The mass of the resulting strontium fluoride precipitate is 22710.4116 g.
To calculate the mass of the resulting strontium fluoride precipitate, we need to know the concentration of each solution and the stoichiometry of the reaction between them.
The balanced equation for the reaction between strontium nitrate and sodium fluoride is:
Sr(NO₃)₂ + NaF --> SrF₂ + 2NO₃-
The initial volume of each solution is given, and we can use stoichiometry to find the initial concentration of each ion in each solution.
First, we need to calculate the initial concentration of strontium ions in the strontium nitrate solution:
[Sr₂+] = [Sr(NO₃)₂] / [NO₃-]
= 2.731 m / 0.4803 mol / 1.8014 mol / 1.0000 mol
= 2.731 / 1.8014
= 0.1506 mol/L
Next, we need to calculate the initial concentration of fluoride ions in the sodium fluoride solution:
[F-] = [NaF] / [F-]
= 3.126 m / 0.010 mol / 0.010 mol / 1.0000 mol
= 3.126 / 0.010
= 31.26 mol/L
Finally, we can use stoichiometry to calculate the initial concentrations of the products and the initial volume of the reaction mixture.
The initial volume of the reaction mixture is the sum of the initial volumes of the two solutions:
V1 + V2 = 210.0 ml
The initial concentrations of the products can be calculated from the balanced equation:
[SrF₂] = [Sr₂+] * [F-] / [Sr(NO₃)₂]
= 0.1506 mol/L * 31.26 mol/L / 2.731 mol/L
= 0.0517 mol/L
[NO₃-] = [NO₃₋] - [Sr(NO₃)₂]
= 0.4803 mol / 2.731 mol / 1.8014 mol / 1.0000 mol
= 0.0143 mol/L
Next, we can use the initial concentrations of the products and the initial volume of the reaction mixture to calculate the initial concentrations of the reactants and the final volume of the reaction mixture.
[Sr₂+] = [SrF₂] + [NO₃-]
= 0.0517 mol/L + 0.0143 mol/L
= 0.0650 mol/L
[Sr(NO₃)₂] = [SrF₂] * [F-] / [Sr₂+]
= 0.0517 mol/L * 31.26 mol/L / 0.0650 mol/L
= 0.1734 mol/L
[NO₃-] = [NO₃-] - [Sr(NO₃)₂]
= 0.0143 mol/L - 0.1734 mol/L
= -0.1591 mol/L
The volume of the reaction mixture is the sum of the volumes of the two solutions:
V1 + V2 = 210.0 ml + 0.0143 mol/L
= 210.0 ml + 0.000143 L
= 210.00143 L
The volume of the reaction mixture is very small due to the large number of moles involved. Therefore, we can express the volume in terms of moles by dividing by the molar volume of each solvent:
V1 + V2 / (V1V2) = 210.00143 / (0.1506 mol/L * 0.0143 mol/L)
= 13074.2 mol/L
The mass of the resulting strontium fluoride precipitate can be calculated by multiplying the volume of the reaction mixture by the initial concentration of strontium ions:
Mass of precipitate = V1 * [Sr₂+]
= 13074.2 mol/L * 0.0650 mol/L
= 8804.4824 mol
The density of strontium fluoride is 2.50 g/mL, so the mass of the precipitate is:
Mass of precipitate = 8804.4824 mol * 2.50 g/mL
= 22710.4116 g
Therefore, the mass of the resulting strontium fluoride precipitate is 22710.4116 g.
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explain why the reaction of an alkyl halide with ammonia gives a low yield of primary amine.
The reaction of an alkyl halide with ammonia gives a low yield of primary amine because the reaction is a nucleophilic substitution reaction and ammonia is a weak nucleophile that is easily outcompeted by solvent molecules or other nucleophiles present in the reaction mixture.
Additionally, the reaction can also lead to the formation of secondary and tertiary amines as well as quaternary ammonium salts, which reduces the yield of primary amine.
The reaction of an alkyl halide with ammonia is a nucleophilic substitution reaction in which the ammonia molecule acts as a nucleophile, attacking the electrophilic carbon atom of the alkyl halide.
However, ammonia is a relatively weak nucleophile compared to other nucleophiles such as hydride ions or alkoxides. This makes the reaction slow and incomplete, resulting in a low yield of primary amine.
Furthermore, the reaction can lead to the formation of secondary and tertiary amines, as well as quaternary ammonium salts, depending on the structure of the alkyl halide and the reaction conditions. These byproducts can further reduce the yield of primary amine.
To increase the yield of primary amine, stronger nucleophiles such as lithium aluminum hydride or sodium azide can be used instead of ammonia, or the reaction conditions can be modified to favor the formation of primary amine.
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how is the molarity of a 0.50 m kbr solution changed when water is added to double its volume?
The molarity of the KBr solution would decrease from 0.50 M to 0.25 M after water is added to double its volume.
When water is added to double the volume of a 0.50 M KBr solution, the molarity of the solution decreases. Molarity is defined as the number of moles of solute per liter of solution.
When water is added, the total volume of the solution increases while the amount of solute (KBr) remains constant. Consequently, the concentration of KBr in the solution decreases.
Since molarity is a measure of concentration, doubling the volume while keeping the same amount of solute reduces the molarity by half. In this case, the molarity of the KBr solution would decrease from 0.50 M to 0.25 M after water is added to double its volume.
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After drawing the Lewis dot structure of HCN, pick the INCORRECT statement from the following. O A. The C-N bond is a double bond. OB. There is a lone pair of electrons on N. OC. There are no lone pairs on H. OD. The C-H bond is a single bond. O E. There are no lone pairs on
After drawing the Lewis dot structure of HCN, the incorrect statement from the options given is D, which says that the C-H bond is a single bond.
After drawing the Lewis dot structure of HCN, we can determine the bonding and nonbonding electron pairs around each atom in the molecule. In the Lewis dot structure of HCN, the carbon atom is in the center and is bonded to both the nitrogen and hydrogen atoms. The nitrogen atom has one lone pair of electrons, and there are no lone pairs on either the carbon or hydrogen atoms.
A. The C-N bond is a double bond. This statement is correct. In the Lewis dot structure of HCN, there are four valence electrons on the carbon atom and five valence electrons on the nitrogen atom. To form a stable molecule, the carbon and nitrogen atoms share two pairs of electrons, forming a double bond.
B. There is a lone pair of electrons on N. This statement is correct. As mentioned earlier, the nitrogen atom has one lone pair of electrons in the Lewis dot structure of HCN.
C. There are no lone pairs on H. This statement is correct. Hydrogen atoms only have one valence electron, which they share with another atom to form a bond. Therefore, there are no lone pairs on the hydrogen atom in HCN.
D. The C-H bond is a single bond. This statement is incorrect. In the Lewis dot structure of HCN, the carbon atom is bonded to the hydrogen atom by a triple bond, which consists of one sigma bond and two pi bonds.
E. There are no lone pairs on C. This statement is correct. In the Lewis dot structure of HCN, there are no lone pairs on the carbon atom.
Therefore, the incorrect statement from the options given is D, which says that the C-H bond is a single bond.
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The pH of a 1.1 M solution of acetic acid (HCH3CO2) is measured to be 2.35. Calculate the acid dissociation constant K, of acetic acid. Round your answer to 2 significant digits. K 8 a x10 X 5 ?
The acid dissociation constant K of acetic acid is approximately 1.8 x 10^-5.
To calculate the K value, first, we need to find the concentration of H+ ions, which can be found using the pH formula:
pH = -log10[H+]
Rearrange the formula to find [H+]:
[H+] = 10^(-pH)
[H+] = 10^(-2.35) ≈ 4.47 x 10^-3 M
Next, we can calculate the concentration of acetate ions (CH3COO-) by subtracting the [H+] from the initial concentration of acetic acid:
[CH3COO-] = 1.1 - 4.47 x 10^-3 ≈ 1.09553 M
Now, we can use the equilibrium expression for the dissociation of acetic acid:
K = [H+][CH3COO-] / [HCH3CO2]
Rearrange the formula to find K:
K = (4.47 x 10^-3)(1.09553) / (1.1 - 4.47 x 10^-3) ≈ 1.8 x 10^-5
Summary: The acid dissociation constant K of a 1.1 M solution of acetic acid with a pH of 2.35 is approximately 1.8 x 10^-5.
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what is the side bond that is of the greatest concern to the salon professional
As a salon professional, one of the biggest concerns is maintaining the side bond of hair extensions. The side bond refers to the bond between the hair extension and the natural hair. If the side bond is weak, the extensions may slip, causing discomfort and possible damage to the natural hair.
Additionally, a weak side bond can lead to the extensions falling out completely, which can be frustrating for clients and damaging to the salon's reputation. To maintain a strong side bond, it is essential to choose high-quality extensions and use a professional-grade bonding agent.
Proper installation techniques, including the use of the correct amount of bonding agent and careful placement of the extensions, can also help ensure a strong side bond. Regular maintenance appointments with a professional stylist can help identify and address any issues with the side bond before they become more significant problems. Overall, maintaining a strong side bond is essential to providing a professional, high-quality hair extension service.
The side bond of greatest concern to the salon professional is the disulfide bond. Disulfide bonds play a crucial role in determining the strength and structure of hair, and they are responsible for maintaining the curl pattern or straightness of hair. These bonds are sensitive to various salon treatments, such as chemical relaxers, perms, and certain coloring processes.
As a salon professional, it's essential to understand how disulfide bonds work to ensure the health of your clients' hair. When applying treatments that involve breaking or reforming disulfide bonds, you must be precise and follow the product instructions carefully. This helps to avoid any potential damage to the hair and maintain the desired results. Additionally, recommending proper aftercare and maintenance products to your clients can further protect their hair and preserve the integrity of the disulfide bonds.
In summary, the disulfide bond is the side bond that salon professionals should be most concerned with, as it plays a significant role in hair strength and structure. Proper knowledge and application of treatments involving these bonds are key to providing clients with the best possible results and maintaining healthy hair.
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g what type of interaction forms when two side chains containing an amino group and a carboxyl group are in close proximity? a. hydrophobic interactions b. hydrogen bond c. salt bridge d. disulfide bridge
The correct answer is c. salt bridge. The salt bridge is also known as an ionic bond or an ionic interaction.
When two side chains containing an amino group and a carboxyl group are in close proximity, they can form a salt bridge. A salt bridge is a type of interaction that occurs between an acidic group (carboxyl group) and a basic group (amino group) when they come close together. The carboxyl group, which has a negative charge at physiological pH, can attract and interact with the positively charged amino group. This electrostatic interaction between the opposite charges forms a salt bridge. The salt bridge is also known as an ionic bond or an ionic interaction.
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Part A Write a balanced equation for the dissociation of the following Brønsted-Lowry acid in water: H2SO4 Express your answer as a chemical equation. Identify all of the phases in your answer.
The dissociation of the Brønsted-Lowry acid, H2SO4, in water can be represented by the following balanced equation: H2SO4 (aq) + 2H2O (l) -> H3O+ (aq) + HSO4- (aq) The resulting ions are in a state of equilibrium in the solution, where the concentrations of the ions are determined by the acid dissociation constant (Ka) of H2SO4.
The equation, H2SO4 is the acid, and it donates two protons (H+) to the water molecules (H2O) to form hydronium ions (H3O+) and hydrogen sulfate ions (HSO4-). The H3O+ ions are the conjugate acid of water, and they are formed through the acceptance of a proton from the H2SO4. The HSO4- ions are the conjugate base of H2SO4, and they are formed through the loss of a proton. The phases in this equation are as follows: H2SO4 (aq) is the acid dissolved in water (l), H2O (l) is the solvent, H3O+ (aq) is the hydronium ion formed, and HSO4- (aq) is the hydrogen sulfate ion formed. Overall, the dissociation of H2SO4 in water is an example of an acid-base reaction, where the acid donates a proton (H+) to the water, and the water acts as a base by accepting the proton. The resulting ions are in a state of in the solution, where the concentrations of the ions are determined by the acid dissociation constant (Ka) of H2equilibrium SO4.
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how many peaks would be expected in your aldol product molecule if a 13 c nmr spectrum were taken? (2 pt
The number of peaks that would be expected in your aldol product molecule if a 13 c nmr spectrum were taken is 8.
What is Dibenzyl acetone?Dibenzyl acetone is an organic compound with the chemical formula C₂0H₂O. It is also known as 4,4'-dibenzylideneacetone or DBA. It is a yellow to brown colored solid and is insoluble in water but soluble in organic solvents such as ethanol and acetone.
Dibenzyl acetone is widely used in organic synthesis as a building block for the preparation of various organic compounds. It is used as a starting material in the synthesis of chiral ligands, pharmaceuticals, and agrochemicals. It is also used as a flavor and fragrance ingredient in the production of perfumes and cosmetics.
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How many peaks would be expected in your aldol product molecule if a 13C NMR spectrum were taken?
aldol product is dibenzyl acetone
