Answer:
Molarity of silver ion: 0.296 M
Explanation:
Reaction:
AgNO₃ + KCl → AgCl↓ + K⁺ + NO₃⁻
From the reaction, we know that the moles of AgCl produced will be the same as the moles of initial silver.
First, let´s calculate the number of moles of AgCl produced:
1.56 g AgCl was produced, that is, (1.56 g AgCl * 1 mol AgCl/143.3 g AgCl) 0.011 moles AgCl.
The moles of silver ion present in the original solution was 0.011 mol. Since this number of moles was present in a 37.1 ml solution, then, in 1000 ml:
moles of silver ion per liter = 1000 ml * 0.011 mol / 37.1 ml = 0.296 mol
Molarity of silver ion = 0.296 M
The molarity of silver ion in the original AgNO3 solution is 0.293 M.
Explanation:To determine the molarity of silver ion in the original AgNO3 solution when 37.1 mL reacts to yield 1.56 g of AgCl, we first need to calculate the moles of AgCl produced using the molar mass of AgCl.
The molar mass of AgCl is 143.3 g/mol. So, the moles of AgCl formed are calculated as follows:
(1.56 g AgCl) / (143.3 g/mol) = 0.010878 mol AgCl
Since AgNO3 reacts with KCl in a 1:1 mole ratio to produce AgCl, the moles of AgNO3 that reacted is also 0.010878 mol. To find the molarity of AgNO3, we divide the moles by the volume of solution in liters:
(0.010878 mol AgNO3) / (0.0371 L) = 0.293 M
Two solutions of sodium acetate are prepared, one having a concentration of 0.1 M and the other having a concentration of 0.01 M. Calculate the pH values when the following concentrations of HCl have been added to each of these solutions: 0.0025 M, 0.005 M, 0.01 M, and 0.05 M.
Answer:
For 0.1 M sodium acetate solution
if concentration of acid is 0.0025 then pH will 6.075
if concentration of acid is 0.005 then pH will 5.775
if concentration of acid is 0.01 then pH will 5.475
if concentration of acid is 0.05 then pH will 4.775
For 0.01 M sodium acetate solution
if concentration of acid is 0.0025 then pH will 5.075
if concentration of acid is 0.005 then pH will 4.775
if concentration of acid is 0.01 then pH will 4.475
if concentration of acid is 0.05 then pH will 3.775
Explanation:
to calculate the pH of a buffer solution we use the following formula
pH = pKa + log [B]/[A] ------------- eq (1)
[B] = concentration of base
[A] = concentration of acid
Given data
[B] = 0.1 M , 0.01M
[A] = 0.0025 M , 0.005 M, 0.01 M, 0.05 M
pKa value for sodium acetate is 4.75
1. First we will calculate the pH values for 0.1 M acetate solution.
If the concentration of acid is 0.0025, then:
[B] = 0.1 M
[A] = 0.0025 M
put these values in eq 1. which is:
pH = pKa + log [B]/[A]
pH = 4.75 + log [0.1]/[0.0025]
pH = 4.75 + log [40]
pH = 4.475 + 1.6
pH = 6.075
If the concentration of acid is 0.005 M, then:
[B] = 0.1 M
[A] = 0.005 M
put these values in eq 1. which is:
pH = pKa + log [B]/[A]
pH = 4.75 + log [0.1]/[0.005]
pH = 4.75 + log [20]
pH = 4.475 + 1.3
pH = 5.775
If the concentration of acid is 0.01, then:
[B] = 0.1 M
[A] = 0.01 M
put these values in eq 1. which is:
pH = pKa + log [B]/[A]
pH = 4.75 + log [0.1]/[0.01]
pH = 4.75 + log [10]
pH = 4.475 + 1
pH = 5.475
If the concentration of acid is 0.05, then:
[B] = 0.1 M
[A] = 0.05 M
put these values in eq 1. which is:
pH = pKa + log [B]/[A]
pH = 4.75 + log [0.1]/[0.05]
pH = 4.75 + log [2]
pH = 4.475 + 0.3
pH = 4.775
2. Now we will calculate the pH values for 0.01 M acetate solution.
If the concentration of acid is 0.0025, then:
[B] = 0.01 M
[A] = 0.0025 M
put these values in eq 1. which is:
pH = pKa + log [B]/[A]
pH = 4.75 + log [0.01]/[0.0025]
pH = 4.75 + log [4]
pH = 4.475 + 0.6
pH = 5.075
If the concentration of acid is 0.005 M, then:
[B] = 0.01 M
[A] = 0.005 M
put these values in eq 1. which is:
pH = pKa + log [B]/[A]
pH = 4.75 + log [0.01]/[0.005]
pH = 4.75 + log [2]
pH = 4.475 + 0.3
pH = 4.775
If the concentration of acid is 0.01 M, then:
[B] = 0.01 M
[A] = 0.01 M
put these values in eq 1. which is:
pH = pKa + log [B]/[A]
pH = 4.75 + log [0.01]/[0.01]
pH = 4.75 + log [1]
pH = 4.475 + 0
pH = 4.475
If the concentration of acid is 0.05 M, then:
[B] = 0.01 M
[A] = 0.05 M
put these values in eq 1. which is:
pH = pKa + log [B]/[A]
pH = 4.75 + log [0.01]/[0.05]
pH = 4.75 + log [0.2]
pH = 4.475 + (-0.7)
pH = 4.475 - 0.7
pH = 3.775
Answer:
a) pH = 4.71
b) pH = 4.704
c) pH = 4.57
d) No buffer here, the pH will be between 2-3
Explanation:
Applying Henderson Hasselbach equation:
pH = pKa + log([A]/[HA])
a) For 0.0025 M:
[A] = 0.1/2 = 0.05 M
[HA] = 0.05 M
After add 0.0025 M of acid:
[A] = 0.05 - 0.0025 = 0.0475 M
[HA] = 0.05 + 0.0025 = 0.0525 M
[tex]pH=4.75+log(\frac{0.0475}{0.0525} )=4.71[/tex]
b) For 0.005 M:
[A] = 0.1/2 = 0.05 M
[HA] = 0.05 M
After add 0.005 M of acid:
[A] = 0.05 - 0.005 = 0.0495 M
[HA] = 0.05 + 0.005 = 0.055 M
[tex]pH=4.75+log(\frac{0.0495}{0.055} )=4.704[/tex]
c) For 0.01 M:
[A] = 0.1/2 = 0.05 M
[HA] = 0.05 M
After add 0.01 M of acid:
[A] = 0.05 - 0.01 = 0.04 M
[HA] = 0.05 + 0.01 = 0.06 M
[tex]pH=4.75+log(\frac{0.04}{0.06} )=4.57[/tex]
d) For 0.05 M:
[A] = 0.1/2 = 0.05 M
[HA] = 0.05 M
After add 0.05 M of acid:
[A] = 0.05 - 0.05 = 0
[HA] = 0.05 + 0.05 = 0.1 M
No buffer here, the pH will be between 2-3
What is the preferred electrical charge of a Sodium ion? O a. +1 O O O b. +2 Oco O O d.-1 O e. 2 O
Konvert the following temperatures from °F to PC: Ta 86°F, (b) -22°F, (C) 50°F, (d) -40°F, (e) 32°F, (f) -459.67°F. Convert each temperature to K.
Answer:
The answers are:
a) 30°C; 303.15K
b) -30°C; 243.15K
c) 50°C; 323.15K
d) -40°C; 233.15K
e) 0°C; 273.15K
f) -273.15 °C ; 0K
Explanation:
To convert the temperature from ° F to ° C we use the following expression:
[tex]C=(F-32)\frac{5}{9}[/tex]
where C es temperature en °C and F is temperature in °F
To obtain the temperature in K we need to add 273.15 to each Celcius temperature
[tex]K=C+273.15[/tex]
The highest temperature recorded in the same city during the past year was 304.89 K. What was the temperature in degrees Celsius?
Final answer:
The temperature of 304.89 Kelvin is equivalent to 31.74°C when converted using the formula C = K - 273.15.
Explanation:
The student has asked about converting the highest temperature recorded in a certain city from Kelvin to degrees Celsius. The formula for converting Kelvin to Celsius is: C = K - 273.15, where C is the temperature in Celsius and K is the temperature in Kelvin. Applying this formula to the given temperature (304.89 K), we get:
C = 304.89 K - 273.15
C = 31.74°C
Hence, the temperature of 304.89 K is equivalent to 31.74°C.
If you feed 100 kg of N2 gas and 100 kg of H2 gas into a
reactor. What is the excess reactant?
Answer:
H₂ gas
Explanation:
The reaction between nitrogen gas and hydrogen gas forms ammonia (the Haber-Bosch process):
N₂ + 3H₂ ⇒ 2NH₃
The excess reactant can be found by comparing the moles of nitrogen and hydrogen. The molar mass of N₂ is 28.00 g/mol and the molar mass of H₂ is 2.02 g/mol.
(100 kg N₂)(1000g/kg)(mol/28.00g) = 3570 mol
(100 kg H₂)(1000g/kg)(mol/2.02g) = 49500 mol
The molar ratio between the reactant N₂ and H₂ is 1N₂:3H₂. The moles of nitrogen required to react with H₂ is:
(49500 mol H₂)(1N₂ / 3H₂) = 16500 mol
The amount of nitrogen required is more than what is available, so nitrogen is the limiting reagent and hydrogen is the excess reagent.
Explain what D and L represent in stereoisomers
Final answer:
The D and L stereochemical descriptors are used to represent the configuration of stereoisomers in monosaccharides. The D- or L- designation is based on the position of the -OH group on the penultimate carbon in the Fisher projection. The D-configuration is commonly found in nature and only dextrorotary amino acids are used by cells to build proteins.
Explanation:
The D and L stereochemical descriptors are used to represent the configuration of stereoisomers, specifically in the context of monosaccharides or sugars. The designation of D or L is based on the position of the -OH group on the second-last carbon (penultimate C) in the Fisher projection. If the -OH group is on the right side, it is assigned D-configuration, and if it is on the left side, it is assigned L-configuration.
These descriptors do not indicate the rotation of plane polarized light, but purely define the configuration. Enantiomers that are D- and L- pairs have the same common name, with the D- or L- designation indicating their configuration. It's important to note that the D- and L- designation does not always correlate with the dextro/levo rotatory nature of the enantiomers in a polarimeter.
For example, D-glucose and L-glucose are enantiomers with the penultimate C defining D- or L-configuration. The D-configuration is commonly found in nature, and only dextrorotary d amino acids (L amino acids) are used by cells to build polypeptides and proteins.
A piece of an unknown metal has a volume of 16.6 mL and a mass of 190.1 grams. The density of the metal is g/mL A piece of the same metal with a mass of 94.6 grams would have a volume of ml. Submit Answer
Answer: The density of the metal is 11.45 g/mL and the volume occupied by 94.6 grams is 8.26 mL
Explanation:
To calculate the density of unknown metal, we use the equation:
[tex]\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}[/tex] ......(1)
Volume of unknown metal = 16.6 mL
Mass of unknown metal = 190.1 g
Putting values in equation 1, we get:
[tex]\text{Density of unknown metal}=\frac{190.1g}{16.6mL}\\\\\text{Density of unknown metal}=11.45g/mL[/tex]
The density of the metal remains the same.
Now, calculating the volume of unknown metal, using equation 1, we get:
Density of unknown metal = 11.45 /mL
Mass of unknown metal = 94.6 g
Putting values in above equation, we get:
[tex]11.45g/mL=\frac{94.6g}{\text{Volume of unknown metal}}\\\\\text{Volume of unknown metal}=8.26mL[/tex]
Hence, the density of the metal is 11.45 g/mL and the volume occupied by 94.6 grams is 8.26 mL
Name two "Storage Polysaccharides" and two "Structural Polysaccharides"
Answer:
Examples of storage polysaccharides - starch and glycogen and structural polysaccharides - cellulose and chitin
Explanation:
Polysaccharides are the complex carbohydrate polymers, composed of monosaccharide units that are joined together by glycosidic bond.
In other words, polysaccharides are the carbohydrate molecules that give monosaccharides or oligosaccharides on hydrolysis.
The examples of storage polysaccharides are starch and glycogen. The examples of structural polysaccharides are cellulose and chitin.
What are the names of the following compounds: FeCl HNO NaSO SO
Answer:
FeCl: Ferric Chloride (also called iron chloride), comes from Fe (ferrum, or iron), and Cl (Chlorine)
HNO: Nitroxyl, from N (Nitrogen), and the acidic nature of a radical ending in -yl.
NaSO: Sodium sulfate, Na (Sodium), S (Sulfur), O (Oxygen).
SO: Sulfur monoxide (Mono-One), O (Oxygen) and S (Sulfur).
A chemist needs to create a series of standard Cu2 (aq) solutions for an absorbance experiment. For the first standard, he uses a pipet to transfer 25.00 mL of a 2.96 M Cu2 (aq) stock solution to a 250.0 mL volumetric flask and adds enough water to dilute to the mark. He then uses a second pipet to transfer 20.00 mL of the second solution to a 100.0 mL volumetric flask and adds enough water to dilute to the mark. Calculate the concentration of the Cu2 (aq) solution in the 100.0 mL volumetric flask.
Answer:
The concentration of the Cu2 in the 100.0 ml volumetric flask is 0.0592 M
Explanation:
In the first dilution, Cu2 was diluted ten times (25 / 250 = 1/10). Then, this dilution was diluted again, but now five times (20 / 100 = 1/5). In total, the solution was diluted 50 times (1/10 * 1/5 = 1/50). The final concentration will be 2.96 M / 50 = 0.0592 M
The quantity of the solute or the substance present in the solution is called the concentration. The concentration of the [tex]\rm Cu_{2}[/tex] in the volumetric flask is 0.0592 M.
What is concentration?Concentration is the molarity of the substance and is given as the ratio of the moles of the solute with the volume in litres.
Given,
The volume of [tex]\rm Cu_{2}[/tex] by first pipet = 25 mlVolume of stock solution = 250 mlThe [tex]\rm Cu_{2}[/tex] is diluted ten times at first,
[tex]\dfrac {25}{250}= \dfrac{1}{10}[/tex]
Given,
Volume of [tex]\rm Cu_{2}[/tex] by second pipet = 20 mlVolume of stock solution = 100 mlThe [tex]\rm Cu_{2}[/tex] is diluted five times the second time,
[tex]\dfrac {20}{100}= \dfrac{1}{5}[/tex]
Total dilution of the solution was done 50 times as,
[tex]\dfrac{1}{10}\times \dfrac{1}{5} = \dfrac{1}{50}[/tex]
The final concentration of the solution will be,
[tex]\dfrac{2.96 \;\rm M}{50} = 0.0592 \;\rm M[/tex]
Therefore, the final concentration is 0.0592 M.
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A technician tares a 100.0 mL volumetric flask at 150.00 g. After adding sodium chloride to the flask it then weighs 158.84 g. Assuming an error of 0.2 mL in the volumetric volume and 0.005 g in the weight, calculate the molar concentration of sodium chloride and its associated standard deviation.
To find the molar concentration of NaCl, subtract the tare weight from the total weight to get the mass of NaCl, calculate moles of NaCl, and divide by the solution volume. To estimate the standard deviation, propagate the errors from the mass and volume measurements according to the rules of error propagation. Specific numerical values for the standard deviation cannot be provided without exact formulas.
Explanation:The question pertains to calculating the molar concentration of sodium chloride (NaCl) in a solution, and its associated standard deviation, given certain experimental measurements and potential error margins. First, the mass of NaCl added to the solution is found by subtracting the tare weight of the volumetric flask from the total weight after NaCl was added, yielding 8.84 g of NaCl. The molecular weight of NaCl is 58.44 g/mol, which allows determination of the moles of NaCl present.
To find the molar concentration, divide the moles of NaCl by the volume of the solution in liters (assuming the 100.0 mL flask volume as ideal, the error in volume would be considered in calculating the standard deviation, not the concentration itself). Then, to address the error margins, propagate the errors from the mass and volume measurements to estimate the standard deviation of the calculated concentration.
Note: Without specific formulas for error propagation and the exact calculation method for standard deviation provided in the question, a detailed numerical solution including the standard deviation calculation cannot be accurately presented. However, this process typically involves the square root of the sum of squared fractional uncertainties of the measurements involved.
What is the focus on the biological organization levels for ecology?
Answer:
The biology of the ecosystem is always studied from the composition of organisms, population and community.
Explanation:
In an ecosystem, various processes are sustained and these processes regulate the environment, as the biotic communities in the ecosystem make up the biosphere. Various life processes, their interaction, movement, and successional development ecosystem and the distribution of organisms are the focus of biological organization levels of ecology. First comes the organism ecology level as the researchers study their adaptations for the structure and physiographic behavior. Second is the population ecology level with a group of interbreeding organisms next level is of Community ecology and ecosystem ecology as they study the different species within the area using a three-dimensional model. After these the ecosystem ecology that is an extension of all the levels in the ecosystem. The natural environment can provide all the needed nutrients and life-supporting factors.Calculate the mass of 1.0 L of helium (He), 1.0 L of chlorine gas (Cl2), and 1.0 L of air (79% N2, 21% O2 by volume) at 25°C and 1 atm total pressure. Explain why a balloon filled with helium rises and why leaks of chlorine gas can be dangerous.
To calculate the mass we use the following formulas:
PV=nRT (1)
and
n = m / M (2)
where:
P - pressure (atm)
V - volume (L)
n - moles
R - gas constant = 0.082 (L × atm) / (mol × K)
T - temperature (°K) (25°C + 273 = 298°K)
m - mass (g)
M - molecular mass (g/mole)
Now we rewrite equation (1):
n = PV / RT
And replace n with m / M from equation (2):
m / M = PV / RT
m = (P × V × M) / (R ×T)
1 L of He will have a mass of:
m = (1 × 1 × 4) / (0.082 × 298) = 0.1637 g
1 L of Cl₂ will have a mass of:
m = (1 × 1 × 71) / (0.082 × 298) = 2.9055 g
1.0 L of air will contain 0.79 L of N₂ and 0.21 L of O₂
0.79 L of N₂ will have a mass of:
m = (1 × 0.79 × 28) / (0.082 × 298) = 0.9052 g
0.21 L of O₂ will have a mass of:
m = (1 × 0.21 × 32) / (0.082 × 298) = 0.2750 g
mass of air = mass of N₂ + mass of O₂
mass of air = 0.9052 + 0.2750 = 1.1802 g
A balloon filed with helium will rise because as you see 1 L of helium is lighter than 1 L of air.
Chlorine gas is dangerous because chlorine is very toxic for human life and more of that is heavier than the air so will diffuse very hard from the area where the leak appeared.
Find the percent by mass of tin in the original sample, assuming that it contains no other reducing agents. A sample of impure tin of mass 0.528 g is dissolved in strong acid to give a solution of Sn2+. The solution is then titrated with a 0.0448 M solution of NO3−, which is reduced to NO(g). The equivalence point is reached upon the addition of 4.03×10−2 L of the NO3− solution.
Final answer:
To find the mass percent of chloride in the original dry sample, you can use the formula: Mass percent of chloride = (mass of chloride / mass of original sample) x 100%. Use the volume of AgNO3 solution used in the titration, the solution's molarity, and the molar mass of chloride to calculate the mass of chloride.
Explanation:
The percent by mass of chloride in the original dry sample can be calculated using the following formula:
Mass percent of chloride = (mass of chloride / mass of original sample) x 100%
In this case, the mass of chloride can be determined by multiplying the volume of AgNO3 solution used in the titration (28 mL) by the molarity of the solution (0.1 M) and the molar mass of chloride (35.453 g/mol).
Then, using the mass of chloride and the mass of the original sample (0.200 g), the percent by mass of chloride in the original dry sample can be calculated.
In a conjugate acid-base pair, the acid typically has one more proton than the base b. one fewer proton than the base. C. two fewer protons than the base. d. the same number of protons as the base. 17 a. TO17 O 0001
Answer:
Statement (a) is true
Explanation:
Conjugate base of an acid is formed from deprotonation of corresponding acid.For an example, consider an acid e.g. [tex]CH_{3}COOH[/tex] (acetic acid)Acid-base equilibrium for acetic acid in aqueous solution is represented as: [tex]CH_{3}COOH+H_{2}O\rightarrow CH_{3}COO^{-}+H_{3}O^{+}[/tex]Here [tex]CH_{3}COO^{-}[/tex] (acetate ion) is the conjugate base of acetic acid.So, clearly, acetic acid has one more proton as compared to acetate ionHence statement (a) is true
In a conjugate acid-base pair, the acid typically has one fewer proton than the base.
Explanation:In a conjugate acid-base pair, the acid typically has one fewer proton than the base. When a proton (H+) is removed from an acid, it forms its conjugate base which has one less proton. For example, water (H2O) is an acid in the conjugate acid-base pair H2O/OH-, where water (H2O) has one more hydrogen ion (H+) than the hydroxide ion (OH-), which is the conjugate base.
The acid-base pairs can be represented as:
H2O/OH-H3PO4/H2PO4-H2SO4/HSO4-NH4+/NH3I have a 5 M stock solution of KCI (For a protocol 100 mM KCl is considered "1X" concentration). If I want to make 10 ml of a 4X concentration KCl solution, how much 5 M stock and how much water do I need to add together?
Answer:
You need 0.8 ml of 5M stock solution and you have to add 9.2 ml of water.
Explanation:
Protocol solution (1X): 100 mM=0.1M
4X: 0.4M
The concentration of a solution is inversely proportional to the volume of a solution, so:
[tex]M_{1}V_{1}=M_{2}V_{2}[/tex]
where:
M1= 5M stock solution
V1= amount of solution we need to collect
M2=4X solution
V2= 10 ml (volumen of 4X solution)
Therefore:
5M×V1=0.4M×10ml
V1={0.4M}{5M}10ml=0.8ml
[tex]5M*V_{1}=0.4M*10ml\\ V_{1}=\frac{0.4M}{5M}10ml=0.8ml[/tex]
To make a 10 ml solution we have to add 9.2 ml of water because V2 es 10 ml.
Which statement is true about obligate anaerobes? View Available Hint(s) Which statement is true about obligate anaerobes? They obtain energy by oxidizing ferrous ions. They will use O2 if it is present, but may obtain energy by fermentation if needed. They use O2 for cellular respiration and cannot grow without it. They are poisoned by O2. They live exclusively by cellular respiration or by anaerobic respiration.
Answer: They are poisoned by O₂
Explanation:
Obligate anaerobes cannot survive in normal concentrations of oxygen. Depending on the species, tolerance varies from 0.5% to 8% oxygen.
Under normal cellular conditions, O₂ turns into O₂⁻ and H₂O₂, toxic to the organism. Obligate anaerobes lack enzymes superoxide dismutase and catalase, capable of turning O₂⁻ and H₂O₂ back into breathable O₂.
The statement that is true about obligate anaerobes is they are poisoned by [tex]O_2[/tex].
Microorganisms known as obligatory anaerobes are incapable of surviving or developing in the presence of oxygen. They cannot detoxify the reactive oxygen species (ROS) created during aerobic respiration because they lack the enzymes catalase and superoxide dismutase. As a result, oxygen is poisonous to them. When cells are exposed to oxygen, toxic byproducts can arise that injure the cells' biological constituents and ultimately cause cell death.
Obligate anaerobes are constrained to anaerobic metabolic pathways, in contrast to facultative anaerobes, which can flip between aerobic and anaerobic metabolism depending on oxygen availability. Typically, they break down organic substances without the need of oxygen through fermentation or anaerobic respiration to produce energy.
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Calculate: (a) the weight (in lbf) of a 30.0 lbm object. (b) the mass in kg of an object that weighs 44N. (c) the weight in dynes of a 15-ton object (not metric tons)
Answer:
a) 965,1 lbf
b) 4,5 kg
c) 1,33 * 10^6 dynes
Explanation:
Mass of an object refers to the amount of mattter it cotains, it can be expressed it gr, kg, lbm, ton, etc.
Weight of an object refers to a force, and is the measurement of the pull of gravitiy on an object. It may be definide as the mass times the acceleration of gravity.
w=mg
In Planet Earth, the nominal "average" value for gravity is 9,8 m/s² (in the International System) or 32,17 ft/s² (in the FPS system).
To solve this problem we'll use the following conversion factors:
1 lbf = 1 lbm*ft/s²
1 N = 1 kg*m/s²
1 dyne = 1 gr*cm/s² and 1 N =10^5 dynes
1 ton = 907,18 kg
1 k = 1000 gr
a) m = 30 lbm
[tex]w = 30 lbm * 32,17 ft/s^{2} = 965, 1 \frac{lmb*ft}{s^{2} } = 965,1 lbf[/tex]
b) w = 44 N
First, we clear m of the weight equation and then we replace our data.
[tex]m = \frac{w}{g} = \frac{44 N}{9,8 \frac{m}{{s}^{2}} } = 4,5 kg[/tex]
c) m = 15 ton
[tex]m=15 ton * \frac{907,18 kg}{1 ton} = 13607,7 kg \\ w = mg = 13607,7 kg * 9,8 m/s2 = 133355,5 N * \frac{10^{5} dynes }{1 N} = 1,33 * 10^{6}dynes [/tex]
The weight of a 30 lbm object is 966 lbf. The mass of an object weighing 44N is 4.49 kg. And, the weight of a 15-ton object is 1.34 x 10¹² dynes.
Explanation:To Calculate: (a) the weight (in lbf) of a 30.0 lbm object we need to use the fact that 1 lbm equals 32.2 lb force (lbf). Therefore, a 30.0 lbm object weight would be 30 lbm * 32.2 = 966 lbf.
In (b), the mass in kg of an object that weighs 44N can be calculated by dividing the weight by Earth's gravity (approximately 9.8m/s²). 44N / 9.8m/s² = 4.49 kg.
Lastly, in (c), to find the weight in dynes of a 15-ton object we first convert the weight to pounds since 1 ton equals 2000 lbs. Then we convert pounds to Newtons (1 lb = 4.44822 N) and finally Newtons to dynes (1 N = 1,000,000 dynes). So, 15 ton * 2000 = 30000lb * 4.44822 N/lb * 1,000,000 dynes/N = 1.34 x 10¹² dynes.
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Glycerol is a syrupy liquid often used in cosmetics and soaps. A 3.25-L sample of pure glycerol has a mass of 4.10 x 10 g.
What is the density of glycerol in g/cm"? Express your answer in grams per cubic centimeter.
Explanation:
It is known that density is the amount of mass present in liter of solution or substance.
Mathematically, Density = [tex]\frac{mass}{volume}[/tex]
It is given that volume is 3.25 L and mass is [tex]4.10 \times 10^{3} g[/tex]. Hence, calculate the density of glycerol as follows.
Density = [tex]\frac{mass}{volume}[/tex]
= [tex]\frac{4.10 \times 10^{3} g}{3.25 L}[/tex]
= [tex]1.26 \times 10^{3} g/L[/tex]
As, 1 L = 1000 [tex]cm^{3}[/tex].
So, [tex]1.26 \times 10^{3} g/L \times \frac{1000 cm^{3}}{1 L}[/tex]
= [tex]1260 \times 10^{6} g/cm^{3}[/tex]
Thus, we can conclude that the density of glycerol is [tex]1260 \times 10^{6} g/cm^{3}[/tex].
write and the integrated rate laws hor zeroth-first- second-order rate laws.
Explanation:
The integrated rate law for the zeroth order reaction is:
[tex][A]=-kt+[A]_0[/tex]
The integrated rate law for the first order reaction is:
[tex][A]=[A]_0e^{-kt}[/tex]
The integrated rate law for the second order reaction is:
[tex]\frac{1}{[A]}=kt+\frac{1}{[A]_0}[/tex]
Where,
[tex][A][/tex] is the active concentration of A at time t
[tex][A]_0[/tex] is the active initial concentration of A
t is the time
k is the rate constant
Answer:
- 0th: [tex]C_A=C_{A0}-kt[/tex]
- 1st: [tex]C_A=C_{A0}exp(-kt)[/tex]
- 2nd: [tex]\frac{1}{C_A}=kt+\frac{1}{C_{A0}}[/tex]
Explanation:
Hello,
For the ideal reaction A→B:
- Zeroth order rate law: in this case, we assume that the concentration of the reactants is not included in the rate law, therefore the integrated rate law is:
[tex]\frac{dC_A}{dt}=-k\\ \int\limits^{C_A}_{C_{A0}} {} \ dC_A= \int\limits^{t}_{0} {-k} \ dt\\C_A-C_{A0}=-kt\\C_A=C_{A0}-kt[/tex]
- First order rate law: in this case, we assume that the concentration of the reactant is included lineally in the rate law, therefore the integrated rate law is:
[tex]\frac{dC_A}{dt}=-kC_A\\ \int\limits^{C_A}_{C_{A0}} {\frac{1}{C_A} } \ dC_A= \int\limits^{t}_{0} {-k} \ dt\\ln(\frac{C_{A}}{C_{A0}} )=-kt\\C_A=C_{A0}exp(-kt)[/tex]
- Second order rate law: in this case, we assume that the concentration of the reactant is squared in the rate law, therefore the integrated rate law is
[tex]\frac{dC_A}{dt}=-kC_A^{2} \\ \int\limits^{C_A}_{C_{A0}} {\frac{1}{C_A^{2} } } \ dC_A= \int\limits^{t}_{0} {-k} \ dt\\-\frac{1}{C_A}+\frac{1}{C_{A0}}=-kt\\\frac{1}{C_A}=kt+\frac{1}{C_{A0}}[/tex]
Best regards.
One solution has a formula C (n) H (2n) O (n) If this material weighs 288 grams, dissolves in weight 90 grams, the solution will have a boiling point of 101.24 ° C. Find the formula
The molecules of this substance When determining the Kb value of water = 0.512 ° C / m and the atomic weight H = 1, C = 12 and O = 16.
Explanation:
The given data is as follows.
Boiling point of water ([tex]T^{o}_{b}) = 100^{o}C[/tex] = (100 + 273) K = 323 K,
Boiling point of solution ([tex]T_{b}) = 101.24^{o}C[/tex] = (101.24 + 273) K = 374.24 K
Hence, change in temperature will be calculated as follows.
[tex]\Delta T_{b} = (T_{b} - T^{o}_{b})[/tex]
= 374.24 K - 323 K
= 1.24 K
As molality is defined as the moles of solute present in kg of solvent.
Molality = [tex]\frac{\text{weight of solute \times 1000}}{\text{molar mass of solute \times mass f solvent(g)}}[/tex]
Let molar mass of the solute is x grams.
Therefore, Molality = [tex]\frac{\text{weight of solute \times 1000}}{\text{molar mass of solute \times mass f solvent(g)}}[/tex]
m = [tex]\frac{288 g \times 1000}{x g \times 90}[/tex]
= [tex]\frac{3200}{x}[/tex]
As, [tex]\Delta T_{b} = k_{b} \times molality[/tex]
[tex]1.24 = 0.512 ^{o}C/m \times \frac{3200}{x}[/tex]
x = [tex]\frac{0.512 ^{o}C/m \times 3200}{1.24}[/tex]
= 1321.29 g
This means that the molar mass of the given compound is 1321.29 g.
It is given that molecular formula is [tex]C_{n}H_{2n}O_{n}[/tex].
As, its empirical formula is [tex]CH_{2}O[/tex] and mass is 30 g/mol. Hence, calculate the value of n as follows.
n = [tex]\frac{\text{Molecular mass}}{\text{Empirical mass}}[/tex]
= [tex]\frac{1321.29 g}{30 g/mol}[/tex]
= 44 mol
Thus, we can conclude that the formula of given material is [tex]C_{44}H_{88}O_{44}[/tex].
In a five-fold serial dilution of a 20 pg/ml solution, all tube dilutions are 1/5. What is the substance concentration in the third tube of this series?
Answer:
0.8 pg/ml
Explanation:
To make the dilutions, you will take 1 ml of the original solution (tube 1) and add 4 ml of solvent. You will now have 20 pg per 5 ml of solution, so your new concentration will be 4 pg/ml (tube 2). Then you will repeat the process, so you will have 4 pg per 5 ml of solution, resulting in a concentration of 0.8 pg/ml (tube 3). The same process will be repeated for tubes 4 and 5.
The reaction of carboxylic acids with alcohol in the presence of an acid catalyst yields Select one: amides O esters O no reaction occurs O aldehydes
Answer:
esters
Explanation:
The -OR group of the alcohol replaces the -OH of the carboxylic acid, forming an ester. See attachment for condensed mechanism.
Which property of metals is explained by the "sea of electrons" within metallic bonding?
A. Malleability
B. All of the Above
C. Electrical conductivity
D. Thermal conductivity
Answer:
B. All of the Above
Explanation:
Sea of electrons -
This model of the metallic bonding helps in explaining the properties like , malleability , ductility ,high electrical conductivity , luster ,high thermal conductivity of the metals in solid state .
The metallic bonding is between metal atoms and the ionic bond links a metal and a non - metal together ,
In case of metallic bonding , bulk of metal atoms are joined .
hence from the question , all the given properties are correct .
What is the pH at each of the points in the titration of 25.00 mL of 0.2000 M
HCl by 0.2000 M NaOH:
i) Before adding NaOH
ii) After adding 24.00 mL NaOH
Answer:
i) pH = 0.6990
ii) pH = 2.389
Explanation:
i) Before adding aqueous NaOH, there are 25.00 mL of 0.2000 M HCl. HCl reacts with the water in the aqueous solution as follows:
HCl + H₂O ⇒ H₃O⁺ + Cl⁻
The HCl and H₃O⁺ are related to each other through a 1:1 molar ratio, so the concentration of H₃O⁺ is equal to the HCl concentration.
The pH is related to the hydronium ion concentration as follows:
pH = -log([H₃O⁺]) = -log(0.2000) = 0.699
ii) Addition of NaOH causes the following reaction:
H₃O⁺ + NaOH ⇒ 2H₂O + Na⁺
The H₃O⁺ and NaOH react in a 1:1 molar ratio. The amount of NaOH added is calculated:
n = CV = (0.2000 mol/L)(24.00 mL) = 4.800 mmol NaOH
Thus, 4.800 mmol of H₃O⁺ were neutralized.
The initial amount of H₃O⁺ present was:
n = CV = (0.2000 mol/L)(25.00 mL) = 5.000 mmol H₃O⁺
The amount of H₃O⁺ that remains after addition of NaOH is:
(5.000 mmol) - (4.800 mmol) = 0.2000 mmol
The concentration of H₃O⁺ is the amount of H₃O⁺ divided by the total volume. The total volume is (25.00 mL) + (24.00 mL ) = 49.00 mL
C = n/V = (0.2000 mmol) / (49.00 mL) = 0.004082 M
The pH is finally calculated:
pH = -log([H₃O⁺]) = -log(0.004082) = 2.389
Green plants use light from the Sun to drive photosynthesis, a chemical reaction in which liquid water and carbon dioxide gas from aqueous glucose and oxygen gas. Calculate the moles of glucose produced by the reaction of 2.40 moles of water. Be sure your answer has a unit symbol, if necessary, and round it to significant digits.
Answer: 0.4 moles of glucose are produced by the reaction of 2.40 moles of water.
Explanation:
Photosynthesis is a phenomenon in which green plants containing chlorophyll use sunlight as a source of energy to convert carbon dioxide and water to form glucose and oxygen.
The balanced chemical equation is:
[tex]6CO_2+6H_2O\overset{sunlight}\rightarrow C_6H_{12}O_6+6O_2[/tex]
According to stoichiometry:
6 moles of water produces = 1 mole of glucose
2.40 moles of water produces = [tex]\frac{1}{6}\times 2.4=0.4[/tex] moles of glucose
Thus 0.4 moles of glucose are produced by the reaction of 2.40 moles of water.
In natural convection heat transfer, the parameter
a) Nusselt Number
b) Rayleigh Number
c) Grashof Number
c) Lewis Number
Answer:
correct answer is option c i.e Grashof Number
Explanation:
The Grashof number is a dimensionless number, which is named after renowned scientist Franz Grashof. The Grashof quantity is defined as the proportion of the buoyant force to viscous force performing on a fluid in a pace boundary layer.
Its function in natural convection is more or less the same as that of Reynolds's number in compelled convection.
To find the formula of a compound composed of iron and carbon monoxide, Fex(CO)y, the compound is burned in pure oxygen, an reaction that proceeds according to the following unbalanced equation.
Fex(CO)y + O2 --> Fe2O3 + CO2
If you burn 1.959 g. of Fex(CO)y and obtain 0.799 g. of Fe2O3 and 2.200 g. of CO2, what is the empirical formula of Fex(CO)y?
Answer:
The empirical formula is: Fe(CO)₅
Explanation:
According the global reaction:
Feₓ(CO)y + O₂ → Fe₂O₃ + CO₂
You should calculate Fe₂O₃ and CO₂ moles, thus:
0,799 Fe₂O₃ grams × [tex]\frac{1 mole}{159.69 Fe2O3 g}[/tex] = 5,00×10⁻³ Fe₂O₃ moles
2,200 CO₂ grams × [tex]\frac{1 mole}{44,01 CO2 g}[/tex] = 5,00×10⁻²CO₂ moles
The ratio between Fe₂O₃ moles and CO₂ moles is 1:10. Thus ratio between x and y must be 1:5 because Fe₂O₃ has 2 irons but CO₂ has just one carbon.
Assuming the formula is Fe₁(CO)₅ the molecular weight is 195,9 g/mol. Thus:
1,959 Fe(CO)₅ grams × [tex]\frac{1 mole}{195,9 Fe(CO)5 g}[/tex] = 1,00×10⁻² Fe(CO)₅ moles
Thus, assuming 1,00×10⁻² moles as basis for calculation, the global reaction is:
1 Fe(CO)₅ + ¹³/₂O₂ → ¹/₂ Fe₂O₃ + 5 CO₂
With this balanced equation the moles produced have sense, thus, the empirical formula is: Fe(CO)₅
I hope it helps!
The empirical formula of the compound Fex(CO)y, formed from 1.959 g of the substance producing 0.799 g of Fe²O³ and 2.200 g of CO², is Fe(CO)⁵.
Explanation:To determine the empirical formula of the compound Fex(CO)y, we must first find the moles of iron (Fe) and carbon monoxide (CO) in the compound. Given that 0.799 g of Fe²O³ and 2.200 g of CO² were produced, we can calculate the number of moles of Fe and C using their molar masses (Fe: 55.85 g/mol, C: 12.01 g/mol, O: 16.00 g/mol).
From Fe²O³, the mass of Fe is 0.799 g x (2 mol Fe / 159.69 g Fe²O³) = 0.0100 mol Fe.
From CO², the mass of C is 2.200 g x (1 mol C / 44.01 g CO²) = 0.0500 mol C.
To find the mole ratio, we use the smallest number of moles as a divisor. Here, it is 0.0100 mol Fe. The ratio of Fe to C in the compound is 0.0100 mol Fe / 0.0100 mol = 1 Fe to 0.0500 mol C / 0.0100 mol = 5 CO.
Therefore, the empirical formula of the compound is Fe(CO)5.
Learn more about empirical formula here:https://brainly.com/question/32125056
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A 36.5 lb child has a Streptococcus infection. Amoxicillin is prescribed at a dosage of 25 mg per kg of body weight b.id What is the meaning of the Latin abbreviation b.i.d.? O every other day O as needed O twice daily O once daily How many hours should pass between each administration? number of hours: 413 How many milligrams of amoxicillin should be given at each administration? mass of amoxicillin: 413 Amoxicillin should be stored between 0 °C and 20 °C. Should the amoxicillin be stored in the freezer or the refrigerator?
Answer:
a) b.i.d: twice daily.
b) 12 hour between each administration.
c) mg amoxicilin/administration = 413.9 mg/administration.
d) should be stored in the refrigerator
Explanation:
mass child = 36.5 Lb * ( 453.592 g/Lb ) = 16556.11 g = 16.556 Kgdosage: 25mg/kg body b.i.d.∴ b.i.d. : refer to twice a day; so 12 hours will pass between each administration of the medication.
⇒ mg amoxicilin/administration = 25 mg/ kg * 16.556 Kg = 413.9 mg amoxicilin.
Amoxicillin should be stored in the refrigerator, since in this section the temperature is kept within the storage range
How many neutrons in an atom of 30p? O a. 17 O b. 30.974 O c. 15 O d. 14 O e. 16
Answer:
The correct option is: c. 15
Explanation:
Phosphorous is a chemical element which belongs to the group 15 of the periodic table and has atomic number 15. It is a highly reactive non-metal of the p-group.
Since, atomic number of an atom is the number of electrons and number of protons for neutral atoms.
So, the number of protons = number of electrons = 15
The atomic mass is obtained by adding the number of neutrons and the protons.
So, number of neutrons + number of protons = 30
So, number of neutrons + 15 = 30
Therefore, the number of neutrons in ³⁰P = 15