Answer:
0.1933 M
Explanation:
Data provided in the question:
Number of Moles of Na₂SO₄ initially taken = 0.145 mol
Volume of solution, V = 750 mL = 0.750 L
Now, the molarity is given as:
Molarity = [tex]\frac{\textup{moles of solute}}{\textup{volume of solution in liter}}[/tex]
or
Molarity = [tex]\frac{0.145}{0.750}[/tex]
or
Molarity = 0.1933 M
The molarity of the solution is 0.1933 M.
Explanation:To calculate the molarity of a solution, you need to know the moles of solute and the volume of solution. In this case, we have 0.145 mol of Na2SO4 and a volume of 750 mL. First, convert the volume to liters by dividing by 1000: 750 mL = 0.75 L. Now, calculate the molarity using the formula: Molarity (M) = moles of solute / volume of solution in liters.
So, Molarity = 0.145 mol / 0.75 L = 0.1933 M
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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.
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)
Describe how you would make 250 ml of a 3 M solution of sodium acetate (NaOAc = 82.03 g/mol). First figure out how much sodium acetate you would need, then describe how you would make the solution if you were given a bottle of solid sodium acetate, a volumetric flask, and DI water.
Answer:
You need to do the following conversion to pass from 3M in 250 mL to g of sodium acetate
[tex]3 M (mol/L)*(1L/1000 mL)*(250 mL)*(82.03 g/1 mol)=61.52 g[/tex]
Explanation:
First, you need to dissolve 61.52 g of solid sodium acetate (MW 82.03 g/mol) in 200 ml of DI water. Then, using a volumetric flask add water to bring the total volume of the solution to 250 mL.
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.
When measuring a solid metal block at constant temperature, which measurement will change in numerical value depending on the location where it is taken? A) weight B) length C) volume D) mass
Answer:
A) weight
Explanation:
The weight is the force that gravity makes in a body, and it is given by the equation:
W = mxg
Where W is the weight, m is the mass, and g is the gravity in the local. So, the length, the volume, and the mass are constant and don't depend on the local that is measured, but the weight depends on the local gravity.
You can obtain a rough estimate of the size of a molecule with the following simple experiment: Let a droplet of oil spread out on a fairly large but smooth water surface. The resulting "oil slick" that forms on the surface of the water will be approximately one molecule thick. Given an oil droplet with a mass of 9.00 × 10−7 kg and a density of 918 kg/m3 that spreads out to form a circle with a radius of 41.8 cm on the water surface, what is the approximate diameter of an oil molecule?
Answer:
The diameter of the oil molecule is [tex]4.4674\times 10^{-8} cm[/tex] .
Explanation:
Mass of the oil drop = [tex]m=9.00\times 10^{-7} kg[/tex]
Density of the oil drop = [tex]d=918 kg/m^3[/tex]
Volume of the oil drop: v
[tex]d=\frac{m}{v}[/tex]
[tex]v=\frac{m}{d}=\frac{9.00\times 10^{-7} kg}{918 kg/m^3}[/tex]
Thickness of the oil drop is 1 molecule thick.So, let the thickness of the drop or diameter of the molecule be x.
Radius of the oil drop on the water surface,r = 41.8 cm = 0.418 m
1 cm = 0.01 m
Surface of the sphere is given as: a = [tex]4\pi r^2[/tex]
[tex]a=4\times 3.14\times (0.418 m)^2=2.1945 m^2[/tex]
Volume of the oil drop = v = Area × thickness
[tex]\frac{9.00\times 10^{-7} kg}{918 kg/m^3}=2.1945 m^2\times x[/tex]
[tex]x= 4.4674\times 10^{-10} m= 4.4674\times 10^{-8} cm[/tex]
The thickness of the oil drop is [tex]4.4674\times 10^{-8} cm[/tex] and so is the diameter of the molecule.
The diameter of an oil molecule can be estimated by considering the oil slick as a cylinder and using the formula for the volume of a cylinder. From the given, we calculate the volume of the oil droplet. Dividing this volume by the area of the slick gives us the height of the cylinder, which represents the diameter of the oil molecule.
Explanation:The oil slick forms a cylindrical shape with a height equal to the diameter of an oil molecule or two times the radius. We know the volume of the oil droplet, oil slick, and the diameter of the oil slick.
The volume of oil droplet (V) = Mass (m) / Density (ρ)
= 9 x 10⁻⁷ kg / 918 kg/m³
= 9.81 x 10⁻¹⁰ m³
The volume of the cylindrical oil slick (V) = πr²h, where r is the radius and h is the height of the cylinder.
So, h (height of the oil molecule) = V / (πr^2)
= 9.81 x 10⁻¹⁰ m³ / (3.14 x (0.418 m)²)
= 1.74 x 10⁻¹⁰ m or 174 pm (picometers)
Since the height of the cylinder is approximately equal to the diameter of an oil molecule, the approximate diameter of an oil molecule is 174 pm.
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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 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!
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.
Draw the resonance structures for SO3 2-
SO3 2- has three possible resonance structures, each with one of the three oxygen atoms having a double bond to the central sulfur atom, while the other two have single bonds. The sulfur atom is surrounded by two bonds and one lone pair of electrons in any given structure.
Explanation:The student asked to draw the resonance structures for SO3 2-. Resonance structures are a way of illustrating the delocalization of electrons within certain molecules, often containing double bonds. In the case of the SO3 2- ion, we have sulfur connected to three oxygen atoms.
In each of the resonance structures, sulfur has an oxidation number of +6 and each oxygen atom has an oxidation number of -2. There are three different resonance structures possible for this ion. This is because one of the oxygen atoms can have a double bond to the sulfur atom while the other two oxygen atoms have single bonds, with sulfur carrying one lone pair. The double bond can be with any of the three oxygen atoms, leading to three possible structures, hence the term resonance structures.
The sulfur atom is surrounded by two bonds and one lone pair of electrons in either resonance structure. Therefore, the electron-pair geometry is trigonal planar, and the hybridization of the sulfur atom is sp².
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why is it necessay to remove the stopper form a
separatoryfunnel when liquid is being drained from it through
thestopcock?
Answer:
It is necessary in order to let air in.
Explanation:
When the stopper is not removed, a vacuum is formed inside the separatory funnel as the liquid drains out. The lower air pressure inside the separatory funnel then causes the liquid to drain improperly.
What is the theoretical yield of bismuth when 198 g of Bi2O3 reacts with excess carbon as shown below? Bi2O3 C(s)>Bi(s) + CO(g) (not balanced)
Answer:
179.4306 g
Explanation:
Given that:
Mass of [tex]Bi_2O_3[/tex] = 198 g
Molar mass of [tex]Bi_2O_3[/tex] = 465.96 g/mol
The formula for the calculation of moles is shown below:
[tex]moles = \frac{Mass\ taken}{Molar\ mass}[/tex]
Thus,
[tex]Moles= \frac{198\ g}{465.96\ g/mol}[/tex]
[tex]Moles\ of\ Bi_2O_3= 0.4293\ mol[/tex]
From the balanced reaction,
[tex]Bi_2O_3+3C\rightarrow 2Bi+3CO[/tex]
1 mole of [tex]Bi_2O_3[/tex] on reaction produces 2 moles of bismuth
So,
0.4293 mole of [tex]Bi_2O_3[/tex] on reaction produces 2 × 0.4293 moles of bismuth
Moles of bismuth = 0.8586 moles
Molar mass of bismuth = 208.9804 g/mol
So, mass of bismuth = Moles × Molar mass = 0.8586 × 208.9804 g = 179.4306 g
Final answer:
The theoretical yield of bismuth in the given reaction is 0.848 moles or 177.2 grams.
Explanation:
To determine the theoretical yield of bismuth (Bi) in the reaction between 198 g of Bi2O3 and excess carbon, we need to balance the equation first. The balanced equation for the reaction is:
Bi2O3 + 3C → 2Bi + 3CO
From the balanced equation, we can see that for every mole of Bi2O3, we get 2 moles of Bi. To calculate the theoretical yield, we need to convert the given mass of Bi2O3 to moles using its molar mass (465.96 g/mol) and then use the stoichiometry to find the moles of Bi.
Mass of Bi2O3 = 198 g
Molar mass of Bi2O3 = 465.96 g/mol
Moles of Bi2O3 = (198 g) / (465.96 g/mol) = 0.424 mol
Moles of Bi = 2 * (0.424 mol) = 0.848 mol
The theoretical yield of bismuth in this reaction is 0.848 moles or you can convert it to grams using the molar mass of bismuth (208.98 g/mol) to get the theoretical yield in grams.
A process stream contains 450 mol/s benzene and 375 mol/s toluene. Calculate the mole fraction of benzene in the process stream.
Answer:
0.5455
Explanation:
The moles of benzene in the process stream in 1 sec = 450 moles
The moles of toluene in the process stream in 1 sec = 375 moles
So, according to definition of mole fraction:
[tex]Mole\ fraction\ of\ benzene=\frac {n_{benzene}}{n_{benzene}+n_{toluene}}[/tex]
Applying values as:
[tex]Mole\ fraction\ of\ benzene=\frac {450}{450+375}[/tex]
Mole fraction of benzene in the process stream = 0.5455
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
A balloon filled with helium gas at 1.00 atm occupies 12.9 L. What volume would the balloon occupy in the upper atmosphere, at a pressure of 0.19 atm and a constant temperature? Enter your answer in the box provided.
Answer:
67,9 L
Explanation:
Boyle's Law indicates that the pressure of a fixed amount of gas at a constant temperature is inversely proportional to the volume of a gas, for a constant amount of gas we can write:
P1V1=P2V2
For the problem:
P1= 1 atm, V1= 12,9 L
P2=0,19 atm, V2=?
Therefore:
V2=P1V1/P2.................... V2=1 atm*12,9L/0,19 atm = 67,9 L
The balloon would occupy a volume of 67,9 L in the upper atmosphere.
Which equation or rule can be used to estimate the pseudo-critical properties of mixtures based on the pure-component critical constants? Select one: a. Newton's correlations b. Soave-Redlich-Kwong equation c. Kay's rule d. van der Waals equation of state e. Peng-Robinson
Answer:
c. Kay's rule
Explanation:
Kay's rule -
The rule is used to determine the pseudo reduced critical parameters of mixture , with the help of using the critical properties of the components of a given mixture .
The equation for Kay's rule is as follows ,
PV = Z RT
Where Z = The compressibility factor of the mixture .
Hence from the given options , the correct answer is Kay's rule .
To estimate the pseudo-critical properties of mixtures based on pure-component critical constants, Kay's rule is used. It estimates the properties based on mole fractions of components, offering a simpler alternative to complex equations like Van der Waals or Peng-Robinson equations.
Explanation:The equation that can be used to estimate the pseudo-critical properties of mixtures based on the pure-component critical constants is Kay's rule. This rule is a simple and common method for calculating the critical properties of mixtures, which assumes that the critical properties are proportional to the mole fractions of the components in the mixture. It is not as complex as the Van der Waals equation of state or the Peng-Robinson equation, which are more accurate but also more complicated.
For ideal solutions, Raoult's Law is often used to determine the vapor pressures and hence the equilibrium states of the components in a solution, given by the equation PA = XAPA, where PA is the partial pressure of component A, XA is the mole fraction of component A, and PAo is the vapor pressure of the pure component A.
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.
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.
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.
Water enters a 4.00-m3 tank at a rate of 6.33 kg/s and is withdrawn at a rate of 3.25 kg/s. The tank is initially half full.
How long does it take for the tank to overflow?
Explanation:
The given data is as follows.
Volume of tank = 4 [tex]m^{3}[/tex]
Density of water = 1000 [tex]kg/m^{3}[/tex]
Since, the tank is initially half-filled. Hence, the volume of water in the tank is calculated as follows.
[tex]\frac{1}{2} \times 4 = 2 m^{3}[/tex]
Also, density of a substance is equal to its mass divided by its volume. Therefore, initially mass of water in the tank is as follows.
Mass = [tex]Density \times initial volume[/tex]
= [tex]1000 \times 2[/tex]
= 2000 kg
Whereas mass of water in tank when it is full is as follows.
Mass = [tex]Density \times final volume[/tex]
= [tex]1000 \times 4[/tex]
= 4000 kg
So, net mass of the fluid to be filled is as follows.
Net mass to be filled = Final mass - initial mass
= 4000 kg - 2000 kg
= 2000 kg
Mass flow rate [tex](m_{in})[/tex] = 6.33 kg/s
Mass flow rate [tex](m_{out})[/tex] = 3.25 kg/s
Time needed to fill tank = [tex]\frac{\text{net mass to be filled}}{\text{net difference of flow rates}}[/tex]
= [tex]\frac{2000 kg}{m_{in} - m_{out}}[/tex]
= [tex]\frac{2000 kg}{6.33 kg/s - 3.25 kg/s}[/tex]
= 649.35 sec
Thus, we can conclude that 649.35 sec is taken by the tank to overflow.
A gas of potassium chlorate molecules KClO3 all decompose into potassium chloride, KCl, and diatomic oxygen, O2. The products and reactants are in a closed container and can all be treated as ideal gases.A. Fill in the smallest possible integers that allows the stoichiometry of the reaction equation to be correct:__ KClO3 → ___ KCl ___ O2B. If there are N molecules of potassium chlorate in the initial state, how many product molecules are there
Answer: A. [tex]2KClO_3\rightarrow 2KCl+3O_2[/tex]
B. [tex]2.5\times N[/tex] molecules of product.
Explanation:
According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.
A. The balanced reaction for decomposition of potassium chlorate is:
[tex]2KClO_3\rightarrow 2KCl+3O_2[/tex]
B. According to avogadro's law, 1 mole of every substance occupies 22.4 L at NTP, weighs equal to the molecular mass and contains avogadro's number [tex]6.023\times 10^{23}[/tex] of particles.
According to stoichiometry :
2 moles of reactant give 5 moles of products
Thus [tex]2\times 6.023\times 10^{23}[/tex] molecules of reactant give [tex]5\times 6.023\times 10^{23}[/tex] molecules of product
Thus N molecules of reactant give= [tex]\frac{5\times 6.023\times 10^{23}}{2\times 6.023\times 10^{23}}\times N=\frac{5}{2}\times N[/tex] molecules of product.
Thus [tex]\frac{5}{2}\times N=2.5\times N[/tex] molecules of product are there.
Final answer:
The balanced chemical equation for the decomposition of potassium chlorate is 2 KClO3 → 2 KCl + 3 O2. If there are N molecules of potassium chlorate in the initial state, the number of product molecules of O2 can be calculated using the ratio of 2:3 in the equation.
Explanation:
The balanced chemical equation for the decomposition of potassium chlorate is:
2 KClO3 → 2 KCl + 3 O2
The stoichiometry of the reaction equation can be balanced by filling in the smallest possible integers as follows:
A. 2 KClO3 → 2 KCl + 3 O2
B. If there are N molecules of potassium chlorate in the initial state, the number of product molecules can be calculated as follows:
The ratio of KClO3 to O2 in the balanced equation is 2:3. So, for every 2 molecules of KClO3 that decompose, 3 molecules of O2 are formed. Therefore, the number of O2 molecules produced is given by the equation:
(3 / 2) N = (3 / 2) × N molecules of O2
A sample of argon (Ar) gas occupies 65.0 mL at 22.0°C and 725 torr. What is the volume of this Ar gas at STP? Enter your answer in the provided box.
Answer: The volume of argon gas at STP is 57.4 mL
Explanation:
STP conditions are:
Pressure of the gas = 1 atm = 760 torr
Temperature of the gas = 273 K
To calculate the volume when temperature and pressure has changed, we use the equation given by combined gas law. The equation follows:
[tex]\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}[/tex]
where,
[tex]P_1,V_1\text{ and }T_1[/tex] are the initial pressure, volume and temperature of the gas
[tex]P_2,V_2\text{ and }T_2[/tex] are the final pressure, volume and temperature of the gas
We are given:
[tex]P_1=725torr\\V_1=65.0mL\\T_1=22^oC=(22+273)K=295K\\P_2=760torr\\V_2=?\\T_2=273K[/tex]
Putting values in above equation, we get:
[tex]\frac{725torr\times 65.0mL}{295K}=\frac{760torr\times V_2}{273K}\\\\V_2=57.4mL[/tex]
Hence, the volume of argon gas at STP is 57.4 mL
Explanation:
The given data is as follows.
[tex]V_{1}[/tex] = 65.0 mL = 0.065 L (as 1 ml = 0.001 L),
[tex]T_{1}[/tex] = [tex]22.0^{o}C[/tex] = (22 + 273) K = 295 K,
[tex]P_{1}[/tex] = 725 torr = 0.954 atm (as 1 torr = 0.00131579 atm),
[tex]V_{2}[/tex] = ?, [tex]T_{2}[/tex] = 273 K,
[tex]P_{2}[/tex] = 1 atm
And, according to ideal gas equation,
[tex]\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}[/tex]
[tex]\frac{0.954 atm \times 0.065 L}{295 K} = \frac{1 atm \times V_{2}}{273 K}[/tex]
[tex]V_{2}[/tex] = 0.0574 L
As, 1 L = 1000 ml. So, 0.0574 L = 57.4 ml.
Thus, we can conclude that the volume of this Ar gas at STP is 57.4 L.
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:
Melamine, C3N3(NH2)3, is used in adhesives and resins. It is manufactured in a two-step process: CO(NH2)2(l) → HNCO(l) + NH3(g) 6 HNCO(l) → C3N3(NH2)3(l) + 3 CO2(g) What mass of melamine, C3N3(NH2)3, will be obtained from 161.2 kg of urea, CO(NH2)2, if the yield of the overall reaction is 76.5 % ?
Answer:
43.13Kg of melamine
Explanation:
The problem gives you the mass of urea and two balanced equations:[tex]CO(NH_{2})_{2}_{(l)}=HNCO_{(l)}+NH_{3}_{(g)}[/tex]
[tex]6HNO_{l}=C_{3}N_{3}(NH_{2})_{3}_{(l)}+3CO_{2}_{(g)}[/tex]
First we need to calculate the number of moles of urea that are used in the reaction, so:
molar mass of urea = [tex]60.06\frac{g}{mol}*\frac{1kg}{1000g}=0.06006\frac{Kg}{mol}[/tex]
The problem says that you have 161.2Kg of urea, so you take that mass of urea and find the moles of urea:
161.2Kg of urea[tex]*\frac{1molofurea}{0.06006Kgofurea}=[/tex]2684 moles of urea
Now from the stoichiometry you have:
2684 moles of urea[tex]*\frac{1molofHNCO}{1molurea}*\frac{1molofmelamine}{6molesofHNCO}[/tex] = 447 moles of melamine
The molar mass of the melamine is [tex]126.12\frac{g}{mol}[/tex] so we have:
[tex]447molesofmelamine*\frac{126.12g}{1molofmelamine}[/tex] = 5637.64 g of melamine
Converting that mass of melamine to Kg:
5637.64 g of melamine *[tex]\frac{1Kg}{1000g}[/tex] = 56.38 Kg of melamine, that is the theoretical yield of melamine.
Finally we need to calculate the mass of melamine with a yield of 76.5%, so we have:
%yield = 100*(Actual yield of melamine / Theoretical yield of melamine)
Actual yield of melamine = [tex]\frac{76.5}{100}*56.38Kg[/tex] = 43.13Kg of melamine
Using stoichiometry, the actual yield of Melamine from a reaction starting with 161.2 kg of urea with an overall yield of 76.5% can be found to be approximately 43.2 kg.
Explanation:The question asks about the actual yield of Melamine, C3N3(NH2)3, from a reaction starting with 161.2 kg of urea, CO(NH2)2, with an overall yield of 76.5%. To understand this prediction, we must use the concept of stoichiometry, which is a method used in chemistry to calculate the quantities of reactants or products in a chemical reaction.
First, calculate the mole of urea used: it is the mass of the urea divided by the molar mass of urea. The molar mass of urea (CO(NH2)2) is approximately 60 g/mole. So, you have 161.2*1000/60 = approximately 2686.67 moles of urea.
From the balanced chemical equation, we can see that one mole of urea produces one mole of HNCO, and six moles of HNCO produce one mole of Melamine. Therefore, in theory, 2686.67/6 = 447.78 moles of Melamine can be produced.
However, the reaction yield is only 76.5%, so the actual yield of the Melamine will be less. Using the percentage yield, we can calculate the actual yield: 447.78 * 0.765 = approximately 342.65 moles.
Finally, to convert from moles to mass, multiply by the molar mass of melamine, which is approximately 126 g/mole: 342.65*126 = approximately 43173.9g or 43.2 kg.
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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]
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.
How do i estimate the density of a solution of two compounds, when im given wt% and the specific gravity?
Answer:
[tex]\rho_s=\frac{\rho_w}{\frac{wt1}{SpGr1}+\frac{wt2}{SpGr2}}[/tex]
The only other data you need is the density of water ρw.
Explanation:
We can start by the volume balance
[tex]V_s=V_1+V_2[/tex]
We can replace the volumes with V=M/ρ
[tex]\frac{M_s}{\rho_s}=\frac{M_1}{\rho_1}+\frac{M_2}{\rho_2}[/tex]
If we divide every term by Ms
[tex]\frac{M_s/M_s}{\rho_s}=\frac{M_1/M_s}{\rho_1}+\frac{M_2/M_s}{\rho_2}[/tex]
By definition, wt=Mi/Msol, so we can replace that in the expression
[tex]\frac{1}{\rho_s}=\frac{wt1}{\rho_1}+\frac{wt2}{\rho_2}[/tex]
Then we have the expression of the density of the solution
[tex]\rho_s=\frac{1}{\frac{wt1}{\rho_1}+\frac{wt2}{\rho_2}}[/tex]
To replace ρ1 and ρ2, you have to multiply the specific gravity of the components and the density of water.
[tex]\rho_s=\frac{1}{\frac{wt1}{SpGr1\rho_w}+\frac{wt2}{SpGr2\rho_w}}\\\\\rho_s=\frac{\rho_w}{\frac{wt1}{SpGr1}+\frac{wt2}{SpGr2}}[/tex]
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.
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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
The vapour pressure of water at 20 C is 2.34 kPa. Given that the heat of vaporisation is 2537.4 kJ/kg, use the Clausius-Clapeyron equation to give the vapour pressure at 40 C.
The Clausius-Clapeyron equation is used to calculate the vapor pressure of water at 40°C given its vapor pressure at 20°C and heat of vaporisation. By rearranging the equation and inserting the known values, the vapor pressure at 40°C can be found, which will be higher than that at 20°C.
Explanation:The Clausius-Clapeyron equation relates the vapor pressure and temperature of a substance to its heat of vaporisation. To calculate the vapor pressure of water at 40°C, given the vapor pressure at 20°C (2.34 kPa) and the heat of vaporisation (2537.4 kJ/kg), we can rearrange the equation to solve for the new vapor pressure:
Clausius-Clapeyron Equation: ln(P2/P1) = -(ΔHvap/R) * (1/T2 - 1/T1)
Where P1 is the initial vapor pressure, P2 is the final vapor pressure, ΔHvap is the molar heat of vaporisation, R is the universal gas constant in J/(mol·K), T1 is the initial temperature in Kelvin, and T2 is the final temperature in Kelvin.
The calculation involves converting both temperatures from Celsius to Kelvin and plugging the values into the equation to solve for P2. After using this equation, we find that the vapor pressure at 40°C is significantly higher than at 20°C.