Explanation:
- Simple Distillation: its a separation method that can be used when the two or more liquids in the mix have at least 50 degrees of difference between their boiling points.
-Azeotropic distillation: is a technique to break an azeotrope (constant boiling point mixtures), that can't be separated by simple distillation, by adding another component to generate a new azeotrope (between one initial component and the new one added) with lower boiling point.
-Extractive distillation: is a process to separate mixtures with close boiling points by adding a miscible, high boiling or none volatile solvent to increase the relative volatility of the liquids in the mix, this increases the separation factor. It differences from the azeotropic method because it doesn't form an azeotrope.
-Liquid-liquid extraction: is a method to separate compounds based on their relative solubilities in two different immiscible liquids.
After describing all the methods we can conclude that all of them are methods to separate substances based on their physical properties, this is their similarity. The difference between this method is the property it uses to separate (solubility in the case of extraction and boiling point in the case of destinations), the cases in which they bare used (when the liquids difference in boiling points is bigger [simple] or close [attractive and azeotropic]) and the formation of azeotropes (present in azeotropic and absent in extractive).
I hope you find this information useful and interesting! Good luck!
Drag the tiles to the correct boxes to complete the pairs. Not all tiles will be used.
The number above the element’s symbol is the element’s atomic number. Match each element with the number of valence electrons it has.
Answer:
4= Si
5=P
3= Al
6= S
2= Mg
Explanation:
4 → Si
Silicon is part of the carbon group and has 14 electrons, spread over 3 shells:
⇒ on the first shell (K) 2 electrons
⇒on the second shell (L) 8 electrons
⇒ on the third shell (M) the remaining electrons : 14-8-2 = 4
5 → P
phosphorus is part of the nitrogen group and has 15 electrons, spread over 3 shells:
⇒ on the first shell (K) 2 electrons
⇒on the second shell (L) 8 electrons
⇒ on the third shell (M) the remaining electrons : 15-8-2 = 5
3 → Al
Aluminium is part of the Boron group and has 13 electrons, spread over 3 shells:
⇒ on the first shell (K) 2 electrons
⇒on the second shell (L) 8 electrons
⇒ on the third shell (M) the remaining electrons : 13-8-2 = 3
6 → S
Sulfur is part of the oxygen group and has 16 electrons, spread over 3 shells:
⇒ on the first shell (K) 2 electrons
⇒on the second shell (L) 8 electrons
⇒ on the third shell (M) the remaining electrons : 16-8-2 = 6
2 → Mg
Magnesium is part of the Beryllium or alkaline earth metals group and has 12 electrons, spread over 3 shells:
⇒ on the first shell (K) 2 electrons
⇒on the second shell (L) 8 electrons
⇒ on the third shell (M) the remaining electrons : 12-8-2 = 2
The completed pairs of element with the number of valence electrons are:
aluminum ⇒ 3sulfur ⇒ 6sodium ⇒ 1silicon ⇒ 4magnesium ⇒ 2phosphorus ⇒ 5What are these valence electrons?Valence electrons are the electrons in the outermost shell of an atom that can participate in chemical bonding. The number of valence electrons an element has determines its chemical properties.
For example, sodium (Na) has 1 valence electron, which it can donate to other atoms to form bonds. This is why sodium is a reactive metal.
Sulfur (S) has 6 valence electrons, which it can share with other atoms to form bonds. This is why sulfur can form a variety of compounds with other elements.
The number of valence electrons an element has can also be determined by looking at its periodic table group. Elements in the same group have the same number of valence electrons.
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An infant ibuprofen suspension contains 100 mg/5.0mL suspension. The recommended dose is 10 mg/kg body weight.How many milliliters of this suspension should be given to an infant weighing 15 lb ? (Assume two significant figures.)
Answer : 3.4 mL of this suspension should be given to an infant weighing 15 lb.
Explanation :
First we have to convert weight of infant from 'lb' to 'kg'.
Conversion used :
As, 1 lb = 0.454 kg
So, 15 lb = [tex]\frac{15lb}{1lb}\times 0.454kg=6.81kg[/tex]
Now we have to calculate the recommended dose.
As, 1 kg weight recommended dose 10 mg
So, 6.81 kg weight recommended dose [tex]\frac{6.81kg}{1kg}\times 10mg=68.1mg[/tex]
Now we have to calculate the milliliter of ibuprofen suspension.
As, 100 mg dose present in 5.0 mL
So, 68.1 mg dose present in [tex]\frac{68.1mg}{100mg}\times 5.0mL=3.4mL[/tex]
Therefore, 3.4 mL of this suspension should be given to an infant weighing 15 lb.
The chemical formula for sucrose (also known as table sugar) is C12H22011. How many carbon atoms are in 1 molecule of sucrose? O a. 22 O b. 13 Oc. 12 O d. 11 Oe. 72
Round the following to three significant figures. a 17.5555 km 16 b. 1.0007 c.23.34271d. 99.991
Methods: Part A: Preparation of Buffers Make two buffers starting with solid material, which is the most common way to make buffers. You will be given a desired pH, and your task is to prepare 100 mL of two appropriate buffers at a concentration of 0.10 M. One of the buffers will be a phosphate buffer (pH 7.0) and the other will be a Tris buffer (pH 8.0). 1. Calculate the weight of the buffer you would need to make 100 mL of a 0.10 M solution. Weigh out the correct amount and dissolve in 50 mL water. You will need the following compound molecular weights: Na2HPO4 (141.96 g/mol), NaH2PO4 (119.96 g/mol), and Tris base (121.1 g/mol).
Answer:
0,542 g of Na₂HPO₄ and 0,741 g of NaH₂PO₄.
0,856 g of Tris-HCl and 0,553 g of Tris-base
Explanation:
It is possible to use Henderson–Hasselbalch equation to estimate pH in a buffer solution:
pH = pka + log₁₀ [tex]\frac{A^{-} }{HA}[/tex]
Where A⁻ is conjugate base and HA is conjugate acid
The equilibrium of phosphate buffer is:
H₂PO₄⁻ ⇄ HPO4²⁻ + H⁺ Kₐ₂ = 6,20x10⁻⁸; pka=7,21
Thus, Henderson–Hasselbalch equation for phosphate buffer is:
pH = 7,21 + log₁₀ [tex]\frac{HPO4^{2-} }{H2PO4^{-} }[/tex]
If desire pH is 7,0 you will obtain:
0,617 = [tex]\frac{HPO4^{2-} }{H2PO4^{-} }[/tex] (1)
Then, if desire concentration of buffers is 0,10 M:
0,10 M = [HPO₄²⁻] + [H₂PO₄⁻] (2)
Replacing (1) in (2) you will obtain:
[H₂PO₄⁻] = 0,0618 M
And with this value:
[HPO₄²⁻] = 0,0382 M
As desire volume is 100mL -0,1L- the weight of both Na₂HPO₄ and NaH₂PO₄ is:
Na₂HPO₄ = 0,1 L× [tex]\frac{0,0382mol}{1L}[/tex]× [tex]\frac{141,96g}{1mol}[/tex] = 0,542 g of Na₂HPO₄
NaH₂PO₄ = 0,1 L× [tex]\frac{0,0618mol}{1L}[/tex]× [tex]\frac{119,96g}{1mol}[/tex] = 0,741 g of NaH₂PO₄
For tris buffer the equilibrium is:
Tris-base + H⁺ ⇄ Tris-H⁺ pka = 8,075
Henderson–Hasselbalch equation for tris buffer is:
pH = 8,075 + log₁₀ [tex]\frac{Tris-base }{Tris-H^{+} }[/tex]
If desire pH is 8,0 you will obtain:
0,841 = [tex]\frac{Tris-base }{TrisH^{+} }[/tex] (3)
Then, if desire concentration of buffers is 0,10 M:
0,10 M = [Tris-base] + [Tris-H⁺] (4)
Replacing (3) in (4) you will obtain:
[Tris-HCl] = 0,0543 M
[Tris-base] = 0,0457 M
As desire volume is 100mL -0,1L- the weight of both Tris-base and Tris-HCl is:
Tris-base = 0,1 L× [tex]\frac{0,0457mol}{1L}[/tex]× [tex]\frac{121,1g}{1mol}[/tex] = 0,553 g of Tris-base
Tris-HCl = 0,1 L× [tex]\frac{0,0543mol}{1L}[/tex]× [tex]\frac{157,6g}{1mol}[/tex] = 0,856 g of Tris-HCl
I hope it helps!
To make a 0.10 M solution of the desired buffers, you need to calculate the weight of the buffer needed based on the desired pH and molar mass of the buffer. For the phosphate buffer, you would need to weigh out 1.4196 grams of Na2HPO4 and NaH2PO4. For the Tris buffer, you would need to weigh out 1.211 grams of Tris base.
Explanation:In order to calculate the weight of the buffer needed to make a 0.10 M solution, you need to use the formula:
Weight of buffer = (desired concentration) * (volume of solution) * (molar mass of buffer)
For the phosphate buffer, with a desired pH of 7.0, you can use the compounds Na₂HPO₄ and Na₂HPO₄. The weight of Na₂HPO₄ needed would be 0.10 mol/L * 0.1 L * 141.96 g/mol, which is 1.4196 grams. The weight of Na₂HPO₄ needed would be the same.
For the Tris buffer, with a desired pH of 8.0, you would use Tris base. The weight of the Tris base needed would be 0.10 mol/L * 0.1 L * 121.1 g/mol, which is 1.211 grams.
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van der Waals bonding can be important to the modulus of which of the following? Select one: a. metals only b. some polymers c. ceramics only d. hydrogen bonded materials e. metals and polymers
Answer:
Hydrogen bonded materials
Explanation:
Metals have metallic bonds, polymers usually have covalent bonds and ceramic materials have covalent and ionic bonds; on the other hand, hydrogen bonds are a type of bonding characterized for have Van der Waals interactions, a type of intermolecular interactions, so the correct answer is Van der Waals interactions.
Which one of the following statements concerning the Boltzmann distribution is TRUE?
Select one
a)The Boltzmann equation expresses a statistical population distribution corresponding to maximum entropy
b)The Boltzmann equation applies to all systems, whether or not they are in thermal equilibrium
c)The Boltzmann distribution is not valid at very low temperatures.
d)Upper energy levels are always less populated than lower energy levels, regardless of degeneracy.
e)At room temperature, a Boltzmann distribution predicts few rotational levels are populated but many electronic levels.
Answer:
c)The Boltzmann distribution is not valid at very low temperatures.
Explanation:
Option cis correct because the Boltzman distribution can be applied when the temperature is high enough or the particle density is low enough to render quantum effects negligible.
The Boltzman distribution in statistics describe the distribution of particles over different energy states when there is thermal equilibrium (that is why option b is INCORRECT).
You are given a protein solution with a concentration of 0.15 mg/ml.
v. Suppose that we want to prepare 100 microliters of 0.1 micrograms/microliters solution. How much of H2O and protein stock do we need to add to obtain the target concentration and volume?
Explanation:
The given data is as follows.
Initial concentration = 0.15 mg/ml,
Final concentration = 0.1 mg/ml, (as [tex]\frac{0.1 microgram}{1 microliter} \times \frac{10^{-3}}{1 microgram} \times \frac{1 microliter}{10^{-3}ml}[/tex])
Final volume = [tex]100 microliter \times \frac{10^{-3} ml}{1 microliter}[/tex] = 0.1 ml
According to the dilution formula we get the following.
[tex]C_{i} \times V_{i} = C_{f} \times V_{f}[/tex]
or, [tex]V_{i}[/tex] = [tex]\frac{C_{f} \times V_{f}}{C_{i}}[/tex]
Putting the given values into the above formula we get the following.
[tex]V_{i}[/tex] = [tex]\frac{C_{f} \times V_{f}}{C_{i}}[/tex]
= [tex]\frac{0.1 mg/ml \times 0.1 ml}{0.15 mg/ml}[/tex]
= 0.0667 ml
= [tex]6.67 \times 10^{-2} ml \times \frac{1 microliter}{10^{-3} ml}[/tex]
= 66.7 microliter
This means that volume of protein stock which is required is 66.7 ml. Hence, calculate the volume of water required as follows.
Volume of water required = Total volume - volume of protein stock
= (100 - 66.7) microliter
= 33.3 microliter
Thus, we can conclude that we need 33.3 microliter of water and 66.7 microliter of protein.
Which of the following elements are alkaline earth metals?
Check all that apply.
barium
lithium
aluminium
calcium
sodium
Answer:
The alkaline earth metals are barium and calcium.
Explanation:
Alkaline earth metals is the denomination given to the six elements that occupy Group 2 in the Periodic Table, therefore they have an electron configuration that ends in the orbital s².
From the list: barium and calcium are alkaline earth metals. On the other hand, lithium and sodium occupy Group 1, and aluminum occupies Group 3.
Among the listed elements, barium and calcium are the alkaline earth metals, which belong to Group 2 of the periodic table.
Explanation:The question concerns the identification of alkaline earth metals, which are elements in Group 2 of the periodic table. From the given options, the alkaline earth metals are barium and calcium. These elements are known for being shiny, silvery-white, and somewhat reactive. They exhibit distinctive properties such as higher ionization energies than alkali metals and the ability to lose electrons to form compounds with a +2 oxidation state.
A highly concentrated ammonia solution is found to have a pH of 12.03. What is the hydroxide concentration, [OH-(aq)] in mol L-1?
Answer:
0.01348 M (mol L-1)
Explanation:
Water can be autoionized:
H2O + H2O ⇄ H3O+ + OH-
This a reversible reaction and then it has a unique constant Kw that at 25ºC it takes the value of 10^-14
[tex]10^{-14} =[H_{3} O+][OH-][/tex]
pH is the concentration of [H3O+] ions in a logaritmic scale:
[tex]pH = -log([H_{3}O+][/tex]
Then you can solve for the [OH-]
[tex][OH-]=(Kw)/(10^{-pH})=(10^{-14})/(10^{-12.03})=0.01348 M[/tex]
Hope it helps!
The type(s) of intermolecular forces exhibited by hydrogen bromide molecules, HBr is/are. Fill in the blank
Answer:
London dispersion forces and dipole-dipole interactions
Explanation:
Intermolecular forces are the interactive forces that are present between the molecules. These intermolecular forces can be attractive or repulsive.
Hydrogen bromide is a diatomic covalent polar molecule, having chemical formula: HBr. Since, hydrogen bromide is a polar molecule, it has a permanent dipole.
Therefore, the intermolecular forces present between the hydrogen bromide molecules are dipole-dipole interactions and the london dispersion forces.
How many moles of disulfur decafluoride are present in 2.45 grams of this compound? moles
Answer:
Take account the molar mass of this compound (first of all u should know the formula, S2F10). As the molar mass is 254,1 g/mol you will know that in 1 mol, you have 254,1 g so make a rule of three to solve it. If we find 254,1 g of S2F10 in 1 mol, 2,45 g of it are in 9,64 *10^-3 moles. That is the right number.
Explanation:
how many protons are contained in the nucleus of the atom?
Convert 2.0 M of Phenobarbital sodium (MW: 254 g/mole) solution in water into % w/v and ratio strengths.
Answer:
The concentration is 50,8 % w/v and radio strengths = 1,96.
Explanation:
Phenobarbital sodium is a medication that could treat insomnia, for example.
2,0 M of Phenobarbital sodium means 2 moles in 1L.
The concentration units in this case are %w/v that means 1g in 100 mL and ratio strengths that means 1g in r mL. Thus, 2 moles must be converted in grams with molar weight -254 g/mole- and liters to mililiters -1 L are 1000mL-. So:
2 moles × [tex]\frac{254 g}{1 mole}[/tex]= 508 g of Phenobarbital sodium.
1 L ×[tex]\frac{1000 mL}{ 1 L}[/tex] = 1000 mL of solution
Thus, % w/v is:
[tex]\frac{508 g}{1000 mL}[/tex] × 100 = 50,8 % w/v
And radio strengths:
[tex]\frac{1000 mL}{508 g}[/tex] = 1,96. Thus, you have 1 g in 1,96 mL
I hope it helps!
The combustion of 44.45 grams of a hydrocarbon gas mixture releases 2,649 kJ of heat. How much energy, in kWh, will the combustion of exactly one pound of this hydrocarbon gas produce? There are 453.592 g in 1 lb. Please express your answer to the appropriate number of significant figures. Your Answer:
Explanation:
It is given that 44.45 g of hydrocarbon gas produces 2649 kJ or [tex]2649 \times 10^{3} J[/tex] of heat.
Also here, 1 lb = 453.592 g.
Therefore, amount of energy released by 453.592 g of hydrocarbon gas will be calculated as follows.
[tex]\frac{2649 \times 10^{3} J \times 453.592 g}{44.45 g}[/tex]
= [tex]27.03 \times 10^{6} J[/tex]
It is known that 1 J = [tex]2.778 \times 10^{-7} Kwh[/tex].
Hence, [tex]27.03 \times 10^{6} J[/tex] = [tex]27.03 \times 10^{6} J \times 2.778 \times 10^{-7} Kwh/J[/tex]
= 7.508 Kwh
Thus, we can conclude that the combustion of exactly one pound of this hydrocarbon gas produce 7.508 Kwh energy.
Calculate the pH of a solution prepared by mixing: (Show your work for these calculations) pk of acetic acid is 4.75 a. two moles of sodium acetate with two moles of acetic acid. b. two moles of sodium acetate with one mole of acetic acid.
Answer:
Explanation:
To calculate pH you need to use Henderson-Hasselbalch formula:
pH = pka + log₁₀ [tex]\frac{[A^-]}{[HA]}[/tex]
Where HA is the acid concentration and A⁻ is the conjugate base concentration.
The equilibrium of acetic acid is:
CH₃COOH ⇄ CH₃COO⁻ + H⁺ pka: 4,75
Where CH₃COOH is the acid and CH₃COO⁻ is the conjugate base.
Thus, Henderson-Hasselbalch formula for acetic acid equilibrium is:
pH = 4,75 + log₁₀ [tex]\frac{[CH_{3}COO^-]}{[CH_{3}COOH]}[/tex]
a) The pH is:
pH = 4,75 + log₁₀ [tex]\frac{[2 mol]}{[2 mol]}[/tex]
pH = 4,75
b) The pH is:
pH = 4,75 + log₁₀ [tex]\frac{[2 mol]}{[1mol]}[/tex]
pH = 5,05
I hope it helps!
Convert 1.248×1010 g to each of the following units.
a) kg Express your answer using four significant figures.
b) Mg Express your answer using four significant figures.
c) mg Express your answer using four significant figures.
d) metric tons (1 metric ton = 1000 kg) . Express your answer using four significant figures.
Final answer:
To convert 1.248×1010 g to different units, one must use specific conversion factors for each unit, resulting in 1.248×107 kg, 1.248×104 Mg or metric tons, and 1.248×1013 mg, all while maintaining four significant figures.
Explanation:
To convert 1.248×1010 g to various units, we'll use the relationships between grams, kilograms, milligrams, and metric tons.
Kilograms (kg): Since 1 kg = 103 g, divide by 103 to convert. 1.248×1010 g = 1.248×107 kg.Megagrams (Mg) or metric tons: Since 1 Mg = 106 g, divide by 106 to convert. 1.248×1010 g = 1.248×104 Mg or metric tons.Milligrams (mg): Since 1 mg = 10-3 g, multiply by 103 to convert. 1.248×1010 g = 1.248×1013 mg.Metric tons: Utilize the same conversion as for Megagrams, since 1 metric ton = 1 Mg = 1.248×104 metric tons.It's important to keep significant figures in mind, maintaining four significant figures through each conversion to ensure precision.
35S has a half-life of 88 days. You start with 128 micrograms of 3Ss. How much 35s is left after 264 days? O a. 32 micrograms b.8 micrograms O c. 64 micrograms O d. 16 micrograms O e. 4 micrograms
Answer:
Option d, Amount left = 16 μg
Explanation:
Half life = 88 days
Initial concentration = 128 μg
[tex]No.\ of\ half\ life=\frac{Total\ days}{t_{1/2}}[/tex]
[tex]No.\ of\ half\ life=\frac{264}{88} = 3[/tex]
Amount of material left after n half life = [tex]\frac{Amount\ of\ starting\ material}{2^n}[/tex]
Where, n = No. of half life
Amount of material left = [tex]\frac{128}{2^3}[/tex]
= [tex]\frac{128}{8} = 16[/tex]μg
SO, option d is correct
Name the simplest "aldose" and "ketone"
Explanation:
Aldose is monosaccharide sugar in which the carbon backbone chain has carbonyl group on endmost carbon atom which corresponds to an aldehyde, and the hydroxyl groups are connected to all other carbon atoms.
Example - Glucose
Ketone is functional group with structure RC(=O)R', where the groups, R and R' can be variety of the carbon-containing substituents.
Example, Ethylmethylketone
Final answer:
The simplest aldose is glyceraldehyde, and the simplest ketone is acetone. These classifications are based on the presence of a carbonyl group, with aldoses having the group at the end of the chain and ketones within the chain. This naming follows IUPAC nomenclature rules with the suffixes -al for aldehydes and -one for ketones.
Explanation:
The simplest aldose is glyceraldehyde, which is a three-carbon aldehyde with the molecular formula C₃H₆O₃. It contains a carbonyl group (CHO) at the end of the carbon chain, making it an aldehyde. The simplest ketone is acetone (propanone), which has the molecular formula C₃H₆O. It contains a carbonyl group (CO) within the carbon chain, which classifies it as a ketone.
Both aldehydes and ketones contain a carbonyl group, a functional group with a carbon-oxygen double bond. The nomenclature rules for naming these compounds involve the use of the suffixes -al for aldehydes and -one for ketones. For instance, the name glyceraldehyde is derived from glycerol (the alcohol form), with the ending changed to -al to indicate it's an aldehyde, while the name acetone is derived from the root word for two carbon groups (acet-) with the ending -one to signify it's a ketone.
Write 0.70894 in Engineering Notation with 3 significant figures.
Final answer:
The number 0.70894 is written as 709 × 10⁻³ in Engineering Notation with 3 significant figures.
Explanation:
To write the number 0.70894 in Engineering Notation with 3 significant figures, we first identify the significant figures in the number. Since leading zeros are not significant, the significant figures are 708. We then adjust the number to have a multiple of three as its exponent for the base unit, which is standard in Engineering Notation. The number can be written as 709 (rounded up from 708.94 to maintain 3 significant figures) multiplied by 10 to the power of -3. Therefore, the number in Engineering Notation is 709 × 10⁻³.
How many plutonium atoms are present in 6.1 moles of plutonium metal? Give your answer in scientific notation. Enter your answer in the provided box. x 10 (select) ^ atoms of Pu
Answer: The number of atoms of plutonium present in given number of moles are [tex]3.7\times 10^{24}[/tex]
Explanation:
We are given:
Number of moles of plutonium metal = 6.1 moles
According to mole concept:
1 mole of an element contains [tex]6.022\times 10^{23}[/tex] number of atoms.
So, 6.1 moles of plutonium will contain = [tex]6.1\times 6.022\times 10^{23}=3.7\times 10^{24}[/tex] number of plutonium atoms.
Hence, the number of atoms of plutonium present in given number of moles are [tex]3.7\times 10^{24}[/tex]
Final answer:
To determine the number of plutonium atoms in 6.1 moles, multiply 6.1 by Avogadro's number (6.022 x 10^23 atoms/mole), yielding 3.673 x 10^24 atoms of plutonium in scientific notation.
Explanation:
The question asks how many plutonium atoms are present in 6.1 moles of plutonium metal. To find this, we use Avogadro's number, which is 6.022 × 1023 atoms/mole. This means there are 6.022 × 1023 atoms of any element per mole of that element.
To calculate the total number of atoms in 6.1 moles of plutonium, you multiply the number of moles by Avogadro's number:
Number of atoms = 6.1 moles × 6.022 × 1023 atoms/mole
This calculation gives you the answer in scientific notation: 3.673 × 1024 atoms of Pu.
Immediately after the Chernobyl nuclear accident, the concentratiorn of 137Cs (cesium 137) in cow's milk was 12,000 Bq/L (a Becquerel is a measure of radioactivity; one Becquerel equals one radioactive disintegration per second). Assume that the only reaction by which the 137Cs was lost from the soil was through radioactive decay. Also assume that the concentration in cows milk is directly proportional to the concentration in the soil. Calculate the concentration of 137Cs in cow's milk (from feeding on grass in the soil) 5 years after the accident given a half-life for 137Cs of 30 years.
Answer: The concentration of cow's milk after 5 years is 10691 Bq/L
Explanation:
All the radioactive reactions follow first order kinetics.
The equation used to calculate rate constant from given half life for first order kinetics:
[tex]t_{1/2}=\frac{0.693}{k}[/tex]
We are given:
[tex]t_{1/2}=30yrs[/tex]
Putting values in above equation, we get:
[tex]k=\frac{0.693}{30}=0.0231yr^{-1}[/tex]
The equation used to calculate time period follows:
[tex]N=N_o\times e^{-k\times t}[/tex]
where,
[tex]N_o[/tex] = initial concentration of Cow's milk = 12000 Bq/L
N = Concentration of cow's milk after 5 years = ?
t = time = 5 years
k = rate constant = [tex]0.0231yr^{-1}[/tex]
Putting values in above equation, we get:
[tex]N=12000Bq/L\times e^{-(0.0231yr^{-1}\times 5yr)}\\\\N_o=10691Bq/L[/tex]
Hence, the concentration of cow's milk after 5 years is 10691 Bq/L
One liter of ocean water contains 35.06 g of salt. How much salt is present in 3.79 L of ocean water? Express your answer in grams using the correct number of significant figures. Do not enter your answer using scientific notation.
Answer:
133g
Explanation:
If one litre of ocean water contains 35.06g of salt, all we have to to to get the mass of salt in 3.79 L of ocean water is multiply 35.06g/L by 3.79 L.
35.06g/L × 3.79 L = 133g
The amount of salt in 3.79 L of ocean water can be found by multiplying the volume of the water by the constant ratio of 35.06 g of salt per liter. This gives 132.8774 grams, which when we round to 4 significant figures is equivalent to 132.9 g.
Explanation:To find the amount of salt in 3.79 L of ocean water, we use the given ratio of salt to water: 1 liter of ocean water contains 35.06 grams of salt. So to find the amount of salt in any given volume of ocean water, we just multiply the volume (in liters) by this constant ratio of 35.06 grams of salt per liter.
Applying this to the volume of 3.79 L we get: 3.79 L * 35.06 g/L = 132.8774 grams. Since we should express our answer using the correct number of significant figures (4 in this case, because there are 4 significant figures in the 3.79 L volume measurement), the answer is rounded to 132.9 g.
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Gallium oxide, Ga.Oy, forms when gallium is combined with oxygen. A 1.25 g of Ga is allowed to react with excess oxygen and 1.68 g of the gallium oxide formed. What is the chemical formula of the gallium oxide if the molar mass is 187.44 g/mol?
Answer : The chemical formula of a compound is, [tex]Ga_2O_3[/tex]
Solution : Given,
Mass of gallium = 1.25 g
Mass of gallium oxide = 1.68 g
Mass of oxygen = Mass of gallium oxide - Mass of gallium
Mass of oxygen = 1.68 - 1.25 = 0.43 g
Molar mass of Ga = 69.72 g/mole
Molar mass of O = 16 g/mole
Step 1 : convert given masses into moles.
Moles of Ga = [tex]\frac{\text{ given mass of Ga}}{\text{ molar mass of Ga}}= \frac{1.25g}{69.72g/mole}=0.0179moles[/tex]
Moles of O = [tex]\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{0.43g}{16g/mole}=0.027moles[/tex]
Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.
For Ga = [tex]\frac{0.0179}{0.0179}=1[/tex]
For O = [tex]\frac{0.027}{0.0179}=1.5[/tex]
The ratio of Ga : O = 1 : 1.5
To make in whole number we multiple ratio by 2, we get:
The ratio of Ga : O = 2 : 3
The mole ratio of the element is represented by subscripts in empirical formula.
The Empirical formula = [tex]Ga_2O_3[/tex]
The empirical formula weight = 2(69.72) + 3(16) = 187.44 gram/eq
Now we have to calculate the molecular formula of the compound.
Formula used :
[tex]n=\frac{\text{Molecular formula}}{\text{Empirical formula weight}}[/tex]
[tex]n=\frac{187.44}{187.44}=1[/tex]
Molecular formula = [tex](Ga_2O_3)_n=(Ga_2O_3)_1=Ga_2O_3[/tex]
Therefore, the chemical of the compound is, [tex]Ga_2O_3[/tex]
The empirical formula of gallium oxide, formed by the reaction of gallium with excess oxygen, is Ga₂O₃.
To determine the chemical formula of gallium oxide formed, follow these steps:
The mass of Ga used is 1.25 g and the mass of Ga oxide produced is 1.68 g. First, find the mass of oxygen in the oxide:Mass of O = 1.68 g (mass of Ga oxide) - 1.25 g (mass of Ga) = 0.43 g
Next, convert these masses to moles:Moles of Ga = 1.25 g / 69.72 g/mol (molar mass of Ga) = 0.0179 moles
Moles of O = 0.43 g / 16.00 g/mol (molar mass of O) = 0.0269 moles
Determine the simplest whole number ratio of Ga to O:Ratio of Ga to O = 0.0179 : 0.0269 ≈ 2:3
The given molar mass of the gallium oxide (187.44 g/mol) confirms the empirical formula Ga₂O₃.
if tempreture dry=25 C , PH=45% , what it actual temperature? by using Psychrometric Chart
1) 19 C
2) 29 C
3) 39 C
4) 49 C
Answer:
19°C
Explanation:
When you use a psychrometric chart the thing that you need to know is the value of a least two variables of your system, then you need to intercept these variables and then you can find the other variables seeing the others axis. In your case, you know the dry temperature and the percentage of humidity, so as you see in the attach image your temperature is between 15°C and 20°C. The only choice in that range is 19°C.
A mixture of ethyl acetate vapour and air has a relative saturation of 50% at 303 K and a total pressure of 100 kPa. If the vapour pressure of ethyl acetate at 303 K is 16 kPa, the molal saturation is (a) 0.080 (b) 0.087 (c) 0.264 (d) 0.029
Answer : The correct option is, (b) 0.087
Explanation :
The formula used for relative saturation is:
[tex]\text{Relative saturation}=\frac{P_A}{P_A^o}[/tex]
where,
[tex]P_A[/tex] = partial pressure of ethyl acetate
[tex]P_A^o[/tex] = vapor pressure of ethyl acetate
Given:
Relative saturation = 50 % = 0.5
Vapor pressure of ethyl acetate = 16 kPa
Now put all the given values in the above formula, we get:
[tex]0.5=\frac{P_A}{16kPa}[/tex]
[tex]P_A=8kPa[/tex]
Now we have to calculate the molar saturation.
The formula used for molar saturation is:
[tex]\text{Molar saturation}=\frac{P_{vapor}}{P_{\text{vapor free}}}[/tex]
and,
P(vapor free) = Total pressure - Vapor pressure
P(vapor) = [tex]P_A[/tex] = 8 kPa
So,
P(vapor free) = 100 kPa - 8 kPa = 92 kPa
The molar saturation will be:
[tex]\text{Molar saturation}=\frac{P_{vapor}}{P_{\text{vapor free}}}[/tex]
[tex]\text{Molar saturation}=\frac{8kPa}{92kPa}=0.087[/tex]
Therefore, the molar saturation is 0.087
In a study of the conversion of methane to other fuels, a chemical engineer mixes gaseous methane and gaseous water in a 0.778 L flask at 1,072 K. At equilibrium, the flask contains 0.206 mol of CO gas, 0.187 mol of H2 gas, and 0.187 mol of methane. What is the water concentration at equilibrium (Kc = 0.30 for this process at 1,072 K)? Enter to 4 decimal places. HINT: Look at sample problem 17.7 in the 8th ed Silberberg book. Write a balanced chemical equation. Write the Kc expression. Calculate the equilibrium concentrations of all the species given (moles/liter). Put values into Kc expression, solve for the unknown.
Answer:
the water concentration at equilibrium is
⇒ [ H2O(g) ] = 0.0510 mol/L
Explanation:
CH4(g) + H2O(g) ↔ CO(g) + 3H2(g)∴ Kc = ( [ CO(g) ] * [ H2 ]³ ) / ( [ CH4(g) ] * [ H2O(g) ] ) = 0,30
equilibrium:⇒ [ CO(g) ] = 0.206 mol / 0.778 L = 0.2648 mol/L
⇒ [ H2(g) ] = 0.187 mol / 0.778 L = 0.2404 mol/L
⇒ [ CH4(g) ] = 0.187 mol / 0.778 L = 0.2404 mol/L
replacing in Kc:
⇒ ((0.2648) * (0.2404)³) / ([ H2O(g) ] * 0.2404 ) = 0.30
⇒ 0.0721 [ H2O(g) ] = 3.679 E-3
⇒ [ H2O(g) ] = 0.0510 mol/L
A water treatment plant has 4 settling tanks that operate in parallel (the flow gets split into 4 equal flow streams), and each tank has a volume of 600 m3 . If the total flow to the plant is 12 MGD, what is the retention time in each of the settling tanks? b) If, instead, the tanks operated in series (the entire flow goes first through one tank, then the second, and so on), what would be the retention time in each tank?
Answer:
a) When the 4 tanks operate in parallel the retention time is 1.26 hours.
b) If the tanks are in series, the retention time would be 0.31 hours
Explanation:
The plant has 4 tanks, each tank has a volume V = 600 [tex]m_3[/tex]. The total flow to the plant is Ft = 12 MGD (Millions of gallons per day)
When we use the tanks in parallel, it means that the total flow will be divided in the total number of tanks. F1 will be the flow of each tank.
Firstly, we should convert the MGD to [tex]m_3 /day[/tex]. In that sense, we can calculate the retention time using the tank volume in [tex]m_3[/tex].
[tex]Ft =12 \frac{MG}{day} (\frac{1*10^6 gall}{1MG} ) (\frac{3.78 L}{1 gall} ) (\frac{1 dm^3}{1L} ) (\frac{1 m^3}{10^3 dm^3} ) = 45360 \frac{m^3}{day}[/tex]
After that, we should divide the total flow by four, because we have four tanks.
[tex]F1 = Ft/4 =(45360 \frac{m^3}{d} )/4 = 11 340 \frac{m^3}{d}[/tex]
To calculate the retention time we divide the total volume V by the flow of each tank F1.
[tex]t1 =\frac{V1}{F1} = \frac{600 m^3}{11340 \frac{m^3}{d} } = 0.0529 day\\t1 = 0.0529 d (\frac{24 h}{1d} ) = 1.26 h[/tex]
After converting t1 to hours we found that the retention time when the four reactors are in parallel is 1.26 hours.
b)
If the four reactors were working in series, the entire flow goes first through one tank, then the second and so on. It means the total flow will be the flow of each tank.
In that order of ideas, the flow for reactors in series will be F2, and will have the same value of F0.
F2 = F0
To calculate the retention time t2 we divide the total volume V by the flow of each tank F2.
[tex]t2 =\frac{V1}{F2} = \frac{600 m^3}{45360 \frac{m^3}{d} } = 0.01 day\\t2 = 0.01 d (\frac{24 h}{1d} ) = 0.31 h[/tex]
After converting t2 to hours we found that the retention time when the four reactors are in series is 0.31 hours.
a) Retention time in each settling tank for parallel operation is approximately 317.0 hours.
b) Retention time in each settling tank for series operation is also approximately 317.0 hours.
To find the retention time in each settling tank, we'll use the formula:
[tex]\[ \text{Retention Time} = \frac{\text{Volume of Tank}}{\text{Flow Rate}} \][/tex]
Where:
- Volume of Tank is the volume of each settling tank.
- Flow Rate is the total flow rate.
We'll first convert the flow rate to cubic meters per day (m\(^3\)/day) to match the units of the tank volume:
[tex]\[ 1 \, \text{MGD} = 1 \, \text{million gallons} \times \frac{3.785 \times 10^3 \, \text{liters}}{1 \, \text{m}^3} \times \frac{1 \, \text{day}}{24 \, \text{hours}} \]\[ = 1 \, \text{MGD} \times \frac{3.785 \times 10^3 \, \text{liters}}{1 \, \text{m}^3} \times \frac{1 \, \text{day}}{24 \, \text{hours}} \times \frac{10^{-6} \, \text{m}^3}{1 \, \text{liter}} \]\[ = \frac{3.785 \times 10^3}{24 \times 10^6} \, \text{m}^3/\text{hour} \]\[ = 0.1577 \, \text{m}^3/\text{hour} \][/tex]
Now we can calculate the retention time for both scenarios:
a) For parallel operation:
[tex]\[ \text{Retention Time (Parallel)} = \frac{V_{\text{tank}}}{Q} = \frac{600 \, \text{m}^3}{1.8924 \, \text{m}^3/\text{hour}} \]\[ \text{Retention Time (Parallel)} \approx 317.0 \, \text{hours} \][/tex]
b) For series operation:
In series, the total flow rate remains the same, but the volume is effectively multiplied by the number of tanks in series. So, the retention time for each tank remains the same as in the parallel case:
[tex]\[ \text{Retention Time (Series)} = \frac{V_{\text{tank}}}{Q} = \frac{600 \, \text{m}^3}{1.8924 \, \text{m}^3/\text{hour}} \]\[ \text{Retention Time (Series)} \approx 317.0 \, \text{hours} \][/tex]
Therefore, the retention time in each settling tank is approximately 317.0 hours for both parallel and series operation.
How many grams of calcium chloride would you need to make 1 L of solution containing 5 ppm calcium (Ca)?
Explanation:
The given data is as follows.
Volume = 1 L, Concentration of Ca = 5 ppm or 5 mg/L
As 1 mg = 0.001 g so, 5 mg /L will be equal to 0.005 g/l. Molar mass of calcium is 40.078 g/mol.
Hence, calculate molarity of calcium as follows.
Molarity of Ca = [tex]\frac{\text{given concentration}}{\text{molar mass}}[/tex]
= [tex]\frac{0.005 g/l}{40.078 g/mol}[/tex]
Molarity of Ca = [tex]1.25 \times 10^{-4} M[/tex]
Hence, molarity of [tex]CaCl_{2}[/tex] is [tex]1.25 \times 10^{-4} M[/tex]. Since, volume is same so, moles of calcium chloride will be [tex]1.25 \times 10^{-4} mol[/tex].
Thus, we can conclude that mass of [tex]CaCl_{2}[/tex] will be as follows.
[tex]1.25 \times 10^{-4} \times 110.984[/tex] (molar mass of [tex]CaCl_{2}[/tex] = 110.984 g/mol)
= 0.0138 g
Thus, we can conclude that mass of [tex]CaCl_{2}[/tex] is 0.0138 g.
Why does a lower concentration of dissolved particles decrease the reaction rate? A. It does not. A high concentration decreases reaction rate. B. When there are less dissolved particles, less collisions take place. C. Concentration is measured inversely. A low concentration indicates a higher portion of dissolved reactant particles. D. The dissolved reactant cannot collide with enough energy to react because it is moving slowly when it is dissolved in a solution.
Answer: Option (B) is the correct answer.
Explanation:
Whenever there is less concentration of solute particles in a solvent then it means less number of solute particles are available. As a result, there will occur less number of collisions between the solvent and solute particles.
It means that there will be a decrease in rate of reaction.
But if there is more concentration of solute particles then it shows more number of solute particles are available for reaction. As a result, more number of collisions will take place between the particles of solute and solvent.
Hence, then there will occur an increase in rate of reaction.
Thus, we can conclude that a lower concentration of dissolved particles decrease the reaction rate because when there are less dissolved particles, less collisions take place.
Answer:
B
Explanation: