Answer:
The density of the solid is 1,316 g/mL
Explanation:
The weight of both Toluene and the solid insoluble is 58,68 g. And the weight of the solid is 32,50 g. Thus, weight of toluene is:
58,68 g - 32,50 g = 26,18 g of Toluene
To know how much volume that toluene occupy you must use density thus:
26,18 g of toluene × ( 1 mL / 0,864 g) = 30,30 mL of toluene
The volume of both Toluene and the solid is 55,00 mL and the volume of toluene is 30,30 mL. Thus, the volume of the solid is:
55,00 mL - 30,30 mL = 24,70 mL
Knowing both volume and weight it is possible to know the density thus:
32,50 g / 24,70 mL = 1,316 g/mL
I hope it helps!
Define "Anomeric" carbon
Anomeric carbon is a stereocenter present in the cyclic structures of carbohydrates (mono or polysaccharides). Being a stereocenter, more exactly an epimer, two diastereosoimers derive from it, designated by the letters α and β; these are anomers, and are part of the extensive nomenclature in the world of sugars.
A1.00 m long beam of stainless steel with a square 2.00 cm x 2.00 cm cross section has a mass of 3.02 kg. What is its density in grams per cubic centimeter? Round your answer to two decimal places.
Answer:
density = 7.55 g/cm^3
Explanation:
we need to get the volume first to get the density , and the volume formula is:
the volume = area * thickness
where :
area = 2 cm * 2 cm = 4 cm^2
and thickness = 100 cm
by substitution:
the volume = 4 * 100 = 400 cm^3
where
density = mass / volume
we have to convert mass from kg to g = 3020 g
by substitution:
density = 3020 g / 400 cm^3
= 7.55 g/cm^3
functional group and bond hybridization of vanillin
Vanillin is the common name for 4-hydroxy-3-methoxy-benzaldehyde.
See attached figure for the structure.
Vanillin have 3 functional groups:
1) aldehyde group: R-HC=O, in which the carbon is double bonded to oxygen
2) phenolic hydroxide group: R-OH, were the hydroxyl group is bounded to a carbon from the benzene ring
3) ether group: R-O-R, were hydrogen is bounded through sigma bonds to carbons
Now for the hybridization we have:
The carbon atoms involved in the benzene ring and the red carbon atom (from the aldehyde group) have a sp² hybridization because they are involved in double bonds.
The carbon atom from the methoxy group (R-O-CH₃) and the blue oxygen's have a sp³ hybridization because they are involved only in single bonds.
A prescription medication requires 5.98 mg per kg of body weight.
A)Convert this quantity to the number of grams required per pound of body weight.
B)Determine the correct dose (in g) for a 191-lb patient. Express your answer with the appropriate units.
To convert from mg/kg to g/lb, use the conversion factors 1000 mg/g and 2.20462 lbs/kg. For a 191-lb patient, the correct dose is approximately 0.51761 g.
Explanation:To convert the prescription medication requirement from milligrams per kilogram (mg/kg) to grams per pound (g/lb), we first need to know the conversion factors between the given units. There are 453.59237 mg in a gram and 2.20462 pounds in a kilogram. Following the conversion steps:
First, we convert 5.98 mg/kg to g/kg by dividing by 1000 (since there are 1000 mg in 1 g):To find the correct dose for a 191-lb patient, we multiply the medication requirement by the patient's weight:
0.00271 g/lb × 191 lbs = approximately 0.51761 g.
Therefore, the correct dose for a 191-lb patient is 0.51761 g.
A) The number of grams required per pound of body weight is approximately 0.00272 grams per pound. B) The correct dose (in g) for a 191-lb patient is approximately 0.519 grams.
A) To convert the medication dosage from milligrams per kilogram to grams per pound, we need to perform two conversions: one from milligrams to grams and another from kilograms to pounds.
1 milligram is equal to 0.001 grams (since there are 1000 milligrams in a gram).1 kilogram is approximately equal to 2.20462 pounds.Starting with the dosage of 5.98 mg per kg, we convert to grams per kilogram: 5.98 mg/kg = 5.98 x 0.001 g/kg = 0.00598 g/kg
Now, we convert from grams per kilogram to grams per pound:
0.00598 g/kg x 2.20462 kg/lb = 0.013175 g/lb
To simplify the calculation, we can round this to a more convenient number, such as 0.00272 g/lb for practical purposes.
B) To determine the correct dose for a 191-lb patient, we multiply the patient's weight in pounds by the dosage in grams per pound:
Dose (in g) = patient's weight (in lb) * dosage (in g/lb)
Dose (in g) = 191-lb x 0.00272 g/lb
Dose (in g) =0.519 g
Therefore, the correct dose for a 191-lb patient is approximately 0.519 grams.
A student is heating a chemical in a beaker with a Bunsen burner.
In a paragraph of at least 150 words, identify the safety equipment that should be used and the purpose of it for the given scenario.
Answer:The student should be wearing a lab coat or maybe an apron to prevent chemicals from spilling or exploding onto their clothes, I do recommend a lab coat better though because it can protect your skin better. Next, make sure while messing with chemicals you are always wearing goggles, if you are not wearing them there is a chance that after touching chemicals you could touch your eyes. And that brings me to washing your hands straight away after messing with chemicals. You could also wear gloves and just take them off when you're done but if you don't have clean hands afterward you could always put the chemicals all over your skin. But in case you do touch your eyes there is always an emergency eyewash station somewhere in the lab room. And if you are to get Chemicals on your skin, in your hair, on your clothes, or to be on fire, there shall be a shower somewhere to get rid of that. But if you read the instructions or listen closely to the teacher you shall have no problem.
Explanation:
I kinda got off topic
When a student is warming a chemical in a container using a special burner, it is very important to focus on safety by using the right safety tools.
What is the safety equipmentFirst, the student needs to wear the right safety clothes like a lab coat, gloves, and goggles to protect themselves from getting splashed or hurt by chemicals. A lab coat stops chemicals from touching the skin, gloves keep the hands safe, and safety goggles protect the eyes from chemicals
and hot things.
Furthermore, using a fume hood is necessary to make sure there is enough fresh air circulating and to remove any dangerous fumes or gases that might be released while heating things up.
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Which of the following reagents and biomolecules are necessary to make a recombinant DNA? a. Restriction enzyme b. DNA fragment to be cloned C. Glucose d. DNA ligase e. Chymotrypsin f. Calcium chloride g. starch h. Ampicillin
Answer:
All of them can be necessary.
Explanation:
In typical DNA cloning, the gene of interest is inserted into a plasmid, this is achieved by using enzymes that "cut and paste" DNA producing a recombinant DNA, considering this we will first need a DNA fragment to be cloned. To "cut and paste" these fragments of DNA we will need restriction enzymes (to cut) and DNA ligase (to paste), this enzyme will recognize the specific target sequence and I'll cut it, another restriction enzyme will also cut the plasmid, then DNA ligase will link the plasmid and target gene together. Now we need to introduce the plasmid into bacteria, to extract it we use glucose as a buffer to maintain the pH-controlled for the plasmid to be stable, so that linear dsDNA (sheared chromosomal DNA) is denatured but closed-circular DNA (plasmid) is not. Once we have our plasmid isolated we can put it into our bacteria (this is called transformation), this is achieved by giving the bacteria a shock that encourages them to take foreign DNA, calcium chloride can improve the results by binding plasmids to lipopolysaccharides in the bacteria. After this shock, some bacteria will accept the plasmid but a portion won't, this is why plasmids typically contain antibiotic resistance genes to allow the bacteria that contain the plasmids to survive after the application of such antibiotic, this means ampicillin is also necessary to isolate our bacteria with recombinant DNA. Finally, you can use these bacteria as "factories" to produce proteins and then obtain them by splitting the bacteria, to achieve this splitting we can use proteases, for example, chymotrypsin. NOw you'll need to purify the proteins you extract one method to do it is using the starch binding domain (SBD) that can be found in some amylolytic enzymes, we can add a recombinant proteins for transferring the starch binding capacity to the target proteins, we will observe both proteins fused to the SBDtag, only the target protein will remain over the starch granules after the wash process.
I hope you find this information useful and interesting! good luck!
For each of the following units and concentration values, mention if they are parts per million (ppm), parts per billion (ppb) or parts per trillion (ppt) to
a. g / Ton
b. mg / L
c. μg / L
d. ng / L (ng = nanogram) and e
e. mg / ton
Answer:
a) ppm
b) ppm
c) ppb
d) ppt
e) ppb
Explanation:
a) You know that 1000 g are 1 kg, and 1000 kg are 1 ton, so (1000)*(1000) g are 1 ton, so 1,000,000 grams are one ton.
b) 1000 mg are 1 g, and 1000 g are 1 liter, so 1,000,000 grams are one liter.
c) You know that 1000 ug are 1 mg, so with the b), we just need to multiply the answer by 1000, so 1,000,000,000 ug are 1 liter.
d) The same as c, 1000 ng are 1 mg. So we are talkinf of ppt.
e) 1000 mg are 1 g. And 1000 g are 1 kg, then 1000 kg are one ton. So 1,000,000,000 mg are one ton.
Determine the molality of a solution of benzene dissolved in toluene (methylbenzene) for which the mole fraction of benzene is 0.176. Give your answer to 2 decimal places
Answer:
2.32 m
Explanation:
So, according to definition of mole fraction:
[tex]Mole\ fraction\ of\ benzene=\frac {n_{benzene}}{n_{benzene}+n_{toluene}}[/tex]
Mole fraction = 0.176
Applying values as:
[tex]0.176=\frac {n_{benzene}}{n_{benzene}+n_{toluene}}[/tex]
[tex]0.176\times ({n_{benzene}+n_{toluene}})={n_{benzene}}[/tex]
So,
[tex]0.176\times n_{toluene}}=0.824\times {n_{benzene}}[/tex]
[tex]{n_{benzene}}=\frac {0.176}{0.824}\times n_{toluene}}[/tex]
[tex]{n_{benzene}}=0.2136\times n_{toluene}}[/tex]
Also, Molar mass of toluene = 92.14 g/mol
Thus,
The formula for the calculation of moles is shown below:
[tex]moles = \frac{Mass\ taken}{Molar\ mass}[/tex]
[tex]Mass=92.14\times n_{toluene}}\ g[/tex]
Also, 1 g = 0.001 kg
So,
[tex]Mass\ of\ toluene=0.09214\times n_{toluene}}\ kg[/tex]
Molality is defined as the moles of the solute present in 1 kg of the solvent.
It is represented by 'm'.
Thus,
[tex]Molality\ (m)=\frac {0.2136\times n_{toluene}}{0.09214\times n_{toluene}}[/tex]
Molality of benzene = 2.32 m
Final answer:
The molality of the solution of benzene dissolved in toluene is 0.00543 mol/kg.
Explanation:
The molality of a solution can be calculated using the mole fraction and the molar mass of benzene and toluene. The mole fraction of benzene can be calculated by dividing the moles of benzene by the total moles of benzene and toluene. The moles of benzene can be determined using the given volumes of benzene and toluene and their respective densities. The molar mass of benzene is 78.11 g/mol. By substituting the values into the formula, we can calculate the molality of the solution to be 0.176 * 78.11 g/mol / (1000 g + 0.867 g/mL * 100 mL + 0.874 g/mL * 300 mL) = 0.00543 mol/kg.
2mL of a serum sample was added to 18mL of phosphate buffered saline (PBS) in Tube 1. 10mL of Tube 1 was added to 40mL of PBS in Tube What is the dilution of serum in Tube 2?
Answer:
Tube 2 has a total dilution of 1:50
Explanation:
We have a 2 ml serum sample added to a 18 mL phosphate buffered saline sample in tube 1. This means now in tube 1 there is 20 mL.
We have a 1:10 (= 2:20) dilution here.
10 ml of this 1:10 diluted tube 1 is taken and added to a 40 mL of PBS in tube 2.
Now we have 50 mL in tube 2.
This is a 10:50 (= 1:5) dilution.
The total dilution is 10x5 = 50
So the total ditultion has a rate 1:50
Tube 2 has a total dilution of 1:50
During the experiment a student precipitated and digested the BaSO4. After allowing the precipitate to settle, they added a few drops of BaCl2 solution, and the previously clear solution became cloudy. Explain what happened.
Answer:
Incomplete precipitation of barium sulfate
Explanation:
The student has precipitated and digested the barium sulfate on his/her side. But on the addition of [tex]BaCl_2[/tex] in the solution, the solution become cloudy. This happened because incomplete precipitation of barium sulfate by the student. When [tex]BaCl_2[/tex] is added, there are still sulfate ions present in the solution with combines with [tex]BaCl_2[/tex] and forms [tex]BaSO_4[/tex] and the formation of this precipitate makes the solution cloudy.
The solution became cloudy after adding BaCl2 because the additional Ba2+ ions reacted with any remaining [tex]SO4^2-[/tex]orming more BaSO4 precipitate, indicating incomplete digestion of the initial BaSO4.
Explanation:During the experiment, the student added a few drops of BaCl2 solution to the clear solution that contained digested BaSO4. BaSO4 is known for its low solubility in water; however, it's soluble in solutions containing ions that can form more soluble compounds with the constituent ions. In the experiment, adding more BaCl2 likely introduced additional Ba2+ ions into the solution. If any unreacted sulfate ions (SO42-) were present, these extra Ba2+ ions could have reacted with them, forming additional BaSO4 precipitate, thus causing the solution to become cloudy. This suggests that the digestion process was not complete, leaving some sulfate ions in the solution which reacted with the added barium ions to form more BaSO4 precipitate.
In water, hydroxides of Group 2 metals a. are all strong bases. b. are all weak bases. c. are all acids. d. are nonelectrolytes
Answer:
The correct answer is: a. are all strong bases
Explanation:
Alkaline earth metals are the chemical elements that belong to the group 2 of the periodic table. The members or elements of this group are all highly reactive metals.
Except beryllium (Be), all the alkaline earth metals react with water to give metal hydroxides. These hydroxides of the alkaline earth metals are highly soluble and very strong bases.
Final answer:
Hydroxides of Group 2 metals in water are all strong bases because they dissociate almost completely into ions, significantly raising the solution's pH by releasing a high concentration of OH- ions.
Explanation:
The question asks about the nature of hydroxides of Group 2 metals when dissolved in water. Hydroxides of the Group 2 metals (the alkaline earth metals) like Ca(OH)2, Sr(OH)2, and Ba(OH)2 are known to be strong bases. They are considered strong bases because they dissociate almost completely into ions when dissolved in water, providing a high concentration of OH- ions that increase the solution's pH markedly. To answer the provided options, a. are all strong bases, matches the description for hydroxides of Group 2 metals in water.
Henry low is Obeyed by a gas when gas has high • Pressure • Temperature • Solubility • Non of the above
Answer:
None of the above
Explanation:
Henry's law -
This law was given by William Henry in the year 1803 , it is also known as the gas law ,
According to Henry's law , the amount of gas which gets dissolved in a liquid is directly related to the partial pressure of gas that is in equilibrium with the liquid , at a constant temperature .
Or it can stated as ,
The gases' solubility in a liquid is directly related to the partial pressure of gas that is in equilibrium with the liquid .
This law is applicable for sparingly soluble gases in liquid solvents .
And the solubility of a gas is independent of temperature and pressure .
Hence ,
the correct option is None of the above .
A polymer P is made up of two monodisperse fractions; fraction A with molecular weight of 1000 g/mole and fraction B with a molecular weight of 100,000 g/mole. The batch contains an equal mole fraction of each fraction. Calculate the number average and the Weight average of polymer P.
Answer:
a)Number average molecular weight is 50, 500 g/mol
b) Weight average molecular weight is 99, 019.8 g/mol
Explanation:
We have a polymer P made up of two monodisperse fractions.
A with molecular weight of MA = 1000 g/mol and B with MB =100000 g/mol.
The batch contains an equal mole fraction of each component A and B.
Let's suppose a total number (Nt) of mols 2 moles. Equal fraction means XA = 0.5 and XB =0.5
Nt = 2 mol
Na = 2*0.5 = 1 mol
Nb = 2*0.5 = 1 mol.
So, we have 1 mol of A, 1 mol of B and 2 moles in total.
a) The number average molecular weight (NAM) is calculate using the mole numbers of each component. In this case, we will multiple each component molecular weight by the number of moles of each one. After that we will sum them and finally to divide by the total number of moles.
NAM = (Na*MA + Nb*MB)/(Nt)
NAM = (1 mol *1000 g/mol + 1*100000 g/mol ) /(2 mol)
NAM = 50500 g/mol
The number average molecular weight for the polymer P is 50,500 g/mol
b) Weight average molecular weight (WAM) is calculated using the mass quantities of each component. Weight mass of A (WA), weight mass of B (WB) are calculate using the moles of A, B and their molecular weights respectively. Total Weight (WT)
WA = Na*MA = 1 mol *1000 g/mol = 1000 g A
WB = Nb*MB = 1mol * 100000 g/mol = 100 000 gB
WT = WA + WB = 101 000 g
Now we will calculate average molecular using weights, we will multiple each component molecular weight by the mass of each one. After that we will sum them and finally to divide by the total mass.
WAM = (WA*MA + WB*MB)/(WT)
WAM = (1000 g *1000 g/mol + 100000 g*100000 g/mol )/(101 000 g)
WAM = 99 019.8 g/mol
The weight average molecular weight for polymer P is 99, 019.8 g/mol
Final answer:
The number average molecular weight (Mn) for the polymer is 50,500 g/mol, while the weight average molecular weight (Mw) is 198,019,802 g/mol, given the equal mole fraction of the two monomer fractions.
Explanation:
The number average molecular weight (Mn) and the weight average molecular weight (Mw) of a polymer comprising two monodisperse fractions.
Firstly, to calculate the Mn, we consider the definition that Mn is the total weight of the polymer divided by the total number of moles. Since each fraction has an equal mole fraction, we take the simple average of the two given molecular weights:
Mn = (1000 g/mol + 100,000 g/mol) / 2 = 50,500 g/molNext, to compute the Mw, which is the sum of the products of the weight contribution of each fraction and its molecular weight squared, divided by the total weight:
Mw = [(1/2) × 1000 g/mol × 1000 g/mol + (1/2) × 100,000 g/mol × 100,000 g/mol] / [(1/2) × 1000 g/mol + (1/2) × 100,000 g/mol]Mw = [1,000,000 + 1 × 10^10] / 50,500Mw = 198,019,802 g/molTo summarize, for a polymer with an equal mole fraction of two fractions with molecular weights of 1000 g/mol and 100,000 g/mol, the number average molecular weight is 50,500 g/mol and the weight average molecular weight is 198,019,802 g/mol.
The distance from Earth to the Moon is approximately 240.000 mi Part C The speed of light is 3.00 x 10 m/s How long does it take for light to travel from Earth to the Moon and back again? Express your answer using two significant figures.
Answer:
2.6 sec
Explanation:
The distance between the Earth and the moon = 240,000 miles
Also,
1 mile = 1609.34 m
So,
Distance between the Earth and the moon = 240,000 × 1609.34 m = 386241600 m
Speed of the light = 3 × 10⁸ m/s
Distance = Speed × Time.
So,
Time = Distance / Speed = 386241600 m / 3 × 10⁸ m/s = 1.3 sec
For back journey = 1.3 sec
So, total time = 2.6 sec
Chemical Equations
Instructions: Solve the following chemical equations.
For the following reaction, calculate how many moles of NO2forms when 0.356 moles of the reactant completely reacts. 2 N2O5(g) ---> 4 NO2(g) + 02(g)
Answer:
0.712 moles of NO₂ are formed.
Explanation:
First, we need to write the balanced equation:
2 N₂O₅(g) ⇄ 4 NO₂(g) + O₂(g)
From the balanced equation, we can see the relationship between the moles of N₂O₅ and the moles of NO₂. Every 2 moles of N₂O₅ that react, 4 moles of NO₂ are formed. Let us apply this relationship to the information given by the problem (0.356 moles of N₂O₅):
[tex]0.356molN_{2}O_{5}.\frac{4molNO_{2}}{2molN_{2}O_{5}} =0.712molNO_{2}[/tex]
39.20 mL of 0.5000 M AgNO3 is added to 270.00 mL
ofwater which contains 5.832 g K2CrO4. A
redprecipitate of Ag2CrO4 forms. What is
theconcentration, in mol/L, of
unprecipitatedCrO42-? Be sure to enter
the correct numberof significant figures. Assume
Ag2CrO4is completely insoluble.
Answer:
concentration of CrO4²⁻ ions in the final solution = 6.53 × 10⁻⁵ mol /L
Explanation:
First we calculate the number of moles of AgNO₃:
number of moles = molar concentration × volume
number of moles = 0.5 × 39.20 = 19.6 mmoles = 0,0196 moles AgNO₃
Then we calculate the number of moles of K₂CrO₄:
number of moles = mass / molar weight
number of moles = 5.832 / 194 = 0.03 moles K₂CrO₄
The chemical reaction will look like this:
2 AgNO₃ + K₂CrO₄ → Ag₂CrO₄ + 2 KNO₃
Now we devise the following reasoning:
if 2 moles of AgNO₃ are reacting with 1 mole of K₂CrO₄
then 0,0196 moles of AgNO₃ are reacting with X moles of K₂CrO₄
X = (0.0196 × 1) / 2 = 0.0098 moles of K₂CrO₄
now the the we calculate the amount of unreacted K₂CrO₄:
unreacted K₂CrO₄ = 0.03 - 0.0098 = 0.0202 moles
now the molar concentration of CrO4²⁻ ions:
molar concentration = number of moles / solution volume (L)
molar concentration = 0.0202 / (39.20 + 270) = 6.53 × 10⁻⁵ mol /L
The following procedure was carried out to determine thevolume
of a flask. The flask was weighed dry and then filled withwater. If
the masses of the empty flask and filled flask were 56.12g and
87.39 g, respectively, and the density of water is 0.9976g/mL,
Calculate the volume of the flask in mL.
Answer: Volume of the flask is 31.34 mL.
Explanation:
Weight of empty flask [tex]w_1[/tex] = 56.12 grams
Weight of flask with water [tex]w_2[/tex] = 87.39 grams
Weight of water [tex]w_3[/tex] = [tex]w_2-w_1[/tex] = (87.39 - 56.12) grams= 31.27 grams
Density of water = 0.9976 g/mL
0.9976 grams are contained in = 1 ml of water
Thus 31.27 grams are contained in = [tex]\frac{1}{0.9976}\times 31.27=31.34[/tex] ml of water.
Thus the volume of the flask is 31.34 mL.
The volume of the flask is calculated by subtracting the mass of the empty flask from the mass of the filled flask and then dividing by the density of water. The resulting volume is approximately 31.36 mL.
To calculate the volume of the flask in mL, we need to use the mass of the water that filled the flask and the known density of water. First, we find the mass of the water by subtracting the mass of the empty flask from the mass of the filled flask. Then, we use the density formula, which is Density = Mass/Volume, to find the volume of the flask.
Here's the step-by-step calculation:
Determine the mass of the water by subtracting the mass of the empty flask from the filled flask: 87.39 g - 56.12 g = 31.27 g.
Using the known density of water (0.9976 g/mL), calculate the volume using the formula Volume = Mass/Density.
Volume of water (Volume of flask) = 31.27 g / 0.9976 g/mL = 31.36 mL.
Therefore, the volume of the flask is approximately 31.36 mL.
What is the volumetric flow rate in L/s of a stream of air (density = 1 kg/m3) at 1 kg/s?
Answer:
Volumetric flow rate: Q = 1000 L/s
Explanation:
Volumetric flow rate, also called the rate of fluid flow, is described as volume of fluid that passes a particular point per unit time. The SI unit of volumetric flow rate is m³/s.
Whereas, mass flow rate is defined as the mass of substance that passes through a point per unit of time. SI unit is kg/s.
Given- mass flow rate: ṁ = 1 kg/s and density: ρ = 1 kg/m³
Therefore, volumetric flow rate can be calculated by
[tex]Q = \frac{\dot{m}}{\rho } = \frac{1 kg/s}{1 kg/m^{3}} = 1 m^{3}/s[/tex]
Since, 1 m³/s = 1000 L/s
Therefore, volumetric flow rate: Q = 1 m³/s = 1000 L/s
A two-liter soft drink bottle can withstand apressure of
5 atm. Half a cup (approximately 120mL) of ethynlalcohol, C2H5OH,
(d=0.789 g/mL) is poured into a soft drink bottleat room
temperature. The bottle is then heated to 100C, changingthe liquid
alcohol to a gas. Will the soft drink bottle withstandthe pressure
or explode?
Explanation:
According to the ideal gas equation, PV = nRT.
where, P = pressure, V = volume
n = no. of moles, R = gas constant
T = temperature
Also, density is equal to mass divided by volume. And, no. of moles equals mass divided by molar mass.
Therefore, then formula for ideal gas could also be as follows.
P = [tex]\frac{mass}{volume \times molar mass} \times RT[/tex]
or, P = [tex]\frac{density}{\text{molar mass}} \times RT[/tex]
Since, density is given as 0.789 g/ml which is also equal to 789 g/L (as 1000 mL = 1 L). Hence, putting the given values into the above formula as follows.
P = [tex]\frac{density}{\text{molar mass}} \times RT[/tex]
= [tex]\frac{789 g/l}{46.06 g/mol} \times 0.0821 L atm/mol K \times 373 K[/tex]
= 525 atm
As two-liter soft drink bottle can withstand a pressure of 5 atm and the value of calculated pressure is 525 atm which is much greater than 5 atm.
Therefore, the soft drink bottle will obviously explode.
Charles' law relates the way two gas properties change when another property remains the same. What are the two changing properties in Charles' law?
Pressure and temperature
Pressure and volume
Pressure, temperature, and volume
Temperature and volume
ik its not B
Answer:
The two changing properties in Charles’ law are temperature and volume.
Explanation:
Charles’ law state the presence of direct relationship between temperature and volume of the system of gas molecules at constant pressure condition. In this law, the expansion of gas has been explained with the increase of temperature.
As the temperature is increased or the system of gas molecules are heated, the gas molecules tend to expand their volume to maintain the pressure same.
So the temperature and volume are directly proportional at constant pressure. Thus the two changing properties of Charles’ law is temperature and volume. The mathematical representation of Charles’ law is
V∝T (at constant Pressure)
[tex]V=kT[/tex]
Here k is the non-zero constant and V and T are volume and temperature respectively.
Choose the pure substance from the list below. osea water sugar air lemonade milk Question 2 (1 point) Saved A chemical change O occurs when methane gas is burned. occurs when paper is shredded. occurs when water is vaporized. occurs when salt is dissolved in water. occurs when powdered lemonade is stirred into water.
Sugar is the pure substance in the list. A chemical change occurs when methane gas is burned, as the chemical composition of methane is altered.
Explanation:In the list provided, the pure substance is sugar. Pure substances are materials that are composed of only one type of particle, and sugar, which is made up of sucrose, meets this criterion. In contrast, substances like sea water, air, lemonade, and milk are mixtures, as they contain more than one kind of particle.
As for the second question, a chemical change happens when the composition of a substance is altered by a chemical reaction. In this scenario, the correct option is 'occurs when methane gas is burned'. Methane's combustion is a chemical reaction that produces water and carbon dioxide, which are chemically distinct from the original methane. Other options like shredding paper, vaporizing water, dissolving salt in water, or stirring powdered lemonade into water are all physical changes, as they don't alter the chemical composition of the substances involved.
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Sugar is considered a pure substance because it maintains the same composition and properties. A chemical change occurs when methane gas is burned, resulting in different substances.
Explanation:From the given list, the pure substance is sugar. A pure substance has a constant composition. All specimens of a pure substance have exactly the same makeup and properties. An example of a pure substance is table sugar, or sucrose, which consists of 42.1% carbon, 6.5% hydrogen, and 51.4% oxygen by mass. It maintains the same physical properties, such as melting point, color, and sweetness, regardless of the source from which it is isolated.
Switching to the second question, the occurrence of a chemical change is evident when methane gas is burned. A chemical change always produces one or more types of matter that differ from the matter present before the change. An instance of this is the combustion or burning of methane which results in carbon dioxide and water. These yielded substances significantly differ from the original methane, hence signifying a chemical change.
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How much power (energy per unit time) can be provided by a 75 m high waterfall with a flow rate of 10,000 L/s? Give answer in kW rate given here is volume per unit time; 10,000 L/s mean that every second 10,000 L of water go through the water fall
Explanation:
It is given that flow rate is 10,000 L/s. As 1 L equals 0.001 [tex]m^{3}[/tex].
Hence, flow rate will be 10 [tex]m^{3}/s[/tex]. Calculate mass of water flowing per second as follows.
Mass flowing per second = density × flow rate
= [tex]1000 kg/m^{3} \times 10 m^{3}/s[/tex]
= [tex]10^{4} kg/s[/tex]
Also, energy provided per second will be as follows.
E = mgh
Putting the given values into the above formula as follows.
E = mgh
= [tex]10^{4} kg/s \times 9.8 m/s^{2} \times 75 m[/tex]
= [tex]735 \times 10^{4} W[/tex]
or, = 7350 kW
Thus, we can conclude that energy per unit time provided will be 7350 kW.
Write the isotopic symbol for the following (show your work) a) An isotope of iodine whose atoms have 78 neutrons b) An isotope of cesium whose atoms have 82 neutrons c) An isotope of strontium whose atoms have 52 neutrons
Final answer:
The isotopic symbol for an isotope of iodine with 78 neutrons is 131I53. The isotopic symbol for an isotope of cesium with 82 neutrons is 137Cs55. The isotopic symbol for an isotope of strontium with 52 neutrons is 90Sr38.
Explanation:
An isotope of iodine with 78 neutrons would have an atomic number of 53. To write the isotopic symbol, we include the atomic number as a subscript and the mass number (atomic number + number of neutrons) as a superscript. Therefore, the isotopic symbol for the iodine isotope with 78 neutrons would be 131I53.
An isotope of cesium with 82 neutrons would have an atomic number of 55. So, the isotopic symbol for the cesium isotope with 82 neutrons would be 137Cs55.
An isotope of strontium with 52 neutrons would have an atomic number of 38. Hence, the isotopic symbol for the strontium isotope with 52 neutrons would be 90Sr38.
Determine the minimum work required by an air compressor. At the inlet the conditions are 150 kg/min, 125 kPa and 33 °C. At the exit, the pressure is 550 kPa. Assume air is an ideal gas with MW 29 g/mol, Cp 3.5R (constant).
Explanation:
The given data is as follows.
MW = 29 g/mol, [tex]C_{p}[/tex] = 3.5 R
Formula to calculate minimum amount of work is as follows.
[tex]W_{s} = C_{p}T_{1}[(\frac{P_{2}}{P_{1}})^{\frac{R}{C_{p}}} - 1][/tex]
= [tex]3.5 \times 8.314 J/k mol \times 306 \times [(\frac{550}{125})^{\frac{1}{3.5}} - 1][/tex]
= 4.692 kJ/mol
Therefore, total work done will be calculated as follows.
Total work done = [tex]m \times W_{s}[/tex]
Since, m = [tex]\frac{150 \times 10^{3}g/min}{29}[/tex]. Therefore, putting these values into the above formula as follows.
Total work done = [tex]m \times W_{s}[/tex]
= [tex]\frac{150 \times 10^{3}g/min}{29} \times 4.692 kJ/min[/tex]
= 24268.96 kJ/min
It is known that 1 kJ/min = 0.0166 kW. Hence, convert 24268.96 kJ/min into kW as follows.
[tex]24268.96 kJ/min \times \frac{0.0166 kW}{1 kJ/min}[/tex]
= 402.86 kW
Thus, we can conclude that the minimum work required by an air compressor is 402.86 kW.
The solubility of acetanilide is 12.8 g in 100 mL of ethanol at 0 ∘C, and 46.4 g in 100 mL of ethanol at 60 ∘C. What is the maximum percent recovery that can be achieved for the recrystallization of acetanilide from ethanol?
A student was given a sample of crude acetanilide to recrystallize. The initial mass of the the crude acetanilide was 171 mg.The mass after recrystallization was 125 mg.
Calculate the percent recovery from recrystallization.
Answer: 72.41% and 26.90% respectively.
Explanation:
At 60°C, you can dissolve 46.4g of acetanilide in 100mL of ethanol. If you lower the temperature, at 0°C, you can dissolve just 12.8g, which means (46.4g-12.8g)=33.6g of acetanilide must have precipitated from the solution.
We can calculate recovery as:
[tex]\%R=\frac{crystalized\ mass}{initial\ mass}*100 =\frac{33.6\ g}{46.4\ g}*100=72.41\%[/tex]
So the answer to the first question is 72.41%.
For the second part just use the same formula, the mass of the precipitate is the final mass minus the initial mass, (171mg-125mg)=46mg.
[tex]\%R=\frac{crystalized\ mass}{initial\ mass}*100 =\frac{46\ mg}{171\ mg}*100=26.90\%[/tex]
So the answer to the second question is 26.90%.
What will be the final pH when 5.865 mL of 3.412 M NaOH is added to 0.5000 L of 1.564 x 10-3 M HCl?
Answer: 12.5
Explanation:
Molarity of a solution is defined as the number of moles of solute dissolved per Liter of the solution.
[tex]Molarity=\frac{moles}{\text {Volume in L}}[/tex]
moles of [tex]NaOH=Molarity\times {\text {Volume in L}}=3.412\times 5.865\times 10^{-3}L=0.02moles[/tex]
moles of [tex]HCl=Molarity\times {\text {Volume in L}}=1.564\times 10^{-3}\times 0.5000L=0.782\times 10^{-3}moles[/tex]
[tex]NaOH+HCl\rightarrow NaCl+H_2O[/tex]
According to stoichiometry:
1 mole of [tex]HCl[/tex] require 1 mole of [tex]NaOH[/tex]
Thus [tex]0.782\times 10^{-3}moles[/tex] will combine with [tex]0.782\times 10^{-3}moles[/tex] of [tex]NaOH[/tex]
Thus [tex](0.02-0.782\times 10^{-3})moles=0.019moles[/tex] of [tex]NaOH[/tex] will be left
Thus Molarity of [tex]OH^-=\frac{0.019}{0.56L}=0.03M[/tex]
[tex]pOH=-\log[OH^-][/tex]
Putting in the values:
[tex]pOH=-\log[0.03][/tex]
[tex]pOH=1.5[/tex]
[tex]pH+pOH=14[/tex]
[tex]pH=14-1.5=12.5[/tex]
Thus final pH will be 12.5.
Draw the Lewis Structure for NaCl
Explanation:
The electronic configuration of sodium with Z = 11 is : 2, 8, 1
The electronic configuration of chlorine with Z = 17 is : 2, 8, 7
The Lewis structure is drawn in such a way that the octet of each atom is complete.
Thus, sodium losses one electron to chlorine and chlorine accepts this electron to form ionic bond.
Thus, the valence electrons are shown by dots in Lewis structure. The structure is shown in image below.
The Lewis structure of NaCl involves sodium losing one electron to become Na+ and chlorine gaining one electron to become Cl-. Draw the ions next to each other to represent the ionic bond. Sodium has no electrons around it while chlorine has a complete octet with the extra electron in brackets.
The Lewis structure of sodium chloride (NaCl), follow these steps:
Identify the valence electrons: Sodium (Na) is in group 1 and has 1 valence electron. Chlorine (Cl) is in group 17 and has 7 valence electrons.Show the transfer of electrons: Sodium will lose its 1 valence electron to achieve a stable electron configuration, becoming a positively charged ion (Na+). Chlorine will gain this electron to complete its octet, becoming a negatively charged ion (Cl-).Represent the ions: Write the Lewis structures of the resulting ions next to each other to indicate the ionic bond:Na: Na+ [ ]
Cl: [ :Cl: ]-
Note that the brackets around the chlorine indicate it has gained an electron and the overall charge of the ion.
The mean for a set of measurements is 4.17 and the standard deviation is 0.14. To show 95% confidence limits, we can write the result of the measurements as 4.17+ -
Answer:
The 95% confidence level is
[tex]4.17 \pm 0.2744[/tex]
Explanation:
If we can apply the central limit theorem, we can approximate this distribution to a normal distribution.
The confidence level (for n=1) is defined as
[tex]X\pm \frac{z*\sigma}{\sqrt{n}}=X\pm z*\sigma[/tex]
For a 95% confidence interval, according to the normal distribution, z=1.96.
Then we have:
[tex]X\pm z*\sigma=4.17 \pm 1.96*0.14=4.17 \pm 0.2744[/tex]
Final answer:
To calculate the 95% confidence interval for a set of measurements with a mean of 3.095 g and a standard deviation of 0.0346 g, you first determine the standard error, then use it to find the margin of error, and finally compute the confidence interval, yielding an estimate that the true mean weight of a penny lies between 3.088 and 3.102 g.
Explanation:
To find the 95% confidence interval for a set of measurements, you utilize statistical principles that involve the mean, standard deviation, and sample size of the data. The basic formula to calculate the confidence interval is mean ± (critical value) * (standard deviation / √n), where √n is the square root of the sample size, and the critical value is determined based on the confidence level. For a 95% confidence level, the critical value (often represented as a z-score in the context of a normal distribution) is approximately 1.96.
Let's assume you're working with a sample size large enough for the central limit theorem to apply, simplifying the computation of the confidence interval to just involving the mean and standard deviation, due to a large sample size making the distribution of sample means approximately normal. Therefore, for the sample of 100 pennies with a mean of 3.095 g and a standard deviation of 0.0346 g, the 95% confidence interval can be calculated as follows:
First, calculate the standard error: SE = 0.0346 / √100 = 0.00346.Then, calculate the margin of error: Margin of Error = 1.96 * SE = 1.96 * 0.00346 = approximately 0.00678.Finally, determine the confidence interval: 3.095 ± 0.00678, which calculates to an interval of approximately 3.088 to 3.102 g.This calculation reveals that we estimate with 95% confidence that the true mean weight of a penny lies between 3.088 and 3.102 g.
Calculate ln (0.345). Report your answer to 3 significant figures.
Answer:
-1.06
Explanation:
0.345 has three significant figures since the zero does not represent a significant figure.
So, usign the calculator, we find ln (0.345) = -1.06421086195
Now, the problem says that you need to report your answer to three significant figures.
So, you should take the first three numbers of the answer:
ln (0.345) = -1.06
The rate of reaction at 550 K is ten times faster than the rate of reaction at 440 K. Find the activation energy from the collision theory. a) 40075.14 J/mol b) 50078.5J/mol c) 44574.5 J/mol d) 43475.5 J/mol
Answer : a) 40075.14 J/mol
Explanation :
According to the Arrhenius equation,
[tex]K=A\times e^{\frac{-Ea}{RT}}[/tex]
or,
[tex]\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}][/tex]
where,
[tex]K_1[/tex] = rate constant at [tex]440K[/tex] = k
[tex]K_2[/tex] = rate constant at [tex]550K[/tex] = 10 k
[tex]Ea[/tex] = activation energy for the reaction = ?
R = gas constant = 8.314 J/mole.K
[tex]T_1[/tex] = initial temperature = 440 K
[tex]T_2[/tex] = final temperature = 550 K
Now put all the given values in this formula, we get :
[tex]\log (\frac{10k}{k})=\frac{Ea}{2.303\times 8.314J/mole.K}[\frac{1}{440K}-\frac{1}{550K}][/tex]
[tex]Ea=40075.14J/mol [/tex]
Therefore, the activation energy for the reaction is 40075.14J/mol.