Answer:
Kₐ = 5.7 x 10⁻⁵
Explanation:
The equilbrium for this acid is
HC₃H₃CO₂ + H2O ⇄ H₃O⁺ + C₃H₃CO₂ ⁻ ,
and the equilibrium constant for acrylic acid is given by the expression:
Kₐ = [ H₃O⁺][ C₃H₃CO₂⁻ ] / [ HC₃H₃CO₂ ]
Since the pH of the 0.23 M solution is known , we can calculate [ H₃O⁺].
The ][ C₃H₃CO₂⁻ ] is equal to [ H₃O⁺] from the above equilibria (1:1)
Finally [ HC₃H₃CO₂ ] is known.
pH = - log [ H₃O⁺]
taking antilog to both sides of the equation
10^-pH = [ H₃O⁺]
Substituting
10^-2.44 = [ H₃O⁺] = 3.6 x 10⁻³
[ C₃H₃CO₂⁻ ] = 3.6 x 10⁻³
Kₐ = ( 3.6 x 10⁻³ ) /0 .23 = 5.7 x 10⁻⁵
To find the Ka of acrylic acid, we use the measured pH to find [H+], set up an ICE table for the dissociation equilibrium, and solve for Ka, which is approximately 5.8 x 10^{-5} when rounded to two significant digits.
Explanation:To calculate the acid dissociation constant (Ka) of acrylic acid, we start by using the pH given. The pH is defined as the negative logarithm of the hydrogen ion concentration ([H+]), which can be represented as pH = -log[H+]. From the pH of 2.44, we find the concentration of hydrogen ions:
[H+] = 10^{-pH} = 10^{-2.44} ≈ 3.63 × 10^{-3} M
Given that acrylic acid (HC3H3CO2) is a weak acid and only partially dissociates in a solution, we can represent its dissociation as follows:
HC3H3CO2(aq) ⇌ H+(aq) + C3H3CO2⁻(aq)
The ICE table approach details the initial concentrations, the change in concentrations, and the final concentrations. For acrylic acid, we get:
Initial: [HC3H3CO2] = 0.23 M, [H+] = 0, [C3H3CO2⁻] = 0Change: [HC3H3CO2] decreases by x, [H+] increases by x, [C3H3CO2⁻] increases by xEquilibrium: [HC3H3CO2] = 0.23-x M, [H+] = x , [C3H3CO2⁻] = xSince the pH corresponds to the hydrogen ion concentration at equilibrium, we also know that x (the increase in [H+]) is equal to 3.63 × 10^{-3} M. Now we can calculate the Ka using the equilibrium expressions:
Ka = \frac{[H+][C3H3CO2⁻]}{[HC3H3CO2]}
Assuming x is small compared to the initial concentration, we can simplify [HC3H3CO2] to be approximately equal to 0.23 M:
Ka ≈ \frac{(3.63 × 10^{-3})^2}{0.23}
Thus, Ka ≈ 5.76 × 10^{-5}. Rounded to two significant digits, Ka ≈ 5.8 × 10^{-5}.
How can you force a solid compound to the bottom of a melting point capillary? a. tap the closed endb. tap the open endc. use a needle
To move a solid compound to the bottom of a melting point capillary tube, one should tap the open end on a hard surface. A needle can be used to adjust the position of the compound if necessary, without compacting it too much.
Explanation:To force a solid compound to the bottom of a melting point capillary tube, you should tap the open end of the capillary tube on a hard surface, such as a bench top. This action utilizes gravity to help settle the compound into the closed end of the tube. If the compound does not move down to the bottom, you can use a long, thin object like a needle to carefully push the compound down without compacting it too tightly, which could affect its melting behavior. Alternatively, tapping the closed end might cause some compacting and is not usually recommended for settling the compound.
While using the capillary tube, it's important to understand capillary action and the characteristics of liquid-glass interactions. For example, if a capillary tube is placed into a beaker of ethylene glycol, the ethylene glycol will rise into the tube by capillary action due to the strong adhesive forces between the polar Si-OH groups on the surface of glass and the molecules of the ethylene glycol, creating a concave meniscus. This is opposite to what happens with a nonpolar liquid like SAE 20 motor oil, which cannot form strong interactions with the polar Si-OH groups, resulting in the oil having a convex meniscus and a reduction in the capillary action.
A sample of gas occupies a volume of 57.9 L at 300K. Use Charles’s Law to calculate the volume (L) when the temperature is 264K. Show the calculation.
Answer:
When the temperature lowers to 264K the volume lowers to 50.95 L
Explanation:
Step 1: Data given
The initial volume of a gas = 57.9 L
The initial temperature = 300 K
The temperature lowers to 264 K
Step 2: Charles's law
V1 / T1 = V2 / T2
⇒ with V1 = the initial volume of the gas = 57.9 L
⇒ with T1 = the initial temperature = 300K
⇒ with V2 the new volume = TO BE DETERMINED
⇒ with T2 = the final temperature = 264K
57.9L /300K = V2 / 264K
V2 = (57.9*264)/300
V2 = 50.95 L
When the temperature lowers to 264K the volume lowers to 50.95 L
If no carbon dioxide is present in your test tube it will be a ____________ color. If a medium amount of CO2 is present, your test tube will be a _______________ color. If a large amount of CO2 is present, your test tube will be __________________.
if there is no carbon dioxide your test tube will be blue
if there is a medium amount of carbon dioxide your test tube is green
if there are high amounts of CO2 it will be
yellow
In the context of bromothymol blue as an indicator, if no CO2 is present, the solution will be blue. If a medium amount of CO2 is present, the solution turns green. A large amount of CO2 would turn the solution yellow due to the acidic conditions created by carbonic acid.
Explanation:The colors in the test tube depend on the identification or indicator test you are performing. However, if we were referring to the bicarbonate buffer system and the formation of carbonic acid, we could consider the pH indicators' reaction to acidity caused by carbon dioxide concentration. One common pH indicator is bromothymol blue. ~ This works as follows:
No carbon dioxide: If there is no carbon dioxide present in your test tube, bromothymol blue will remain blue. This indicates a basic or neutral solution. Medium CO2: If a medium amount of CO2 is present, bromothymol blue will change to green. This occurs because an increase in carbonic acid results in a more acidic solution, causing the color change. Lots of CO2: If a large amount of CO2 is present, bromothymol blue will turn yellow, indicating a more acidic environment due to a further increase in carbonic acid.
It's important to remember, this answer assumes the use of bromothymol blue, a common pH indicator. Different indicators may produce different color changes.
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Draw a structure for an alcohol that exhibits a molecular ion at M+ = 74 and that produces fragments at m/z = 59, m/z = 56 and m/z = 45.
Answer:
Please refer to the attachment below for answer and explanation.
Explanation:
Please refer to the attachment below for answer and explanation.
A solution is made by adding 0.350 g Ca(OH)2(s), 45.0 mL of 1.00 M HNO3, and enough water to make a final volume of 75.0 mL. Assuming that all of the solid dissolves, what is the pH of the final solution?
Final answer:
To find the pH of the final solution after reacting Ca(OH)2 and HNO3, calculate moles of reactants, determine moles needed for the reaction, and use the formula for a strong base to find the final pH.
Explanation:
Step 1: Calculate the moles of HNO3 used: 0.045 L × 1.00 mol/L = 0.045 mol.
Step 2: Determine the moles of Ca(OH)2 needed to react with the HNO3: 2 moles of HNO3 react with 1 mole of Ca(OH)2, so 0.045 mol HNO3 requires 0.0225 mol Ca(OH)2.
Step 3: Use the volume and molarity of the Ca(OH)2 to find the pH using the formula for a strong base: pH = 14 - pOH. From the given data, we can calculate pOH = -log(0.672) = 0.173. Therefore, pH = 14 - 0.173 = 13.827.
The pH of the final solution is 0.324.
To determine the pH of the final solution, follow these steps:
Calculate moles of Ca(OH)₂:
Molar mass of Ca(OH)₂ is 74.10 g/mol. Moles of Ca(OH)₂:
(0.350 g) / (74.10 g/mol) = 0.00472 mol
Calculate moles of HNO₃:
Molarity (M) = moles/volume in L. For HNO₃, moles:
(1.00 M) x (0.0450 L) = 0.0450 mol
Determine reaction and limiting reagent:
The balanced equation for the reaction is:
Ca(OH)₂ + 2 HNO₃ ➞ Ca(NO3)₂ + 2 H₂O
Moles of HNO₃ required to react with Ca(OH)₂:
0.00472 mol Ca(OH)₂ x 2 = 0.00944 mol HNO₃
Excess HNO₃:
0.0450 mol - 0.00944 mol = 0.03556 mol
Calculate H⁺ concentration and pH:Final volume of solution = 75.0 mL = 0.0750 L. H⁺ concentration:
[H+] = 0.03556 mol / 0.0750 L = 0.474 MpH = -log[H⁺] :pH = -log(0.474) = 0.324Identify which of the concentration expressions can also be used to describe a solution with a concentration of 1 mg/mL of solute. Assume the density of the solution is 1.00 g/mL
When presented with a concentration of 1 mg/mL, one can express it through molarity (if the molar mass of the solute is known) or mass-volume percent concentration. In this case, the solution's mass-volume percent concentration would be 0.1%.
Explanation:The two concentration expressions that can be used to describe a solution with a concentration of 1 mg/mL are molarity and mass-volume percent concentration.
Molarity is defined as the amount of solute in moles divided by the volume of the solution in liters. Assuming the molar mass of the solute is known, the molarity can be calculated using the provided concentration.
Mass-volume percent concentration (% m/v) is another applicable concept. It is defined as the mass of solute in grams divided by the volume of the solution in milliliters, multiplied by 100%. In your case, as the solution is 1 mg/mL, it translates to a mass-volume percent concentration of 0.1%.
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Final answer:
To identify a concentration expression compatible with 1 mg/mL for a solution with density 1.00 g/mL, molarity can be used if the molar mass of the solute is known. Molality is not directly applicable without knowing the weight of the solvent separately.
Explanation:
When considering a solution with a concentration of 1 mg/mL of solute and a density of 1.00 g/mL, it's important to note that molality cannot be used directly, because it requires the weight of the solvent (not the solution), and it's expressed in kilograms. The question pertains to a solution that has its density equivalent to that of water, thus allowing for certain concentration expressions like molarity to provide a good approximation to molality for dilute aqueous solutions.
To convert from mg/mL to molarity (M), we need the molar mass of the solute. For example, if we have potassium bromide (KBr), which has a molar mass of 119.0 g/mole, a 1 mg/mL solution would be 1 g/L or 1000 mg/L. To find the molarity, divide the concentration in grams per liter (1 g/L) by the molar mass of KBr, yielding approximately 0.0084 M. This would be similar to the molality for a dilute KBr solution because the molality and molarity converge when the density is close to that of water (1.00 g/mL).
You are making homemade ice cream and can use either 200 g Ice Melt (CaCl2) or 200 g rock salt (NaCl) to lower the freezing point of 7.00 kg of ice (assume the ice is pure water). Which salt would help you eat your ice cream faster (i.e. lower the freezing temperature of the ice)?
Calcium chloride (CaCl2) is more effective at lowering the freezing point of ice than sodium chloride (NaCl) because it produces more ions when it dissociates, leading to greater freezing point depression.
To determine which salt would help you eat your homemade ice cream faster by lowering the freezing point of ice more effectively, we need to consider the colligative properties of each salt. The property of interest here is freezing point depression, which is a colligative property that states the freezing point of a solution is lower than that of the pure solvent due to the presence of solute particles.
Calcium chloride (CaCl2) is more effective at lowering the freezing point than sodium chloride (NaCl) because it dissociates into three ions (one Ca2+ and two Cl-), while NaCl dissociates into only two ions (one Na+ and one Cl-). Since the extent of freezing point depression is dependent on the number of solute particles in solution, CaCl2 will lower the freezing point more than NaCl for the same mass.
Given that the freezing points of saturated solutions are −22°C for NaCl and −30°C for CaCl2, using 200 g of CaCl2 would result in a lower freezing temperature compared to the same amount of NaCl, thus allowing your ice cream to freeze quicker.
If 6 moles of A and 2 moles of B are reacted, what is the maximum number of moles of C that can be formed
Given the balanced equation 2A + 3B = 3C, with 6 moles of A and 2 moles of B, A is the limiting reactant. The maximum moles of C formed is 9 (Option D).
First, let's identify the limiting reactant by comparing the mole ratios of A and B in the balanced equation:
[tex]\[2A + 3B \rightarrow 3C\][/tex]
The mole ratio between A and B is 2:3. If you have 6 moles of A and 2 moles of B, you can compare the actual ratio (6 moles A to 2 moles B) with the ratio in the balanced equation (2 moles A to 3 moles B).
The actual ratio is 6:2, which simplifies to 3:1. The balanced ratio is 2:3. The limiting reactant will be the one with the smaller ratio, so in this case, A is the limiting reactant.
Now, let's determine the maximum number of moles of C that can be formed using the stoichiometry of the balanced equation:
For every 2 moles of A, 3 moles of C are formed.
Since you have 6 moles of A (the limiting reactant), you can use the ratio to find the moles of C:
[tex]\[\text{Moles of C} = \left( \frac{3 \text{ moles C}}{2 \text{ moles A}} \right) \times 6 \text{ moles A} = 9 \text{ moles C}\][/tex]
Therefore, the correct answer is:
D. 9 moles
The probable question may be:
Consider the chemical equation: 2A+3B=3C. If 6 moles of A and 2 moles of B are reacted, what is the maximum number of moles of C that can be formed?
A. 4 moles
B. 2 moles
C. 3 moles
D. 9 moles
Which atom(s) in the molecule CH3OCH3 will display tetrahedral electron-pair geometry? (Select all that apply)a. Ob. Cc. Hd. none of the above
In a tetrahedral molecular geometry, a central atom is located at the center with four substituents that are located at the corners of a tetrahedron.
The number of bonds in the structure is as follows:-
5 bond10 electronsThe angle between the elements is 109.5.
The geometry is how the bonds and the lone pairs of electrons are distributed around the atom, and the angles formed by them occur to minimize the repulsive forces of the bonds and the lone pairs.
For example, the molecule of CO₂ is linear, because the central atom C has no lone pairs (the valence shell of it has 4 electrons, so it shares all of them to be stable), and does only two bonds. But the molecule of H₂O is angular, because the central atom O has two lone pairs (its valence shell has 6 electrons, so it only shares 2 electrons to be stable), and the repulsive forces are minimized with an angle of 104.45°.
To the geometry be tetrahedral, the atom must have 4 bonds and no lone pairs, and it only happens at the two C in the molecule given.
H only does one bond, and O does 2 bonds and has 2 lone pairs.
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In the molecule CH3OCH3, the carbon (C) atoms will display a tetrahedral electron-pair geometry due to the sp³ hybridization commonly found in organic compounds.
Explanation:In the molecule CH3OCH3, the atoms that will display tetrahedral electron-pair geometry are the carbon (C) atoms. This is because carbon atoms, particularly when involved in organic compounds like this one, often form bonds using sp³ hybridization, which results in a tetrahedral shape. The tetrahedral geometry of carbon occurs when it forms four bonds, often with hydrogen and other carbon atoms.
This conformation is one of the most common in organic chemistry and is noteworthy for the fact that all the bonds are 109.5° apart - the ideal bond angle in a true tetrahedral geometry.
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A 0.1014 g sample of a purified CHO compound was burned in a combustion apparatus and produced 0.1486 g CO2 and 0.0609 g of H2O. What is the empirical formula of this CHO compound? Enter as C#H#O#, e.g. C2H3O2
Answer: the empirical formula is CH2O
Explanation:Please see attachment for explanation
A small sphere of initial volume V is filled with n moles of helium at initial pressure and temperature P and T.
Complete Question:
A small sphere of initial volume V is filled with n moles of helium at initial pressure and temperature and T. Which of the following statements is true?
a) The volume decreases to V/2, and the pressure increases to 4P when the temperature is T/2
b) n/2 moles of gas are removed, the volume is decreased to V/2, and the pressure decreases to P/4 with a drop in temperature of T/2
c) n moles of gas are added, the total sample is heated to 2T, and the pressure drops to P/2 when the volume increases to 8V
d) The amount of gas is doubled to 2n, the pressure is doubled to 2P, and the volume is doubled to 2V, with a corresponding temperature drop to T/2
Answer:
c
Explanation:
Let's consider the helium as an ideal gas, so it can be studied by the ideal gas law, which states:
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. Because R is constant:
PV/nT = R. Thus the initial state must be equal to the final state.
So, let's check the statements:
a) Let's indicate the final state as P₂, V₂, n₂ and T₂. So, if T₂ = T/2:
PV/nT = P₂V₂/n₂T₂
PV/nT = P₂V₂/n₂(T/2)
PV/nT = 2P₂V₂/n₂T
So, if V₂ = V/2 and P₂ = 4P:
PV/nT = 2*(V/2 * 4P)/n2T
PV/nT = 4VP/n2T
Which is not correct!
b) Now, if T₂ = T/2:
PV/nT = 2P₂V₂/n₂T
If n/2 is removes, n₂ = n/2. And, V₂ = V/2 and P₂ = P/4:
PV/nT = 2*(V/2 * P/4)/(n/2)*T
PV/nT = 4*(V/2 *P/4)/nT
PV/nT = PV/2nT
Which is not corret!
c) Now, if V₂ = 8V:
PV/nT = P₂*8V/n₂T₂
And n₂ = n +n = 2n, T₂ = 2T and P₂ = P/2:
PV/nT = (P/2)*8V/2n*2T
PV/nT = 8*(PV)/2*2n*2T
PV/nT = 8*(PV)/8*(nT)
PV/nT = PV/nT
So, it's correct!
d) Now, T₂ = T/2, n₂ = 2n, P₂ = 2P, and V₂ = 2V:
PV/nT = 2P*2V/2n*(T/2)
PV/nT = 4PV/nT
Which is not correct!
What is the partial pressure of oxygen gas in a mixture which contains 540 mmHg He, 203 303 Pa N2 if the total pressure of the mixture is 5.00 atmospheres?
Answer:
Partial pressure O₂ = 2.284 atm
Explanation:
This is problem with unit conversion. Let's convert everything to atm
760 mmHg ___ 1 atm
540 mmHg ___ (540 / 760) = 0.710 atm
101325 Pa ____ 1 atm
203303 Pa ____ (203303 / 101325) = 2.006 atm
Total pressure of the mixture = Sum of partial pressure of each gas
0.710 atm (He) + 2.006 (N₂) + Partial Pressure O₂ = 5 atm
Partial pressure O₂ = 5 atm - 2.006 atm - 0.710 atm = 2.284 atm
An object has a mass of 0.0124kg and a volume of 1893mm3. What is its density in grams per cubic centimeter?
The object's density is calculated by converting the mass into grams (12.4 grams) and the volume into cubic centimeters (1.893 cm3), and then dividing mass by volume. The resulting density is approximately 6.56 grams per cubic centimeter.
Explanation:The density of an object is calculated by dividing its mass by its volume. We also need to convert the units to be compatible. Since your mass is in kilograms we will convert it to grams, 0.0124 kg is the same as 12.4 grams. Your volume is in cubic millimeters (mm3), and we need it in cubic centimeters (cm3). 1 cm3 is equivalent to 1,000 mm3. Therefore, 1893mm3 is equivalent to 1.893 cm3. Now we simply calculate the density: 12.4g / 1.893cm3 = approximately 6.56 grams per cubic centimeter.
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Unit conversions.(a) Convert 10,000 dynes to units of lbm ∙ ft/s^2 and lbf(b) Convert 0.2 atm to units of Kpa and lbf/in^2(c) Convert 37 °C (physiological temperature) to units of °F and K.(d) Convert 50 in 2 ∙ lbm/s^2 units of joules and cal.
Answer:
(a) 0.72 lbm· ft/s²
(b) 20.3 kPa, 2.94 lbf / in²
(c) 98.6 ºF, 310 K
(d) 1.5 x 10⁻² J, 6.1 x 10⁻² cal
Explanation:
Our strategy here will be to find the conversion factors for the quantities we are asked in each part, and perform the calculations.
(a) 10,000 dynes to lbm ·ft/s²
here we are asked to convert the force of 10,000 dynes to lbm ·ft/s². Recall that F= ma ( m= mass, a = acceleration), thus
10,000 dynes = 10 g cm/s²
converting the force
10,000 g cm/s² x (1 lbm/454 g) x (1 ft / 30.48 cm ) /s² = 0.72 lbm· ft/s²
(b)
1 atm = 101.33 pa
0.2 atm x ( 101.33 kPa ) = 20.3 kPa
1 atm = 14.7 lbf / in²
0.2 atm x ( 14.7 lbf / in² /atm ) = 2.94 lbf / in²
(c) The formula for the conversion from ºC to ºF is:
ºF = 9/5 ºC +32
ºF = 9/5 ( 37ºC) + 32 = 98.6 ºF
K = ºC + 273
K = (37 + 273) K = 310 K
(d) 50 in²·lbm/s² to joules and calories
Since the unit in² ·lbm/s² is not that common, lets convert it using their definition.
These are energy units, and we know the energy is the force times distance. In turn force is mass times acceleration so that the units of energy are mass time distance per time squared.
Joules is the unit of energy in the metric system.
50 in² lbm/s² = 50 in²x ( 2.54 cm/in x 1m /100cm)² x (1lbm x 0.454 Kg/lbm)/s²
= 1.5 x 10⁻² Kg m²/² = 1.5 x 10⁻² J
To convert to cal it wilñl be easier to use the value in joules just calculated:
1.5 x 10⁻² J x (4.184 cal/J) = 6.1 x 10⁻² cal
Final answer:
This response provides detailed conversions for different units in physics including dynes, lbm, atm, and temperature measurements. It explains how to convert between various units and provides step-by-step examples for each conversion.
Explanation:
Unit conversions:
(a) To convert 10,000 dynes to lbm ∙ ft/s² and lbf:
1 dyne = 1 g cm/s² = 10⁻⁵ N, therefore, 10,000 dynes = 0.1 N. Using 1 lb = 453.59 g, 1 ft = 0.3048 m, 1 lb = 4.448 N, the conversion is: 10,000 dynes = 0.1 N = 0.0225 lb or 0.1 N = 0.0225 lbf.
(b) To convert 0.2 atm to Kpa and lbf:
1 atm = 101.3 Kpa, therefore, 0.2 atm = 20.26 Kpa. Also, 1 atm = 14.7 lb/in², so 0.2 atm = 2.94 lb/in².
(c) Converting 37 °C to °F and K:
To convert °C to °F: °F = (°C × 9/5) + 32. So, 37 °C = 98.6 °F. To obtain K from °C, use K = °C + 273. Thus, 37 °C = 310 K.
What is the coefficient for H2O when the equation ? Ca(OH)2(aq) + ? H3PO4(aq) → ? Ca3(PO4)2(s) + ? H2O(ℓ) is balanced using the smallest possible integers?
To balance the equation Ca(OH)2(aq) + H3PO4(aq) → Ca3(PO4)2(s) + H2O(ℓ), you need to place a coefficient of 3 in front of Ca(OH)2, 6 in front of H3PO4, and 3 in front of H2O.
Explanation:To balance the equation Ca(OH)2(aq) + H3PO4(aq) → Ca3(PO4)2(s) + H2O(ℓ), we can start by balancing the calcium (Ca) atoms. There are three calcium atoms on the right side (Ca3(PO4)2) and only one on the left side (Ca(OH)2). To balance the calcium atoms, we need to place a coefficient of 3 in front of Ca(OH)2.
Next, we can balance the hydrogen (H) atoms. There are 6 hydrogen atoms on the left side (2 from Ca(OH)2 and 4 from H3PO4), so we need to place a coefficient of 6 in front of H3PO4.
Finally, we can balance the oxygen (O) atoms. There are 4 oxygen atoms on the left side (from Ca(OH)2) and 6 oxygen atoms on the right side (3 from Ca3(PO4)2 and 3 from H2O). To balance the oxygen atoms, we need to place a coefficient of 3 in front of H2O.
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A 5.000 gram sample of lenthionine contains 0.6357 g carbon and 0.1070 g hydrogen, and the remainder is sulfur. The molar mass of lenthionine is 188.4 g/mol. Determine the chemical formula for lenthionine.
Answer: C2H4S5
Explanation:
Since the total mass is 5.000g
Mass of sulphur = 5.000-(0.6357+0.1070)
Mass of sulphur = 4.2573g
Using Empirical relation
C= 0.6357 H= 0.1070 S= 4.2573
Divide through by their molar mass to obtain the smallest ratio
C= 0.6357/12 H=0.1070/1 S=4.2573/32
C= 0.053 H= 0.1070 S= 0.133
Divide through by the smallest ratio (0.053)
C=0.053/0.053 H=0.1070/0.053 S=0.133/0.053
C=1 H=2 S=2.5
1:2:2.5 ,multiply through by 2 ,to obtain whole numbera
2:4:5
Therefore the empirical formula is C2H4S5. Thus only gives the ratio
Molecular formula is the chemical formula .
(Empirical formula) n = molecular formula
(C2H4S5)n = molar mass
[(12×2) + ( 1×4) +(32×5)]n = 188.4
188n=188.4
n= 1
Molecular formula = (C2H4S5)×1
Therefore the chemical formula of
lenthionine is C2H4S5
A sample of hydrogen effuses through a porous container about 9 times faster than an unknown gas. Estimate the molar mass of the unknown gas.
Answer:
164 g/mol
Explanation:
According to Graham's law, the rate of effusion of a gas (r) is inversely proportional to the square root of its molar mass (M).
rH₂/rX = √[M(X)/ M(H₂)]
(rH₂/rX)² = M(X)/ M(H₂)
M(X) = (rH₂/rX)² × M(H₂)
M(X) = (9)² × 2.02 g/mol
M(X) = 164 g/mol
The molar mass of the unknown gas is 164 g/mol.
The molar mass of the unknown gas is 64 g/mol.
According to Graham's law:The rate of effusion of a gas (r) is inversely proportional to the square root of its molar mass (M).
It is given by:
[tex]rH_2/rX = \sqrt{[M(X)/ M(H_2)]} \\\\(rH_2/rX)^2 = M(X)/ M(H_2)\\\\M(X) = (rH_2/rX)^2 * M(H₂)\\\\M(X) = (9)^2 * 2.02 g/mol\\\\M(X) = 164 g/mol[/tex]
Thus, the molar mass of the unknown gas is 164 g/mol.
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Which would have a greater impact on the freezing point depression of ice, CrF3 or CaF2? Explain briefly.
Answer: [tex]CrF_3[/tex] has greater impact on the freezing point depression of ice.
Explanation:
Depression in freezing point is given by:
[tex]\Delta T_f=i\times K_f\times m[/tex]
[tex]\Delta T_f=T_f-T_f^0[/tex] = Depression in freezing point
i= vant hoff factor
[tex]K_f[/tex] = freezing point constant
m= molality
[tex]\Delta T_f=i\times K_f\times \frac{\text{mass of solute}}{\text{molar mass of solute}\times \text{weight of solvent in kg}}[/tex]
a) i = 4 for [tex]CrF_3[/tex] as it dissociates to give 4 ions in water.
[tex]CrF_3\rightarrow Cr^{3+}+3F^-[/tex]
b) i = 3 for [tex]CaF_2[/tex] as it dissociates to give 3 ions in water.
[tex]CaF_2\rightarrow Ca^{2+}+2F^-[/tex]
As the vant hoff factor is higher for [tex]CrF_3[/tex] , it has greater impact on the freezing point depression of ice.
Final answer:
CrF3 would have a greater impact on the freezing point depression of ice than CaF2 because it dissociates into more particles, four compared to CaF2 which dissociates into three.
Explanation:
In determining which substance would have a greater impact on the freezing point depression of ice, we need to consider the number of particles each substance releases into solution when dissolved. CrF3 will ionize to form one Cr3+ ion and three F- ions, for a total of four particles. Conversely, CaF2 will dissociate to form one Ca2+ and two F- ions, resulting in three particles. According to colligative properties, CrF3 will therefore have a greater effect on freezing point depression than CaF2 because it produces a larger number of dissociated particles.
A solution is prepared at that is initially in chlorous acid , a weak acid with , and in potassium chlorite . Calculate the pH of the solution. Round your answer to decimal places.
The question is incomplete, here is the complete question:
A solution is prepared at 25°C that is initially 0.075 M in chlorous acid [tex](HClO_2)[/tex] , a weak acid with [tex]K_a=1.1\times 10^{-2}[/tex], and 0.34 M in potassium chloride [tex](KClO_2)[/tex] . Calculate the pH of the solution. Round your answer to 2 decimal places.
Answer: The pH of the solution is 2.62
Explanation:
The chemical equation for the reaction of chlorous acid and potassium hydroxide follows:
[tex]HClO_2+KOH\rightarrow KClO_2+H_2O[/tex]
To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:
[tex]pH=pK_a+\log(\frac{[salt]}{[acid]})[/tex]
[tex]pH=pK_a+\log(\frac{[NaHCO_3]}{[H_2CO_3]})[/tex]
We are given:
[tex]pK_a[/tex] = negative logarithm of acid dissociation constant of chlorous acid = 1.96
[tex][KClO_2]=0.34M[/tex]
[tex][HClO_2]=0.075M[/tex]
pH = ?
Putting values in above equation, we get:
[tex]pH=1.96+\log(\frac{0.34}{0.075})\\\\pH=2.62[/tex]
Hence, the pH of the solution is 2.62
Under certain conditions, the substance hydrobromic acid can be broken down to form hydrogen and bromine.
If 25.0 grams of hydrobromic acid react to form 0.3 grams of hydrogen, how many grams of bromine must simultaneously be formed?
2. Under certain conditions, the substance ammonium chloride can be broken down to form ammonia and hydrogen chloride.
If 29.1 grams of ammonium chloride react to form 9.3 grams of ammonia, how many grams of hydrogen chloride must simultaneously be formed?
Answer:
1. 24.7g of Br2
2. 19.9g of HCl
Explanation:Please see attachment for explanation
When ammonia is mixed with hydrogen chloride (HCl), the white solid ammonium chloride (NH4Cl) is produced. Suppose 10.0 g ammonia is mixed with the same mass of hydrogen chloride. What substances will be present after the reaction has gone to completion, and what will their masses be?
Final answer:
After 10.0 g of ammonia reacts with 10.0 g of hydrogen chloride, 31.5 g of ammonium chloride is produced with leftover hydrogen chloride remaining.
Explanation:
When ammonia is mixed with hydrogen chloride (HCl), there is a chemical reaction in which ammonium chloride (NH4Cl) is formed. This reaction can be represented by the balanced chemical equation: NH3 (g) + HCl (g) → NH4Cl (s). The molar mass of ammonia (NH3) is approximately 17 g/mol and that of hydrogen chloride (HCl) is approximately 36.5 g/mol. When 10.0 g of each reactant is mixed, stoichiometry is used to determine the products and their masses after the reaction goes to completion.
Form the given masses, we can calculate the moles of each reactant: 10.0 g NH3 ÷ 17 g/mol ≈ 0.588 mol NH3 and 10.0 g HCl ÷ 36.5 g/mol ≈ 0.274 mol HCl. Since the reaction occurs in a 1:1 molar ratio, ammonia is the limiting reactant because we have less moles of it compared to moles of HCl. Therefore, all the ammonia will react with HCl to produce 0.588 mol of NH4Cl. The mass of the produced NH4Cl can be found by multiplying the moles of NH4Cl by its molar mass (approximately 53.5 g/mol), so 0.588 mol × 53.5 g/mol ≈ 31.5 g of NH4Cl. There will be no remaining ammonia. Since HCl is in excess, after the reaction the system will contain 31.5 g of NH4Cl and some unreacted HCl.
The "nitrogen rule" of mass spectrometry requires a compound containing an odd number of nitrogens to have an odd-mass molecular ion and a compound containing an even number of nitrogens to have an even-mass molecular ion.
What is the molecular formula of the CHN-containing compound imidazole, M+ = 68?
Answer:
C₃H₄N₂ is the molecular formula
Explanation:
The molecular ion peak is obtained at 68 units, which is the molecular mass of the compound. The even number of nitrogen atoms gives even mass of the molecule. In this problem, the mass is even so there is an even number of nitrogen atoms. It can be either two or four, from which it must be two nitrogen, as four nitrogen atoms will give higher mass. Now, we know that 28 out of 68 units belongs to nitrogen
The number of carbon atoms can also be determined in the same manner. Only one or two carbon atoms will leave much of the remaining units (28 or 16 units respectively) to hydrogen atoms. This leaves the option of three carbon atoms as any number higher than it will give molecular mass more than 68 units.
For hydrogen atoms, only 4 units are remaining this means four hydrogen atoms are present in the molecule. This is possible, if we consider that the compound is cyclic and contains two double bonds.
PS: If we consider two carbon atoms then there will be 16 units left which means sixteen hydrogen atoms. The valency of two carbon and two nitrogen atoms will only allow a maximum of eight hydrogen atoms in the molecule (giving only 60 units).
Imidazole, with a molecular ion of 68, has the molecular formula C3H4N2, determined through understanding the application of the nitrogen rule in mass spectrometry.
Explanation:The molecular formula for the compound imidazole, which contains carbon (C), hydrogen (H), and nitrogen (N), and has a molecular ion of 68, is C3H4N2. According to the nitrogen rule in mass spectrometry, nitrogen contributes either +1 or -1 to the mass, depending on whether the number of nitrogens in the compound is odd or even. Imidazole has an even number of nitrogens, hence the even-mass molecular ion of 68. The makeup of the remaining 68 mass units results from the carbon and hydrogen in the molecule: 3 carbons (each with a mass of 12, for a total of 36) and 4 hydrogens (each with a mass of 1, for a total of 4), and 2 nitrogens (each with a mass of 14, for a total of 28). So, the total mass (36 + 4 + 28) equals 68, aligning with the M+ value.
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value of Δ H ∘ rxn for the equation NH 3 ( g ) + 2 O 2 ( g ) ⟶ HNO3 ( g ) + H 2 O ( g )
Answer: - 894.6 kJ/mol.
Explanation:
Hess law is states that the changes in enthalpies in a chemical reactions is independent of the pathway between the initial and final states.
∆H is the change in the sum of the internal energy of a system.
We are to find the Value of ΔH°(rxn) for the equation:
NH3 (g) + 2 O2 (g) ⟶ HNO3 (g) + H2O (g). ----------------------------------(**).
From the series of equations given;
==> 4NH3 (g) + 5O2 (g) -------->4 NO(g) + 6H2O (l). ∆H = -1166.0 kJ/mol.--------------------------------------(1).
===> 2NO(g) + O2 (g) ------> 2NO2 (g). ∆H = -116.2 kJ/mol.---------------(2).
===> 3NO2 (g) + H2O (l) ---------> 2HNO3 (aq) + NO (g). ∆H = -137.3 kJ/mol.-------------------------------------(3).
The first thing to do is to multiply equation (2) by 3. Also, multiply equation (3) by 2. This will give us equation (4) and (5) respectively.
6NO + 3 O2 ----------------> 6NO2. ∆H= 3 × (-116.2 kJ/mol) = -348.6 kJ/mol. ------------------------------------(4).
6NO2 + 2 H2O ----------------> 4HNO3 + 2 NO. ∆H= 2 × (-137.3kj/mol) = -274.6 kJ/mol ---------------------------(5).
Next, add equations (4) and (5) to give;
4NO +3O2 +2H2O -------------> 4HNO3. ∆H = -623.2 kJ/mol. -----(6).
Add this equation to the equation (1) from above, we have;
4NH3 + 8O2 --------------> 4HNO3 + 4H2O. ∆H= -1789.2 kJ/mol. --------(7).
Then, divide the equation (7) above by 2 to give us back the equation (**).
NH3 (g) + 2 O2 (g) ⟶ HNO3 (g) + H2O (g). ∆H= -894.6 kJ/mol.
Δ H^∘ (rxn)= - 894.6 kJ/mol.
Sulfurous acid is a polyprotic acid. Write balanced chemical equations for the sequence of reactions that sulfurous acid can undergo when it's dissolved in water.
Answer:
The reactions are:
H₂SO₃ + H₂ O ⇄ HSO₃⁻ + H₃O⁺
HSO₃⁻ + H₂O ⇄ SO₃⁻² + H₃O⁺
Explanation:
This is the sulfurous acid → H₂SO₃.
It is a weak diprotic acid and it is diprotic because it can release 2 protons to increase H₃O⁺ from water. As every weak acid, the reactions are in equilibrium, that's why it has two Ka. As it can release protons, it is an acid from Bronsted Lowry and an acid from Arrhenius.
H₂SO₃ + H₂ O ⇄ HSO₃⁻ + H₃O⁺ Ka₁
HSO₃⁻ → this is called acid sulfite
HSO₃⁻ + H₂O ⇄ SO₃⁻² + H₃O⁺ Ka₂
2H₂O ⇄ H₃O⁺ + OH⁻ Kw
(Don't forget to put water equation, always)
Sulfurous acid, a diprotic acid, ionizes in two stages when dissolved in water. The first stage produces the hydrogen sulfite ion, HSO3¯, while the second stage further ionizes this ion to form the sulfite ion, SO3²-. The first ionization is more prominent than the second, reflecting the moderate strength of sulfurous acid.
Explanation:Sulfurous acid, being a polyprotic acid, ionizes in two stages when dissolved in water. In the first step, sulfurous acid gives up one proton (H⁺) to produce the hydrogen sulfite ion, HSO3¯:
H2SO3(aq) ⟶ H⁺(aq) + HSO3¯(aq)
In the second stage, the hydrogen sulfite ion can further ionize to give up another proton to form the sulfite ion, SO3²-:
HSO3¯(aq) ⟶ H⁺(aq) + SO3²-(aq)
It's important to note that the ionization is much more significant in the first stage than the second, due to sulfurous acid being a moderately strong acid.
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Student measured the absorbance of the substance X depending on its concentration in solution. She used 0.2 cm cuvette. Her calibration curve is a straight line with the slope 2.3·103 M-1. The extinction coefficient of substance X in this solution is:
Answer:
The extinction coefficient is 1.15 x 10⁴ M⁻¹.cm⁻¹ (value is not rounded off)
Explanation:
According to Lambert-Beer law
[tex]A = ε.b.C[/tex]
Here, A is absorbance, ε is extinction coefficient, b is the length of the cuvette and C is the molar concentration of substance X.
This equation is used for the relation between concentration and absorbance of electromagnetic radiation absorbing species. It is a linear equation and can be used for making a calibration curve, which is used for the analysis of an unknown concentration solution. The slope of this curve according to the equation is the product of extinction coefficient (M⁻¹.cm⁻¹) and the length of the cuvette in cm.
In this problem, the slope is provided, which can be mathematically represented as:
[tex]slope = ε.b[/tex]
[tex]2.3 X 10^{3}M^{-1} = ε.(0.2 cm)[/tex]
[tex]ε = 1.15 X 10^{4} M^{-1}.cm^{-1}[/tex] (not rounded off)
[tex]ε = 1 X 10^{4} M^{-1}.cm^{-1}[/tex] (rounded off)
A chemist added an excess of sodium sulfate to a solution of a soluble barium compound to precipitate all of the barium ion as barium sulfate, BaSO4. How many grams of barium ion are in a 441-mg sample of the barium compound if a solution of the sample gave 403 mg BaSO4 precipitate? What is the mass percentage of barium in the compound?
Answer : The mass percentage of barium in the compound is, 53.8 %
Explanation : Given,
Mass of barium compound = 441 mg
Mass of barium sulfate = 403 mg = 0.403 g (1 mg = 0.001 g)
The balanced chemical reaction will be:
[tex]Ba^{2+}(aq)+Na_2SO_4(aq)\rightarrow BaSO_4(s)+2Na^+(aq)[/tex]
First we have to calculate the moles of [tex]BaSO_4[/tex]
[tex]\text{Moles of }BaSO_4=\frac{\text{Mass of }BaSO_4}{\text{Molar mass of }BaSO_4}[/tex]
Molar mass of [tex]BaSO_4[/tex] = 233.38 g/mole
[tex]\text{Moles of }BaSO_4=\frac{0.403g}{233.38g/mole}=0.001727mole[/tex]
Now we have to calculate the moles of barium ion.
From the balanced chemical reaction, we conclude that
As, 1 mole of barium sulfate produced from 1 mole of barium ion
So, 0.001727 mole of barium sulfate produced from 0.001727 mole of barium ion
Now we have to calculate the mass of barium ion.
[tex]\text{ Mass of }Ba^{2+}=\text{ Moles of }Ba^{2+}\times \text{ Molar mass of }Ba^{2+}[/tex]
Molar mass of barium = 137.3 g/mol
[tex]\text{ Mass of }Ba^{2+}=(0.001727moles)\times (137.3g/mole)=0.2371g[/tex]
Now we convert the mass of barium ion from gram to mg.
Conversion used : (1 g = 1000 mg)
Mass of barium ion = 0.2371 g = 237.1 mg
Now we have to calculate the mass percentage of barium in the compound.
Mass percent of barium = [tex]\frac{237.1mg}{441mg}\times 100=53.8\%[/tex]
Thus, the mass percentage of barium in the compound is, 53.8 %
Some instant cold packs contain ammonium nitrate and a separate pouch of water. When the pack is activated by squeezing to break the water pouch, the ammonium nitrate dissolves in water and the pack gets cold. The heat of solution for ammonium nitrate is 25.4 kJ/mol.
a. Is the dissolution of ammonium nitrate endothermic or exothermic?
b. A cold pack contains 135.0 g of water and 50.0 g of ammonium nitrate. What will be the final temperature of the activated cold pack, if the initial temperature is 25.0 degree C? (Assume that the specific heat of the solution is the same as that for water, 4.184 J/g degree C and no heat is lost).
The dissolution of ammonium nitrate is an endothermic process. In terms of calculating the final temperature of an activated cold pack containing water and ammonium nitrate, you would use a specific formula with the given data. The details of this calculation were outlined in the detailed answer.
Explanation:a. The dissolution of ammonium nitrate is an endothermic process. This is because it absorbs heat from its surroundings in the process of dissolving, leading to a drop in temperature. Hence, endothermic reactions cool their surroundings. The given heat of solution for ammonium nitrate is positive (+25.4 kJ/mol), which is characteristic of endothermic reactions.
b. To calculate the final temperature of the activated cold pack, you would need to use the formula q = mcΔT, where q is the heat absorbed or released, m is the mass, c is the specific heat, and ΔT is the change in temperature (final-initial). Given data can be used to calculate the heat absorbed by the reaction, then using that value along with the given initial temperature and specific heat, we can find the final temperature. Please note that the calculations involved may require quite a few steps and it may be best to consult your chemistry teacher or textbook for the specific mathematical operation.
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Draw the structure with the molecular formula C6H14O that most likely produced the below IR and 1H NMR. The 13C NMR data is 78.3 ppm, 43.9 ppm, 27.3 ppm, 13.3 ppm.
Answer:
The answer is 2,3-dimethylbutan-2-ol and the structure is attached below.
Explanation:
Although we are not provided with ¹H-NMR spectrum and IR spectrum but still we can elucidate the ¹³C-NMR data and finalize a plausible structure.
First of all we look at the molecular formula, we can conclude from the formula that the structure given is saturated in nature because the hydrogen deficiency index of this formula is zero. Hence, we can say that there is no double bond either between Carbon atoms or between carbon and oxygen atom. This can also be proved by the absence of peaks in downfield as unsaturated compounds and carbonyl compounds give value above 100 and 200 ppm respectively.
Secondly, we can also conclude that among the six carbon atom two pairs of them are having same electronic environment because we are having only 4 signals hence we can conclude that two pairs have same chemical shift values.
Also, after making every possinble isomer of given molecular formula the structure of 2,3-dimethylbutan-2-ol was found to be the most accurate structure.
Despite the absence of certain spectra, analysis of the ¹³C-NMR data suggests a saturated structure. The most fitting isomer is identified as 2,3-dimethylbutan-2-ol based on chemical shift values.
While lacking specific data from the ¹H-NMR and IR spectra, an insightful elucidation of the ¹³C-NMR data allows for a plausible structural determination.
Examining the molecular formula, the absence of a hydrogen deficiency index suggests a saturated structure, ruling out double bonds or carbonyl groups. The lack of peaks in downfield further supports this, as unsaturated and carbonyl compounds typically exhibit values above 100 and 200 ppm, respectively.
Considering the six carbon atoms, the presence of only four signals indicates two pairs with identical chemical shift values, implying equivalent electronic environments.
Upon exploring various possible isomers for the given molecular formula, the most fitting structure emerged as 2,3-dimethylbutan-2-ol. This conclusion aligns with the observed NMR data and satisfies the saturation criteria.
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As an EPA scientist studying catalytic converters and urban smog, you want to find Kc for the following reaction: 2NO2(g) ⇋ N2(g) + 2 O2(g) Kc = ?. Use the following data to find the unknown Kc (1) ½ N2(g) + 1/2 O2(g) ⇋ NO(g) Kc = 4.8x10-10 (2) 2NO2(g) ⇋ 2NO(g) + O2(g) Kc = 1.1x10-5
Answer:
Kc = 4.774 * 10¹³
Explanation:
the desired reaction is
2 NO₂(g) ⇋ N₂(g) + 2 O₂(g)
Kc =[N₂]*[O₂]² /[NO₂]²
Since
1/2 N₂(g) + 1/2 O₂(g) ⇋ NO(g)
Kc₁= [NO]/(√[N₂]√[O₂]) → Kc₁²= [NO]²/([N₂][O₂])
and
2 NO₂(g) ⇋ 2 NO(g) + O₂(g)
Kc₂= [NO]²*[O₂]/[NO₂]² → 1/Kc₂= [NO₂]²/([NO]²[O₂])
then
Kc₁²* (1/Kc₂) = [NO]²/([N₂]*[O₂]) *[NO₂]²/([NO]²[O₂]) = [NO₂]²/([N₂]*[O₂]²) = 1/Kc
Kc₁² /Kc₂ = 1/Kc
Kc= Kc₂/Kc₁² =1.1*10⁻⁵/(4.8*10⁻¹⁰)² = 4.774 * 10¹³
Final answer:
To find the Kc for the reaction 2NO²(g) ⇌ N²(g) + 2O²(g), we reverse and manipulate the given equilibria, taking reciprocals and squaring as necessary, and then multiplying the constants to get the overall Kc.
Explanation:
To determine the equilibrium constant Kc for the given reaction 2NO²(g) ⇌ N²(g) + 2O²(g), we can manipulate the given equilibria (1) and (2) to yield the desired reaction.
First, we reverse the equilibrium (1), 1/2 N²(g) + 1/2 O²(g) ⇌ NO(g), whose Kc is 4.8 x 10-10. When an equation is reversed, the equilibrium constant becomes the reciprocal of the original. Thus, the new Kc for NO(g) ⇌ 1/2 N²(g) + 1/2 O²(g) is 1 / (4.8 x 10-10). To find the Kc for N²(g) + O²(g) ⇌ 2NO(g), we need to multiply the new reaction by two, which means squaring the Kc.
Second, we consider reaction (2), which is 2NO²(g) ⇌ 2NO(g) + O²(g), with a Kc of 1.1 x 10-5.
By adding the modified equilibrium (1) with the reaction (2), we get the desired equilibrium: 2NO(g) + O2(g) ⇌ 2NO2(g). We then multiply the Kc values of these individual steps to obtain the overall Kc for the desired reaction. This is because the equilibrium constant of the overall reaction is the product of the equilibrium constants of the individual steps.
In summary:
Reverse equilibrium (1) Kc: 1 / (4.8 x 10-10)Square this value to get Kc for N²(g) + O²(g) ⇌ 2NO(g)Use the Kc for 2NO²(g) ⇌ 2NO(g) + O²(g), which is 1.1 x 10-5Multiply these Kc values for the overall reaction KcThe solubility product of calcium fluoride (CaF2(s); fluorite) is 310-11 at 25C. Could a fluoride concentration of 1.0 mg L-1 be obtained in water that contains 200 mg L-1 of calcium?
The given question is incomplete. The complete question is as follows.
The solubility product of calcium fluoride () is at 25 degrees C. Will a fluoride concentration of 1.0 mg/L be soluble in a water containing 200 mg/L of calcium?
Explanation:
Reaction equation for the given chemical reaction is as follows.
[tex]CaF_{2} \rightleftharpoons Ca^{2+} + 2F^{-}[/tex]
Therefore, expression for [tex]K_{sp}[/tex] will be as follows.
[tex]K_{sp} = [Ca^{2+}][F^{-}]^{2}[/tex]
=
Also, moles of per liter = \frac{\text{mass of F^{-} per L}}{\text{molar mass of F}}[/tex]
= [tex]\frac{1.0 \times 10^{-3}}{19.0}[/tex]
= [tex]5.263 \times 10^{-5} mol[/tex]
Hence, [tex][F^{-}] = \frac{\text{moles of F^{-}}{volume}[/tex]
= [tex]\frac{5.263 \times 10^{-5}}{1}[/tex]
= M
Now, moles of per L = \frac{\text{mass of Ca^{2+} per L}}{\text{molar mass of Ca}}[/tex]
= [tex]\frac{200 \times 10^{-3}}{40.1}[/tex]
= M
Also, [tex][Ca^{2+}] = \frac{moles of Ca^{2+}}{volume}[/tex]
= [tex]\frac{4.988 \times 10^{-3}}{1}[/tex]
= M
Hence, ionic product =
= [tex](4.988 \times 10^{-3}) \times (5.263 \times 10^{-5})^{2}[/tex]
= [tex]1.38 \times 10^{-11}[/tex]
As, the ionic product is less than the [tex]K_{sp}[/tex], this means that the fluoride will be soluble in water containing the calcium.