Final answer:
The empirical formula for this compound is CH₂O.
Explanation:
To find the empirical formula of a compound with given mass percentages, we first assume a sample size of 100 grams. This makes it easy to convert mass percent to grams directly. For the compound containing 40.0% C, 6.71% H, and 53.28% O, we would have 40.0 grams of carbon, 6.71 grams of hydrogen, and 53.28 grams of oxygen.
Next, we convert these masses to moles using the molar mass of each element (Carbon: 12.01 g/mol, Hydrogen: 1.008 g/mol, Oxygen: 16.00 g/mol).
Carbon: 40.0 g ÷ 12.01 g/mol = 3.33 moles of CHydrogen: 6.71 g ÷ 1.008 g/mol = 6.66 moles of HOxygen: 53.28 g ÷ 16.00 g/mol = 3.33 moles of OTo determine the simplest integer ratio of the elements, divide the moles of each element by the smallest number of moles calculated. In this case, all values come down to 1, which gives us the simple ratio of 1:2:1. Thus, the empirical formula is CH₂O.
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 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.
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.
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).
Which of the following are true about balanced chemical reactions? (Select all that apply)
a
Atoms of each element are equal on both sides.
b
All chemical formulas have a coefficient greater than one.
c
It is better to change subscripts to balance equations.
d
Coefficients are added to balance the equation.
e
Chemical equations must be balanced to satisfy the Law of Conservation of Mass.
f
Balanced chemical equations help to determine the number of moles needed in the reaction.
Answer: D. There must be as equal number of atoms of each element on both sides of the equation.
Explanation: I just took the test on Plato and this is correct
1. Where can you find safety data sheets (MSDS) for the chemicals used in the lab ?
2. Which region in the IR spectrum could be used to distinguish between butanoic acid and 2-butanone?
(a) 3200-3600 cm⁻¹
(b) 1600 cm⁻¹
(c) 1680-1750 cm⁻¹
(d) 2500-3300 cm⁻¹
Answer:
Option d
Explanation:
Safety data sheets (MSDS) contains information regarding safe handling practices of hazardous properties of chemicals used, remedies on exposure.
By rule, MSDS should be provided by suppliers of the chemicals.
physical and chemical properties, technical data, trade and common names, hazards, remedies on exposure and safe handling practices on chemicals to users and those involved in their handling and transportation.
2-butanone is a ketone having carbonyl group (C=O) whereas butanoic acid is a carboxylic acid.
IR frequency of ketonic C=O is 1750 – 1680 cm⁻¹.
Butanoic acid also has C=O group but peak is observed at 3000 – 2500 cm⁻¹.which corresponds to Carboxylic Acid O-H Stretch. Therefore, if frequency 3000 – 2500 cm⁻¹. frequency can be used to distinguish between between butanoic acid and 2-butanone.
Therefore, among given, option d is correct.
Safety Data Sheets (SDS) can be found at http://www.msds.com. The IR spectrum region 1680-1750 cm⁻¹ distinguishes between butanoic acid and 2-butanone through different behavior of their carbonyl groups.
Explanation:Locating Safety Data Sheets and Distinguishing IR Spectrum Regions
Safety Data Sheets (SDS), formerly known as Material Safety Data Sheets (MSDS), for chemicals used in the lab can often be found on dedicated websites like http://www.msds.com or directly from the chemical manufacturer's website. These sheets are crucial for understanding the handling, risks, and disposal procedures of chemicals.
To distinguish between butanoic acid and 2-butanone using infrared (IR) spectroscopy, you would look at differing absorbance regions that are characteristic of their functional groups. Butanoic acid has a carboxyl group that typically shows absorbance in the O-H stretch region (2500-3300 cm⁻¹) and the C=O stretch of acids (1700-1750 cm⁻¹). In contrast, 2-butanone has a carbonyl group that absorbs in the C=O stretch region (1680-1750 cm⁻¹), but without the broad O-H stretching band. Thus, the region that could be used to distinguish between butanoic acid and 2-butanone is option (c) 1680-1750 cm⁻¹, which captures the different behavior of the carbonyl group in each compound.
What volume (in mL) of 6 M acetic acid would have to be added to 500mL of a solution of 0.20M sodium acetate in order to achieve a pH = 5.0? The pKa of acetic acid is 4.75.
Answer:
9.3 mL of 6 M acetic acid needs to be added to 500 mL of a solution of 0.20 M sodium acetate to a achieve a pH of 5.0
Explanation:
This problem can be solved by the Henderson-Hasselbalch equation. It's formula is:
[tex]pH=pKa+log(\frac{[CH_3COO^-]}{[CH_3COOH]})[/tex]
The molar concentration can be replaced by the moles of the solute as the volume of the buffer will be the same for both species
[tex]pH=pKa+log(\frac{n_{CH_3COO^-}}{n_{CH_3COOH}})[/tex]
Placing the given data:
[tex]5.0=4.75+log(\frac{0.1}{n_{CH_3COOH}})[/tex]
[tex]n_{CH_3COOH}=(\frac{0.1}{10^{5.0-4.75}})\\\\n_{CH_3COOH}=(\frac{0.10}{1.778})\\\\ n_{CH_3COOH}=0.056moles[/tex]
The volume required to obtain the above-calculated moles can be determined by the molarity of acetic acid
[tex]M_{CH_3COOH}=\frac{n_{CH_3COOH}}{V_{CH_3COOH}(L)}\\\\V_{CH_3COOH}=\frac{n_{CH_3COOH}}{M_{CH_3COOH}}\\\\V_{CH_3COOH}=\frac{0.056}{6.0}\\\\V_{CH_3COOH}=0.0093L\\\\or\\\\V_{CH_3COOH}=9.3mL[/tex]
The above volume of acetic acid can be added to 500 mL of acetate to form 5.0 pH buffer
The volume of 6 M acetic acid needed is (0.356/6)x(500+V(A-)).
Explanation:To calculate the volume of 6 M acetic acid needed to achieve a pH of 5.0, we need to use the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
In this case, the ratio of sodium acetate ([A-]) to acetic acid ([HA]) should be 1:1 to achieve a pH of 5.0.
Using the equation, we can rearrange it to find the concentration of acetic acid:
pH = pKa + log([A-]/[HA])
[HA] = [A-] x 10^(pH - pKa)
Plugging in the values, we get:
[HA] = 0.20 M x 10^(5.0 - 4.75) = 0.20 M x 10^0.25 = 0.20 M x 1.78 = 0.356 M
Now, we can calculate the volume of 6 M acetic acid needed:
[HA] x V(HA) = [HA] x V(A-)
0.356 M x V(HA) = 0.356 M x (500 mL + V(A-))
V(HA) = 0.356 M x (500 mL + V(A-)) / 6 M
V(HA) = (0.356/6)x(500+V(A-)) mL
Since the ratio of sodium acetate to acetic acid is 1:1, the volume of 6 M acetic acid needed is equal to the volume of 0.20 M sodium acetate added to the solution. Therefore, the volume of 6 M acetic acid needed is (0.356/6)x(500+V(A-)).
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Arrange the following alkyl bromides in order from most reactive to least reactive in an SN₂ reaction: 1-bromo-2-methylbutane, 1-bromo-3-methylbutane, 2-bromo-2-methylbutane, and 1-bromopentane.
Answer:
1- bromopentane 1- bromo-3-methylbutane 1-bromo2-methylbutane 2-bromo-2-methylbutane
Explanation:
SN square reaction is a concentrated reaction. All the bond making and bond braking occur in one step
The nucleophile attack the back side of the carbon that bears the halide and replace it
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.
All methods of chromatography operate on the same basic principle that Select one: a. one component of the mixture will chemically react with the mobile phase b. the one component of the mixture will be completely insoluble in the mobile phase c. the components of the mixture will destribute unequally between mobile and stationary phase d. one component of the mixture will chemically react with the stationary phase
Answer:C
Explanation:
Chromatography separates compounds by taking advantage of their polarity. The stationary phase is generally very polar. The mobile phase can be pure hexane or various ratios of hexane with a polar eluent added. The more polar the compound, the more it interacts with the stationary phase and won’t move very far up the plate compared to the non-polar or less polar compounds that interact more with the non-polar hexane.
Final answer:
All methods of chromatography operate based on the principle that components of a mixture will distribute unequally between the mobile and stationary phase, due to differences in intermolecular forces and affinities.
Explanation:
The question asks about the basic principle on which all methods of chromatography operate. The correct answer is c. The components of the mixture will distribute unequally between the mobile and stationary phases. In chromatography, the separation of components is achieved because different components have different affinities for the stationary and mobile phases. A component with a higher affinity for the stationary phase will move more slowly through the chromatography system compared to a component with a higher affinity for the mobile phase. This differing affinity is often due to differences in the strength of intermolecular forces between the components and the phases. For example, in liquid chromatography, if a compound has strong intermolecular forces with the stationary phase (e.g., through hydrogen bonding or van der Waals forces), it will tend to be 'adsorbed' onto the stationary phase and thus move through the system more slowly compared to compounds that have weaker interactions and therefore travel with the mobile phase.
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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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.
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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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
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 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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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.
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.
What is the density of a piece of aluminum that has a mass of 270 grams and a volume of 100.0 cubic centimeters? If this piece was cut in half what would be the density of one of the pieces?
The density of the original piece of aluminum is 2.7 grams per cubic centimeter. If one piece is cut in half, the density of each piece would remain the same.
Explanation:In order to find the density of an object, you divide the mass of the object by its volume. The mass of the aluminum piece is 270 grams and its volume is 100.0 cubic centimeters. So, the density would be 270 grams divided by 100.0 cubic centimeters, which equals 2.7 grams per cubic centimeter.
If the piece of aluminum is cut in half, the mass of each piece would be half of the original mass, so each piece would have a mass of 135 grams. Since the volume of each piece remains the same, 100.0 cubic centimeters, the density of one of the pieces would still be 2.7 grams per cubic centimeter.
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What is the mass of water necessary to generate 11.2 L of hydrogen gas if calcium metal reacts with water at standard temperature and pressure (STP)?
Answer:
The mass of water is 18 g
Explanation:
The reaction of calcium with water can be represented in the equation below:
Ca + 2H₂O --------->Ca(OH)₂ +H₂
1 Mole of gas at STP = 22.4L
From the displacement reaction above, calculate the mass of water that will produce 22.4L of hydrogen gas at STP.
Mass of 2H₂O = 2(2x1 + 16) = 2X18 = 36 g/mol
Using proportional analysis;
36 g of 2H₂O produced 22.4 L of H₂, then
what mass of 2H₂O will produce 11.2L of H₂ ?
Mathematically,
22.4 L ----------------------------------> 36g
11.2 L -----------------------------------> ?
Cross and multiply, to obtain the expression below
= (11.2 X 36)/22.4
= 18 g
Therefore, the mass of water is 18 g
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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.
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.
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
In most chemical reactions the amount of product obtained is
Answer:
called theoretic yield
In most chemical reactions, the amount of product obtained is typically less than the theoretical yield due to various reasons such as incomplete reactions, loss of product during isolation, or the reversible nature of some reactions.
Explanation:In most chemical reactions, the amount of product obtained is usually less than the theoretical yield which is the amount predicted by a stoichiometric calculation based on the number of moles of all reactants present. This is because some reactants may not react to form the product, some products may be lost during the isolation step, or the reaction may not go to completion because it is reversible. For instance, if you mix 10 grams of hydrogen with 10 grams of oxygen in a sealed container, you won't get 20 grams of water, but less, because not all the hydrogen and oxygen will react.
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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
What is the trivial solution for a set of homogeneous equations? When is the solution not trivial. Give examples.
With respect to the categories of assets, liabilities, and stockholders' equity presented on the balance sheet (statement of financial position), what are U.S. GAAP and IFRS differences?
Explanation:
GAAP is a generally accepted accounting principle in U.S. it refers to common sets of accepted accounting principle, standards, procedures that the companies and its accountants must follow in order to compile their financial statement.
IFRS are sets of international accounting standards That specify how the financial statements will disclose different types of transactions and other activities. The International Accounting Standards Board (IASB) issues IFRS which defines precisely how accountants are required to maintain and record their accounts. In an attempt to have an universal accounting system, IFRS was developed so that business and accounts can be interpreted from industry to industry, and country to country.
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.
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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Be sure to answer all parts. In winemaking, the sugars in grapes undergo fermentation by yeast to yield CH3CH2OH (ethanol) and CO2. During cellular respiration, sugar and ethanol are "burned" to water vapor and CO2. (a) Using C6H12O6 for sugar, calculate ΔH o rxn of fermentation and of respiration (combustion). Fermentation = kJ Respiration = kJ (b) Write a combustion reaction for ethanol. Include the physical states of each reactant and product. (c) Which releases more heat from combustion per mole of C, sugar or ethanol?
a) ΔH of Fermentation = - 2816 kJ/mol
ΔH of Respiration = - 1409.2 kJ/mol
b)C₂H₅OH (l) + 3O₂ (g) → 3H₂O(g) + 2CO₂(g)
c) Combustion per mole of sugar
Let's solve for each part one by one:
a)
The fermentation of sugar is given as follows:C₆H₁₂O₆ (s) + 6O₂ (g) → 6O₂ (g) + 6H₂O (l)
ΔHrxn = ΔHformation (products) - ΔHformation (reactants)
[tex]= (6 mol * -393.5kJ/mol)+ (6mol * -285.8kJ/mol) - (1 mol * -1260kJ/mol + 6mol * 0)\\\\= -2 816 kJ/mol[/tex]
The heat of combustion of ethanol is given as follows:C₂H₅OH (l) + 3O₂ (g) → 3H₂O(g) + 2CO₂(g)
Let's assume that 1 mole of ethanol is burnt. The heat of reaction, ΔHrxn is given by this equation:
ΔHrxn = ΔHformation (products) - ΔHformation (reactants)
[tex]= (2 mol* - 393.5kJ/mol) + ( 3mol * -285.8kJ/mol) - [ (1mol * -235 kJ/mol) + ( 3mol * 0.0000kJ)]\\\\= -1644 - (-235.2)\\\\= - 1409.2 kJ/mol[/tex]
The negative sign indicates that the process of combustion is exothermic.
b) The combustion reaction for ethanol can be given as:
C₂H₅OH (l) + 3O₂ (g) → 3H₂O(g) + 2CO₂(g)
c) From the comparisons of the ΔHrxn, sugar produces more energy.
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Final answer:
The enthalpy change for fermentation and respiration (combustion) can't be calculated without additional data on standard enthalpies of formation. The combustion reaction for ethanol is balanced, and to compare the heat released from sugar or ethanol combustion, we would need their specific heats of combustion in kJ/mol.
Explanation:
The question asks us to calculate the change in enthalpy (ΔHorxn) for both fermentation and respiration (combustion) processes and write a balanced chemical equation for the combustion of ethanol. To answer these, we start by looking at each process.
Fermentation
The balanced chemical equation for the fermentation of glucose to ethanol (CH3CH2OH) and carbon dioxide (CO2) is given by:
C6H12O6 → 2 CH3CH2OH + 2 CO2
We cannot calculate ΔHorxn for fermentation without the standard enthalpy of formation for each species.
Respiration (Combustion)
The combustion of glucose can be represented as:
C6H12O6 + 6 O2 → 6 CO2 + 6 H2O
Again, to calculate ΔHorxn, we need the standard enthalpies of formation for each compound.
Combustion Reaction for Ethanol
The balanced combustion reaction of ethanol is:
C2H5OH(l) + 3 O2(g) → 2 CO2(g) + 3 H2O(l)+ 29.7 kJ/g
This reaction shows that for each mole of ethanol burned, carbon dioxide and water are produced, and heat is released.
Comparison of Heat Released
To compare which releases more heat per mole of carbon, sugar or ethanol, we would need to know the specific heat of combustion for each substance in kJ/mol.