Answer:
In the second step, the chemist must measure 8.3 ml of concentrated acid
Explanation:
The concentration of the final solution can be obtained using the pH value:
pH = -log[H] = 0.60
[H] = 10^(-0.60) = 0.25 M
Then, the final concentration of HCl will be 0.25 M because HCl is a monoprotic acid, which means that HCl only has one hydrolyzable proton. Therefore: [HCl] = [H].
The number of moles of HCl in the final solution will be equal to the number of moles present in the volume taken from the stock solution:
n° of moles in the volume taken from stock solution = n° moles in the final solution.
The number of moles can be calculated as follows:
number of moles = concentration * volume
Then:
Ci * Vi = Cf * Vf
where
Ci = concentration of the stock solution
Vi = volume taken from the stock solution
Cf = concentration of the final solution
Vf = volume of the final solution
Replacing with the data, we can obtain Vi:
6.0 M * Vi = 0.25 M * 200.0 ml
Vi = 8.3 ml
Draw an arrow-pushing mechanism for the elimination reaction between NaOH in ethanol and each of the following haloalkanes: (E1/E2)
(i) 1-bromobutane;
(ii) 2-bromo-2-methylpentane.
Answer:
The mechanisms for the elimination reactions between NaOH in ethanol and the halogenoalkanes are demonstrated in the figure attached.
Explanation:
(i) 1-bromobutane will suffer elimination to for an alkene. The mechanism will be E2, which means that the attack and the elimination will occur simultaneously. This is the preferred mechanism because the bromine is in a primary carbon.
(ii) 2-bromo-2-methylpentane will suffer elimination to for an alkene. The mechanism will be E1, which means that the attack and the elimination will occur in two different steps. The bromine will be eliminated in the first step with the formation of a carbocation and in a second step the double bond will be formed after the anionic attack. This is the preferred mechanism because the bromine is in a terciary carbon which is able to stabilize the carbocation formed.
Calculate the molarity of a solution that contains 3.11 mol of NaNO3 dissolved in 2.50 L. Enter your answer in the provided box.
Answer:
Molarity of a solution that contains 3.11 mol of NaNO3 is 1,24 M
Explanation:
We understand molarity as the number of moles of solute that are contained in 1 L of solution, then if in a solution of 2.50 L we have 3.11 moles, it remains to calculate how many moles do we have in 1 liter.
2,50 L .......... 3,11 moles
1 L .................. x
X = ( 1 L x 3,11 moles) / 2,50 L = 1,24
Assume that aniline is insoluble in water. The normal boiling points of aniline and water are 184.1°C and 100°C respectively. At a pressure of I atm, the boiling point of a mixture of aniline and water is a) 184.1°C b) 100°C c) less than 100°C d) greater than 100°C but less than 184.1°C.
Answer:
At a pressure of I atm, the boiling point of a mixture of aniline and water is b) 100°C.
Explanation:
Assuming that aniline is insoluble in water there will be no interaction between liquids. There will be no positive interactions that can increase the boiling point or negative interactions that can decrease the boiling point. Thus the boiling points of each substance will not be affected. Since the boiling point of water is 100ºC, the mixture will start to evaporate at 100ºC.
Answer:
B.)
Explanation:
One side of a plane wall is held at 200°C while the other side is exposed to a convective environment having T10°C and h 100 W/m2 °C. The wall has a k - 2.6 W/m °C and is 30 cm thick. Calculate the heat flux through the wall. What is the temperature on the surface of the cold side of the wall? (Remember this problem!! wll come up frequently later in the course.)
Answer:
[tex]\frac{\dot Q}{A} = 151.33 W/m^2[/tex]
[tex]T_c = 25.153°C[/tex]
Explanation:
Given data:
one side wall temperature 200°C
other side wall temperature 10°C
h = 100 W/m^2 °C
k = 2.6W/m °C
wall thickness L = 30 cm
we know that heat flux is given as
[tex]\frac{\dot Q}{A} = \frac{ T_A - T\infty}{\frac{L}{K} + \frac{1}{h}}[/tex]
[tex]\frac{\dot Q}{A} = \frac{ 20- 10} {\frac{0.30}{2.6} + \frac{1}{100}}[/tex]
[tex]\frac{\dot Q}{A} = 151.33 W/m^2[/tex]
[tex]1515.33 W/m^2 = \frac{ T_A - T_c}{\frac{L}{K}}[/tex]
solving for temperature for cold surface is given as
[tex]T_c = -1515.33 \times \frac{0.3}{2.6} + 200[/tex]
[tex]T_c = 25.153°C[/tex]
The following data were measured for the reaction BF3(g)+NH3(g)→F3BNH3(g): Experiment [BF3](M) [NH3](M) Initial Rate (M/s) 1 0.250 0.250 0.2130 2 0.250 0.125 0.1065 3 0.200 0.100 0.0682 4 0.350 0.100 0.1193 5 0.175 0.100 0.0596 Part A What is the rate law for the reaction? What is the rate law for the reaction? rate=k[BF3]2[NH3] rate=k[BF3][NH3] rate=k[BF3][NH3]2 rate=k[BF3]2[NH3]2
The rate law for the reaction BF3(g)+NH3(g)→F3BNH3(g) based on the input data is rate = k[BF3][NH3], where k is the rate constant, [BF3] and [NH3] denote the molar concentrations of BF3 and NH3, indicating a first order relationship for both reactants.
Explanation:In order to determine the rate law for the reaction BF3(g)+NH3(g)→F3BNH3(g), we have to look at how the initial rate changes with respect to the change in concentration of the reactants. Using the given data, we can compare experiments where only one reactant's concentration is changed while the concentration of the other reactant remains constant. Upon analysis, we see that when the concentration of BF3 doubles, the rate also doubles, suggesting a first order relationship. Similarly, when the concentration of NH3 doubles, the rate doubles, indicating a first order relationship for NH3 as well. Hence, given that both BF3 and NH3 are first order, the rate law for the reaction should be: rate = k[BF3][NH3].
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The rate law for the reaction BF3(g) + NH3(g) → F3BNH3(g) is rate = k[BF3][NH3].
Explanation:
The rate law for the reaction BF3(g) + NH3(g) → F3BNH3(g) can be determined by analyzing the effect of changing concentrations on the initial rate of the reaction. By comparing the rates of reaction for different experiments, we can observe how changing the concentrations of the reactants affects the rate. In this case, it is clear that the rate of reaction is directly proportional to the concentration of BF3 and NH3, so the rate law for the reaction is rate = k[BF3][NH3].
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Round 129.752416 to the requested number of significant figure a. 3 significant figures c. 6 significant figures_22 b. 4 significant figures d. 7 significant figures
To round 129.752416 to the requested number of significant figures:
- For 3 significant figures, round down to 129.7.
- For 4 significant figures, round up to 129.8.
- For 6 and 7 significant figures, keep the given digits.
To round 129.752416 to the requested number of significant figures:
For 3 significant figures, we look at the digit next to the third significant figure. If it is 5 or greater, we round up, otherwise, we round down. In this case, the digit next to the third significant figure is 2, which is less than 5, so the rounded number is 129.7.For 4 significant figures, we follow the same process as before but include the digit next to the fourth significant figure. The digit next to the fourth significant figure is 4, which is less than 5, so the rounded number is 129.8.For 6 significant figures, we include all the digits given. The rounded number is 129.752416.For 7 significant figures, we include all the digits given. The rounded number is 129.752416.A slice of Swiss cheese contains 47 mg of sodium. (a) What is this mass in grams? .047 (b) What is this mass in ounces? (16 oz = 453.6 g) 0.0016 X (c) What is this mass in pounds? (1 lb 453.6 g) 0.0001x
Based on conversion ratios, 47 mg of sodium would be:
a. 0.047 grams. b. 0.00166 ounces.c. 0.000104 pounds.Mass in grams1g is 1,000 grams.
47 mg would be:
= 0.47 / 1,000
= 0.047 grams
Mass in Ounces16 ounces is 453.6 grams so 0.047 grams would be:
= (0.047 x 16) / 453.6
= 0.00166 ounces.
Mass in Pounds1 pound is 453.6 grams so 0.047 grams would be:
= 0.047 / 453.6
= 0.000104 pounds
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To convert 47 mg of sodium to grams, divide by 1000. To convert grams to ounces, use the conversion factor 1 oz = 28.35 g. To convert grams to pounds, use the conversion factor 1 lb = 453.6 g.
Explanation:To convert 47 mg to grams, divide by 1000:
47 mg = 47/1000 = 0.047 grams
To convert grams to ounces, use the conversion factor:
0.047 g x (1 oz/28.35 g) = 0.00166 oz
To convert grams to pounds, use the conversion factor:
0.047 g x (1 lb/453.6 g) = 0.0001035 lb
Write a balanced half-reaction for the reduction of permanganate ion (Mno) to manganese ion Mn?) in acidic aqueous solution. Be sure to add physical state symbols where appropriate. 0-0 Cb X 5 2
The reduction of permanganate ion to manganese ion in acidic solution is represented by the balanced half-reaction: MnO₄⁻(aq) + 8H⁺(aq) + 5e⁻ → Mn²⁺(aq) + 4H₂O(l). It shows the permanganate ion gaining 5 electrons to become a manganese(II) ion, with the protons reacting with permanganate to form water.
Explanation:The reduction of the permanganate ion (MnO⁴⁻) to the manganese(II) ion (Mn²⁺) in acidic solution can be illustrated with the following balanced half-reaction:
MnO₄⁻(aq) + 8H⁺(aq) + 5e⁻ → Mn²⁺(aq) + 4H₂O(l)
This equation indicates that the permanganate ion gains 5 electrons (is reduced) in an acidic solution to result in a manganese(II) ion. The 8 protons react with permanganate to form 4 molecules of water, which is also reflected in the equation.
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The reduction half-reaction for permanganate to manganese ion in an acidic solution is MnO4- + 8H+ + 5e- → Mn2+ + 4H2O(l), where oxygen and hydrogen are balanced by adding water and hydrogen ions, respectively, and the charge is balanced by adding electrons.
Explanation:To write a balanced half-reaction for the reduction of the permanganate ion (MnO4-) to manganese ion (Mn2+) in an acidic aqueous solution, we start with the half-reaction:
MnO4- + 8H+ + 5e- → Mn2+ + 4H2O(l)
In this balanced equation, the oxygen atoms are balanced by adding water molecules (H2O) to the right side, and the hydrogen atoms are balanced by adding hydrogen ions (H+) to the left side. Additionally, 5 electrons (5e^-) are added to the left side to ensure that the overall charge is balanced between the reactant and product sides of the equation.
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A compressed cylinder of gas contains 45.6 mol of N2 gas at a pressure of 3.75 x 105 Pa and a temperature of 23.6°C. What volume of gas has been released into the atmosphere if the final pressure in the cylinder is 5.67 x 105 Pa? Assume ideal behavior and that the gas temperature is unchanged.
Answer: 0.102 Liters
Explanation
According to the ideal gas equation:
[tex]PV=nRT[/tex]
P = Pressure of the gas = [tex]3.75\times 10^5 Pa[/tex] = 3675 atm (1 kPa= 0.0098 atm)
V= Volume of the gas = ?
T= Temperature of the gas = 23.6°C = 296.6 K [tex]0^00C=273K[/tex]
R= Gas constant = 0.0821 atmL/K mol
n= moles of gas = 45.6
[tex]V=\frac{nRT}{P}=\frac{45.6\times 0.0821\times 296.6}{3675}=0.302L[/tex]
Boyle's Law: This law states that pressure is inversely proportional to the volume of the gas at constant temperature and number of moles.
[tex]P\propto \frac{1}{V}[/tex] (At constant temperature and number of moles)
[tex]P_1V_1=P_2V_2[/tex]
where,
[tex]P_1[/tex] = initial pressure of gas = [tex]3.75\times 10^5 Pa[/tex]
[tex]P_2[/tex] = final pressure of gas = [tex]5.67\times 10^5 Pa[/tex]
[tex]V_1[/tex] = initial volume of gas = 0.302 L
[tex]V_2[/tex] = final volume of gas = ?
[tex]3.75\times 10^5 \times 0.302=5.67\times 10^5\times V_2[/tex]
[tex]V_2=0.199L[/tex]
The final volume has to be 0.199 L, thus (0.302-0.199) L= 0.102 L must release into the atmosphere.
Therefore the answer is 0.102 L
Chloroform flows through a 4.26 inch inside-diameter pipe at the rate of 3.60 gallons per minute. What is the average velocity of chloroform in the pipe? Number ft/s The specific gravity of chloroform is 1.49. What is the mass flow rate of the liquid for the conditions described above?
Answer:
1) 0,081 ft/s
2) 0,746 lb/s
Explanation:
The relation between flow and velocity of a fluid is given by:
Q=Av
where:
Q, flow [ft3/s] A, cross section of the pipe [ft2]v, velocity of the fluid [ft/s]1)
To convert our data to appropiate units, we use the following convertion factors:
1 ft=12 inches
1 ft3=7,48 gallons
1 minute=60 seconds
So,
[tex]Q=\frac{3,60 gallons}{1 min}*\frac{1min}{60 s}*\frac{1ft3}{7,48gallons}=0,00802 \frac{ft3}{s}[/tex]
As the pipe has a circular section, we use A=πd^2/4:
[tex]d=4,26 inch *\frac{1ft}{12 inch}=0,355ft\\ A=\pi \frac{0,355^{2} }{4}=0,0989ft2[/tex]
Finally:
Q=vA......................v=Q/A
[tex]v=\frac{0,00802ft^{3} /s}{0,0989ft^{2} }=0,081ft/s[/tex]
2)
The following formula is used to calculate the specific gravity of a material:
SG = ρ / ρW
where:
SG = specific gravity, ρ = density of the material [lb/ft3]ρW = density of water [lb/ft3] = 62.4 lbs/ft3then:
ρ = SG*ρW = 1,49* 62,4 lb/ft3 = 93 lb/ft3
To calculate the mass flow, we just use the density of the chloroform in lb/ft3 to relate mass and volume:
[tex]0,00802 \frac{ft3}{s}*\frac{93lb}{1ft3}=0,746lb/s[/tex]
Draw the seven constitutional isomers of the cycloalkane with the formula C6H12 Draw the two stereoisomers of 1,3-dibromocyclobutane
To draw the seven constitutional isomers of the cycloalkane with the formula C6H12, we need to consider different arrangements of carbon atoms in a cyclic structure
To draw the seven constitutional isomers of the cycloalkane with the formula C6H12, we need to consider different arrangements of carbon atoms in a cyclic structure with the given molecular formula:
1. Cyclohexane:
H H H
| | |
H--C--C--C--H
| | |
H H H
2. Methylcyclopentane:
H H H
| | |
H--C--C--C--C--H
| | |
H H H
3. Dimethylcyclobutane:
H H H
| | |
H--C--C--C--C--H
| | | |
H H H H
4. Ethylcyclopentane:
H H H
| | |
H--C--C--C--C--H
| | |
H H H
5. 1,1-Dimethylcyclobutane:
H H H
| | |
H--C--C--C--C--H
| | | |
H H H H
6. 1,2-Dimethylcyclopentane:
H H H
| | |
H--C--C--C--C--C--H
| | |
H H H
7. 1,3-Dimethylcyclobutane:
H H H
| | |
H--C--C--C--C--H
| | |
H H H
For the two stereoisomers of 1,3-dibromocyclobutane, we can have:
1. Cis-1,3-dibromocyclobutane:
Br H
| |
H--C--C--C--Br
| |
H H
2. Trans-1,3-dibromocyclobutane:
Br Br
| |
H--C--C--C--H
| |
H H
These structures represent the seven constitutional isomers of cycloalkane C6H12 and the two stereoisomers of 1,3-dibromocyclobutane.
If one mole of a substance has a mass of 56.0 g, what is the mass of 11 nanomoles of the substance? Express your answer in nanograms using the correct number of significant figures. Do not enter your answer using scientific notation.
The mass of 11 nanomoles of the substance is 616 nanograms.
Given that one mole of a substance has a mass of 56.0 g, we can calculate the mass of 11 nanomoles of the substance as follows:
1 mole = 56.0 g
1 nanomole = 56.0 g / 1,000,000,000
=[tex]5.60 * 10^{-8} g[/tex]
Now, to find the mass of 11 nanomoles:
Mass = [tex]11 nanomoles * 5.60 * 10^{-8}g/nanomole[/tex]
Calculating this gives us:
Mass =[tex]6.16 * 10^{-7} g[/tex]
To express the answer in nanograms, we need to convert grams to nanograms:
1 g = 1,000,000,000 ng
So, [tex]6.16 * 10^{-7} g = 6.16 * 10^2 ng = 616 ng[/tex]
Therefore, the mass of 11 nanomoles of the substance is 616 nanograms.
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To find the mass of 11 nanomoles of a substance, multiply the number of nanomoles by the molar mass of the substance. In this case, the mass is 616.0 ng.
Explanation:To calculate the mass of 11 nanomoles of a substance, we need to know the molar mass of the substance. If one mole of the substance has a mass of 56.0 g, then the molar mass is 56.0 g/mol. To find the mass of 11 nanomoles, we can use the conversion factor:
11 nmol * 56.0 g/mol = 616.0 ng
Therefore, the mass of 11 nanomoles of the substance is 616.0 nanograms (ng).
Know the general characteristics of the following; mixtures, solutions, compounds, molecules, and acids and bases.
Explanation:
A mixture is material which is composed up of two or more substances and these substances are physically combined. The identities of each specie in the mixture are retained.
A solution is a type of the homogeneous mixture which is composed of two or more substances. The specie which is dissolved into the another substance is known as solute and in which it is dissolved is known as a solvent.
A compound is the substance which is formed when two or more elements are bonded together chemically.
A molecules are electrically neutral and is a combination of of two or more atoms which are held together by means of chemical bonds.
Acids are the species which furnish hydrogen ions in the solution or is capable of forming bonds with electron pair species as they are electron deficient species.
Bases are the species which furnish hydroxide ions in the solution or is capable of forming bonds with electron deficient species as they are electron rich species.
In order to study hydrogen halide decomposition, a researcher fills an evacuated 1.11 L flask with 0.822 mol of HI gas and allows the reaction to proceed at 428°C: 2HI (g) ⇋ H2(g) + I2(g)
At equilibrium, the concentration of HI = 0.055 M. Calculate Kc. Enter to 4 decimal places.
HINT: Look at sample problem 17.6 in the 8th ed Silberberg book. Write a Kc expression. Find the initial concentration. Fill in the ICE chart. Put the E (equilibrium) values into the Kc expression.
Answer:
Kc = 168.0749
Explanation:
2HI(g) ↔ H2(g) + I2(g)initial mol: 0.822 0 0
equil. mol: 2(0.822 - x) x x
∴ [ HI ]eq = 0.055 mol/L = 2(0.822 - x) / (1.11 L )
⇒ 1.644 - 2x = 0.055 * 1.11
⇒ 1.644 = 2x + 0.06105
⇒ 2x = 1.583
⇒ x = 0.7915 mol equilibrium
⇒ [ H2 ] eq = 0.7915mol / 1.11L = 0.7130 M = [ I2 ] eq
⇒ Kc = ([ H2 ] * [ I2 ]) / [ HI ]²
⇒ Kc = ( 0.7130² ) / ( 0.055² )
⇒ Kc = 168.0749
The equilibrium constant (Kc) for the reaction where 2HI decomposes into H2 and I2 at a given temperature can be calculated using the ICE method. After setting up the Kc expression and solving for x, the concentration changes of HI, H2, and I2 can be used to find Kc, which is approximately 0.1176.
Explanation:To calculate the equilibrium constant (Kc) for the decomposition of hydrogen iodide (HI) into hydrogen gas (H₂) and iodine gas (I₂), we must first write the Kc expression for the reaction:
2HI (g) ⇌ H₂ (g) + I₂ (g)
The Kc expression is Kc = [H₂][I₂] / [HI]^2.
Next, we use an ICE (Initial, Change, Equilibrium) table to keep track of the concentrations throughout the reaction:
Initial: [HI] = 0.822 mol / 1.11 L = 0.7409 M (since the flask is evacuated, [H₂] and [I₂] initially are 0)Change: At equilibrium, [HI] has decreased by x to (0.7409 - 2x) M, since HI decomposes into one mole of H₂ and one mole of I₂ for every two moles of HI that decompose.Equilibrium: [HI] = 0.055 M, so 0.7409 - 2x = 0.055 M. [H₂] and [I₂] both increase by x to x M given the stoichiometry of the reaction.To calculate x, solve the equation 0.7409 - 2x = 0.055, yielding x = 0.34295 M. Now we know at equilibrium:
[HI] = 0.055 M[H₂] = x = 0.34295 M[I₂] = x = 0.34295 MInserting these values into the Kc expression we get:Kc = (0.34295)(0.34295) / (0.055)^2
Simplifying gives us:Kc = 0.1176 to 4 decimal places.
what is conjugate base for h20?
Answer:
OH-
Explanation:
H2O can act as base or acid.
When H2O acts as a base the conjugate acid is H3O+
When H2O acts as an acid the conjugate base is OH-
How much magnesium sulfate heptahydrate is required to prepare 200 mL of 0.05 M solution? 10 g 2.46 g 0 0.05 g 12.38 24.65 g
Final answer:
To prepare a 0.05 M magnesium sulfate heptahydrate solution with a volume of 200 mL, you will need 2.46 grams of magnesium sulfate heptahydrate.
Explanation:
To calculate the amount of magnesium sulfate heptahydrate required to prepare 200 mL of 0.05 M solution, we need to use the formula:
moles = molarity x volume
First, we convert the volume from milliliters to liters:
200 mL x (1 L/1000 mL) = 0.2 L
Next, we substitute the given values into the formula:
moles = 0.05 mol/L x 0.2 L = 0.01 mol
Finally, we calculate the mass of magnesium sulfate heptahydrate using its molar mass:
moles = mass (g) / molar mass (g/mol)
0.01 mol = mass (g) / 246.48 g/mol
mass (g) = 0.01 mol x 246.48 g/mol = 2.46 g
Which of the following arises from interaction of dipoles? 3. a) lonic bond b) Covalent bond c) Metallic bond d) Hydrogen bond
Answer:
Hydrogen bond
Explanation:
Hydrogen bond is a type of dipole dipole interaction.
Hydrogen bond is present between the hydrogen atom attached to an electronegative atom and other electronegative atom. Compounds in which hydrogen is attached with N, F and O, show hydrogen bonding.
As N, F and O are more electronegative than H, so a positively charge is develop on H atom and negative charge is develop on electronegative atom.
Hydrogen bond is a weaker bond. It is weak as compared to ionic bond and covalent bond but stronger than van der Waals interaction.
Ionic bond is formed by the complete transfer and gain of electrons. It is a electrostatic attraction between positively charged and negatively charged species. Fox example bonding in NaCl
Covalent bond is formed by the sharing of electrons between two atoms. For example bonding in Cl_2.
Metallic bond is present between mobile electrons and positively charged kernels. It exist in metals only.
Answer:
d) Hydrogen bond
Explanation:
Hydrogen bonding is a special type of the dipole-dipole interaction and it occurs between hydrogen atom that is bonded to highly electronegative atom which is either fluorine, oxygen or nitrogen atom.
Partially positive end of the hydrogen atom is attracted to partially negative end of these atoms which is present in another molecule. It is strong force of attraction between the molecules.
1. Separation of amino acids by ion-exchange Chromatography. Mixtures of amino acids can be analyzed by first separating the mixture into its components through ion-exchange chromatography. Amino acids placed on a cation-exchange resin containing sulfate (-SO3-) groups flow down the column at different rates because of two factors that influence their movement (1) ionic attraction between the sulfonate residues on the column and positively charged functional groups on the amino acids, and (2) hydrophobic interactions between amino acid side chains and the strongly hydrophobic backbone of the polystyrene resin. For each pair of amino acids listed, determine which will be eluted first from the cation-exchange column by a ph 7.0 buffer.
a.) Asp and Lys
b.) Arg and Met
c.) Glu and Val
d.) Gly and Val
e.) Ser and Ala
a.) Asp and Lys
Asp will elute first from the column because it has less positively charged functional groups than Lys.
b.) Arg and Met
Met will elute first from the column because it has less positively charged functional groups than Lys.
c.) Glu and Val
Glu will elute first from the column because it has more negativity functional groups than Lys and will be not be much retained by the -SO₃⁻ groups from the ion-exchange coloumn.
d.) Gly and Val
Gly will elute first from the column because Lys have a longer alkyl chain which will be attracted by the strongly hydrophobic backbone for the resin.
e.) Ser and Ala
Ser will be eluted first from the column because Ala alkyl chain will be more attracted by the strongly hydrophobic backbone for the resin. Ser have an -OH group which will decrease the hydrophobicity of the alkyl chain and will not be so much retained on the column.
Charlotte is driving at 70.4 mi/h and receives a text message. She looks down at her phone and takes her eyes off the road for 4.54 5. How far has Charlotte traveled in feet during this time? distance:
The distance travelled by charlotte is 7269ft/mi.
Given:
Charlotte's speed = 70.4 mi/h.
Time she takes her eyes off the road = 4.54 seconds.
To calculate the distance Charlotte has traveled in feet while looking at her phone, it is important to convert her speed from miles per hour (mi/h) to feet per second (ft/s), and then use that speed to calculate the distance.
Convertion of speed to feet per second:
1 mile = 5280 feet
1 hour = 3600 seconds
Speed in ft = (70.4 (mi/h) × 5280(ft/mi)) / (3600(s/h))
Speed in ft = 103.25ft.
Calculation of distance:
Distance =Speed / Time
Distance = 103.25 * 70.4
Distance = 7269 ft/mi
Therefore, the distance travelled by charlotte is 7269ft/mi.
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Charlotte traveled approximately 469.761 feet while looking at her phone for 4.54 seconds by driving at a speed of 70.4 mi/h.
Explanation:Calculating Distance Traveled While DistractedCharlotte is driving at 70.4 mi/h. To find out how far she has traveled in feet while looking at her phone for 4.54 seconds, we need to convert the speed from miles per hour to feet per second and then multiply by the time in seconds.
First, we convert the speed from miles per hour to feet per second using the conversion factors 1 mile = 5280 feet and 1 hour = 3600 seconds:
70.4 mi/h × 5280 ft/mi × 1/3600 h/s = 103.46667 ft/s.
Now, we multiply the speed in feet per second by the time she is distracted:
103.46667 ft/s × 4.54 s = 469.761 ft.
Charlotte has therefore traveled approximately 469.761 feet while looking down at her phone for 4.54 second
A drug decomposes by a first order mechanism, with a half-life of 5.00 years. Calculate how long it will take for 80% of the drug to decompose
Answer : The time taken for the decomposition of drug will be 11.6 years.
Explanation :
Half-life = 5.00 years
First we have to calculate the rate constant, we use the formula :
[tex]k=\frac{0.693}{5.00}[/tex]
[tex]k=0.139\text{ years}^{-1}[/tex]
Now we have to calculate the time taken.
Expression for rate law for first order kinetics is given by:
[tex]t=\frac{2.303}{k}\log\frac{a}{a-x}[/tex]
where,
k = rate constant
t = time taken = ?
a = let initial amount of the reactant = 100 g
a - x = amount left after decay process = 100 - 80 = 20 g
Now put all the given values in above equation, we get
[tex]t=\frac{2.303}{0.139}\log\frac{100}{20}[/tex]
[tex]t=11.6\text{ years}[/tex]
Therefore, the time taken for the decomposition of drug will be 11.6 years.
An interpretation of the results of many tests is called A) an experiment. B) a prediction D) a theory
Final answer:
An interpretation of the results of many tests is called a theory. It is a well-substantiated explanation of an aspect of the natural world, based on extensive research and experimentation.
Explanation:
An interpretation of the results of many tests is called a theory. A theory is a well-substantiated explanation of an aspect of the natural world. It is a scientific explanation that has been repeatedly tested and supported by many experiments.
For example, the theory of evolution is a well-supported explanation of how species have evolved over time. It is based on extensive research and experimentation.
It is important to note that a theory in science is different from the everyday use of the word, which often refers to a guess or speculation. In science, a theory is a well-tested and supported explanation.
Part A In a particular experiment at 300 ∘C, [NO2] drops from 0.0138 to 0.00886 M in 374 s . The rate of disappearance of NO2 for this period is ________ M/s. In a particular experiment at 300 , drops from 0.0138 to 0.00886 in 374 . The rate of disappearance of for this period is ________ . −6.06×10−5 6.60×10−6 7.57×104 2.64×10−5 1.32×10−5
Final answer:
The rate of disappearance of NO2 in the experiment at 300 °C is approximately 1.32 × 10^-5 M/s, calculated by the change in concentration over time.
Explanation:
The rate of disappearance of NO2 in the experiment at 300 °C is calculated by determining the change in concentration over the change in time. The initial concentration of NO2 is 0.0138 M and it drops to 0.00886 M over 374 s. The change in concentration (Δ[NO2]) is 0.0138 M - 0.00886 M = 0.00494 M. The rate of disappearance is then Δ[NO2]/Δt = 0.00494 M/374 s.
After calculating this, we get the rate of disappearance of NO2 to be approximately 1.32 × 10-5 M/s, which is one of the possible choices provided in the question.
Conversion of mass to moles A continuous feed to a separation unit is 1,000 kg/h of 45 wt% methanol and 55 wt% water, whose molecular weights are 32 and 18, respectively. Compute: (a) feed rate in lbmol/h, and (b) composition in mole fractions.
Answer:
Total feed rate = 98.3 lbmol/h
methanol mole fraction = 0.315
water mole fraction = 0.685
Explanation:
First of all, it is needed to calculate the feed mass of methanol and water in kg/h.
For methanol:
[tex]m_{methanol} = m\%wt_{methanol}/100 = (1000kg/h)(45\%)/100[/tex]
[tex]m_{methanol} = 450kg/h[/tex]
For water:
[tex]m_{water} = m\%wt_{water}/100 = (1000kg/h)(55\%)/100[/tex]
[tex]m_{water} = 550 kg/h[/tex]
Now, change from mass units (kg/h) to moles units (kmol/h and lbmol/h) using simple conversion factors:
For methanol:
[tex]n_{methanol} = (450\frac{kg}{h})(\frac{1 kmol}{32 kg} )[/tex]
[tex]n_{methanol} = 14.1kmol/h[/tex]
For water:
[tex]n_{water} = (550\frac{kg}{h})(\frac{1 kmol}{18 kg} )[/tex]
[tex]n_{water} = 30.6kmol/h[/tex]
Change units from kmol/h to lbmol/h
For methanol:
[tex]n_{methanol} = (14.1\frac{kmol}{h})(\frac{1 lbmol}{0.454 kmol} )[/tex]
[tex]n_{methanol} = 31.0 lbmol/h[/tex]
For water:
[tex]n_{water} = (30.6\frac{kg}{h})(\frac{1 lbmol}{0.454 kmol} )[/tex]
[tex]n_{water} = 67.3 lbmol/h[/tex]
Sum moles of methanol and water in lbmol/h to compute the total feed rate:
[tex]n = 31.0 lbmol/h + 67.3 lbmol/h[/tex]
[tex]n = 98.3 lbmol/h[/tex]
Divide both methanol and water moles feed rates by total feed rate:
For methanol:
[tex]X_{methanol} = \frac{31.0 lbmol/h}{98.3 lbmol/h}[/tex]
[tex]X_{methanol} = 0.315[/tex]
For water:
[tex][X_{water} = \frac{67.3 lbmol/h}{98.3 lbmol/h}[/tex]
[tex]X_{water} = 0.685[/tex]
End
The weight percent of concentrated H2SO4, molar mass=980 g/mol, is 960% and its density is 184 g/ml. What is the molarity of concentrated H2SO4? 9.79 M 12.0 M 18.0M 532M 245 M avigator AN 20 Backspace HKLIL
Answer:
C H2SO4 = 9.79 M
Explanation:
molarity (M) ≡ # dissolved species / V slnH2SO4 ↔ H3O+ + SO4-∴ %w/w H2SO4 = 960% = g H2SO4 / g sln * 100
⇒ 9.6 = g H2SO4 / g sln
calculation base: 1000 g sln
⇒ g H2SO4 = 9600g
⇒ mol H2SO4 = 9600 g H2SO4 * ( mol H2SO4/ 980g H2SO4 ) = 9.796 mol H2SO4
⇒ V sln = 1000g sln / 1000g/L = 1 L sln
∴ ρ H20 ≅ 1000 Kg/m³ = 1000 g/L
⇒ C H2SO4 = 9.796 mol H2SO4 / 1 L sln
⇒ C H2SO4 = 9.796 M
The measured cube has the following information: mass = 3.4800 g height = length - width = 1:00 cm • Determine volume of cube V- • Determine calculate density of cube. D-
Answer:
Volume of cube = [tex]L^3[/tex] = [tex]1 cm^3[/tex]
Density = [tex] \frac{m}{V} = 3.48 \frac{g}{cm^3}[/tex]
Two iron oxide samples are given to you where one is red and the other is black. You perform a chemical analysis and you find that the red sample has a Fe/O mass ratio of 2.327 and the black has a Fe/O mass ratio of 3.491. You suspect the red sample is simple rust or Fe2O3. What is the chemical formula for the black sample?
Answer:
Chemical formula for the black iron oxide sample is FeO
Explanation:
Given:
Mass ratio of black iron oxide sample, Fe/O = 3.491
To determine:
The chemical formula
Calculation:
The mass ratio is Fe:O = 3.491 : 1
Mass of Fe = 3.491 g
Mass of O = 1.000 g
Atomic mass of Fe =55.85 g/mol
Atomic mass of O = 16.00 g/mol
[tex]Moles\ Fe = \frac{3.491g}{55.85g/mol} =0.625\\\\Moles\ O = \frac{1.000g}{16.00g/mol} =0.0625\\[/tex]
Therefore, the molar ratio of Fe:O = 1:1
Hence, chemical formula is FeO
How would you prepare 250.0 mL of 0.00200 M Na2S2O3? Describe what you would do in the lab. Include amounts and types of glassware and equipment that you would use.
Answer:
You have to weigh 0.079 g of Na₂S₂0₃ and dissolve it in water at a final volume of 250 ml.
Explanation:
A 0.00200 M Na₂S₂O₃ solution has 0.002 mol of Na₂S₂O₃ per liter of solution. As we know that 1 L = 1000 ml, and that the molecular weight of Na₂S₂O₃ IS 158 g/mol, we can calculate the mass of Na₂S₂O₃ to weigh as follows:
mass= [tex]\frac{0.002 mol Na2S2O3}{1000 ml solution}[/tex] x [tex]\frac{158 g}{1 mol Na2S2O3}[/tex] x 250 ml solution
mass= 0.079 g
To prepare the solution, we have to weigh in a beaker with 0.079 g of Na₂S₂O₃ by employing an analytical balance. Then, we have to dissolve the mass in a volume of aproximately 230 ml water. For this, we measure the 230 ml of water in a graduated cylinder, we add the volume to the beaker with the mass and we agitate until total disolution. Finally, we tranfer the total amount of the solution in the glass to a volumetric flask with a capacity of 250 ml. We add water until we reach the capacity, and then we homogenize the solution.
I'm kinda stumped here :/
Suppose now that you wanted to determine the density of a small crystal to confirm that it is phosphorus. From the literature, you know that phosphorus has a density of 1.82 g/m^3 . How would you prepare 20.0 mL of the liquid mixture having that density from pure samples of CHCl^3 ( d= 1.492 g/mL) and CHBr^3( d= 2.890 g/mL)? (Note: 1 mL = 1 cm^3 .)
Answer:
To prepare 20,0 mL of the liquid mixture you should mix 15,3 mL of CHCl₃ with 4,7 mL of CHBr₃
Explanation:
Here you have two variables: The volume of both CHCl₃ (X) and CHBr₃ (Y). To find these two variables you must have, at least, two equations.
You know total volume is 20,0 mL. Thus:
X + Y = 20,0 mL (1)
The other equation is:
[tex]\frac{X}{20,0mL\\}[/tex] × 1,492 g/mL + [tex]\frac{Y}{20,0 mL}[/tex] × 2,890 g/mL = 1,82 g/mL (2)
If you replace (1) in (2):
[tex]\frac{X}{20,0mL\\}[/tex] × 1,492 g/mL + [tex]\frac{20,0 mL - X}{20,0 mL}[/tex] × 2,890 g/mL = 1,82 g/mL
Solving:
X = 15,3 mL
Thus, using (1):
20,0 mL - 15,3 mL = Y = 4,7 mL
Thus, to prepare 20,0 mL of the liquid mixture you must mix 15,3 mL of CHCl₃ with 4,7 mL of CHBr₃.
I hope it helps!
Many buffers are polyprotic, such as carbonic acid. Bicarbonate (HCO3) has a pKa of 10.33; while Carbonic acid (HCO3) has a pKa of 6.35. Use the Henderson-Hasselbalch equation to mathematically determine which form of this buffer predominates at pH 7.4 (homeostasis pH of blood). Besides your mathematical answer, state the form that predominates too.
Answer:
Carbonic acid/bicarbonate; bicarbonate
Explanation:
H₂CO₃ + H₂O ⇌ H₃O⁺ + HCO₃⁻; pKₐ₁ = 6.35
HCO₃⁻ + H₂O ⇌ H₃O⁺ + CO₃²⁻; pKₐ₂ = 10.33
[tex]\text{pH}& = &\text{pK}_{\text{a}} + \log\dfrac{[\text{A}^{-}]}{\text{[HA]}}\\\\7.4 & = &\text{pK}_{\text{a}} + \log\dfrac{[\text{A}^{-}]}{\text{[HA]}}[/tex]
The best buffer is one for which pKₐ ≈ pH.
6.35 is closer to 7.4, so the carbonic acid/bicarbonate form of the buffer predominates
The pH of the blood is higher (more basic) than the pKₐ of carbonic acid, so its basic form (bicarbonate, HCO₃⁻) predominates.
Final answer:
The bicarbonate form of carbonic acid predominates in the buffer system at pH 7.4.
Explanation:
The Henderson-Hasselbalch equation is used to determine the pH of a buffer solution. In this case, we are trying to determine which form of the carbonic acid buffer predominates at pH 7.4, which is the homeostasis pH of the blood.
The Henderson-Hasselbalch equation is:
pH = pKa + log ([base] / [acid])
Given that the pKa of carbonic acid is 6.35, we can plug in the values to calculate:
pH = 6.35 + log (0.024 / 0.0012)
Simplifying the equation, we get:
pH = 6.35 + log (20)
pH = 6.35 + 1.3010
pH = 7.65
So, at pH 7.4, the bicarbonate (HCO3-) form of carbonic acid predominates in the buffer system.
A city in Laguna generates 0.96 kg per capita per day of Municipal Solid Waste (MSW). Makati City in Metro Manila generates 1.9 kg per capita per day of MSW. If both cities have population of 20000 inhabitants, determine the following
i. amount of MSW generated in each city
ii. number of trucks needed to collect the MSW twice weekly, if each truck has a
capacity of 4.4 tons and operates 5 days per week. Two loads per day at 75%
capacity is the travel quota per day
iii. -the number and volume of MSW in tons that enter the landfill
Answer:
i) amount of MSW generated:
Laguna: 19200 Kg/day
Makati: 38000 Kg/day
ii) number of trucks to collect twice weekly:
Laguna: 3 trucks
Makati: 5 trucks
iii) volume of MSW in tons that enter landfill/week:
Laguna: 147.84 ton/week
Makati: 292 ton/week
Explanation:
i) Laguna: 0.96 Kg person / day of MSW * 20000 = 19200 Kg MSW / day
⇒ Laguna: 19200 Kg/day * ( 7day/ week ) = 134400 Kg/week
Makati: 1.9 Kg person / day of MSW * 20000 = 38000 Kg MSW / day
⇒ Makati: 38000 Kg/day * ( 7day/week ) = 266000 Kg/week
ii) truck capacity = 4.4 ton * ( Kg / 0.0011 ton ) = 4000 Kg
⇒ quote/day = 4000 Kg * 0.75 = 3000 Kg
⇒ loads/day = 2 * 3000 kg = 6000 Kg
⇒ operate/week = 5 * 6000 Kg = 30000 Kg
∴ Laguna: number of trucks needed/week= 134400 / 30000 = 4.48 ≅ 5 trucks
⇒ number of trucks to collect twice weekly = 5 / 2 = 2.5 ≅ 3 trucks
∴ Makati : number of trucks needed/week = 266000 / 30000 = 8.86 ≅ 9 trucks
⇒ number of trucks to collect twice weekly = 9 / 2 = 4.5 ≅ 5 trucks
iii) enter landfill/week:
Laguna: 134400Kg MSW/week * ( 0.0011 ton/Kg ) = 147.84 ton/week
Makati: 266000Kg MSW/week * ( 0.0011 ton/Kg ) = 292 ton/week