Answer:
0.8 pg/ml
Explanation:
To make the dilutions, you will take 1 ml of the original solution (tube 1) and add 4 ml of solvent. You will now have 20 pg per 5 ml of solution, so your new concentration will be 4 pg/ml (tube 2). Then you will repeat the process, so you will have 4 pg per 5 ml of solution, resulting in a concentration of 0.8 pg/ml (tube 3). The same process will be repeated for tubes 4 and 5.
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!
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.
The complete combustion of 1.00 kg of coal releases about 3.0 x 10E7 J of energy. The conversion of 1.00 kg of mass into energy is equivalent to the burning of how many metric tonnes of coal?
Answer: 0.001 metric tonnes
Explanation:
According to Einstein equation, Energy and mass are inter convertible.
[tex]E=mc^2[/tex]
E= Energy
m = mass
c= speed of light
Given : Complete combustion of 1.00 kg of coal releases about [tex]3.0\times 10^7 J[/tex] of energy.
Given mass: 1 kg
Converting kg to metric ton using the conversion factor:
1 kg=0.001 metric tonnes
Thus 1 kg of coal would also be equal to 0.001 metric tonnes.
Answer:
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Explanation:
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.
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.
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
Solve for x, where M is molar and s is seconds.
x=(2.4×103M−2s−1)(0.38M)3
Answer:
x = 130 Ms⁻¹
Explanation:
The equation is:
x = (2.4 x 10³ M⁻²s⁻¹)(0.38 M)³
The expression raised to the power of 3 is simplified:
x = (2.4 x 10³ M⁻²s⁻¹)(0.054872 M³)
The two values are multiplied together and units canceled:
x = 130 Ms⁻¹
Answer:
x = 131.69Ms⁻¹
Explanation:
x = (2.4×10³M⁻²s⁻¹) (0.38M)³
Upon expanding, we have;
x = (2.4 × 10³M⁻²s⁻¹) (0.38M) * (0.38M) * (0.38M)
x = (2.4 × 10³M⁻²s⁻¹) (0.054872 M³)
x = 2.4 * 0.054872 * 10³ * M⁻²s⁻¹ * M³
x = 0.13169 * 10³ * M⁻²s⁻¹ * M³
x = 131.69 * M⁽⁻²⁺³ ⁾* s⁻¹
x = 131.69 * Ms⁻¹
x = 131.69Ms⁻¹
Name the source of lactose and name the type of enzyme that is required to digest it
Answer:
Lactose is found in animal milk is lactase is the enzyme that is required to digest it.
Explanation:
Lactose is a disaccharide made of glucose and galactose units. This sugar is naturally present in animal milk and it is digested (broken in its units) by the lactase enzyme.
Some people are lactose intolerant because their bodies is not able to produce the enzyme. If they ingest dairy products it may cause health issues. Nowadays it is possible to purchase the lactase enzyme and ingest it with a dairy food to avoid any health effects.
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.
A second order reaction of the type A+B>P was carried out in a solution that was initially .075 mol dm-3 in A and .03 mol dm-3 in B. After 1 hour, the concentration of A had fallen to .02 mol dm-3. a. Calculate the rate constant. b. What is the half life of the reactant? Answers: a. 16.2 dm3mol-hr-, 4.5E-3 dm3mol-s- b. 5.1E3s, 2.1E3 s
Answer:
The rate constant of the reaction is [tex]1.0185\times 10^{-2} dm^3/ mol s[/tex].
[tex]1.31\times 10^3 s[/tex] is the half life of the reactant.
Explanation:
A+B → P
Integrated rate law for second order kinetics is given by:
[tex]\frac{1}{A}=kt+\frac{1}{A_0}[/tex]
A = concentration left after time t = [tex]0.02 mol /dm^3[/tex]
[tex]A_o[/tex] = Initial concentration = [tex]0.075 mol /dm^3[/tex]
t = 1 hour = 3600 seconds
[tex]\frac{1}{0.02 mol /dm^3}=k\times 3600 s+\frac{1}{0.075 mol /dm^3}[/tex]
[tex]k=\frac{1}{3600 s}\times (\frac{1}{0.02 mol /dm^3}-\frac{1}{0.075 mol /dm^3)}[/tex]
[tex]k = 1.0185\times 10^{-2} dm^3/ mol s[/tex]
Half life for second order kinetics is given by:
[tex]t_{\frac{1}{2}}=\frac{1}{k\times a_0}[/tex]
[tex]t_{\frac{1}{2}}=\frac{1}{1.0185\times 10^{-2} dm^3/ mol s\times 0.075 mol /dm^3}=1.31\times 10^3 s[/tex]
The rate constant of the reaction is [tex]1.0185\times 10^{-2} dm^3/ mol s[/tex].
[tex]1.31\times 10^3 s[/tex] is the half life of the reactant.
Which of the following is true about the following chemical reaction at equilibrium? PICK ALL THAT APPLY.
C6H12O6 + 6O2 6CO2 + 6H2O + Energy
A. The concentration of H2O equals the concentration of O2.
B. The rate of the forward reaction, reactants to products, equals the rate of the reverse reaction, products to reactants.
C. The C-C bonds hold more energy in C6H12O6 than the C-O bonds in CO2. Hint: think teter-toter.
D. The concentration of O2 is changing at equilibrium.
Answer:
Option B and C are correct
Explanation:
Chemical reaction at equillibrium means: the rate of forward reaction equals the rate of reverse reaction. This means no product will be consumed, neither formed.
A) The concetration of H2O equals the concentration of O2. This is false.
At the equillibrium the concetnrations are constant. But not necessarily equal.
B) The rate of the forward reaction, reactants to products, equals the rate of the reverse reaction, products to reactants. This is true. IT's the definition of a the equillibrium.
C) The C-C bonds hold more energy in C6H12O6 than the C-O bonds in CO2. Hint: think teter-toter.
This is to, C-C bonds hold more energy than C-O, that's why C6H12O6 is burned, energy is released.
D) The concentration of O2 is changing at equilibrium. This is false. The concentration of O2 is constant.
Upon decomposition, one sample of magnesium fluoride produced 2.15 kg of magnesium and 3.36 kg of fluorine. A second sample produced 1.00 kg of magnesium. How much fluorine (in grams) did the second sample produce?
Answer:
The second sample will produce 1563 grams of fluorine (F2)
Explanation:
The reaction will be MgF2 → Mg + F2
The stoichiometry ratio of MgF2 and F2 is 1 : 1.
That means for 1 mole of MgF2 consumed there is 1 mole of F2 produced.
The first sample produces 2.15 kg of magnesium and 3.36 kg of fluorine
The second sample produced 1 kg of magnesium and x kg of fluorine
This we can show in the following equation =
2.15kg / 3.36 kg = 1kg / x
2.15/3.36 = 0.63988
0.63988 = 1/ x
x= 1/0.63988 = 1.563 kg
1.563kg = 1563 grams
The second sample will produce 1563 grams of fluorine (F2)
What are the Lewis definitions of an acid and a base? In what way are Bronsted definitions? they more general than the In terms of orbitals and electron arrangements, what must be present for a molecule or an ion to act as a Lewis acid (use HT and BF3 as examples)? What must be present for a molecule or ion to act as a Lewis base (use OH and NH3 as examples)?
Explanation:
Lewis definition of Acids and Bases
Chemical species which are capable of accepting electron pairs or donating protons are called Lewis acid.
Chemical species which are capable of donating electron pairs or accepting protons are called Lewis base.
Bronsted definition of acids and bases
Chemical species which are capable of donating H+ are called Bronsted acids.
Chemical species which are capable of accepting H+ are called Bronsted bases.
So all Bronsted acids are Lewis acids but all Lewis acids are not Bronsted acids.
For a chemical species to behave as Lewis acid, they must have:
Incomplete octetDouble bondVacant d-orbitalsFor example, in BF3, octet of boron is incomplete, so it can accept a pair of electron and behaves as Lewis acid.
For a chemical species to behave as Lewis base, they must have:
lone pair of electronsFor example, NH3 and OH, both N and O have lone pairs of electrons, hence behave as Lewis base.
In Lewis theory, an acid is an electron pair acceptor and a base is an electron pair donor. Bronsted definitions are a subset of this concept, and are therefore less general. Lewis acids require an empty orbital, while Lewis bases need a lone pair of electrons.
Explanation:The Lewis definition of an acid is a species that can accept an electron pair, whereas a Lewis base is a species that has an electron pair available for donation to a Lewis acid. A clear example of a Lewis acid is BF3 since it can accept an electron pair due to its empty orbital. On the other hand, a Lewis base, such as OH-, has a lone pair of electrons that it can donate.
In regards to the Bronsted definitions, they define an acid as a proton donor and a base as a proton acceptor but they represent a subcategory within Lewis theory and are therefore less general.
As for the electron arrangements, a molecule or ion can act as a Lewis acid if it has an empty orbital to accept an electron pair. On the contrary, for a molecule or ion to behave as a Lewis base, the presence of a lone pair of electrons is required, as seen in NH3.
Learn more about Lewis Acids and Bases here:https://brainly.com/question/34706043
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Question 5 (1 point) You need to prepare 1.0 L of solution containing 1.0 mM MgCl2 and 0.15 M NaCl form stock solutions of 1.0 M MgCl2 and 0.50 M NaCl. You will: use 1.0 mL MgCl2, 0.30 L NaCl, fill up with water to 1 L Ouse 100 mL MgCl2, 300 ml NaCl, fill up with water to 1 L use 1.0 mL MgCl2, 30.0 ml NaCl, fill up with water to 1 L use 0.0010 L MgCl2, 0.030 L NaCl, fill up with water to 1 L use 100 ml MgCl2, 300 ml NaCl, fill up with water to 1L
Answer:
Use 1.0 mL MgCl2, 0.30 L NaCl, fill up with water to 1 L
Explanation:
The dilution equation relates the concentration C₁ and volume V₁ of an undiluted solution to the concentration C₂ and volume V₂ as follows:
C₁V₁ = C₂V₂
We want to solve for V₁, the amount of stock solution that is required for the dilution to 1.0L.
V₁ = (C₂V₂) / C₁
For the MgCl₂ stock solution, the following volume is required:
V₁ = (C₂V₂) / C₁ = (1.0mM)(1.0L) / (1.0M) = 1 mL
For the NaCl stock solution, the following volume is required:
V₁ = (C₂V₂) / C₁ = (0.15M)(1.0L) / (0.50M) = 0.30 L
The volumes V₁ are the diluted until the total volume reaches V₂.
Consider the following reaction: 2 NO(g) + 2H2(g) → N2(g) + 2 H2O(g) The rate law for this reaction is first order in H2 and second order in NO. What would happen to the rate if the initial concentration of NO tripled while all other factors stayed the same? The rate will increase by a factor of 9. The rate will decrease by a factor of 3. The rate will double. The rate will triple. The rate will remain constant.
Answer: The rate will increase by a factor of 9.
Explanation:
Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.
[tex]2NO(g)+2H_2(g)\rightarrow N_2(g)+2H_2O(g)[/tex]
Given: Order with respect to [tex]NO[/tex] = 2
Order with respect to [tex]H_2[/tex] = 1
Thus rate law is:
[tex]Rate=k[NO]^2[H_2]^1[/tex]
k= rate constant
It is given that the initial concentration of NO is tripled while all other factors stayed the same
[tex]Rate'=k[3\times NO]^2[H_2]^1[/tex]
[tex]Rate'=k[3]^2[NO]^2[H_2]^1[/tex]
[tex]Rate'=k\times 9[NO]^2[H_2]^1[/tex]
[tex]Rate'=9\times Rate[/tex]
Thus the rate will increase by a factor of 9.
The correct answer is that the rate will increase by a factor of 9.
To understand why the rate increases by a factor of 9 when the concentration of NO is tripled, we need to consider the rate law for the reaction, which is given as first order in H2 and second order in NO. The rate law can be expressed as:
[tex]\[ \text{Rate} = k[\text{NO}]^2[\text{H2}] \][/tex]
where [tex]\( k \)[/tex] is the rate constant, and [NO] and [H2] are the concentrations of NO and H2, respectively.
Since the reaction is first order with respect to H2, the rate is directly proportional to the concentration of H2. However, the reaction is second order with respect to NO, which means the rate is proportional to the square of the concentration of NO.
If the concentration of NO is tripled (let's denote the initial concentration of NO as [NO]_initial, then the new concentration is [tex]3[NO]_initial)[/tex], the new rate can be calculated as follows:
[tex]\[ \text{New Rate} = k(3[\text{NO}]_{\text{initial}})^2[\text{H2}] \][/tex]
[tex]\[ \text{New Rate} = k \cdot 9 \cdot [\text{NO}]_{\text{initial}}^2 \cdot [\text{H2}] \][/tex]
[tex]\[ \text{New Rate} = 9 \cdot k[\text{NO}]_{\text{initial}}^2[\text{H2}] \][/tex]
Comparing the new rate to the initial rate:
[tex]\[ \text{Initial Rate} = k[\text{NO}]_{\text{initial}}^2[\text{H2}] \][/tex]
we see that the new rate is 9 times the initial rate because the concentration of NO has been tripled, and due to the second-order dependence on NO, the rate is multiplied by the square of the factor by which the concentration of NO has increased [tex](3^2 = 9).[/tex]
Therefore, the rate will increase by a factor of 9 when the initial concentration of NO is tripled, while keeping all other factors constant.
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.
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.
Report the answer with the correct number of significant figures: a. 12.3 mm x 15.853 mm = b. 72.98 · 15.830 ml = c. 12.3 mm + 15.853 mm = d. 172.3 cm - 15.853 cm - 12.5 pts
Explanation:
The rule apply for the multiplication and division is :The least number of significant figures in any number of the problem determines the number of significant figures in the answer.
The rule apply for the addition and subtraction is :The least precise number present after the decimal point determines the number of significant figures in the answer.
a) 12.3 mm × 15.853 mm :
[tex] 12.3 mm\times 15.853 mm :194.9919 mm^2\approx 195 mm^2[/tex]
Three significant figures.
b) 72.98 · 15.830 ml
[tex]72.98\times 15.830 ml=1,155.2734 ml\approx 1,155 mL[/tex]
Three significant figures.
c) 12.3 mm + 15.853 mm
[tex]12.3 mm + 15.853 mm = 28.153 mm\approx 28.1 mm[/tex]
Three significant figures.
d) 172.3 cm - 15.853 cm
[tex] 172.3 cm - 15.853 cm= 156.447 cm\approx 156.4 cm[/tex]
Four significant figures.
Final answer:
The student's Mathematics question involves calculating with significant figures. When multiplying or dividing values, the result is rounded to the number of significant figures in the least precise value, while addition and subtraction are rounded to match the least number of decimal places.
Explanation:
The subject of this question is Mathematics, specifically related to significant figures in calculations, which is a concept typically covered in High School mathematics or physics courses.
Report the answer with the correct number of significant figures:
(a) To multiply 12.3 mm by 15.853 mm, we should round our answer to the same number of significant figures as the number with the fewest significant figures. In this case, 12.3 has three significant figures. Therefore, the result should also be reported with three significant figures.
(b) When multiplying 72.98 by 15.830 ml, the number with the fewest significant figures is 72.98, which has four significant figures. The answer should be reported with four significant figures.
(c) When adding 12.3 mm to 15.853 mm, we look for the least precise value in terms of decimal places, which is 12.3 mm (it is precise to the tenths place). Therefore, our result should also be rounded to the tenths place.
(d) In the subtraction 172.3 cm - 15.853 cm - 12.5 cm, since 12.5 cm has the least number of decimal places (one decimal place), our final result should also be rounded to one decimal place.
When it comes to significant figures, the rule for multiplication and division is to match the result to the number with the fewest significant figures. For addition and subtraction, we round the result to the least number of decimal places in the values used.
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
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.
A reaction at -6.0 °C evolves 786. mmol of sulfur tetrafluoride gas. Calculate the volume of sulfur tetrafluoride gas that is collected. You can assume the pressure in the room is exactly 1 atm. Be sure your answer has the correct number of significant digits. volume: 1 X I
Answer:
V of Sulfur tetrafluoride is 17.2 L
Explanation:
Given data;
T = -6°C = 267K [1° C = 273 K]
n = 786 mmol of SF4 which is 0.786 mol
P = 1 atm
from ideal gas law we have
PV = nRT
where n is mole, R is gas constant, V is volume
[tex]V = \frac{nRT}{P}[/tex][tex]V = \frac{0.786 mol \times 0.082 atmL/mol K \times* 267K}{1atm} = 17.2 L[/tex]
V of Sulfur tetrafluoride is 17.2 L
In a demonstration of strong electrolytes, weak electrolytes, and non-electrolytes, Professor Popsnorkle used a lightbulb apparatus that showed solution conduction of electricity by the brightness of the lightbulb. He tested H2SO4(aq), NH3 and a 1M sugar solution. What was the result?
Answer:
The result was that H₂SO₄ is a strong electrolyte, NH₃ is a weak electrolyte and sugar is a non-electrolyte.
Explanation:
The experiment will show that the H₂SO₄ solution will light the lightbulb brightly, the NH₃ solution will light it weakly and the sugar solution will not light it at all.
This happens because to light the lightbulb we need free charges moving in the solution. Therefore, the substance that produce more ions will be the best electrolyte.
H₂SO₄ → 2H⁺ + SO₄²⁻
NH₃ + H₂O → NH₄⁺ + OH⁻
Sugar will not generate ions
Given the densities of the following pure liquids, what volume of each is necessary to make 250 mL of a 350 mM solution? a. Ethanol - 0.789 g/cm? b. Acetone -0.791 g/cm c. Formic acid - 1.220 g/cm3 d. tert-Butylamine - 0.696 g/cm?
Answer :
(a) The volume of ethanol liquid needed are, 2.4 L
(b) The volume of ethanol liquid needed are, 1.9 L
(c) The volume of ethanol liquid needed are, 1.6 L
(d) The volume of ethanol liquid needed are, 1.7 L
Explanation : Given,
Volume of solution = 250 mL = 0.250 L (1 l = 1000 mL)
Concentration of solution = 350 mM = 0.350 M (1 mM = 0.001 M)
First we have to calculate the moles of solution.
[tex]\text{Moles of solution}=\text{Concentration of solution}\times \text{Volume of solution}=0.350M\times 0.250L=0.0875mole[/tex]
(a) For ethanol liquid :
To we have to calculate the mass of ethanol.
[tex]\text{Mass of ethanol}=\frac{\text{Moles}}{\text{Molar mass of ethanol}}=\frac{0.0875mole}{46g/mole}=0.0019g[/tex]
Now we have to calculate the volume of ethanol.
[tex]Volume=\frac{Mass}{Density}=\frac{0.0019g}{0.789g/cm^3}=0.0024cm^3=0.0024mL=2.4L[/tex]
(b) For acetone liquid :
To we have to calculate the mass of acetone.
[tex]\text{Mass of acetone}=\frac{\text{Moles}}{\text{Molar mass of acetone}}=\frac{0.0875mole}{58g/mole}=0.0015g[/tex]
Now we have to calculate the volume of acetone.
[tex]Volume=\frac{Mass}{Density}=\frac{0.0015g}{0.791g/cm^3}=0.0019cm^3=0.0019mL=1.9L[/tex]
(c) For formic acid liquid :
To we have to calculate the mass of formic acid.
[tex]\text{Mass of formic acid}=\frac{\text{Moles}}{\text{Molar mass of formic acid}}=\frac{0.0875mole}{46g/mole}=0.0019g[/tex]
Now we have to calculate the volume of formic acid.
[tex]Volume=\frac{Mass}{Density}=\frac{0.0019g}{1.220g/cm^3}=0.0016cm^3=0.0016mL=1.6L[/tex]
(d) For tert-Butylamine liquid :
To we have to calculate the mass of tert-Butylamine.
[tex]\text{Mass of tert-Butylamine}=\frac{\text{Moles}}{\text{Molar mass of tert-Butylamine}}=\frac{0.0875mole}{73g/mole}=0.0012g[/tex]
Now we have to calculate the volume of tert-Butylamine.
[tex]Volume=\frac{Mass}{Density}=\frac{0.0012g}{0.696g/cm^3}=0.0017cm^3=0.0017mL=1.7L[/tex]
To determine the volume of each pure liquid necessary to make a 350 mM solution with a volume of 250 mL, calculate the required mass of each substance and then use the density to find the volume.
Explanation:To determine the volume of each pure liquid necessary to make a 350 mM solution with a volume of 250 mL, we need to use the formula:
Volume (mL) = (mass (g) / density (g/cm³)) * 1000
For each pure liquid, calculate the mass of the substance needed by multiplying its molarity with the desired volume in moles:
Mass (g) = molarity (mol/L) * volume (L) * molar mass (g/mol)
Then, use the formula for volume to find the necessary volume of each pure liquid:
Volume (mL) = (mass (g) / density (g/cm³)) * 1000
Calculate the molarity of 0.400 mol of Na, S in 1.30 L of solution. molarity: Calculate the molarity of 23.9 g of MgS in 843 mL of solution. molarity:
Final answer:
To calculate the molarity of a solution, divide the moles of solute by the liters of solution.
Explanation:
To calculate the molarity of a solution, we need to know the amount of solute and the volume of the solution.
Molarity (M) = moles of solute / liters of solution
In the first question, we have 0.400 mol of Na2S in 1.30 L of solution. So, the molarity would be:
M = 0.400 mol / 1.30 L = 0.308 M
In the second question, we have 23.9 g of MgS in 843 mL of solution. First, we need to convert grams of MgS to moles using the molar mass of MgS (40.3 g/mol). Then, we need to convert mL to L. Finally, we can calculate the molarity:
M = (23.9 g / 40.3 g/mol) / (843 mL / 1000 mL/L) = 0.712 M
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.
What is the main difference between laboratory scale distillation and industrial distillation?
Answer: Distillation is a process in which we use the boiling point, condensation of the substances to separate them from each other, it is a physical separation not the chemical because physical property of substance is used in it.
Now, the distinction in between a laboratory distillation and an industrial distillation is that, the distillation process in laboratory distillation works in batches, while in the industrial distillation it occurs in continuous manner, industrial distillation is the large scale distillation.
Final answer:
The main difference between laboratory scale distillation and industrial distillation lies in the scale and complexity of the process.
Explanation:
The main difference between laboratory scale distillation and industrial distillation lies in the scale and complexity of the process. Laboratory scale distillation is typically used for smaller quantities and is carried out using simple apparatus. It is commonly used for research, quality control, and small-scale production.
On the other hand, industrial distillation is conducted on a much larger scale and involves more complex equipment. It is used in industries such as oil refining and chemical manufacturing, where large quantities of materials need to be processed. Industrial distillation may include fractionating columns and other specialized equipment to separate components from mixtures.
What is the difference between natural capital and natural resources?
Answer: A resource which we get naturally from the enviornment such as, air water, soil etc. is known as natural resources. Natural resources can be categories on the basis of renewability and it is of two types; renewable and non-renewable.
A natural capital is the stock of natural resources; which includes soil, air and all the living organisms present on this planet
Final answer:
Natural resources are raw materials found in nature used for economic gain, while natural capital includes these resources plus the ecosystem services that sustain life. Whereas natural resources can be renewable or nonrenewable, natural capital focuses on the sustainable management of these resources and ecosystem services.
Explanation:
The difference between natural capital and natural resources is found in their definitions and applications. Natural resources are materials or substances that occur in nature and can be used for economic gain; they are the raw inputs we find in our environment that have not been altered by human hands. Examples of natural resources include water, minerals, timber, and fossil fuels. They are essential for producing goods and services, but they vary in their abundance and renewability. Renewable resources, such as solar energy and timber, can be replenished over time, while nonrenewable resources, like fossil fuels and minerals, are finite and diminish with use.
On the other hand, natural capital encompasses not just the natural resources themselves, but also the ecosystem services that sustain and fulfill human life. Natural capital refers to the world's stocks of natural assets which include geology, soil, air, water and all living organisms. There is a growing recognition of the need to manage natural capital sustainably to avoid depleting these vital assets. The concept of natural capital broadens the scope of what is considered valuable in the natural environment beyond just the extractive resources.
It's important to note that while all natural capital includes natural resources, not all natural resources serve as natural capital. This distinction lies in the broader ecological value natural capital provides, benefiting economic processes as well as the environment in ways that maintain or enhance our stock of natural assets.
Write 23,665,700 in Scientific Notation with 4 significant figures.
Answer: The given number in scientific notation is [tex]2.367\times 10^{7}[/tex]
Explanation:
Scientific notation is the notation where a number is expressed in the decimal form. This means that the number is always written in the power of 10 form. The numerical digit lies between 0.1.... to 9.9.....
If the decimal is shifting to right side, the power of 10 is negative and if the decimal is shifting to left side, the power of 10 is positive.
We are given:
A number having value = 23,665,700
Converting this into scientific notation, we get:
As, the decimal is shifting to left side, the power of 10 will be positive.
[tex]\Rightarrow 23,665,700=2.367\times 10^{7}[/tex]
Hence, the given number in scientific notation is [tex]2.367\times 10^{7}[/tex]
Calculate AS° for the reaction below: N2(g)+202(g) 2NO2(g) where ASo for N2(g), O2(g), & NO2(g), respectively, is 191.5, 205.0, & 240.5 J/mol-K -156.0 J/K 156.0 J/K 120.5 J/K -120.5 J/K O OOO
Answer:
ΔS Rx = -120, 5 J/K
Explanation:
The ΔS in a reaction is defined thus:
ΔS Rx = ∑ n S°products - ∑ m S°reactants
For the reaction:
N₂(g) + 2 O₂(g) → 2NO₂(g)
ΔS Rx = 2 mol × 240,5 [tex]\frac{J}{mol.K}[/tex] - [ 1 mol × 191,5 [tex]\frac{J}{mol.K}[/tex] + 2 mol × 205,0 [tex]\frac{J}{mol.K}[/tex]]=
ΔS Rx = -120, 5 J/K
A negative value in ΔS means a negative entropy of the process. Doing this process entropycally unfavorable.
I hope it helps!
The concentration of dye in Solution A is 20.422 M. A serial dilution is performed to make Solutions B and C. In the first dilution, 2 mL of Solution A is diluted with 12 mL water to make Solution B. Then, 2 mL of Solution B is then diluted with 1 mL of water to make Solution C. What is the concentration of dye in Solution C? Provide your response to three significant figures in units of molarity.
Answer:
1.95 M
Explanation:
Considering
Moles of A = Moles of B
Or,
[tex]Molarity_{A}\times Volume_{A}=Molarity_{B}\times Volume_{B}[/tex]
Given that:
[tex]Molarity_{A}=20.422\ M[/tex]
[tex]Volume_{A}=2\ mL[/tex]
[tex]Volume_{B}=2+12=14\ mL[/tex]
[tex]Molarity_{B}=?\ M[/tex]
So,
[tex]20.422\times 2=Molarity_{B}\times 14[/tex]
Molarity of solution B = 2.92 M
Again Considering,
Moles of B = Moles of C
Or,
[tex]Molarity_{B}\times Volume_{B}=Molarity_{C}\times Volume_{C}[/tex]
Given that:
[tex]Molarity_{B}=2.92\ M[/tex]
[tex]Volume_{B}=2\ mL[/tex]
[tex]Volume_{C}=2+1=3\ mL[/tex]
[tex]Molarity_{C}=?\ M[/tex]
So,
[tex]2.92\times 2=Molarity_{C}\times 3[/tex]
Molarity of solution C = 1.95 M
The concentration of dye in Solution C is 1.945 M after performing the serial dilutions.
Explanation:To find the concentration of dye in Solution C after serial dilutions, we can use the dilution equation which states that the concentration times the volume before dilution (C1V1) is equal to the concentration times the volume after dilution (C2V2), or C1V1 = C2V2.
For Solution B, the initial concentration is 20.422 M (C1) and the initial volume is 2 mL (V1). This solution is diluted with 12 mL water, so the final volume (V2) is 2 mL + 12 mL = 14 mL. Using the dilution equation, we calculate the concentration of Solution B (C2).
(20.422 M)(2 mL) = (C2)(14 mL)
C2 = (20.422 M)(2 mL) / (14 mL) = 2.917 M
Next, we perform a second dilution to create Solution C. We take 2 mL of Solution B and add 1 mL of water, giving a final volume of 3 mL. Again, we use the dilution equation to find the concentration of Solution C.
(2.917 M)(2 mL) = (C3)(3 mL)
C3 = (2.917 M)(2 mL) / (3 mL) = 1.945 M
Therefore, the concentration of dye in Solution C is 1.945 M, rounded to three significant figures.