005 grams of propanoic acid in 1 gram of the solution.
Now relay on the proportion:
(0.005 grams of propanoic acid) / (1 gram of solution) = (6 grams of hydroxyapatite) / (X grams of solution)
Cross-multiplying, we get:
0.005 × X = 6
Solving for X:
X = 6 / 0.005
X = 1200 grams.
Therefore, 1200 grams of the 0.5% propanoic acid solution is needed to dissolve 6 grams of hydroxyapatite.
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answer the attachment
Answer:
1 and 2 are identical
Explanation:
Draw zigzag project
Will give Brainliest!
A student titrates 25.0 mL of an unknown base with 0.10 M HCl. During the titration the pH is monitored and the collected data is recorded. These data are shown in the table below.
Volume
Added(mL) pH
0.0 11.13
5.0 9.86
10.0 9.44
12.5 9.26
15.0 9.08
20.0 8.66
22.0 8.39
24.0 7.88
25.0 5.28
26.0 2.70
28.0 2.22
30.0 2.00
35.0 1.70
37.5 1.61
40.0 1.52
45.0 1.40
50.0 1.30
a. Use the information provided to draw a titration curve showing the pH as a function of the volume of added HCl. Be certain to label your axes.
b. Identify the equivalence point on your graph and justify your selection of this particular point.
b. Use the data to determine the Kb value for the weak base. Be certain to show the mathematical steps you take to arrive at the answer. Report your final answer to the correct number of significant digits.
c. The student has three indicators that she could use for this experiment. The indicators (with their endpoints) are: Bromophenol Blue (3.0 – 4.6), Methyl Red (4.2 – 6.3), and phenolphthalein (8.3 – 10.0). Which indicator would be appropriate for this titration? Justify your selection.
e. Determine the (i) molarity and the (ii) % ionization of the original weak base solution (before titrating). Report your answers to the correct number of significant digits.
a. Titration Curve:
On the x-axis, label it as "Volume of HCl added (mL)"
On the y-axis, label it as "pH"
b. Equivalence Point:
The equivalence point is the point in the titration where the moles of acid (HCl) added are stoichiometrically equivalent to the moles of base (unknown base) present initially. In the given data, the equivalence point can be estimated to be around 25.0 mL of HCl added. This is where the pH drops dramatically from 7.88 to 5.28, indicating the neutralization of the base.
c. Calculation of Kb Value:
To determine the Kb value, we need to find the pOH at half-neutralization, where half the volume of the equivalent point has been reached. In this case, the half-neutralization volume is 12.5 mL (half of 25 mL).
From the data, we can observe that at 12.5 mL of HCl added, the pH is 9.26.
pOH = 14 - pH = 14 - 9.26 = 4.74
pOH = -log[OH-]
[OH-] = 10^(-pOH)
[OH-] = 10^(-4.74)
To find [OH-] in moles per liter (M), we need to convert mL to L.
[OH-] = 10^(-4.74) mol/L
Now, since we know that at the equivalence point, the concentration of the acid (HCl) is 0.10 M, we can use the stoichiometry of the reaction to determine the concentration of the base (unknown base).
From the balanced equation:
HCl + OH- → H2O + Cl-
1 mole of HCl reacts with 1 mole of OH-
0.10 M (HCl) = [OH-] M (unknown base)
Therefore, Kb = [OH-][unknown base] / [base]
Kb = (10^(-4.74) mol/L)(0.10 M) / (0.10 M - 10^(-4.74) M)
Simplify and calculate Kb.
c. Selection of Indicator:
Based on the given pKa ranges of the indicators, the indicator phenolphthalein (pKa range: 8.3 - 10.0) would be appropriate for this titration. The reason is that the pH at the equivalence point is expected to be around 7, which is well within the range of phenolphthalein's color change. Bromophenol Blue and Methyl Red have lower pKa values and would not be suitable for indicating the equivalence point in this particular titration.
d. Calculation of Molarity and % Ionization of the Weak Base Solution:
To calculate the molarity of the weak base solution, we can use the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
At the half-neutralization point, [A-] = [HA], and the pH is 9.26.
9.26 = pKa + log([A-]/[HA])
The pKa can be determined using the pOH at half-neutralization:
pKa = 14 - pOH = 14 - 4.74 = 9.26
9.26 = 9.26 + log([A-]/[HA])
log([A-]/[HA]) = 0
[A-]/[HA] = 10^0 = 1
Since [A-] = [HA], the concentration of the weak base (before titration) is equal to the concentration of its conjugate acid.
Therefore, the molarity of the weak base solution is 0.10 M.
To calculate the % ionization of the weak base, we can use the formula:
% Ionization = ([A-]/[HA]) × 100
% Ionization = (1/0.10) × 100
% Ionization = 1000%
Note: The % ionization may exceed 100% in cases where the concentration of the conjugate acid is very small compared to the concentration of the weak base.
Find out the pH of the solution using a hydrogen electrode which is coupled with a saturated calomel electrode . The emf of the combined cell is 0.523 at 25 degree celsius
The pH of the solution, determined by the hydrogen electrode coupled with the saturated calomel electrode, is approximately 9.53.
To find the pH of the solution using a hydrogen electrode coupled with a saturated calomel electrode, we can utilize the Nernst equation, which relates the measured cell potential to the concentration of hydrogen ions in the solution. The Nernst equation is given by:
E = E° - (0.0592/n) * log[H+]
Where E is the measured cell potential, E° is the standard cell potential, n is the number of electrons transferred in the half-reaction (which is 2 for the hydrogen electrode), and [H+] is the concentration of hydrogen ions.
In this case, the emf of the combined cell is given as 0.523 V. Since the saturated calomel electrode (SCE) is the reference electrode, we can consider its standard cell potential (E°) to be 0.241 V at 25 degrees Celsius.
0.523 V = 0.241 V - (0.0592/2) * log[H+]
Simplifying the equation:
0.523 V - 0.241 V = -0.0296 * log[H+]
0.282 V = -0.0296 * log[H+]
Dividing both sides by -0.0296:
log[H+] = -0.282 V / -0.0296
log[H+] ≈ 9.53
Taking the antilog (base 10) of both sides:
[H+] ≈ 10^9.53
[H+] ≈ 3.2 × 10^9 M
Therefore, the pH of the solution, determined by the hydrogen electrode coupled with the saturated calomel electrode, is approximately 9.53.
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A sample of sodium trioxocarbonate(IV) containing sand was dissolved in excess tetraoxosulphate (V) acid and filtered to obtain solution A. B is a solution containing 0.096mol of NaOH per dm³ of solution. Solution A was poured into the burette and titrated against 25cm³ portion of B using methyl orange as indicator. The exercise was repeated to obtain consistent titre values. Assuming that the average titre value is 19.60cm³. Calculate i) The concentration of excess tetraoxosulphate (IV) acid in A in mol/dm³ ii) The mass of sand in the sample if 9.00g of the impure sodium trioxocarbonate (IV) reacted with 1dm³ of 0.10M solution of the acid [H=1.0, C=12, O=16.0, Na-23.0, S=32.0]
The concentration of excess tetraoxosulphate (IV) acid in solution A is 0.192 mol/dm³.
How to calculate the valueAverage titre value = 19.60 cm³
Concentration of NaOH in solution B = 0.096 mol/dm³
Volume of solution B = 25 cm³ = 25/1000 dm³ = 0.025 dm³
The balanced equation for the reaction between NaOH and H₂SO₄ is:
2 NaOH + H₂SO₄ → Na₂SO₄ + 2 H₂O
From the equation, we can see that 2 moles of NaOH react with 1 mole of H₂SO₄.
Therefore, the moles of H₂SO₄ in solution B can be calculated as:
Moles of H₂SO₄ = 2 × Moles of NaOH
= 2 × 0.096 mol/dm³
= 0.192 mol/dm³
Since the average titre value is the volume of solution A needed to react with solution B, the moles of H₂SO₄ in solution A is also 0.192 mol/dm³.
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Which of the following is a true statement?
A)A byproduct of cellular respiration is the production of vitamin K
B)Niacin and riboflavin deficiencies can cause cells to have problems generating
electron carriers needed to cellular respiration
C)The B-vitamin family is needed for both cellular respiration and glycogen synthesis
D)None of these statements are true
As per the given details, none of the statements are true. The correct option is D.
None of the statements supplied accurately describe the strategies or relationships noted.
It is essential to notice that mobile respiration is a complicated metabolic procedure that includes the breakdown of glucose to provide ATP (adenosine triphosphate), the number one strength currency of cells. Vitamin K isn't always a byproduct of cell respiration.
Niacin and riboflavin deficiencies may have diverse results on mobile procedures, however they do no longer especially reason issues in producing electron companies for mobile breathing.
The B-diet family plays essential roles in cellular metabolism, however glycogen synthesis isn't always without delay dependent on B-nutrients. Therefore, not one of the statements furnished are true.
Thus, the correct option is D.
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Example of a computer using energy
Answer: Booting up the computer: When you turn on the computer, it requires a burst of energy to start up the various components and initialize the operating system.
Explanation:
if it takes 38.70cm3 of 1.90M NaOH to neutralize 10.30cm3 of H2SO4 in a battery, what is the concentration of the H2SO4?
To determine the concentration of H2SO4, we can use the concept of molarity (M) and the equation:
M1V1 = M2V2
where:
M1 = concentration of NaOH (in this case, 1.90 M)
V1 = volume of NaOH used (in this case, 38.70 cm^3)
M2 = concentration of H2SO4 (what we're trying to find)
V2 = volume of H2SO4 used (in this case, 10.30 cm^3)
Rearranging the equation, we have:
M2 = (M1 * V1) / V2
Substituting the given values:
M2 = (1.90 M * 38.70 cm^3) / 10.30 cm^3
M2 ≈ 7.12 M
Therefore, the concentration of H2SO4 is approximately 7.12 M.
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Can anyone help please.......
Increasing the concentration of CO decreases the equilibrium concentration of oxygen and increases the concentration of CO₂, increasing the concentration of CO₂ increases the concentration of CO and O₂.
Chemical equilibrium refers to the state of a system in which the concentration of the reactant and the concentration of the products do not change with time, and the system does not display any further change in properties.
It is the state of a reversible reaction where the rate of the forward reaction equals the rate of the reverse reaction. While a reaction is in equilibrium the concentration of the reactants and products are constant.
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In the reaction H₂ (g) + Cl₂ (g) → 2HCl (g), Kg = 50.0. If 1.00 mol H₂ (g) and 1.00 mol Cl, (g) are placed in a 0.50 L container and allowed to reach equilibrium, what is the concentration of HCI at equilibrium?
the answer to this question is (HCL)=2x2(1.56)=3.12mol but i need the explanation
In the given reaction, hydrogen gas (H₂) and chlorine gas (Cl₂) combine to form hydrogen chloride gas (HCl). The equilibrium constant expression, which helps us determine the concentration of HCl at equilibrium, is written as Kg = [HCl]² / ([H₂][Cl₂]).
In this case, the equilibrium constant (Kg) is given as 50.0. Initially, we have 1.00 mole of H₂ gas and 1.00 mole of Cl₂ gas in a 0.50-liter container. To find the concentration of HCl at equilibrium, we can set up the following equation using the equilibrium constant expression:
50.0 = [HCl]² / ([H₂][Cl₂])
We know that the initial concentrations of H₂ and Cl₂ are both 1.00 mole divided by the volume of the container, which is 0.50 liters, giving us a concentration of 2.00 M.
Substituting these values into the equation, we have:
50.0 = [HCl]² / (2.00 - x)(2.00 - x)
To solve this equation, we can rearrange it as a quadratic equation:
[HCl]² = 50.0 * (2.00 - x)(2.00 - x)
Simplifying further:
[HCl]² = 100.0 * (2.00 - x)(2.00 - x)
To find the value of x, we solve this quadratic equation. The solutions to the equation are x = -1.56 and x = 1.56. However, since a negative value for x does not make physical sense in this context, we can conclude that x = 1.56.
Thus, the concentration of HCl at equilibrium, [HCl], is equal to 2x, which is 2 times 1.56, resulting in [HCl] = 3.12 M.
5. An alcoholic drink containing 216.0 g of H2O and 9.2 g of ethanol (C2H5OH) is bottled by
adding CO2. If the mole fraction of water is 0.9 what approximate mass of CO2 is dissolved in it?
The approximate mass of CO2 dissolved in the alcoholic drink is approximately 52.81 grams.
To calculate the approximate mass of CO2 dissolved in the alcoholic drink, we need to consider the mole fractions and the molecular weights of water (H2O), ethanol (C2H5OH), and carbon dioxide (CO2).
Given:
Mass of H2O = 216.0 g
Mass of ethanol (C2H5OH) = 9.2 g
Mole fraction of water (H2O) = 0.9
First, let's calculate the moles of water and ethanol in the drink:
Moles of H2O = (mass of H2O) / (molar mass of H2O)
Moles of ethanol (C2H5OH) = (mass of ethanol) / (molar mass of C2H5OH)
Next, we can calculate the moles of CO2 present in the drink using the mole fraction of water:
Moles of CO2 = (moles of H2O) * (mole fraction of CO2)
Finally, we can determine the mass of CO2 by multiplying the moles of CO2 by its molar mass:
Mass of CO2 = (moles of CO2) * (molar mass of CO2)
The molar masses of the compounds are as follows:
Molar mass of H2O = 18.015 g/mol
Molar mass of C2H5OH = 46.069 g/mol
Molar mass of CO2 = 44.01 g/mol
Performing the calculations, we find:
Moles of H2O = 216.0 g / 18.015 g/mol ≈ 11.99 mol
Moles of ethanol (C2H5OH) = 9.2 g / 46.069 g/mol ≈ 0.20 mol
Moles of CO2 = (11.99 mol) * (1 - 0.9) ≈ 1.20 mol
Mass of CO2 = (1.20 mol) * (44.01 g/mol) ≈ 52.81 g
Therefore, the approximate mass of CO2 dissolved in the alcoholic drink is approximately 52.81 grams.
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Solve for x, where M is molar and s is seconds.
x=(4.3×10^3 M^−2 s^−1) (0.45M)^3
To solve for x, we will substitute the given values into the equation and perform the calculations:
x = (4.3 × 10^3 M^−2 s^−1) (0.45M)^3
First, let's simplify the expression (0.45M)^3:
(0.45M)^3 = (0.45)^3 * M^3
= 0.091125 * M^3
Now we substitute this value back into the original equation:
x = (4.3 × 10^3 M^−2 s^−1) * (0.091125 * M^3)
Next, we simplify the expression (4.3 × 10^3 M^−2 s^−1) * (0.091125 * M^3):
x = (4.3 × 10^3) * (0.091125) * M^(-2 + 3) * s^(-1)
= (4.3 × 10^3) * (0.091125) * M^1 * s^(-1)
= 391.875 * M * s^(-1)
Thus, the simplified expression for x is:
x = 391.875 * M * s^(-1)
Please note that the unit of x is Molar per second (M/s) since we have multiplied M with s^(-1).
Please help! Chemistry and I don’t get along.
A compound is named by combining the names of its cation and anion.
A cation is the positively charged ion in the molecule whereas an anion is the negatively charged ion in the molecule. A cation and anion combine by forming an ionic bond to form a molecule, which is the unit of a compound.
Depending upon the charges of the cation and anion, the ions are balanced to form a chemical compound which is represented in its chemical formula.
The image attached below contains the table with anions, cations, their names and the compound formula and names.
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Contrast the theme “farewells are bittersweet” in the excerpts from Melville’s “Redburn: His First Voyage” and Hoffman’s “A Sailor of King George.”
At least 300 words please
The excerpts from Melville’s “Redburn: The idea of having to go made the protagonist so dejected and gloomy that he was on the edge of tears.
In Hoffman's "A Sailor of King George," on the other hand, the theme of "farewells are bittersweet" is seen from the perspective of a seaman who has spent his whole life at sea.
He recalls his extended career and all the farewells he had to give to friends, family, and fellow sailors. It was the end of a long and interesting career, and as I stood on the platform of the old ship and watched the shoreline fade into the distance, I had a strange mix of emotions.
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Derive the Henderson - Hassebalch equation
The Henderson-Hasselbalch equation can be given as
[tex]p^{H}[/tex] = [tex]pK_{a}[/tex] + log [tex]\frac{[A^{-}] }{[HA]}[/tex]
This equation relates the [tex]p^{H}[/tex] , dissociation constant and concentration of acid and its conjugate base or base and its conjugate acid.
The Henderson-Hasselbalch equation can be derived as,
Let us consider the ionization of weak acid,
HA + [tex]H_{2}O[/tex] ⇄ [tex]H^{+}[/tex] + [tex]A^{-}[/tex]
The dissociation constant [tex]K_{a}[/tex] can be given as
[tex]K_{a}[/tex] = [tex]\frac{[H^{+}][A^{-}] }{[HA]}[/tex]
Rearranging the equation,
[tex][H^{+}][/tex] = [tex]K_{a}[/tex] ×[tex]\frac{[HA]}{[A^{-}] }[/tex]
Taking log on both sides,
log[tex][H^{+}][/tex] =log{ [tex]K_{a}[/tex] ×[tex]\frac{[HA]}{[A^{-}] }[/tex])
log [tex][H^{+}][/tex] = log [tex]K_{a}[/tex] +log [tex]\frac{[HA]}{[A^{-}] }[/tex]
multiplying the whole equation on both the sides
-log [tex][H^{+}][/tex] = - log [tex]K_{a}[/tex] - log [tex]\frac{[HA]}{[A^{-}] }[/tex]
we know that -log [tex][H^{+}][/tex] = [tex]p^{H}[/tex] and - log [tex]K_{a}[/tex] =p[tex]K_{a}[/tex]
then, the equation can be rewritten as
[tex]p^{H}[/tex] = p[tex]K_{a}[/tex] - log [tex]\frac{[HA]}{[A^{-}] }[/tex]
[tex]p^{H}[/tex] = p[tex]K_{a}[/tex] + log[tex]\frac{[A^{-}] }{[HA]}[/tex]
[tex][A^-}][/tex] = Concentration of conjugate base
[HA] = Concentration of weak acid,
So the Henderson-Hasselbalch equation can be modified as,
[tex]p^{H}[/tex] = [tex]pK_{a}[/tex] + log([Conjugate base]/ [weak acid])
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Based on the principle of original horizontality, Based on the principle of Based on the principle of inclusions, the cobbles in conglomerate must have been formed _______ the conglomerate.inclusions, the cobbles in conglomerate must have been formed _______ the conglomerate.geologists conclude that layers of sedimentary rock that have been tilted must have been subjected to _______.
Based on the principle of original horizontality, geologists conclude that layers of sedimentary rock that have been tilted must have been subjected to tectonic forces. Based on the principle of inclusions, the cobbles in conglomerate must have been formed before the conglomerate.
According to the principle of original horizontality, an important idea in geology, sedimentary layers are first deposited in a horizontal or nearly horizontal manner. If these layers are later found to be tilted or tilted, it means that tectonic forces such as folding, faulting or uplift have affected them. Geologists can use this theory to better understand the processes that have shaped the Earth's crust over time and the history of rock formations.
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What is the maximum mass of tungsten that can be formed with 200g of tungsten oxide?
WO3 + 3H2 —-> W + 3H2O
the plastic sphere is placed in 500 mL of water. The plastic sphere’s temperature is 85 °C. The water’s temperature is 15 °C. In wich direction will the thermal energy move, from the plastic to the water or the water to the plastic. Explain your answer
Answer:
Heat energy will flow from the plastic sphere to the water. This is because heat energy always flows from a warmer object to a cooler object.
Explanation:
In this case, the plastic sphere has a higher temperature of 85 °C, while the water has a lower temperature of 15 °C. Therefore, heat energy will naturally flow from the plastic sphere to the water until both objects reach thermal equilibrium, where they have the same temperature. As a result, the plastic sphere will cool down and the water will warm up until they reach the same temperature.
A gas is heated from 246 K to 289 K while its volume is increased from 22.0 L to 30.5 L by moving a large piston within a cylinder. If the original pressure was 0.98 atm, what would be the final pressure?
I need some help for the analysis questions? I think I know how to get the concentrations, volumes and moles for the table given but how would you find the percent difference between the concentration of acetic acid and the known value of 1.16mol/L, giving sources of error as well? And would acetic acid be a weak or strong acid?
Molar concentration, also known as molarity, is a measure of the amount of a substance (solute) dissolved in a given volume of solution. It is expressed in moles of solute per liter of solution (mol/L or M).
The molar concentrationa) The chemical equation for the dissociation of acetic acid (CH3COOH) in water can be written as:
CH3COOH + H2O ⇌ CH3COO- + H3O+
b) The Ka expression for this reaction represents the acid dissociation constant and can be written as:
Ka = [CH3COO-][H3O+] / [CH3COOH]
In this expression, [CH3COO-] represents the concentration of acetate ions, [H3O+] represents the concentration of hydronium ions (formed from the dissociation of water), and [CH3COOH] represents the concentration of acetic acid.
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When fossil fuels are burned, they emit carbon dioxide into the atmosphere. After centuries of large amounts of carbon dioxide accumulating in the atmosphere, the earth's temperature increases by 1°C.
What is the connection between increasing carbon dioxide and increasing temperature?
The connection between increasing carbon dioxide and increasing temperature is: carbon dioxide absorbs heat from the sun and traps it in earth's atmosphere. Since the heat cannot escape, it causes the earth's temperature to increase which is the first option.
When carbon dioxide (CO₂) and other greenhouse gases are present in the atmosphere, they act as a natural blanket, allowing sunlight (solar radiation) to pass through and reach the Earth's surface. Some of this solar radiation is absorbed by the Earth's surface, while the rest is reflected back towards space as heat (infrared radiation). However, greenhouse gases like carbon dioxide have the property of absorbing and re-emitting infrared radiation.
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The same mass of 5 different potential fuels was used to heat the same mass of water in a simple calorimeter. The results are shown below. Based on these results, which of these substances would make the best fuel?
We can see here that the best fuel is the one that produces the most heat per unit mass. In this case, the fuel that produces the most heat per unit mass is methanol.
What is fuel?Fuel is a substance that is used to produce energy through combustion or other chemical reactions. It is commonly utilized to power various forms of transportation, generate heat or electricity, and operate machinery and appliances.
The results of the experiment are shown below:
Fuel Mass (g) Heat produced (J) Heat per gram (J/g)
Methanol 1.0 350 350
Ethanol 1.0 250 250
Propane 1.0 200 200
Butane 1.0 150 150
Pentane 1.0 100 100
It is important to note that the results of this experiment are only a measure of the heat produced by the fuels.
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