Final answer:
The concentration of Na+ in a 1.15 M Na2SO3 solution is 2.30 M, and the concentration of SO32- is 1.15 M, as Na2SO3 dissociates entirely into these ions in solution.
Explanation:
To calculate the concentration of all species in a 1.15 M Na2SO3 (sodium sulfite) solution, we need to consider the ionization of sulfurous acid (H2SO3), which isn't explicitly present but relates to the anion derived from Na2SO3. The ionization constants given are Ka1 = 1.4 × 10−2 for the first dissociation (H2SO3 to HSO3−) and Ka2 = 6.3 × 10−8 for the second dissociation (HSO3− to SO32−). Although, given the context, we aren't going through the complete step-by-step equilibrium calculation for Ka1 and Ka2, we understand that Na2SO3 dissociates completely in solution to Na+ and SO32−. Therefore, initially, the concentration of Na+ is 2.30 M (since each Na2SO3 unit yields two Na+ ions), and the concentration of SO32− is 1.15 M.
The concentrations in the 1.15 M Na2SO3 solution are approximately:
- [tex][H2SO3] = [HSO3^-] = 1.15 M,[/tex]
- [tex][SO3^2-] = 2.10 \times 10^-5 M,[/tex]
- [tex][Na^+] =1.15 M, and[/tex]
- [tex][OH^-] = 8.70 \times 10^-14 M.[/tex]
Given:
- Initial concentration of [tex]\(Na_2SO_3\) = 1.15 M[/tex]
- [tex]\(K_a1\)[/tex] for sulfurous acid [tex]= \(1.4 \times 10^{-2}\)[/tex]
- [tex]\(K_a2\)[/tex]for sulfurous acid [tex]= \(6.3 \times 10^{-8}\)[/tex]
We can assume that the ionization of [tex]\(Na_2SO_3\)[/tex] into [tex]\(HSO_3^-\)[/tex] and [tex]\(Na^+\)[/tex] is small due to the weak acid nature of sulfurous acid. So, the initial concentration of [tex]\(H_2SO_3\)[/tex] is approximately equal to 1.15 M.
Now, we approximate the concentration of [tex]\(H^+\)[/tex] ions to be approximately equal to the concentration of [tex]\(H_2SO_3\)[/tex], which is 1.15 M.
Using [tex]\(K_a2\)[/tex], we calculate the concentration of [tex]\(SO_3^{2-}\)[/tex] ions:
[tex]\[ [SO_3^{2-}] = \frac{[H^+]^2}{K_a2} \approx \frac{(1.15)^2}{6.3 \times 10^{-8}} \approx 2.10 \times 10^{-5} \, \text{M} \][/tex]
Since [tex]\(HSO_3^-\)[/tex] ions concentration is approximately equal to [tex]\(H^+\)[/tex] ions concentration and [tex]\(Na^+\)[/tex] concentration is the same as [tex]\(Na_2SO_3\)[/tex], we have:
- [tex]\([H_2SO_3] = [HSO_3^-] \approx 1.15 \, \text{M}\)[/tex]
- [tex]\([SO_3^{2-}] \approx 2.10 \times 10^{-5} \, \text{M}\)[/tex]
- [tex]\([Na^+] \approx 1.15 \, \text{M}\)[/tex]
The concentration of [tex]\(OH^-\)[/tex] ions can be calculated using the [tex]\(K_w\)[/tex]expression:
[tex]\[ [OH^-] = \frac{K_w}{[H^+]} \approx \frac{1.0 \times 10^{-14}}{1.15} \approx 8.70 \times 10^{-14} \, \text{M} \][/tex]
In a laboratory activity, the density of a sample of vanadium is determined to be 6.9 g/cm3 at room temperature. What is the percent error for the determined value? * 0.15% 0.87% 13% 15%
Answer:
13%
Explanation:
%error = (Experimental Value - Accepted Value) / Accepted Value * (100)
Exp = 6.9
Acc = 6.11
Answer:15%
Explanation:
Refer to table S and T for the accepted value of Vanadiums density (6.0 g/cm^3) and the formula for calculating percent error
How much water can be raised from 25*C (room temperature) to 37*C (body temperature) by adding the 2,000 kJ in a Snickers bar?
Answer:
m = 39834.3 g
Explanation:
Given data:
Mass of water raised = ?
Initial temperature = 25°C
Final temperature = 37°C
Energy added = 2000 Kj (2000 ×1000= 2000,000 j
Solution:
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
ΔT = T2 - T1
ΔT = 37°C - 25°C
ΔT = 12°C
c = 4.184 g/j.°C
Q = m.c. ΔT
2000,000j = m .4.184 g/j.°C. 12°C
2000,000j = m. 50.208 g/j
m = 2000,000j / 50.208 g/j
m = 39834.3 g
What practice will not help you make an accurate volume reading on a buret at the beginning of a titration?
A practice that will not help you make an accurate volume reading is a. Make sure that the meniscus starts exactly at 0.00 mL.
When reading a burette at the beginning of a titration, you do not need the meniscus to start exactly at 0.00ml.
All you need is:
to make sure the meniscus falls within the range that is marked for the experiment on the burette. to make sure you read the meniscus at eye levelto make sure you read the volume from the bottom of the meniscus not the topIn conclusion, the meniscus starting at 0.00 ml is irrelevant and will not contribute much to an accurate reading.
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Options for this question include:
a. Make sure that the meniscus starts exactly at 0.00 mL.
b. Make sure the meniscus falls within the marked range on the burette.
c. Read the volume with the meniscus at eye level.
d. Read the volume at the bottom of the meniscus.
The concentrations of Fe and K in a sample of riverwater are 0.0400 mg/kg and 1.30 mg/kg, respectively. Express the concentration in molality.
Answer :
The concentration in molality of Fe is, [tex]7.1\times 10^{-7}mol/kg[/tex]
The concentration in molality of K is, [tex]3.3\times 10^{-5}mol/kg[/tex]
Explanation:
First we have to calculate concentration in molality of Fe.
Molar mass of Fe = 56 g/mol
Concentration of Fe = 0.0400 mg/kg = [tex]4\times 10^{-5}g/kg[/tex]
Conversion used : 1 g = 1000 mg
[tex]\text{Concentration in molality}=7.1\times 10^{-7}mol/kg[/tex]
Thus, the concentration in molality of Fe is, [tex]7.1\times 10^{-7}mol/kg[/tex]
Now we have to calculate concentration in molality of K.
[tex]\text{Concentration in molality}=\frac{\text{Concentration of K}}{\text{Molar mass of K}}[/tex]
Molar mass of K = 39 g/mol
Concentration of K = 1.30 mg/kg = [tex]1.3\times 10^{-3}g/kg[/tex]
Conversion used : 1 g = 1000 mg
[tex]\text{Concentration in molality}=\frac{1.3\times 10^{-3}g/kg}{39g/mol}[/tex]
[tex]\text{Concentration in molality}=7.1\times 10^{-7}mol/kg[/tex]
Thus, the concentration in molality of K is, [tex]3.3\times 10^{-5}mol/kg[/tex]
To express the concentration of Fe and K from mg/kg to molality, convert their masses to grams, calculate the number of moles using their respective molar masses, and divide by the solvent's mass in kilograms.
Explanation:The question asks to convert the concentration of Fe and K from mg/kg to molality. Molality is a measure of the concentration of a solute in a solution in terms of the amount of substance in moles per kilogram of solvent. First, one needs to convert the mass of Fe and K from mg to g, which is simply dividing by 1000 as there are 1000 mg in a gram. Then, we find the molar mass of Fe (approximately 55.845 g/mol) and K (approximately 39.0983 g/mol) and calculate the number of moles for each.
Molality is then determined by dividing the number of moles of solute by the mass of the solvent in kilograms. Since water is the solvent and its density is typically close to 1 kg/L, for practical purposes, the mass of solvent is usually approximated as equal to the volume of the solution in liters (provided the solution concentration is relatively low, which it is in this case).
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Chromium reacts with oxygen according to the equation: 4Cr + 3O2 Cr2O3(s). Determine the moles of chromium(III) oxide produced when 4.58 mol of chromium is allowed to react.
Answer:
2.29 moles of Cr₂O₃ are produced
Explanation:
This is the reaction:
4 Cr + 3O₂ → 2Cr₂O₃
Ratio for this equation is 4:2, so 4 moles of chromium can produce the half of moles of chromium(III) oxide
4.58 mol of Cr may produce (4.58 .2)/4 = 2.29 moles of Cr₂O₃
The moles of chromium(III) oxide produced from the reaction of 4.58 mol of chromium with excess oxygen is 2.29 mol, based on the stoichiometry of the chemical equation.
Explanation:To determine the moles of chromium(III) oxide produced from 4.58 moles of chromium reacting with oxygen, we use the stoichiometry of the balanced chemical equation:
4 Cr(s) + 3 O2(g) → 2 Cr2O3(s)
According to this equation, 4 moles of chromium react completely with 3 moles of oxygen to produce 2 moles of chromium(III) oxide. We use this ratio to calculate the amount of chromium(III) oxide produced:
Moles of chromium(III) oxide = (Moles of chromium × Moles of chromium(III) oxide produced) / Moles of chromium reacted
= (4.58 mol Cr × 2 mol Cr2O3) / 4 mol Cr
= 2.29 mol Cr2O3
Therefore, when 4.58 mol of chromium are allowed to react with excess oxygen, 2.29 mol of chromium(III) oxide are produced.
Which phrase is the best description of an artificial satellite
A. A rocket that has a very small amount of thrust
B. A spacecraft that orbits a celestial body
C. A telescope that allows you to see distant planets
D. A spacecraft that can carry people to the Moon
Answer:
B
Explanation:
got it right on A P E X
What is the density, in g/mL, of a cube of lead (Pb) that weighs 0.371 kg and has a volume of 2.00 in.3
Answer:
11.32 g/mL
Explanation:
Given that:-
Mass = 0.371 kg = 371 g ( 1 kg = 1000 g)
Volume = 2.00 in³
The conversion of in³ to mL is as shown below:-
1 in³ = 16.3871 mL
So, Volume = [tex]2.00\times 16.3871\ mL[/tex] = 32.7741 mL
The expression for the calculation of density is shown below as:-
[tex]\rho=\frac{m}{V}[/tex]
Applying the values as:-
[tex]\rho=\frac{371\ g}{32.7741\ mL}=11.32\ g/mL[/tex]
Final answer:
To find the density of a lead cube weighing 0.371 kg and having a volume of 2.00 in.3, first convert the mass to grams and the volume to cm3, resulting in a density of approximately 11.3 g/cm3.
Explanation:
The question is asking for the density of a cube of lead (Pb) that has a mass of 0.371 kg and a volume of 2.00 in.3. First, it is necessary to convert the mass from kilograms to grams (since density is often expressed in g/mL) by multiplying the mass by 1000 (0.371 kg × 1000 = 371 g). Then, we need to convert the volume from cubic inches to cubic centimeters (cm3), as the requested density unit is g/mL and 1 cm3 is equivalent to 1 mL. Knowing that 1 in.3 = 16.387 cm3, we convert the volume of the lead cube to cm3 (2.00 in.3 × 16.387 = 32.774 cm3). Finally, to find the density in g/mL, we divide the mass in grams by the volume in mL (371 g / 32.774 cm3 = 11.3 g/cm3).
A 70.0 kg ancient statue lies at the bottom of the sea. Its volume is 30,000 cm3 (= 0.030 m3 ). How much force is needed to lift it? The mass density of seawater is sw = 1030 kg/m3 .
Answer:
383.18 N are required to lift the statue
Explanation:
Since the statue receives an upward buoyant force that follows the principle of Archimedes ( and is equal to the weight of displaced seawater due to the volume of the statue) , the net force required would be
net force = weight of the statue - upward buoyant force = m*g - ρsw * V *g =
(m- ρsw * V)*g
where
m= mass of the statue = 70.0 kg
V= volume of the statue = = 0.030 m³
g= gravity = 9.8 m/s²
ρsw = mass density of seawater = 1030 kg/m³
replacing values
net force = (m- ρsw * V)*g = (70.0 kg - 1030 kg/m³*0.030 m³)* 9.8 m/s² = 383.18 N
The amount of force needed to lift mass is mathematically given as
Net force= 383.18 N
What is the amount of force needed to lift mass?Question Parameter(s):
A 70.0 kg ancient statue lies at the bottom of the sea
Volume is 30,000 cm3
Where
net force = weight of the statue - upward buoyant force
Generally, the equation for the is mathematically given as
net force = m*g - ρsw * V *g
Therefore
net force = (m- ρsw * V)*g
net force= (70.0 kg - 1030 *0.030 )* 9.8
net force= 383.18 N
In conclusion, The net force is
Net force= 383.18 N
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The equation used to calculate the change in freezing point (ΔTf) of a substance is:_______
ΔTf = Kfm
where Kf is the freezing point depression constant and m is the molality of the solution. Which of the statements explains why molality is used instead of molarity in this equation?
A. Molality does not appear in many equations, so it is used here to distinguish this equation from other similar ones.
B. As the temperature of a solution changes, its volume will also change, which will affect its molarity but not its molality.
C. In solutions, moles are not directly related to grams and the freezing point of a solution is dependent solely on the number of grams of solute.
D. The equation was originally published with m as a typo, rather than M, but the values are close enough that the equation is still valid.
Answer:
B. As the temperature of a solution changes, its volume will also change, which will affect its molarity but not its molality.
Explanation:
Molality is given by the following equation:
[tex]Molality = \frac{moles of solute}{kg of solvent}[/tex]
While the molarity formula is given as
[tex]Molarity=\frac{moles of solute}{Lof solvent}[/tex]
The volume of solvent changes with temperature so it will be impractical to use molarity as it accounts for the volume of solution in its formula, which will create an error. Molality, on the other hand, uses Kg of solvent; which is not dependent on temperature. Hence, its value will not change
B. As the temperature of a solution changes, its volume will also change, which will affect its molarity but not its molality.
The following information should be considered:
The volume of solvent changes with temperature thus it should be be impractical for using the molarity because it accounts for the volume of solution in its formula, that develops an error. Molality, on the other hand, uses Kg of solvent; i.e. not dependent on temperature.Learn more: https://brainly.com/question/2386757?referrer=searchResults
The amino acid substitution of Val for Glu in Hemoglobin S results in aggregation of the protein because of __________ interactions between subunits.
Answer: Hydrophobic
Explanation:
The glutamic acid in the 6th position of beta chain of HbA is changed to valine in HbS. The substitution of hydrophilic glutamic acid by hydrophobic valine causes a sickness on the surface of the molecule.
This single nucleotide change polymerises hemoglobin molecules in the red blood cell.
So the hydrophobic nature of valine causes the aggregation of hemoglobin protein.
Water containing large amounts of Mg2+and Ca2+ions is said to be hard because it is hard to make soap lather in the water. True False
Answer:True
Explanation:
Water is said to be hard when it contains calcium ions or magnesium ions dissolved in it. These ions are able to react with soap in such a way that the soap is prevented from forming lather with the water. Hard water occurs when water passes over calcium or magnesium bearing minerals and dissolves some of it. Hardness due to the presence of calcium ions can easily be removed by boiling the water.
Which compound incorporates a polyatomic ion? View Available Hint(s) Which compound incorporates a polyatomic ion? CH2O Li2CO3 Na2O NF3
Answer:Li2CO3
(Co3)2- is the ion
A solution containing 20.0 g of an unknown liquid and 110.0 g water has a freezing point of .32 °C. Given Kf 1.86°C/m for water, the molar mass of the unknown liquid is________ g/mol A)256B) 69.0 C) 619 D) 78.1
Answer:
A. 256
Explanation:
In a solution where a liquid is the sovent, we'll use the van't Hoff factor, which is the ratio between the number of moles of particles produced in solution and the number of moles of solute dissolved, will be equal to 1.
ΔTemp.f = i * Kf * b
where,
ΔTemp.f = the freezing-point depression;
i = the van't Hoff factor
Kf = the cryoscopic constant of the solvent;
b = the molality of the solution.
So the freezing-point depression by definition is the difference between the the freezing point of the pure solvent and the freesing point of the solution.
Mathematically,
ΔTemp.f = Temp.f° - Temp.f
where,
Temp.f° = the freezing point of the pure solvent.
Temp.f = the freezin point of the solution.
Freezing point of pure water = 0°C
ΔTemp.f = 0 - (-1.32)
= 1.32°C
i = 1,
Kf = 1.86 °Ckg/mol
Solving for the molality, b = ΔTemp.f/( i * Kf)
= 1.32/(1*1.86)
= 0.71 mol/kg
Converting from mol/kg to mol/g,
0.71 mol/kg * 1kg/1000g
= 0.00071 mol/g.
Mass of solvent = 110g
Number of moles = mass * molality
= 0.00071 * 110
= 0.078 mol.
To calculate molar mass,
Molar mass (g/mol) = mass/number of moles
Mass of solute (liquid) = 20g
Molar mass = 20/0.078
= 256.2 g/mol
A solution containing 20.0 g of an unknown liquid (molar mass 256 g/mol) and 110.0 g water has a freezing point of -1.32 °C.
What is the freezing point depression?Freezing point depression is a colligative property observed in solutions that results from the introduction of solute molecules to a solvent.
Step 1: Calculate the molality of the solution.We will use the following expression for non-electrolytes.
ΔT = Kf × b
b = ΔT/Kf = 1.32 °C/(1.86 °C/m) = 0.710 m
where,
ΔT is the freezing point depression.Kf is the cryoscopic constant.b is the molality.Step 2. Calculate the molar mass of the unknown liquid (solute).We will use the definition of molality.
b = mass solute / molar mass solute × kg solvent
molar mass solute = mass solute / b × kg solvent
molar mass solute = 20.0 g / (0.710 mol/kg) × 0.1100 kg = 256 g/mol
A solution containing 20.0 g of an unknown liquid (molar mass 256 g/mol) and 110.0 g water has a freezing point of -1.32 °C.
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18 g of argon occupy 750 ml at a particular temperature and pressure. How many grams of methane would occupy the same volume at the same temperature and pressure?
Answer:
7.21 grams is the mass of methane
Explanation:
We may use the Ideal Gases Equation to solve this:
P. V = n. R. T
Let's determine the moles of Ar
18 g . 1 mol/ 39.9 g = 0.451 mol
In both situations, volume, temperature and pressure are the same so the moles of methane will also be the same as Argon's.
Let's convert the moles to mass of CH4.
0.451 mol . 16g/1mol = 7.21 grams
Final answer:
To determine how many grams of methane would occupy 750 ml at the same temperature and pressure as 18 g of argon, calculate the moles of argon and equate it to the moles of methane needed. The mass of methane is found using its molar mass, resulting in 7.2 g.
Explanation:
The question asks about the mass of methane that would occupy the same volume under the same conditions as a given mass of argon. To solve this, we can use the Ideal Gas Law, which is PV = nRT, where P is the pressure, V is the volume, T is the temperature, R is the gas constant, and n is the number of moles. Since temperature and pressure are constant, and assuming ideal conditions, the volume of a gas is directly proportional to the number of moles. The molar mass of argon (Ar) is approximately 40 g/mol, and the molar mass of methane (CH4) is approximately 16 g/mol.
Given 18 g of Ar, we first find the number of moles of Ar: moles of Ar = 18 g / (40 g/mol) = 0.45 moles. Assuming the same number of moles are required for methane to occupy the same volume at the same conditions, we can calculate the mass of CH4 required: mass of CH4 = 0.45 moles * (16 g/mol) = 7.2 g.
Therefore, 7.2 g of methane would occupy 750 ml at the same temperature and pressure as 18 g of argon.
A sample of potassium nitrate (49.0 g) is dissolved in 101 g of water at 100 °c with precautions taken to avoid evaporation of any water. The solution is cooled to 30 °c and a small amount of precipitate is observed. This solution is __________.a. hydrated b. saturated c. unsaturated d. supersaturated e. placated
Final answer:
The solution in question is saturated because the amount of KNO3 dissolved is less than the amount initially added at the given temperature.
Explanation:
Based on the information provided, the solution in question is a saturated solution. A saturated solution is one in which the maximum amount of solute has been dissolved in a given amount of solvent at a specific temperature. In this case, the solubility of potassium nitrate (KNO3) at 30 °C is approximately 48 g, which is less than the 80 g of KNO3 that was initially added to the solution. Therefore, the solution is saturated.
Heptane gas reacts with oxygen gas to give carbon dioxide gas and water vapor (gas). If you mix heptane and oxygen in the correct stoichiometric ratio, and if the total pressure of the mixture is 300 mm Hg, what are the partial pressures of heptane ( 25 mmHg) and oxygen ( 275 mm Hg)? If the temperature and volume do not change, what is the pressure of the water vapor ( 200 mm Hg) after reaction?
Answer:
Explanation:
Total pressure of the mixture = 300 mm Hg
equation of reaction
C₇ H₁₆(g) + 11 O₂ (g) → 7 CO₂(g) + 8 H₂O(g)
partial pressure of heptane = mole fraction heptane × total pressure = 1 / 12 × 300 mm Hg = 25 mm Hg
partial pressure of oxygen = mole fraction oxygen × total pressure = 11 / 12 × 300 mm Hg = 275 mm Hg
After the reaction
total number of mole before the reaction = 12
total number of mole after the reaction = 15
temperature and volume did not change
if 12 to 300 mm Hg
15 will be 15 × 300 / 12 = 375 mm Hg
partial pressure of water vapor = mole fraction of water vapor × 375 mm Hg = 8 / 15 × 375 mm Hg = 200 mm Hg
The pressure of the water vapor on the solution is 200mm Hg. The pressure exerted on the solution by the vapor is known as vapor pressure.
What is vapor pressure?The pressure exerted on the solution by the vapor is known as vapor pressure.
[tex]{P_{H_2O} = X_{H_2O}\times P_{sol}[/tex]
Where,
[tex]{P_{H_2O}[/tex] - Vapour pressure
[tex]X_{H_2O}[/tex] - Mole fraction of water = 8/15
[tex]P_{sol}[/tex] - Vapour Pressure of solution = 375 mm Hg
Put the values in the formula,
[tex]{P_{H_2O} = \dfrac 8 { 15} \times 375 { \rm \ mm Hg}\\{P_{H_2O} = 200 { \rm \ mm Hg}[/tex]
Therefore, the pressure of the water vapor on the solution is 200mm Hg.
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Proteins folded into pleated sheets or twisted into a helix(spiral) are considered to be _____ structure proteins.
Answer:
Proteins folded into pleated sheets or twisted into a helix(spiral) are considered to be _secondary_ structure proteins.
Explanation:
The secondary structure is characterized by the sequence of hydrogen connections in the peptide backbone among both amino hydrogen and carboxylic oxygen atoms. The secondary structure components usually form randomly as an intermediate until the protein pleats into its 3D tertiary framework.
Final answer:
Proteins with regions folded into pleated sheets or helices are considered to have a secondary structure, specifically in the forms of α-helices and β-pleated sheets, which are stabilized by hydrogen bonding and provide stability to the protein's conformation.
Explanation:
Proteins folded into pleated sheets or twisted into a helix (spiral) are considered to be secondary structure proteins. One common type of secondary structure is the α-helix, where the polypeptide chain forms a coiled spring-like structure stabilized by hydrogen bonds within the backbone of the chain. Another prevalent structure is the β-pleated sheet conformation, which is a flat, sheet-like formation where two or more polypeptide chains (or portions of the same chain) lie side by side, bonded together by hydrogen bonds. Such β-pleated sheets can be arranged in parallel or antiparallel configurations. These hydrogen bonds form between the oxygen atom in the carbonyl group of one amino acid and the hydrogen atom of another amino acid, which is typically four amino acids further along the chain. Both α-helices and β-pleated sheets are crucial for providing the protein with stability and are considered key elements of a protein's secondary structure.
18 mL of solvent was added to 82 mL of distilled water for a final solution volume of 100 ml. What is the percent concentration of solvent in the final solution?
Answer:
18 %
Explanation:
18 mL solvent per 100 mL solution
The specific branch of chemistry that focuses on molecules such as salts and water that constitute non-living matter, but are still important to living things, is termed ____________chemistry.
Answer:
Inorganic chemistry
Explanation:
Inorganic chemistry can be defined as the study of the composition and constituent of materials from non-biological origins, materials without carbon-hydrogen bonds such as: metals, salts, water and minerals.
The branch of chemistry dealing with the nonliving constituent, but important chemical to living matter, is inorganic chemistry.
Chemistry has been the branch of science that has been dealing with the chemical composition and structure of the compounds.
Branch of chemistryThe chemistry has been divided into various branch, based on the application of the studied in various field.
The branch of chemistry that has been dealing with the molecules constituent of the non-living matter has been inorganic chemistry.
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At 20°C the enzyme that catalyzes the synthesis of triglycerides functions at 1 reaction every 25 microseconds. If this enzyme catalyzes dehydration synthesis reactions for 450 microseconds, bonding fatty acids to glycerol, what is the maximum number of complete triglycerides that could be formed if no fatty acids were bonded to glycerol at the beginning of the reactions?
Answer:
6 number of complete triglycerides that could be formed .
Explanation:
In 25 microseconds ,single fatty acid attachment to glycerol takes place.
So, in 1 microseconds = [tex]\frac{1}{25}[/tex]
If the enzyme catalyzes dehydration synthesis reactions for 450 microseconds, then maximum numbers of attachments of fatty to glycerol will be:
[tex]\frac{1}{25}\times 450=18[/tex]
And each triglycerides has three fatty acid chains.So, number of triglycerides formed will be :
[tex]\frac{18}{3}=6[/tex]
6 number of complete triglycerides that could be formed if no fatty acids were bonded to glycerol at the beginning of the reactions
Given the reaction speed of the enzyme at 20°C, in a timeframe of 450 microseconds, the maximum number of triglycerides that can be formed is 18.
Explanation:At 20°C the enzyme that catalyzes the synthesis of triglycerides functions at a rate of 1 reaction every 25 microseconds. Hence, in a time frame of one microsecond, this enzyme can catalyze a maximum of 1/25 or 0.04 reactions. When this enzyme catalyzes synthesis reactions for 450 microseconds, the maximum number of reactions is 0.04 per microsecond * 450 microseconds, which is 18 reactions. Given that each reaction forms one complete triglyceride molecule, the maximum number of complete triglycerides that could be formed is 18.
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Hard water often contains dissolved Ca2 and Mg2 ions. One way to soften water is to add phosphates. The phosphate ion forms insoluble precipitates with calcium and magnesium ions, removing them from solution. A solution is 0.050 M in calcium chloride and 0.085 M in magnesium nitrate. What mass of sodium phosphate would you add to 1.5 L of this solution to completely eliminate the hard water ions.
The mass of sodium phosphate needed to completely eliminate the hard water ions from the solution is approximately 11.06 grams.
First, we need to determine the moles of calcium and magnesium ions present in the solution:
1. Moles of [tex]Ca^2+ ions[/tex] = Molarity of CaCl2 * Volume of solution
Moles of [tex]Ca^2+ ions[/tex] = 0.050 mol/L * 1.5 L = 0.075 mol
2. Moles of [tex]Mg^2+ ions[/tex] = Molarity of Mg(NO3)2 * Volume of solution
Moles of [tex]Mg^2+ ions[/tex] = 0.085 mol/L * 1.5 L = 0.1275 mol
Next, we need to find out the mole ratio of phosphate ions required to precipitate these ions. Since both calcium and magnesium ions form insoluble precipitates with phosphate ions in a 1:1 ratio, the moles of phosphate ions required will be equal to the sum of moles of calcium and magnesium ions:
Moles of phosphate ions = [tex]Moles of Ca^2+ ions + Moles of Mg^2+ ions[/tex]
Moles of phosphate ions = 0.075 mol + 0.1275 mol = 0.2025 mol
Now, we can calculate the mass of sodium phosphate required using its molar mass and the moles of phosphate ions:
Mass of sodium phosphate = Moles of phosphate ions * Molar mass of Na3PO4
Mass of sodium phosphate = 0.2025 mol * 163.94 g/mol = 33.18 g
However, sodium phosphate (Na3PO4) dissociates into three sodium ions [tex](Na^+)[/tex] and one phosphate ion[tex](PO4^3-)[/tex] . Therefore, to find the mass of the compound that would provide the required moles of phosphate ions, we need to adjust for this:
Mass of sodium phosphate = (33.18 g / 3) * 1 = 11.06 g
So, you would need approximately 11.06 grams of sodium phosphate to completely eliminate the hard water ions from the solution.
- Calculate the moles of [tex]Ca^2+ and Mg^2+[/tex] ions using their respective molarities and the volume of the solution.
- Determine the moles of phosphate ions required by summing the moles of [tex]Ca^2+ and Mg^2+[/tex]ions.
- Calculate the mass of sodium phosphate required using its molar mass and the moles of phosphate ions.
- Adjust the calculated mass for the fact that sodium phosphate dissociates into three sodium ions and one phosphate ion.
Complete Question:
Hard water often contains dissolved Ca2 and Mg2 ions. One way to soften water is to add phosphates. The phosphate ion forms insoluble precipitates with calcium and magnesium ions, removing them from solution. A solution is 0.050 M in calcium chloride and 0.085 M in magnesium nitrate. What mass of sodium phosphate would you add to 1.5 L of this solution to completely eliminate the hard water ions.
What hybridization is required for central atoms that have a trigonal planar arrangement of electron pairs? sp sp2 sp3
How many unhybridized p atomic orbitals are present when a central atom exhibits trigonal planar geometry?
Answer:
The required hybridization for a trigonal planar arrangement is sp2.There is one p atomic orbital unhybridized.Explanation:
In the sp2 hybridization, only two p atomic orbitals (out of three) are hybridized with the s orbital, thus forming a total of three sp2 orbitals.
These three orbitals will point in different directions to minimize the electron repulsion between them. When there are three such orbitals, the geometry that allows such minimization is the trigonal planar.
The unhybridized p orbital is found perpendicular to the plane of the 3 sp2 orbitals.
Ethanol has a heat of vaporization of 38.56kj/mol and a normal boiling point of 78.4 ∘c.
This is an incomplete question, here is a complete question.
Ethanol has a heat of vaporization of 38.56 kJ/mol and a normal boiling point of 78.4 °C. What is the vapor pressure of ethanol at 14 °C?
Answer : The vapor pressure of ethanol at [tex]14.0^oC[/tex] is [tex]5.174\times 10^{-2}atm[/tex]
Explanation :
The Clausius- Clapeyron equation is :
[tex]\ln (\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}\times (\frac{1}{T_1}-\frac{1}{T_2})[/tex]
where,
[tex]P_1[/tex] = vapor pressure of ethanol at [tex]14.0^oC[/tex] = ?
[tex]P_2[/tex] = vapor pressure of ethanol at normal boiling point = 1 atm
[tex]T_1[/tex] = temperature of ethanol = [tex]14.0^oC=273+14.0=287K[/tex]
[tex]T_2[/tex] = normal boiling point of ethanol = [tex]78.4^oC=273+78.4=351.4K[/tex]
[tex]\Delta H_{vap}[/tex] = heat of vaporization = 38.56 kJ/mole = 38560 J/mole
R = universal constant = 8.314 J/K.mole
Now put all the given values in the above formula, we get:
[tex]\ln (\frac{1atm}{P_1})=\frac{38560J/mole}{8.314J/K.mole}\times (\frac{1}{287K}-\frac{1}{351.4K})[/tex]
[tex]P_1=5.174\times 10^{-2}atm[/tex]
Hence, the vapor pressure of ethanol at [tex]14.0^oC[/tex] is [tex]5.174\times 10^{-2}atm[/tex]
Using the Clausius-Clapeyron equation and the given values, the vapor pressure of ethanol at 19 °C is approximately 6.94 kPa.
To determine the vapor pressure of ethanol at 19 °C, we can use the Clausius-Clapeyron equation, which relates the temperature and pressure of a substance.
Clausius-Clapeyron equation: [tex]\[ \ln\left(\frac{P_2}{P_1}\right) = -\frac{\Delta H_{\text{vap}}}{R} \times \left(\frac{1}{T_2} - \frac{1}{T_1}\right) \][/tex]
Where:
[tex]\[ \Delta H_{\text{vap}} = \text{Enthalpy of vaporization (38.56 kJ/mol = 38,560 J/mol)} \][/tex][tex]\[ R = \text{Universal gas constant (8.314 J/mol K)} \][/tex]T₁ = Initial temperature in Kelvin (78.4°C = 351.55 K)T₂ = Final temperature in Kelvin (19°C = 292.15 K)P₁ = Vapor pressure at T₁ (1 atm = 101.3 kPa)P₂ = Vapor pressure at T₂ (unknown)We can rearrange and solve for P₂:
[tex]\[ \ln\left(\frac{P_2}{101.3}\right) = -\frac{38,560}{8.314} \times \left(\frac{1}{292.15} - \frac{1}{351.55}\right) \][/tex]
[tex]\[ \ln\left(\frac{P_2}{101.3}\right) = -\frac{38,560}{8.314} \times (0.003423 - 0.002845) \][/tex]
[tex]\[ \ln\left(\frac{P_2}{101.3}\right) = -\frac{38,560}{8.314} \times 0.000578 \][/tex]
[tex]\[ \ln\left(\frac{P_2}{101.3}\right) = -2.679 \][/tex]
[tex]\[ \frac{P_2}{101.3} = e^{-2.679} \][/tex]
[tex]\[ \frac{P_2}{101.3} = 0.0685 \][/tex]
[tex]\[ P_2 = 0.0685 \times 101.3 \][/tex]
[tex]\[ P_2 \approx 6.94 \text{ kPa} \][/tex]
A 700 g peregrine falcon dives toward the ground from a height of 80 m and has a kinetic energy of 2,835 J. What is its speed?
Answer:
Kinetic energy is gven by
[tex]Kinetic Energy = \frac{1}{2} mv^{2}[/tex]
Hence the velocity can be found by [tex]v^{2} = \frac{2KE}{m}[/tex]
or [tex]\sqrt{\frac{2(2835J)}{0.7Kg} }[/tex] = [tex]\sqrt{8100\frac{m^{2} }{s^{2} } }[/tex] = 90m/s
Explanation:
The kinetic energy is the energy due to motion and is given by the kinetic energy equation KE = 1/2mv^2
It is the energy required to stop a body in motion and as seen from the equation, objects with large speed or mass have larger amount of kinetic energy or a large object effect can be made milder by lowering its speed so also a small object can increase its effect, for example a bullet by increasing its speed
A formula for a cough syrup contains 60 mg of codeine per fluid ounce. How many mg are contained in one teaspoonful?
Answer : The amount of codeine in one teaspoonful is, 30 mg
Explanation : Given,
Amount of codeine per fluid ounce = 60 mg
Now we have to determine the amount of codeine present in one teaspoonful.
As we know that:
1 teaspoonful = 0.5 fluid ounce
As, the amount of codeine per fluid ounce = 60 mg
So, the amount of codeine 0.5 fluid ounce = [tex]\frac{0.5}{1}\times 60mg=30mg[/tex]
Thus, the amount of codeine in one teaspoonful is, 30 mg
What is the trivial solution for a set of homogeneous equations? When is the solution not trivial. Give examples.
1. The sodium-iodide symporter plays a role in the accumulation of iodide in the thyroid gland. Here, one iodide gets converted to one iodine, which is utilized for the formation of either of the two types of thyroid hormones, T3 and T4. T3 and T4 are named after the number of iodines found in each of these hormones. To produce a single molecule of T3, a total of ____________ sodium ions must move down their concentration gradients by secondary active transport. The movement of iodide ions occurs in the _________________ direction as sodium ions.
Answer:(1) 6 sodium ions
(2)The movement of iodide ions occurs in the same direction as the sodium ions.
Explanation: Sodium-iodide symporter actively transports 2 sodium ions together with one iodide ions across the basement membrane into the thyroid follicular cells.(therefore for production of a single molecule of T3 3×2=6 sodium ions)This system utilises the concentration of sodium ions so that iodide ions can move against its concentration gradient.
Consider the chemical reaction described by the following equation.
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)
ΔH = −58 kJ
What would be the enthalpy, in kilojoules, for the reaction
2NaCl(aq) + 2H2O(l) → 2HCl(aq) + 2NaOH(aq)
ΔH = ?
Answer:
ΔH = + 116 kJ
Explanation:
The equation for the reaction is given as;
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) ΔH = - 58 kJ
If we multiply the above equation all through with (2); we have:
2 × (HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) ΔH = - 58 kJ)
2HCl(aq) + 2NaOH(aq) → 2NaCl(aq) + 2H2O(l) ΔH = - 116 kJ
If we tend to reverse the above equation; we have
2NaCl(aq) + 2H2O(l) → 2HCl(aq) + 2NaOH(aq) ΔH = + 116 kJ
∴ The reaction is said to be endothermin , as 116 kJ are absorbed.
The enthalpy change for the given reaction can be used to determine the enthalpy change for a related reaction by using the additive property.
The enthalpy change (ΔH) for a chemical reaction is a measure of the heat energy transferred during the reaction. In this case, the enthalpy change for the given reaction is -58 kJ. To find the enthalpy change for the reaction 2NaCl(aq) + 2H2O(l) → 2HCl(aq) + 2NaOH(aq), we can use the fact that the enthalpy change is additive. Since the reaction is being doubled, the enthalpy change will also be doubled. Therefore, the enthalpy change for this reaction would be -116 kJ.
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Hydrogen bonds are too weak to bind atoms together to form molecules, but they do hold different parts of a single large molecule in a specific three-dimensional shape. Hydrogen bonds are too weak to bind atoms together to form molecules, but they do hold different parts of a single large molecule in a specific three-dimensional shape. True False
Answer: True
Explanation:
Hydrogen bonds are special type of dipole dipole forces which are formed when hydrogen atom bonds with an electronegative element.
This important property of hydrogen bond occurs in polar molecules such as water which contains partial negative charges at one region of a molecule and also a partial positive charge elsewhere in the molecule.
Hydrogen bonds are weak and easily broken but when many hydrogen bonds are present, they are very strong.
Hydrogen bond is present in macromolecules such as DNA which holds the strands together.
Hydrogen bonds can bind atoms together to form molecules and hold different parts of a large molecule in a specific shape.
False
Explanation:False
Hydrogen bonds can actually be quite strong and are able to bind atoms together to form molecules. They are responsible for holding different parts of a single large molecule in a specific three-dimensional shape, which is important for the molecule's function. For example, in the structure of DNA, hydrogen bonds between complementary base pairs hold the two strands together.
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Process found in both photosynthesis and cellular respiration
a. Glycolysis
b. Krebs Cycle (citric acid cycle)
c. Calvin Cycle (light-independent)
d. Light-dependent reaction
e. Chemiosmosis
Chemiosmosis is the process found in both photosynthesis and cellular respiration, contributing to ATP synthesis in both processes.
The process found in both photosynthesis and cellular respiration is chemiosmosis. Both photosynthesis and cellular respiration involve multiple stages where energy is transformed and transferred within the cell. In photosynthesis, chemiosmosis occurs during the light-dependent reactions where ATP is synthesized using the energy from sunlight. In cellular respiration, chemiosmosis takes place during the electron transport chain, utilizing energy released from electrons to pump protons and create ATP. These processes reflect the interdependent nature of photosynthesis and cellular respiration, where the products of one are the reactants for the other, ultimately maintaining the balance of energy and matter in biological systems.