Answer:
0.8 mL.
Explanation:
You need to know that the equivalence in mcg to mg is 1000 mcg are 1 mg; so 500 mcg are 0.5 mg. Now you know that there are 0.5 mg per 2 mL, so if you divide all by 2, you will know that you have 0.25 mg per mL. Now you applied a rule of three: ((0.2 mg)(1 ml))/(.25 mg) = 0.8 mL. So the nurse need to administer 0.8 mL to provide 0.2 mg of dose.
Perform the following calculation, giving the answer in the correct number of significant digits. (1.145 x 109 g/mol) (0.0035 mol)/(8.57 x 10 g/mL)=
Answer:
The answer is 0.005097 mL.
Explanation:
Significant figures : The figures in a number which express the value of the magnitude of a quantity to a specific degree of an accuracy is known as significant digits.
Given :
[tex]x=\frac{(1.145 x 109 g/mol)\times (0.0035 mol)}{(8.57 x 10 g/mL)}[/tex]
[tex]x=\frac{0.4368 g}{8.57\times 10 g/mL}=0.005097053 mL[/tex]
[tex]x=0.005097053 mL\approx 0.005097 mL[/tex]
You are given a solution that is 518 mM lactose. You need to make up 4.5 L of 16.7 mM solution. What volume do you need to transfer from your stock?
Answer:
The volume you need to transfer from the stock solution is 0.145 l
Explanation:
Since the number of moles of lactose in the volume of stock solution that you transfer will be the same as the number of moles of lactose in the final solution, you can use this expression:
number of moles in volume to transfer = number of moles in the final solution
Since number of moles = concentration * volume (if the concentration is expressed in molarity), then:
Ci * Vi = Cf * Vf
where:
Ci = concentration of the stock solution.
Vi = volume of the stock solution to be transferred.
Cf = concentration of the final solution
Vf = volume of the final solution
Then, replacing with the data:
518 mM * Vi = 16.7 mM * 4.5 l
Vi = 16.7 mM * 4.5 l / 518 mM
Vi = 0.145 l or 145 ml
Notice that any concentration unit can be used, as long as the units of the concentration of the stock and final solution are the same.
What is the mass of a proton in Gigagrams? ) 1.67 10-39Gg C ) 1.67 10-33 Gg D) 1.67 10-30G
Answer: The mass of proton is [tex]1.67\times 10^{-33}Gg[/tex]
Explanation:
Proton is one of the sub-atomic particle which is present in the nucleus of the atom.
We know that:
Mass of proton = [tex]1.67\times 10^{-27}kg[/tex]
To convert this into Giga grams, we use the conversion factor:
[tex]1Gg=1\times 10^6kg[/tex]
Converting the given value into Giga grams, we get:
[tex]\Rightarrow 1.67\times 10^{-27}kg\times \frac{1Gg}{1\times 10^6kg}=1.67\times 10^{-33}Gg[/tex]
Hence, the mass of proton is [tex]1.67\times 10^{-33}Gg[/tex]
The oxidation of the sugar glucose, C6H12O6, is described by the following equation. C6H12O6(s) + 6 O2(g) → 6 CO2(g) + 6 H2O(l) ΔH = −2802.5 kJ/mol The metabolism of glucose gives the same products, although the glucose reacts with oxygen in a series of steps in the body.
(a) How much heat in kilojoules can be produced by the metabolism of 18.1 g of glucose?
(b) How many Calories can be produced by the metabolism of 18.1 g of glucose?
Answer:
(a) 282 kJ
(b) 67.4 Calories
Explanation:
(a) The molar enthalpy, ΔH = −2802.5 kJ/mol, means that the heat produced by the reaction is 2802.5 kJ per mol of glucose.
We can multiply the enthalpy by the number of moles of glucose to get the heat produced by the metabolism. Grams of glucose will be converted to moles using the molar mass of glucose (180.156 g/mol):
(18.1 g)(mol/180.156g)(2802.5 kJ/mol) = 282 kJ
(b) Using the result we obtained above, kJ will be converted to Calories using the conversion factor of 4.184J = 1 cal. Calorie with a capital C is the same as a kilocalorie.
(282 kJ)(1 cal/4.184J) = 67.4 kcal = 67.4 Calories
Answer:
For a: The amount of heat produced for given amount of glucose is -283.05 kJ
For b: The amount of heat produced for given amount of glucose is -67648.9 Cal
Explanation:
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
Given mass of glucose = 18.1 g
Molar mass of glucose = 180.16 g/mol
Putting values in above equation, we get:
[tex]\text{Moles of glucose}=\frac{18.1g}{180.16g/mol}=0.101mol[/tex]
The given chemical reaction follows:
[tex]C_6H_{12}O_6(s)+6O_2(g)\rightarrow 6CO_2(g)+6H_2O(l);\Delta H=-2802.5kJ/mol[/tex]
For a:By Stoichiometry of the reaction:
When 1 mole of glucose is reacted, the amount of heat released is 2802.5 kJ
So, when 0.101 moles of glucose is reacted, the amount of heat released is [tex]\frac{2802.5}{1}\times 0.101=283.05kJ[/tex]
Hence, the amount of heat produced for given amount of glucose is -283.05 kJ
For b:To convert the heat produced in kilo joules to calories, we use the conversion factor:
1 kJ = 239 Cal
So, [tex]-283.05kJ\times (\frac{239Cal}{1kJ})=-67648.9Cal[/tex]
Hence, the amount of heat produced for given amount of glucose is -67648.9 Cal
Which model of election configuration is the most accurate? The quantum mechanical atomic model The Rutherford atomic model The Bohr atomic model O Plum pudding atomic model The quantum gravity atomic model
Answer:
Quantum mechanical atomic model.
Explanation:
The first model of electronic configuration was given by Bohr's model.
The most accurate model of electronic configuration is the quantum mechanical atomic model.
Bohr's model has various limitations:
1. It does not explain the Zeeman effect and stark effect.
2. It is not valid for multi-electron system.
3. Heisenberg uncertainty principle is not followed by this model
The quantum mechanical atomic model explains all the four quantum numbers for the electronic configuration of an atom in the periodic table.
The quantum mechanical atomic model considered the Heisenberg uncertainty principle.
Answer:
Quantum mechanical atomic model.
Explanation:
The atom model given by quantum mechanics is the most modern, precise and complex, based on the mathematical form of the atomic structure.
Quantum theory states that matter has properties associated with waves, which is why the atom model was based on this theory. The so-called “Uncertainty Principle” states that the electron has no exact position in the electrosphere, no definite speed and direction. This is why the Bohr atom, with electrons spinning in circular orbits, is surpassed by the quantum model.
What is the easiest way to make 250.0 mL of each of the following solutions if you begin with 250.0 ml. of 1.00M NaCl? a) 0.100 M NaCl b) 0.0100 M Naci c) 0.00100 M NaCl d) 0.000100 M NaCl e) 0.0000100 M Naci En Melation of
Answer:
The easiest way to make these solutions is preparing serial dilutions with dilution factor of 10.
Explanation:
In chemistry, a dilution is the way to reducing the concentration of a chemical adding more solvent.
To prepare the first 0,100 M NaCl from 1,00 M NaCl:
You must take an aliquot of 25,0 mL of 1,00 M NaCl and dilute until 250,0 mL. Thus, the concentration will decrease ten times until the expected concentration. The dilution factor is 10.
To prepare the second 0,0100 M of NaCl:
From 0,100 M NaCl take an aliquot of 25 mL an dilute until 250,0 mL. Thus, the concentration will decrease ten times.
So, to prepare each solution you must take the last one and dilute it in a dilution factor of 10.
I hope it helps!
Vehicle gasoline has many slightly different compositions depending on local laws. However, on average it has a density of 0.737 kg/L. This morning you pumped 7.2 gallons of gasoline in order to fill up your fuel tank. How much mass have you added to your car, in kilograms?
1 gal = 3.785 L
Use appropriate Sig Figs in answer!
Answer: 20.1 kg.
Explanation:
To calculate mass of a substance, we use the equation:
[tex]\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}[/tex]
We are given:
Density of gasoline = [tex]0.737kg/L[/tex]
Volume of gasoline = [tex]7.2gallons=7.2\times 3.785L=27.252L[/tex] (Conversion factor: [tex]1gal=3.785L[/tex] )
Putting values in above equation, we get:
[tex]0.737kg/L=\frac{\text{Mass of gasoline}}{27.252L}\\\\\text{Mass of gasoline}=20.084kg[/tex]
The rule apply for multiplication and division is :
The least precise number present after the decimal point determines the number of significant figures in the answer, thus the answer musty have three significant digits.
Hence, the mass added is 20.1 kg.
Be sure to answer all parts. Write a balanced equation (including physical states) for the following reaction: Sodium carbonate, upon heating, produces sodium oxide and carbon dioxide.
Answer:
I think it would be:
NaCO3 (s)-->Na2O (s) + CO2 (g)
The balanced chemical equation when Sodium carbonate heated to produces sodium oxide and carbon dioxide is as,
[tex]Na_{2} CO_{3} (s)[/tex] → [tex]Na_{2} O (s) + CO_{2} (g)[/tex]
The above reaction is thermal decomposition reaction.
What is thermal decomposition reaction?The chemical decomposition caused by the heat is called thermal decomposition reaction.It is also known as thermolysis.Example: [tex]Na_{2} CO_{3} (s)[/tex] → [tex]Na_{2} O (s) + CO_{2} (g)[/tex]What is decomposition reaction?The reaction in which chemical species or reactant break down into smaller part.learn about decomposition reaction,
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In 1983, a pilot miscalculated the fuel requirement for Air Canada Fight 143 from Montreal to Edmonton. Halfway through the flight, the plane ran out of fuel because of an English-metric conversion error. Suppose that based on the distance and conditions, the flight would require at least 2.2 x 104 kg of fuel. The plane started off having 5, 756 liters in the tank when the pilot arrived. If there are 803 g of jet fuel in 1 liter, determine how much fuel (in liters) would be needed for a successful trip.
Answer:
27,397.23 L would be needed for a successful trip.
Explanation:
The problem gives us the density (ρ) of the fuel, by telling us that there are 803 g of fuel in 1 L, in which case:
ρ=[tex]\frac{mass}{Volume}=\frac{803g}{1L} =803\frac{g}{L}[/tex]
The required mass of fuel is 2.2 * 10⁴ kg, we can convert this value into g:
2.2 * 10⁴ kg * [tex]\frac{1000g}{1kg}[/tex] = 2.2 * 10⁷ g
We calculate the required volume (V), using the mass and density:
[tex]803 g/L = \frac{2.2*10^{7}g }{V} \\V=\frac{2.2*10^{7}g }{803g/L}\\ V=27397.26 L[/tex]
Thus 27,397.23 L would be needed for a successful trip.
A researcher prepares a buffer of acetic acid and sodium acetate with a pH of 5.0.5.0. The total concentration of both components in the buffer is 250 mM,250 mM, and acetic acid has a pKapKa of 4.75.4.75. What are the concentrations of acetic acid and sodium acetate in the buffer? concentration of acetic acid: mM concentration of sodium acetate: mM How many moles of acetic acid and sodium acetate are in 2 L2 L of the buffer? amount of acetic acid: mol amount of sodium acetate: mol How many grams of acetic acid and sodium acetate are in 2 L2 L of the buffer? The molar mass of acetic acid is 60.05 g/mol;60.05 g/mol; the molar mass of sodium acetate is 82.03 g/mol.82.03 g/mol. mass of acetic acid: g mass of sodium acetate:
1.Concentration of acetic acid is 89.99 mM. Concentration of sodium acetate is 160.0 mM . 2.0.180 moles of acetic acid. 0.32 moles of sodium acetate. 3. Mass of acetic acid in 2 L solution is 10.81 g. Mass of sodium acetate in 2 L solution is 26.25 g.
The amount of matter in an item is indicated by its mass, which is a fundamental physical attribute of matter. It is a scalar quantity that is typically measured in kilogrammes (kg) or grammes (g). Mass is separate from weight, which is the gravitational force exerted on an object and can change depending on the strength of the gravitational field.
1.pH = pKa + log [Sodium acetate] / [Acetic acid]
5.0 = 4.75 + log [Sodium acetate] / [Acetic acid]
log [Sodium acetate] / [Acetic acid] = 5.0 - 4.75
= 0.25
[Sodium acetate] = 1.778 x [Acetic acid]
[Sodium acetate] + [Acetic acid] = 250 mM
1.778 x [Acetic acid] + [Acetic acid] = 250 mM
2.778 x [Acetic acid] = 250 mM
[Acetic acid] = 250 mM / 2.778
= 89.99 mM
[Sodium acetate] = 1.778 x [Acetic acid]
= 1.778 x 89.99 mM
= 160.0 mM
2.Concentration of acetic acid = 89.99 mM
= 0.08999 M
Concentration of sodium acetate = 160.0 mM
= 0.16 M
2 x 0.08999 = 0.180 moles of acetic acid.
2 x 0.16 = 0.32 moles of sodium acetate.
3.Mass of acetic acid = no of moles x molar mass
= 0.180 mol x 60.05 g /mol
= 10.81 g
Mass of sodium acetate = no of moles x molar mass
= 0.32 mol x 82.03 g /mol
= 26.25 g
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Final answer:
Using the Henderson-Hasselbalch equation, the concentrations of acetic acid and sodium acetate in the buffer solution are calculated to be approximately 89 mM and 161 mM, respectively. When considering a 2 L volume of this buffer, there would be approximately 0.178 moles of acetic acid and 0.322 moles of sodium acetate, which translates to 10.69 grams of acetic acid and 26.25 grams of sodium acetate.
Explanation:
To calculate the concentrations of acetic acid and sodium acetate in the buffer, we use the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
Given:
pH = 5.0pKa = 4.75Total concentration ([A-] + [HA]) = 250 mMLet's calculate [A-] and [HA] (where [A-] is the concentration of sodium acetate and [HA] is the concentration of acetic acid):
5.0 = 4.75 + log([A-]/[HA])
=> log([A-]/[HA]) = 0.25
=> [A-]/[HA] = 10^0.25
=> [A-]/[HA] ≈ 1.78
Since [A-] + [HA] = 250 mM, we can set up a system of equations:
[A-] = 1.78[HA]
[A-] + [HA] = 250 mM
Solving for [HA] and [A-] gives us:
[HA] ≈ 89 mM
[A-] ≈ 161 mM
Next, we calculate the amount of acetic acid and sodium acetate in 2 L of the buffer:
amount of acetic acid = [HA] × 2 L = 0.089 mol/L × 2 L = 0.178 mol
amount of sodium acetate = [A-] × 2 L = 0.161 mol/L × 2 L = 0.322 mol
Finally, to calculate the mass of acetic acid and sodium acetate in 2 L of the buffer, we use their molar masses:
mass of acetic acid = 0.178 mol × 60.05 g/mol = 10.69 g
mass of sodium acetate = 0.322 mol × 82.03 g/mol = 26.25 g
How many milliliters of a 1:5000 w/v solution of the preservative benzalkonium chloride can be made from 125 mL of a 0.2% w/v solution of the preservative? a. 1250 mL b. 2500 mL C. 625 mL d. 1875 ml
Final answer:
To determine the volume of a 1:5000 w/v solution that can be made from 125 mL of a 0.2% w/v solution of benzalkonium chloride, we calculate the mass of the solute in the initial solution and divide it by the concentration of the desired dilute solution, giving us 1250 mL.
Explanation:
To calculate the target volume of a 1:5000 w/v solution from a 0.2% w/v solution of benzalkonium chloride, we first need to understand the concentration terms. A 0.2% w/v solution means 0.2 g of solute is present in every 100 mL of solution. For a 1:5000 w/v solution, there is 1 g of solute in 5000 mL of solution, or essentially y, 0.0002 g/mL (1 g / 5000 mL).
We then figure out how much benzalkonium chloride is in the initial 125 mL of the 0.2% solution. Since 0.2% w/v is equivalent to 0.2 g/100 mL, we have 0.2 g in 100 mL, hence in 125 mL, we'll have 0.25 g (since 125 mL is 1.25 times 100 mL).
To work out how much of the 1:5000 solution we can make with 0.25 g, we use the concentration of the 1:5000 solution, which is 0.0002 g/mL. Dividing the total available mass of solute (0.25 g) by the concentration (0.0002 g/mL), we get 1250 mL. Thus, the answer is a. 1250 mL.
Tris is a weak base. The basic form of tris can be written tris0 and the protonated (conjugate acid) form can be written tris. The pKa of tris is 8.1. a. What is the ratio of the basic and the acidic forms of tris at pH 6.1? b. What is the useful buffering range of tris?
Answer:
a) [Tris0] : [Tris] = 1 : 100
b) Range = 7.1 to 9.1
Explanation:
a) Calculation of ratio of the basic and the acidic forms of tris
pH of a buffer is calculate using Henderson-Hasselbalch equation
[tex]pH = pKa+log\frac{Salt}{Acid}[/tex]
Conjugate acid of Tris dissociated as
[tex]Tris \leftrightharpoons Tris0 + H^+[/tex]
For tris,
Salt or Basic form = tris0
Acid or Acidic form = Tris
pKa = 8.1
pH = 6.1
[tex]pH = pKa+log\frac{Tris0}{Tris}[/tex]
[tex]6.1 = 8.1+log\frac{Tris0}{Tris}[/tex]
[tex]log\frac{Tris0}{Tris} = -2[/tex]
[tex]\frac{Tris0}{Tris} = antilog (-2)[/tex]
[tex]\frac{Tris0}{Tris} = 10^{-2}[/tex]
[Tris0] : [Tris] = 1 : 100
b) Range of Tris
Range of any buffer is:
From (pKa -1) to (pKa+1)
So, range of Tris is:
From (8.1 - 1) to (8.1 +1)
or from 7.1 to 9.1
We put 50 g of ice of 0 °C in a beaker containing 450 g of beer of 20 °C, mix, and wait for heat exchange within the beaker. What will be the final temperature?
Answer : The final temperature will be, 292 K
Explanation :
In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.
[tex]q_1=-q_2[/tex]
[tex]m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)[/tex]
where,
[tex]c_1[/tex] = specific heat of ice = [tex]2.05J/g.K[/tex]
[tex]c_2[/tex] = specific heat of beer = [tex]4.2J/g.K[/tex]
[tex]m_1[/tex] = mass of ice = 50 g
[tex]m_2[/tex] = mass of beer = 450 g
[tex]T_f[/tex] = final temperature = ?
[tex]T_1[/tex] = initial temperature of ice = [tex]0^oC=273+0=273K[/tex]
[tex]T_2[/tex] = initial temperature of beer = [tex]20^oC=273+20=293K[/tex]
Now put all the given values in the above formula, we get:
[tex]50g\times 2.05J/g.K\times (T_f-273)K=-450g\times 4.2J/g.K\times (T_f-293)K[/tex]
[tex]T_f=291.971K\approx 292K[/tex]
Therefore, the final temperature will be, 292 K
Define "Homoglycan" and "Heteroglycan" and give an example of each
Answer:
Polysaccharides, also known as glycans, are the polymers of carbohydrate. These polymers are composed of the monosaccharide monomeric units that are joined together by glycosidic bonds.
A polysaccharide molecule composed of the same type of monosaccharide units is known as homopolysaccharide or homoglycan.
Example: Cellulose, glycogen, and cellulose.
Whereas, a polysaccharide molecule composed of more than type of monosaccharide is known as heteropolysaccharides or heteroglycans.
Examples: Hyaluronic acid and heparin
Which of the following does not describe a measurement standard? Group of answer choices Measurement standards avoid ambiguity. A standard need not agree with a previously defined size. Confusion is eliminated when the correct measurement is applied. Measurement standards must be unchanging.
Answer:
Confusion is eliminated when the correct measurement is applied
Explanation:
A measurement standard is often described as the reference to which other measurements are judged or based on. Most times, measurement standard are used to determine how accurate an experimental measurement is.
A measurement standard may not eliminate confusion when the correct measurement is applied. If the scientist does not agree with the measurement standard being used, then problem will set in.
The statement 'A standard need not agree with a previously defined size' does not describe a measurement standard because consistency with previously defined sizes is crucial for accuracy. Measurement standards ensure comparability of data across various domains, driving advancements in both science and technology.
Explanation:The statement that does not describe a measurement standard is 'A standard need not agree with a previously defined size.' Measurement standards must indeed be consistent with previously defined sizes to ensure accuracy and eliminate confusion. Measurement standards are crucial to avoid ambiguity and to make sure that experimental data from different labs can be compared accurately. By having unchanging standards, like for the size of a meter, we ensure that a unit of measure remains constant over time, which is indispensable for consistency in scientific experiments and technology development.
Furthermore, the correct measurement standards are essential for the growth of technology as they allow for precise and reliable data. The relationship between science and technology is symbiotic; they drive each other's advancement. As technology improves, new measurement techniques become possible, leading to more detailed observations and more accurate scientific experiments, which in turn can further advance technology.
An ideal gas with MW = 30 is flowing at a steady rate through a circular pipe (cross-section area of 57.8 cm^2) at a temperature of 1.60 × 10^2 °C and pressure of 49.3 psig. The volumetric flow rate is 940 L/min, what is the molar flow rate (mol/s)?
Answer:
n = 1.48 mol/s
Explanation:
ideal gas:
PV = RTn⇒ n = PV/RT
∴ T = 1.60 E2 °C = 160°C = 433 K
∴ R = 0.082 atm.L/K.mol
∴ P = 49.3 psig * ( atm/ 14.6959psig ) = 3.35 atm
∴ V = 940 L/min
⇒ n = (( 3.35 )*( 940 )) / (( 0.082 )*( 433 ))
⇒ n = 88.813 mol/min * ( min / 60s )
⇒ n = 1.48 mol/s
What is the concentration, in millimolar, of a 1 mg/ml solution of NaCl (molecular weight = 58 Da)?
Explanation:
Relation between molarity and molar mass is as follows.
Molarity = [tex]\frac{mass}{\text{molar mass}} \times \frac{1000}{V (in ml)}[/tex]
It is given that mass is 1 mg/ml which is also equal to [tex]10^{-3}[/tex] g.
Molecular mass = 58 Da = 58 g/mol
Volume = 1 ml
Therefore, calculate molarity as follows.
Molarity = [tex]\frac{mass}{\text{molar mass}} \times \frac{1000}{V (in ml)}[/tex]
= [tex]\frac{10^{-3}g}{58 g/mol} \times \frac{1000}{1 ml}[/tex]
= 0.0172 molar
It is known that 1 molar equals 1000 millimolar.
So, 0.0172 molar = [tex]0.0172 molar \times \frac{1000 millimolar}{1 molar}[/tex]
= 17.2 millimolar
Thus, we can conclude that the concentration of given solution is 17.2 millimolar.
Butane is burned as a lighter fluid in disposable lighters. Write a balanced equation for the complete oxidation reaction that occurs when butane (C4H10) burns in air. Use the smallest possible integer coefficients.
Answer:
The balanced equation for the complete oxidation reaction uses 2 moles of butane and 13 moles of Oxygen.
Explanation:
The oxidation of butane is shown in the following picture. As it is a complete combustion, the products of this reaction are carbon dioxide and water.
Butane (C₄H₁₀) combusts in the presence of oxygen (O₂) to produce carbon dioxide (CO₂) and water (H₂O). The balanced reaction is: 2C₄H₁₀(g) + 13O₂(g) -> 8CO₂(g) + 10H₂O(g). This is a combustion reaction, which is exothermic and releases heat and light.
Explanation:The question is asking for the balanced combustion reaction of butane, which is burned as a lighter fluid in disposable lighters. In the presence of sufficient oxygen, butane (C₄H₁₀) combusts to produce carbon dioxide (CO₂) and water (H₂O). The reaction can be represented and balanced as follows:
2C₄H₁₀(g) + 13O₂(g) -> 8CO₂(g) + 10H₂O(g)
This reaction is an example of a combustion reaction, where a hydrocarbon, like butane, burns in the air to produce carbon dioxide and water. It's also an exothermic reaction, releasing energy in the form of heat and light, which is what makes butane useful as a lighter fluid.
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In a five-fold serial dilution of a 20 ug/mL solution, all tube dilutions are 1/5. What is the substance concentration in the third tube of this series?
Answer:
In the third tube, the concentration is 0.16 ug/mL
Explanation:
In the first step, the solution is diluted by 5. Then, the concentration will be
20 ug/mL / 5 = 4 ug/mL
Then, in the second step this 4 ug / ml solution is diluted by a factor of five again:
4 ug /ml / 5 = 0.8 ug/mL
This solution is then diluted again by 5 and the concentration in the third tube will be then:
0.8 ug/mL / 5 = 0.16 ug/mL
Another way to calculate this is to divide the original concentration by the dilution factor ( 5 in this case) elevated to the number of dilutions. In this case:
Concentration in the third tube = 20 ug/mL / 5³ = 0.16 ug/mL
Identify the sample and analyte in each of the scenarios.
(a) Determination of the lead content in paint. lead paint
(b) Analysis of the nitrate content of soil near a local water source. soil nitrate
(c) Measurement of the citric acid found in a lime.
Identify the following as either sample or analyte.
(1) lead
(2) paint chips
(3) soil
(4) nitrate
(5) lime wedge
(6) citric acid
Answer:
a) Analyte: lead. Sample: paint.
b) Analyte: nitrate. Sample: soil.
c) Analyte: citric acid. Sample: Lime
1) Lead: Analyte.
2) Paint chips: Sample.
3) Soil: Sample.
4) Nitrate: Analyte.
5) Lime wedge: Sample.
6) Citric acid: Analyte.
Explanation:
A sample is a portion of material selected from a larger quantity of material while an analyte is the chemical of the system that will be analysed.
Thus:
a) Analyte is lead while you must take a sample of paint to analyze this lead.
b) Analyte is the nitrate while sample must be soil.
c) Analyte is citric acid and lime is the sample
1) Lead: Analyte.
2) Paint chips: Sample.
3) Soil: Sample.
4) Nitrate: Analyte.
5) Lime wedge: Sample.
6) Citric acid: Analyte.
Final answer:
In scenarios (a) to (c), the samples are paint chips, soil, and lime wedge, respectively, and the analytes are lead, nitrate, and citric acid, respectively. Item identifications are lead and nitrate as analytes, and paint chips, soil, and lime wedge as samples.
Explanation:
In analytical chemistry, the analyte is the substance whose chemical content is being measured or analyzed, and the sample is the material or product that contains the analyte. Here are the identifications for each scenario:
(a) The sample is the paint chips and the analyte is lead.(b) The sample is the soil and the analyte is nitrate.(c) The sample is the lime wedge and the analyte is citric acid.To clarify, here are the identifications for each item listed:
(1) Analyte: lead(2) Sample: paint chips(3) Sample: soil(4) Analyte: nitrate(5) Sample: lime wedge(6) Analyte: citric acidThe partial pressure of CO2 gas above the liquid in a carbonated drink is 0.51 atm. Assuming that the Henry's law constant for CO2 in the drink is that same as that in water, 3.7 x 10-2 mol/L atm, calculate the solubility of carbon dioxide in this drink. Give your answer to 3 decimal places.
Answer: The molar solubility of carbon dioxide gas is 0.002 M
Explanation:
Henry's law states that the amount of gas dissolved or molar solubility of gas is directly proportional to the partial pressure of the liquid.
To calculate the molar solubility, we use the equation given by Henry's law, which is:
[tex]C_{CO_2}=K_H\times p_{liquid}[/tex]
where,
[tex]K_H[/tex] = Henry's constant = [tex]3.7\times 10^{-2}mol/L.atm[/tex]
[tex]p_{CO_2}[/tex] = partial pressure of carbonated drink = 0.51atm
Putting values in above equation, we get:
[tex]C_{CO_2}=3.7\times 10^{-2}mol/L.atm\times 0.51atm\\\\C_{CO_2}=1.887\times 10^{-2}mol/L=0.002M[/tex]
Hence, the molar solubility of carbon dioxide gas is 0.002 M
Answer:
0.019 mol/L
Explanation:
Given data
Partial pressure of CO₂ above the liquid in a carbonated drink (P): 0.51 atmHenry's law constant for CO₂ (k): 3.7 × 10⁻² mol/L . atmSolubility of carbon dioxide in the drink (C): ?According to Henry's law, the solubility of a gas in a liquid is proportional to its partial pressure.
C = k × P
C = (3.7 × 10⁻² mol/L . atm) × 0.51 atm
C = 0.019 mol/L
Use the References to access important val A piece of an unknown metal has a volume of 19.7 mL and a mass of 427.0 grams. The density of the metal is g/mL A piece of the same metal with a mass of 64.9 grams would have a volume of Submit Answer
Answer:
The density of the metal is 21.675 g/mL.
The volume of metal with mass of 64.9 grams is 2.994 mL.
Explanation:
Density is defined as mass present in unit volume of the substance.
[tex]Density=\frac{Mass}{Volume}[/tex]
Mass of unknown metal = m = 427.0 g
Volume of the unknown metal = v = 19.7 mL
Density of the unknown metal = d
[tex]d=\frac{m}{v}=\frac{427.0 g}{19.7 mL}=21.675 g/mL[/tex]
Mass of the same unknown metal = M = 64.9 g
Volume of same unknown metal with mass M = V
Density of the metal = d = 21.675 g/mL
[tex]21.675 g/ mL=\frac{64.9 g}{V}[/tex]
V = 2.994 mL
A cylinder with a movable piston records a volume of 12.6L when 3.0 mol of oxygen is added. The gas in the cylinder has a pressure of 5.83 atm. The cylinder develops a leak and the volume of gas is now recorded to be 12.1 L at the same pressure. How many moles of oxygen are lost?
Answer : The moles of oxygen gas lost are 2.88 mole
Explanation :
According to the Avogadro's Law, the volume is directly proportional to the number of moles of the gas at constant pressure and temperature.
[tex]V\propto n[/tex] (At constant temperature and pressure)
or,
[tex]\frac{V_1}{V_2}=\frac{n_1}{n_2}[/tex]
where,
[tex]V_1[/tex] = initial volume of oxygen gas = 12.6 L
[tex]V_2[/tex] = final volume of oxygen gas = 12.1 L
[tex]n_1[/tex] = initial moles of oxygen gas = 3.0 mole
[tex]n_2[/tex] = final moles of oxygen gas = ?
Now put all the given values in the above formula, we get the final moles of oxygen gas.
[tex]\frac{12.6L}{12.1L}=\frac{3.0mole}{n_2}[/tex]
[tex]n_2=2.88mole[/tex]
Therefore, the moles of oxygen gas lost are 2.88 mole
A boy on the beach holds a spherical balloon filled with air. At 7:00 am, the temperature on the beach is 35°F and the balloon has a diameter of 20 cm. Six hours later, the outside temperature is 100°F. Assume that the air is an ideal gas, that the atmospheric pressure remains constant, and that no air was lost or added to the balloon. What is the final diameter of the balloon (in cm)?
Answer:
The final diameter of the balloon is 20.84 cm.
Explanation:
To answer this problem, because both the number of moles (because no air was lost or added to the balloon) and the pressure remain constant, we can use Charles' law, or the law of volumes:
V₁T₂=V₂T₁
Where V is Volume (in L) and T is temperature (in K). The subscripts 1 and 2 refer to different states of the gas: for this problem, let's say that ₁ refers to the state of the gas at 7:00 am and that ₂ refers to the state 6 hours later.
To calculate the volume occupied by the gas we use the formula for the volume of a sphere:
[tex]V=\frac{4}{3}*\pi (\frac{d}{2})^{3}[/tex]
Thus the volume of the balloon at 7:00 am is 4188.79 cm³
Then we convert °F into K:
(35°F + 459.67) * 5/9 = 274.82 K = T₁(100°F + 459.67) * 5/9 = 310.93 K = T₂Now we use Charles' law to calculate V₂:
V₂=V₁T₂/T₁
V₂= [tex]\frac{4188.79cm^{3}*310.93K}{274.82 K}=4739.18cm^{3}[/tex]
Lastly we calculate the final diameter of the balloon:
[tex]4739.18cm^{3}=\frac{4}{3}*\pi (\frac{d}{2})^{3}\\3554.38cm^{3}=\pi (\frac{d}{2})^{3}\\1131.40cm^{3}=(\frac{d}{2})^{3}\\10.42cm=\frac{d}{2}\\20.84=d[/tex]
Suppose the average concentration of copper in water is measured to be 2.5 x 10" M. a. Express this concentration in mg/L. b. Express this concentration in ug/L. c. Does this concentration of copper exceed the freshwater acute criteria maximum concentration of 65 ppb?
Answer:
a) C Cu = 1588.6 mg/L
b) C Cu = 1588600 μg/L
c) this concentration exceeds the acute freshwater criteria:
∴ C Cu = 1588.6 mg/L >> 0.065 mg/L
Explanation:
∴ C Cu = 2.5 E-2 mol/L
∴Mw Cu = 63.546 g/mol
a) C Cu = 2.5 E-2 mol/L * 63.546 g/mol = 1.5886 g/L
⇒ C Cu = 1.5886 g/L * ( 1000 mg/g ) = 1588,6 mg/L
b) C Cu = 1588.6 mg/L * ( μg / 0.001 mg ) = 1588600 μg/L
c) C Cu = 65 ppb * ( ppm / 1000ppb ) = 0.065 ppm = 0.065 mg/L
∴ ppm ≡ mg/L
⇒ C Cu = 1588.6 mg/L >> 0.065 mg/L; this concentration exceeds the acute freshwater criteria
If 72.1 mL of 0.543 M H2SO4 completely titrates 39.0 mL of KOH solution, what is the molarity of the KOH solution? a. 0.317 M b. 0.502 M c. 1.00 M d. 2.01 M
Answer: The correct answer is Option d.
Explanation:
To calculate the molarity of base, we use the equation given by neutralization reaction:
[tex]n_1M_1V_1=n_2M_2V_2[/tex]
where,
[tex]n_1,M_1\text{ and }V_1[/tex] are the n-factor, molarity and volume of acid which is [tex]H_2SO_4[/tex]
[tex]n_2,M_2\text{ and }V_2[/tex] are the n-factor, molarity and volume of base which is KOH
We are given:
[tex]n_1=2\\M_1=0.543M\\V_1=72.1mL\\n_2=1\\M_2=?M\\V_2=39.0mL[/tex]
Putting values in above equation, we get:
[tex]2\times 0.543\times 72.1=1\times M_2\times 39.0\\\\M_2=2.01M[/tex]
Hence, the correct answer is Option d.
The concentration of the base is 2.01 M
What is neutralization?A neutralization reaction is a reaction that occurs between an acid and a base to yield salt and water only;
This reaction occurs as follows; H2SO4 + 2KOH ----->K2SO4 + H2O
From the titration formula;
concentration of acid CA = 0.543 M
volume of acid VA = 72.1 mL
concentration of baseCB = ?
volume of base VB= 39.0 mL
Number of moles of acid = 1
Number of moles of base = 2
Given that;
CAVA/CBVB = NA/NB
CAVANB = CBVBNA
CB = 0.543 M * 72.1 mL * 2/39.0 mL * 1
CB = 2.01 M
Learn more about molarity:https://brainly.com/question/4219437?
Show the conversion factor from Patolbf/ft2is 0.02089.
Explanation:
1 Pascal = 1 N/m²
To convert Pa to lbf/ft²
So, the conversion of N to pound force (lbf) is shown below as:
1 N = 0.224809 pound force (lbf)
The conversion of m² to ft² is shown below:
1 m² = 10.7639 ft²
So,
[tex]1\ Pa=\frac {1\ N}{1\ m^2}=\frac {0.224809\ lbf}{10.7639\ ft^2}
1 Pa = 0.02089 lbf / ft²
Hence proved.
Rearrange the following equation and solve for x when the hydrogen ion concentration is 0.01 M. pH=4+logx (Recall that pH=−log[H+]. Use the hints if you do not obtain one of the answers.) View Available Hint(s) Rearrange the following equation and solve for when the hydrogen ion concentration is 0.01 . (Recall that . Use the hints if you do not obtain one of the answers.) x=2 x=0.40 x=0.1 x=1×10−2
To solve the equation pH = 4 + log(x) and find the value of x when the hydrogen ion concentration is 0.01 M, rearrange the equation by subtracting 4 from both sides. Substitute the value of pH with its equivalent -log(0.01) and simplify the equation. The value of x is 0.0001.
Explanation:To solve the equation pH = 4 + log(x) and find the value of x when the hydrogen ion concentration is 0.01 M, we need to rearrange the equation. Subtracting 4 from both sides, we get log(x) = pH - 4. Using the fact that pH = -log[H+], we substitute the value of pH with its equivalent -log(0.01). Simplifying the equation, we have log(x) = -log(0.01) - 4. Taking the antilog (10^x) of both sides, we get x = 10^(-log(0.01) - 4). Evaluating this expression gives us x = 0.0001.
A balloon filled with helium has a volume of 4.6 x 103 L at 31°C. What volume will the balloon occupy at 56°C if the pressure surrounding the balloon remains constant? Be sure to answer all parts. Enter your answer in scientific notation. x 10 (select) ^L
Answer:
5x10^3 L
Explanation:
Charle's Law states that the volume of a fixed amount of gas maintained at constant pressure is directly proportional to the absolute temperature of the gas, for a constant amount of gas we can write:
[tex]\frac{V1}{T1}=\frac{V2}{T2}[/tex]
As the pressure of the balloon doesn't change, we can use Charle's Law to solve the problem. Firs we change the given temperatures to absolute temperature units ( °K), using the following relations:
°K=273,15+°C
Therefore:
V1=4.6 x 10^3 L, T1=273,15+31=304,15°K
V2=?, T2=273,15+56=329,15°K
[tex]V2=\frac{V1T2}{T1}=\frac{4.6*10^3*329,15}{304,15}=4978,1L=5*10^3L[/tex]
The new volume of the balloon is 5x10^3 L.
The volume of a helium-filled balloon at 56°C under constant pressure can be calculated using Charles's Law by converting temperatures to Kelvins and using the volume-temperature direct proportionality relationship.
Explanation:The subject of this question involves the relationship between volume and temperature of a gas under constant pressure, described by Charles's Law. This law states that the volume of an ideal gas is directly proportional to its temperature (in Kelvins) when pressure is held constant. To use Charles's Law for calculating the new volume of a balloon, first, convert the temperatures from Celsius to Kelvins (K = °C + 273.15). The initial temperature is 31°C, which is 304.15K, and the final temperature is 56°C, which is 329.15K.
Apply Charles's Law using the formula V1/T1 = V2/T2. Let V1 be the initial volume (4.6 x 10³ L), T1 be the initial temperature (304.15K), V2 be the final volume, and T2 be the final temperature (329.15K). Solving for V2 gives:
V2 = (V1 x T2) / T1 = (4.6 x 10³ L x 329.15K) / 304.15K
Calculating this will give us the new volume V2 which we express in scientific notation.
Which of the following statistical metrics is an appropriate measure of the variability of data around the mean? standard deviation O mode width O median
Answer:
Variability of data around the mean is best expressed through the standard deviation.
Explanation:
We have some statistical metrics, and we need to find the best measure of the variability of data around the mean.
First variable is mode. Mode of a set of data values is the value that appears most often. It means the most frequent data.
Second variable is Median. The median is a simple measure of central tendency. It is located around 50% of data range.
Third variable is standard deviation. The standard deviation is found by taking the square root of the average of the squared deviations of the values subtracted from their average value. It means, it will show the dispersion of data around the average value.
According to the definition presents for the three variables, Standard deviation is the most appropriate measure of variability of data around the mean.
Answer:
Correct answer is: Variability of data around the mean is best expressed through the standard deviation.
Explanation:
Got it correct on Plato Hope This helped :D