Answer:
[tex]T_{sol}=-15.9\°C[/tex]
Explanation:
Hello,
In this case, we can analyze the colligative property of solutions - freezing point depression - for the formed solution when ethylene glycol mixes with water. Thus, since water freezes at 0 °C, we can compute the freezing point of the solution as shown below:
[tex]T_{sol}=T_{water}-i*m*Kf[/tex]
Whereas the van't Hoff factor for this solute is 1 as it is nonionizing and the molality is:
[tex]m=\frac{mol_{solute}}{kg\ of\ water}=\frac{45.0g*\frac{1mol}{62g} }{85.0g*\frac{1kg}{1000g} } =8.54m[/tex]
Thus, we obtain:
[tex]T_{sol}=0\°C+(-8.54m*1.86\frac{\°C}{m} )\\\\T_{sol}=-15.9\°C[/tex]
Best regards.
The freezing point of a solution prepared from 45.0 g ethylene glycol and 85.0 g of water is -15.9 °C.
What is freezing point depression?Freezing point depression is a drop in the temperature at which a substance freezes, caused when a smaller amount of another, non-volatile substance is added.
Step 1: Calculate the molality of the solution.We will use the definition of molality.
b = mass of solute / molar mass of solute × kg of solvent
b = 45.0 g / 62.07 g/mol × 0.0850 kg = 8.53 m
Step 2: Calculate the freezing point depression (ΔT).We will use the following expression, where Kf is the cryoscopic constant of water.
ΔT = Kf × b = 1.86 °C/m × 8.53 m = 15.9 °C
Step 3: Calculate the freezing point of the solution.The freezing point of pure water is 0°C.
T = 0°C - 15.9 °C = -15.9 °C
The freezing point of a solution prepared from 45.0 g ethylene glycol and 85.0 g of water is -15.9 °C.
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A local barista serves coffee at 85 C. You add ice to the coffee to cool it to 55 C. Assume that an ice cube is 24g and -18.5 degrees Celsius. Hiw many ice cubes would you need to add to your 355mL cup of coffee to bring it to 55 degrees Celsius?.. The specific heat of ice is 2.95J/g degrees Celsius, the specific heat is 4.184 J/g degrees Celsius, and the specific heat of fusion of water is 334 J/g. Remember that an ice cube will need to be warmed to 0 degrees Celsius, will melt, and then the newly melted water will be warmed to 55 degrees Celsius.
A .1
B .3
C .4
D .2
Answer:
B. 3
Explanation:
To decrease the temperature of your coffee from 85°C to 55°C your system need to absorb energy. This energy will be absorbed from the addition of some ice.
How many energy must be absorbed? You can use:
Q = C×m×ΔT
Where Q is heat (Energy), C is specific heat of your solution (4.184J/g°C), m is its mass (mass of 355mL of coffee = 355g) and ΔT is change in temperature (85°C-55°C = 30°C)
Replacing, your ice needs to absorb:
Q = C×m×ΔT
Q = 4.184J/g°C×355g×30°C
Q = 44559.6J
The energy that is taken from an ice cube to change its temperature from -15°C to 55°C is:
Energy from -15°C to 0°C (C of ice = 2.95J/g°C):
Q = C×m×ΔT
Q = 2.95J/g°C×24g×15°C
Q = 1062J
Now the energy taken to pass the ice from solid to liquid is:
Q = ΔHf×m
Q = 334J/g×24g
Q = 8016J
And the energy to increase the temperature of 0°C to 55°C of 24g of ice:
Q = 4.184J/g°C×24g×55°C
Q = 5522.9J
And the total energy that 1 ice cube needs is:
Q = 1062J + 8016J + 5522.9J
Q = 14600.9J
But you need 44559.6J to decrease the temperature of your coffee, that is:
44600J / 14600.9J = 3.05
≈ 3 ice cubes to decrease the temperature of the coffee.
Right solution:
B. 3Answer:
3
Explanation:
I’m not positive but I think it’s correct
If D+2 would react with E-1, what do you predict to be the formula?
Answer:
DE2
Explanation: for every one D+2 you need two E-1 because +2=-2
You mix 500.0 mL of 0.250 M iron(III) chloride solution with 425.0 mL of 0.350 M barium chloride solution. Assuming the volumes are additive, what is the molarity of chloride ion in the mixture.
Answer:
[tex]M=0.727M[/tex]
Explanation:
Hello,
In this case, since iron (III) chloride (FeCl3) and barium chloride (BaCl2) are both chloride-containing compounds, we can compute the moles of chloride from each salt, considering the concentration and volume of the given solutions, and using the mole ratio that is 1:3 and 1:2 for the compound to chlorine:
[tex]n_{Cl^-}=0.50L*0.250\frac{molFeCl_3}{L}*\frac{3molCl^-}{1molFeCl_3}=0.375molCl^- \\\\n_{Cl^-}=0.425L*0.350\frac{molBaCl_2}{L}*\frac{2molCl^-}{1molBaCl_2}=0.2975molCl^-[/tex]
So the total mole of chloride ions:
[tex]N_{Cl^-}=0.2975mol+0.375mol=0.6725molCl^-[/tex]
And the total volume by adding the volume of each solution in L:
[tex]V=0.500L+0.425L=0.925L[/tex]
Finally, the molarity turns out:
[tex]M=\frac{0.6725molCl^-}{0.925L}\\ \\M=0.727M[/tex]
Best regards.
Suppose you have a container filled with air at 212 oF. The volume of the container 1.00 L, the pressure of air is 1.00 atm. The molecular composition of air is 79% N2 and 21% O2 for simplification. Calculate the mass of air and moles of O2 in the container.
Answer:
[tex]m_{air}=0.947g[/tex]
[tex]n_{O_2} =0.00686molO_2[/tex]
Explanation:
Hello,
In this case, we can firstly use the ideal gas equation to compute the total moles of the gaseous mixture (air) with the temperature in Kelvins:
[tex]T=212\°F=100\°C=373.15K\\\\n=\frac{PV}{RT}=\frac{1.00atm*1.00L}{0.082\frac{atm*L}{mol*K}*373.15K}\\ \\n=0.0327mol[/tex]
In such a way, since the molar mass of air is 28.97 g/mol, we can compute the mass of air with a single mass-mole relationship:
[tex]m_{air}=0.0327mol*\frac{28.97g}{1mol} =0.947g[/tex]
Finally, knowing that the 21% of the 0.0327 moles of air is oxygen, its moles turn out:
[tex]n_{O_2}=0.0327mol*\frac{0.21molO_2}{1mol} =0.00686molO_2[/tex]
Best regards.
Which of the following functional groups is formed from the condensation of carboxylic acids???
a. acid anhydride
b. acid halide
c. amide
d. ester
e. ether
Answer:
a
Explanation:
its made up of carbon and hydrogen
Interpret the following equation for a chemical reaction using the coefficients given:
Cl2(g) + F2(g) 2ClF(g)
On the particulate level:
________ of Cl2(g) reacts with ______ of F2(g) to form______ of ClF(g).
On the molar level:
______ of Cl2(g) reacts with______ of F2(g) to form______ of ClF(g).
Answer and Explanation:
Given the following chemical equation:
Cl₂(g) + F₂(g) ⇒ 2ClF(g)
The coefficients are: 1 for Cl₂, 1 for F₂ and 2 for ClF. The coefficients indicate the number of units of each ompound that participates in the reaction. It gives the proportion of reactants and products in the reaction. These units can be molecules or moles. In this reaction, we can say:
On the particulate level: 1 molecule of Cl₂(g) reacts with 1 molecule of F₂(g) to form 2 molecules of ClF(g).
On the molar level: 1 mol of Cl₂(g) reacts with 1 mol of F₂(g) to form 2 mol of ClF(g).
A certain reaction has the following general form. aA → bB At a particular temperature and [A]0 = 2.80 ✕ 10−3 M, concentration versus time data were collected for this reaction, and a plot of 1/[A] versus time resulted in a straight line with a slope value of +3.40 ✕ 10−2 L mol−1 s−1. (a) Determine the rate law, the integrated rate law, and the value of the rate constant for this reaction. (Rate expressions take the general form: rate = k . [A]a . [B]b.) rate law:
Answer:
Rate law: [tex]r=k[A]^2[/tex]
Integrated rate law: [tex]\frac{1}{[A]}=kt+ \frac{1}{[A]_0}[/tex]
Rate constant: [tex]k=3.40x10^{-2}\frac{L}{mol*s}[/tex]
Explanation:
Hello,
In this case, since the slope is obtained by plotting 1/[A] and it has the units L/(mol*s) or 1/(M*s), we can infer the reaction is second-order, therefore, its rate law is:
[tex]r=k[A]^2[/tex]
The integrated rate law:
[tex]\frac{1}{[A]}=kt+ \frac{1}{[A]_0}[/tex]
That is obtained from the integration of:
[tex]\frac{d[A]}{dt}=-k[A]^2[/tex]
And of course, since the slope equals the rate constant, its value is:
[tex]k=3.40x10^{-2}\frac{L}{mol*s}[/tex]
Regards.
When 3-methylpent-2-ene is treated with mercury(II) acetate in methanol and the resulting product isreacted with NaBH4, what is the primary organic compound which results
Answer:
3-methylpentan-3-ol
Explanation:
In this case, we have an "Oxymercuration reaction". With this in mind, we will have to add an "OH" to the most substituted carbon of the double bond and we will obtain 3-methylpentan-3-ol. To understand how this molecule is produced we have to check the mechanism:
The mercury(II) acetate ([tex]Hg(OAC)_2[/tex]) is an ionic substance. So, this substance can be ionized into his ions and we will have the cation [tex]HgOAc^+[/tex] and the anion [tex]AcO^-[/tex]. The cation will attack the double bond and vice-versa to produce a "cyclic intermediate". Then a water molecule will attack the most substituted carbon and the cyclic compound would be broken producing a new bond C-O with a positive charge in the oxygen. Then a deprotonation step takes place and finally, the [tex]NaBH_4[/tex] would reduce the compound to produce the final alcohol.
See figure 1
I hope it helps!
In each of the three reactions between NaOH and HCl, the sign of q for the water was positive. This means the the sign of q for the reaction was ______ and the reaction was ______.
Answer:
This means the the sign of q for the reaction was _NEGATIVE _____ and the reaction was _EXOTHERMIC_____.
Explanation:
In calorimetry, when heat is absorbed by the solution, the q-value of the solution will have a positive value. This means that the reaction will produce heat for the solution to absorb and thus the q-value for the reaction will be negative. This is an exothermic reaction.
Whereas, when heat is absorbed from the solution, the q-value for the solution will have a negative value. This means that the reaction will absorb heat from the solution and so the reaction is endothermic, and q value for the reaction is positive.
So, from the question, since the q-value of water is positive, it means that heat is absorbed by the solution and the reaction will produce a negative value of q and it's an exothermic reaction because the reaction produces heat for the solution.
For the product of the reaction below, which proton is removed irreversibly by NaNH2 base, thus preventing any isomerization of the alkyne bond in the product?
Answer:
The Highly acidic proton joined to one of the carbon in the ALKYNE bond.
(Kindly Check the attachment for the drawing because the solution will need us to draw).
Explanation:
So, let us start by defining some major key terms in this particular Question given above;
(1). ISOMERIZATION: isomerization can simply be defined as the kind is of chemical rearrangement whichay lead to the breaking and the formation of new bonds.
(2). NaNH2 BASE: Sodium amide is a Chemical compound which has a Molar mass of 39.01 g/mol and Heat capacity (C) of 66.15 J/mol K. It is also known as sodamide. It is a good nucleophile.
(3). ALKYNE BOND: it is a C-C joined together by three bonds.
The chemical reaction given in the Question is given in the attachment too.
Therefore, The Highly acidic proton joined to one of the carbon in the ALKYNE bond is removed irreversibly by NaNH2 base.
Briefly in your own words, describe the face-centered cubic unit cell in such a way that someone reading only your description, would be able to make a reasonable sketch of the structure.
Answer:
Face-centered Cubic Unit Cell (FCC) is consists of 8 atoms at the corners (one atom at each corner) and 6 atoms are present at the face-center (one atom at each face). Each eight atoms at corners contribute 1/8 and six atoms at faces contribute 1/2 in each unit cell.
If the equilibrium constant of the reaction is 0.85, then which statement is true if the mass of A is 10.5 grams; the density of B is 0.82 g/ml; the concentration of C is 0.64 M; and the concentration of D is 0.38 M.
A(s) + 3 B(l) _____ 2(aq) + D(aq)
Pick the correct statement about this system.
A. Q < K and reaction shifts left
B. Q > K and reaction shifts left
C. Q > K and reaction shifts right
D. Q = K and reaction does not shift
E. Q < K and reaction shifts right
Answer:
E. Q < K and reaction shifts right
Explanation:
Step 1: Write the balanced equation
A(s) + 3 B(l) ⇄ 2(aq) + D(aq)
Step 2: Calculate the reaction quotient (Q)
The reaction quotient, as the equilibrium constant (K), only includes aqueous and gaseous species.
Q = [C]² × [D]
Q = 0.64² × 0.38
Q = 0.15
Step 3: Compare Q with K and determine in which direction will shift the reaction
Since Q < K, the reaction will shift to the right to attain the equilibrium.
Predict the product of the following Wittig reaction. Be sure your answer accounts for stereochemistry, where appropriate. If multiple stereoisomers form, be sure to draw all products using appropriate wedges and dashes.
1. PPh3
5-iodo-1-phenyl-1-pentanone →
2. n- BuLi
Answer:
Final product: cyclopent-1-en-1-ylbenzene
Explanation:
In this case, we have a Wittig reaction. The addition of [tex]PPh_3[/tex] and n-Buli will produce the "Ylide compound". First, we will have an Sn2 reaction in which the iodide is replaced by triphenylphosphine. Then the base n-Buli will remove a hydrogen atom to form a double bond (Ylide compound). Then the double bond will be delocalized to produce a carbanion. This carbanion, will attack the carbon in the carbonyl group generating a negative charge in the oxygen. Then the negative charge will attack the phosphorous atom to produce a cyclic structure. Finally, the cyclic structure is broken to produce the alkene (cyclopent-1-en-1-ylbenzene).
See figure 1
I hope it helps!
State five difference between ionic compound and covalent compound
Answer:
Compound are defined as the containing two or more different element .
(1) Ionic compound and (2) Covalent compound.
Explanation:
Covalent compound :- covalent compound are the sharing of electrons two or more atom.
Covalent compound are physical that lower points and compared to ionic .
Covalent compound that contain bond are carbon monoxide (co), and methane .
Covalent compound are share the pair of electrons.
Covalent compound are bonding a hydrogen atoms electron.
Ionic compound a large electrostatic actions between atoms.
Ionic compound are higher melting points and covalent compound.
Ionic compound are bonding a nonmetal electron.
Ionic electron can be donate and received ionic bond.
Ionic compound bonding kl.
Provide the reagents necessary to carry out the following conversion. Group of answer choices KMnO4, NaOH,H2O KMnO4, H3O , 75oC H2SO4, heat 1. mCPBA 2. H3O none of these
Answer:
KMnO4,H3O^+,75°C
Explanation:
The conversion of cyclohexene to trans-1,2-cylohexanediol is an oxidation reaction. Alkenes are oxidized in the presence of potassium permanganate and acids to yield the corresponding diols.
These diols may also be called glycols. They are molecules that contain two -OH(hydroxyl) groups per molecule. The reaction closely resembles the addition of the two -OH groups of hydrogen peroxide to an alkene.
The bright color of potassium permanganate disappears in this reaction so it can be used as a test for alkenes.
How many moles of aqueous sodium ions and sulfide ions are formed when 2.05 mol of sodium sulfide dissolved in water
Explanation:
We have to put out the balanced chemical equation before proceeding. The equation for the dissociation of sodium sulfide in water is given as;
Na₂S --> 2Na⁺ + S²⁻
From the stoichiometry of the reaction we can tell that;
1 mol of Na₂S produces 2 moles of Na⁺ and 1 mol of S²⁻.
Sodium ion:
1 mol of Na₂S = 2 mol of Na⁺
2,05 = x
upon solving for x;
x = 2 * 2.05 = 4.10 moles
Sulfide ion:
1 mol of Na₂S = 1 mol of S²⁻
2.05 = x
upon solving for x;
x = 1 * 2.05 = 2.05 moles
Answer:
HEY ITS RIGHT ABOVE!!!
Explanation:
i did 1.85 instead of 2.05 and it was right... im so sorry
The addition of 0.242 L of 1.92 M KCl to a solution containing Ag+ and Pb2+ ions is just enough to precipitate all of the ions as AgCl and PbCl2. The total mass of the resulting precipitate is 65.08 g. Find the mass of PbCl2 and AgCl in the precipitate. Calculate the mass of PbCl2 and AgCl in grams.
Answer:
Mass PbCl₂ = 50.24g
Mass AgCl = 14.84g
Explanation:
The addition of Cl⁻ ions from the KCl solution results in the precipitation of AgCl and PbCl₂ as follows:
Ag⁺ + Cl⁻ → AgCl(s)
Pb²⁺ + 2Cl⁻ → PbCl₂(s)
If we define X as mass of PbCl₂, moles of Cl⁻ from PbCl₂ are:
Xg × (1mol PbCl₂/ 278.1g) × (2moles Cl⁻ / 1 mole PbCl₂) = 0.00719X moles of Cl⁻ from PbCl₂
And mass of AgCl will be 65.08g-X. Moles of Cl⁻ from AgCl is:
(65.08g-Xg) × (1mol AgCl/ 143.32g) × (1mole Cl⁻ / 1 mole AgCl) = 0.45409 - 0.00698X moles of Cl⁻ from AgCl
Moles of Cl⁻ that were added in the KCl solution are:
0.242L × (1.92mol KCl / L) × (1mole Cl⁻ / 1 mole KCl) = 0.46464 moles of Cl⁻ added.
Moles Cl⁻(AgCl) + Moles Cl⁻(PbCl₂) = Moles Cl⁻(added)
0.45409 - 0.00698X moles + (0.00719X moles) = 0.46464 moles
0.45409 + 0.00021X = 0.46464
0.00021X = 0.01055
X = 0.01055 / 0.00021
X = 50.24g
As X = Mass PbCl₂
Mass PbCl₂ = 50.24gAnd mass of AgCl = 65.08 - 50.24
Mass AgCl = 14.84gThe masses of the compounds in the precipitate can be found my knowing
the number of moles of chloride ion contributed by each compound.
The mass of PbCl₂ in the precipitate is approximately 49.24 gThe mass of AgCl in the precipitate is approximately 15.84 gReasons:
The given parameter are;
Volume of KCl solution added = 0.242 L
Concentration of KCl solution = 1.92 M KCl
The ions in the solution to which KCl is added = Ag⁺ and Pb²⁺ ions
Precipitates formed = AgCl and PbCl₂
The mass of the precipitate = 65.08 g
Required:
The mass of PbCl₂ and AgCl in the precipitate
Solution;
Number of moles of chloride ions in a mole of PbCl₂ = 2 moles
Number of moles of chloride ions in a mole of AgCl = 1 mole
Let X represent the mass of PbCl₂ in the precipitate, we have;
The mass of AgCl in the precipitate = 65.08 g - X
[tex]\mathrm{Number \ of \ moles \ of \ PbCl_2} = \dfrac{X \, g}{278.1 \, g} =\mathbf{ \dfrac{X }{278.1}}[/tex]
Number of moles of chloride ions from PbCl₂ is therefore;
[tex]\mathrm{Number \ of \ moles \ of \ Cl^- from \ PbCl_2} =\mathbf{ 2 \times \dfrac{X }{278.1} \ moles \ of \ Cl^-}[/tex]
[tex]\mathrm{Number \ of \ moles \ of \ AgCl \ in \ the \ precipitate} = \dfrac{65.08 -X }{143.32}[/tex]
[tex]\mathrm{Number \ of \ moles \ of \ Cl^- from \ AgCl} = \mathbf{ \dfrac{65.08 -X }{143.32}} \ moles \ of \ Cl^-[/tex]
The number of moles of chloride ions from one mole of KCl = 1 mole
Number of moles of chloride ions from 0.242 L of 1.92 M KCl is therefore;
0.242 L × 1.92 moles/L = 0.46464 moles
Number of moles of chloride ions from KCl = 0.46464 moles
[tex]0.46464 \ moles \ from \ KCl = \overbrace{ \dfrac{ 2 \times X }{278.1} + \dfrac{65.08 -X }{143.32}} \ moles \ in \ PbCl_2 \ and \ AgCl[/tex]
Which gives;
[tex]\displaystyle \frac{192}{896089} \cdot X + \frac{1627}{3583} = \frac{1452}{3125}[/tex]
Therefore;
[tex]\displaystyle X = \frac{\frac{1452}{3125} - \frac{1627}{3583} }{ \frac{192}{896089} } = \frac{105864850549}{2149800000} \approx \mathbf{ 49.24}[/tex]
The mass of PbCl₂ in the precipitate, X ≈ 49.24 g
The mass of AgCl in the precipitate = 65.08 g - 49.24 g ≈ 15.84 g
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The nutrition label on the back of a package of hotdogs (purchased within the US) indicates that one hotdog contains 100 calories. How many calories does a hotdog actually have?
A. 1,000
B. It depends on how many hotdogs you eat
C. 100
D. 10
E. 100,000
Answer:
C. 100
Explanation:
Biochemical researches and studies have found out that an average health hotdog has a calorie of between 100 and 150 which is usually dependent on the additives.
Since the nutrition label on the back of a package of hotdogs (purchased within the US) indicates that one hotdog contains 100 calories then it truly contains such amount of calories. The standard number of calories present in a hotdog is independent of the amount eaten by individuals.
At 298 K, Kc is 2.2×105 for the reaction F(g)+O2(g)⇌O2F(g) . What is the value of Kp at this temperature? Express your answer using two significant figures.
Answer:
The value of Kp at this temperature is 9.0*10³
Explanation:
The equilibrium constant Kp describes the relationship that exists between the partial pressures of the reactants and products, while Kc represents the relationship that exists between the concentrations of the reactants and products that participate in the reaction.
The general relationship between the constants Kp and Kc results:
Kp=Kc*[tex](R*T)^{moles of product - moles of reagent}[/tex]
In this case:
Kc= 2.2*10⁵R = gas constant = 0.0821 [tex]\frac{atm*L}{mol*K}[/tex] T = Kelvin temperature = 298 K moles of gaseous products - moles of gaseous reactants = 1 - 2 = -1Replacing:
Kp=2.2*10⁵*[tex](0.0821*298)^{-1}[/tex]
Solving:
Kp≅9.0*10³
The value of Kp at this temperature is 9.0*10³
what is 1 +1 (a) 11 (b) 3 (c) 6 (d) 2
Answer:
(d) 2
Explanation:
Lets say that there are 2 apples or 1+1 if you count them you would do 1,2 so 2 would be the final answer
Select the correct answer. A certain reaction has this form: aA bB. At a particular temperature and [A]0 = 2.00 x 10-2 Molar, concentration versus time data were collected for this reaction and a plot of ln[A]t versus time resulted in a straight line with a slope value of -2.97 x 10-2 min-1. What is the half-life of this reaction? A. 23.33 seconds B. 0.043 minutes C. 0.0043 seconds D. 23.33 minutes E. 1680 minutes
Answer:
[tex]\large \boxed{\text{D. 23.34 min}}[/tex]
Explanation:
1. Find the order of reaction
Use information from the graph to find the order.
If a plot of ln[A] vs time is linear, the reaction is first order and the slope = -k.
2. Find the half-life
[tex]k = \dfrac{\ln2}{ t_{\frac{1}{2}}}\\\\k = \text{-slope} = -(-2.97 \times 10^{-2} \text{ min}^{-1}) =2.97 \times 10^{-2} \text{ min}^{-1} \\ t_{\frac{1}{2}} =\dfrac{\ln 2}{k} = \dfrac{\ln 2}{2.97 \times 10^{-2}\text{ min}^{-1}} =\textbf{23.34 min}\\\\\text{The half-life is $\large \boxed{\textbf{23.34 min}}$}[/tex]
The half life of the reaction is 23.33 minutes.
We know that for a first order reaction;
ln[A]t = ln[A]o - kt
A plot of ln[A]t against time (t) will yield a straight line graph with a slope of -k.
From the question, the slope is -2.97 x 10-2 min-1.
So, -2.97 x 10-2 min-1 = - k
k = 2.97 x 10-2 min-1
The half life of a first order reaction is obtained from;
t1/2 = 0.693/k
t1/2 = 0.693/2.97 x 10-2 min-1
t1/2 = 23.33 minutes
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A gas mixture containing N2 and O2 was kept inside a 2.00 L container at a temperature of 23.0°C and a total pressure of 1.00 ATM the partial pressure of oxygen was 0.722 ATM how many grams of nitrogen are present in the gas mixture
Answer:
0.641 g of Nitrogen are present in the mixture.
Explanation:
We use the Ideal Gases Law, to solve this question.
For the mixture:
P mixture . V mixture = mol mixture . R . T
We convert the T° to K → 23°C + 273 = 296 K
R = Ideal gases constant → 0.082 L.atm/mol.K
1 atm . 2L = mol mixture . 0.082 L.atm/mol.K . 296K
2 atm.L / ( 0.082 mol /L.atm) . 296 = 0.0824 moles
We know that sum of partial pressure = 1
Partial pressure N₂ + Partial pressure O₂ = 1
1 - 0.722 atm = Partial pressure N₂ → 0.278 atm
We apply the mole fraction concept:
Partial pressure N₂ / Total pressure = Moles N₂ / Total moles
Moles N₂ = (Partial pressure N₂ / Total pressure) . Total moles
Moles N₂ = (0.278 atm / 1 atm) . 0.0824 mol → 0.0229 moles
We convert the moles to mass → 0.0229 mol . 28 g/mol = 0.641 g
641 mg
A hot lump of 27.4 g of aluminum at an initial temperature of 69.5 °C is placed in 50.0 mL H2O initially at 25.0 °C and allowed to reach thermal equilibrium. What is the final temperature of the aluminum and water, given that the specific heat of aluminum is 0.903 J/(g·°C)? Assume no heat is lost to surroundings.
Answer:
[tex]\large \boxed{29.7 \,^{\circ}\text{C}}[/tex]
Explanation:
There are two heat transfers involved: the heat lost by the aluminium and the heat gained by the water.
According to the Law of Conservation of Energy, energy can neither be destroyed nor created, so the sum of these terms must be zero.
Let the Al be Component 1 and the H₂O be Component 2.
Data:
For the Al:
[tex]m_{1} =\text{27.4 g; }T_{i} = 69.5 ^{\circ}\text{C; }\\C_{1} = 0.903 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$}[/tex]
For the water:
[tex]m_{2} =\text{50.0 g; }T_{i} = 25.0 ^{\circ}\text{C; }\\C_{2} = 4.184 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$}[/tex]
Calculations
(a) The relative temperature changes
[tex]\begin{array}{rcl}\text{Heat lost by Al + heat gained by water} & = & 0\\m_{1}C_{1}\Delta T_{1} + m_{2}C_{2}\Delta T_{2} & = & 0\\\text{27.4 g}\times 0.903 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$} \times\Delta T_{1} + \text{50.0 g} \times 4.184 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$}\Delta \times T_{2} & = & 0\\24.74\Delta T_{1} + 209.2\Delta T_{2} & = & 0\\\end{array}[/tex]
(b) Final temperature
[tex]\Delta T_{1} = T_{\text{f}} - 69.5 ^{\circ}\text{C}\\\Delta T_{2} = T_{\text{f}} - 25.0 ^{\circ}\text{C}[/tex]
[tex]\begin{array}{rcl}24.74(T_{\text{f}} - 69.5 \, ^{\circ}\text{C}) + 209.2(T_{\text{f}} - 25.0 \, ^{\circ}\text{C}) & = & 0\\24.74T_{\text{f}} - 1719 \, ^{\circ}\text{C} + 209.2T_{\text{f}} -5230 \, ^{\circ}\text{C} & = & 0\\233.9T_{\text{f}} - 6949\, ^{\circ}\text{C} & = & 0\\233.9T_{\text{f}} & = & 6949 \, ^{\circ}\text{C}\\T_{\text{f}}& = & \mathbf{29.7 \, ^{\circ}}\textbf{C}\\\end{array}\\\text{The final temperature is $\large \boxed{\mathbf{29.7 \,^{\circ}}\textbf{C}}$}[/tex]
Check:
[tex]\begin{array}{rcl}27.4 \times 0.903 \times (29.7 - 69.5) + 50.0 \times 4.184 (29.7 - 25.0)& = & 0\\24.74(-39.8) +209.2(4.7) & = & 0\\-984.6 +983.2 & = & 0\\-985 +983 & = & 0\\0&=&0\end{array}[/tex]
The second term has only two significant figures because ΔT₂ has only two.
It agrees to two significant figures
suppose you are titrating vinegar, which is an acetic acid solution
Answer:
0.373 M
Explanation:
The balanced equation for the reaction is given below:
HC2H3O2 + NaOH —> NaC2H3O2 + H2O
From the balanced equation above, the following were obtained:
Mole ratio of the acid, HC2H3O2 (nA) = 1
Mole ratio of the base, NaOH (nB) = 1
Next, we shall write out the data obtained from the question. This include:
Volume of base, NaOH (Vb) = 32.17 mL
Molarity of base, NaOH (Mb) = 0.116 M
Volume of acid, HC2H3O2 (Va) = 10 mL
Molarity of acid, HC2H3O2 (Ma) =..?
The molarity of the acid solution can be obtained as follow:
MaVa/MbVb = nA/nB
Ma x 10 / 0.116 x 32.17 = 1
Cross multiply
Ma x 10 = 0.116 x 32.17
Divide both side by 10
Ma = (0.116 x 32.17) /10
Ma = 0.373 M
Therefore, the concentration of the acetic acid is 0.373 M.
The correct IUPAC name for the following compound is
Answer:
1-cyclopentylhexan-2-one
Explanation:
1-cyclopentylhexan-2-one
Given these data in a study on how the rate of a reaction was affected by the concentration of the reactants,
Experiment [A] [B] [C] Rate (mol L‑1 hr‑1 )
1 0.200 0.100 0.600 5.0
2 0.200 0.400 0.400 80.0
3 0.600 0.100 0.200 15.0
4 0.200 0.100 0.200 5.0
5 0.200 0.200 0.400 20.0
From this data, what is the numerical value of the rate constant, (k), for this reaction (value that would be found using the same units used in the data above)?
a. 2083
b. 694
c. 417
d. 2500
e. 83.3
Answer:
d. 2500
Explanation:
In a kinetic study with 3 different reactants, you change concentrations of the reactants to see how this concentration affects rate of reaction. General law is:
v = k [A]ᵃ [B]ᵇ [C]ⁿ
If you see 1 and 3 experiments, the concentration of C change from 0.600M to 0.200M but reaction rate doesn't change, thus n=0:
v = k [A]ᵃ [B]ᵇ [C]⁰
v = k [A]ᵃ [B]ᵇ×1
Now, reaction 2 and reaction 4 change B from 0.400M to 0.200M having the other reactants constant. When B is duplicated, rate increase 4 times. That means b = 2:
v = k [A]ᵃ [B]ᵇ
v = k [A]ᵃ [B]²
Finally, if you see 3 and 4 reactions, A change from 0.200M to 0.600M and the reaction rate change from 15.0 to 5.0, That means if the concentration of A is triplicated, reaction rate will be triplicated to. Thus a=1:
v = k [A]ᵃ [B]²
v = k [A] [B]²
Relpacing this equation in any experiment (Experiment 5, for example):
20.0 = k [0.200] [0.200]²
2500 = k
That means right answer is:
d. 2500cetylene gas is often used in welding torches because of the very high heat produced when it reacts with oxygen gas, producing carbon dioxide gas and water vapor. Calculate the moles of oxygen needed to produce of water.
Answer:
0.60 mol
Explanation:
There is some info missing. I think this is the original question.
Acetylene gas is often used in welding torches because of the very high heat produced when it reacts with oxygen gas, producing carbon dioxide gas and water vapor. Calculate the moles of oxygen needed to produce 1.5 mol of water.
Step 1: Given data
Moles of water required: 1.5 mol
Step 2: Write the balanced equation
C₂H₂(g) + 2.5 O₂(g) ⇒ 2 CO₂(g) + H₂O(g)
Step 3: Calculate the moles of oxygen needed to produce 1.5 mol of water
The molar ratio of O₂ to H₂O is 2.5:1. The moles of oxygen needed to produce 1.5 mol of water are (1/2.5) × 1.5 mol = 0.60 mol
what is the net ionic equation with its physical states? (NH4)2CO3(aq)+Ca(ClO4)2(aq)⟶CaCO3(s)+2NH4ClO4(aq)
Answer: The net ionic equation is [tex]CO_3^{2-}(aq)+Ca^{2+}(aq)\rightarrow CaCO_3(s)[/tex]
Explanation:
A double displacement reaction is one in which exchange of ions take place. The salts which are soluble in water are designated by symbol (aq) and those which are insoluble in water and remain in solid form are represented by (s) after their chemical formulas.
[tex](NH_4)_2CO_3(aq)+Ca(ClO_4)_2(aq)\rightarrow CaCO_3(s)+2NH_4ClO_4(aq)[/tex]
The equation can be written in terms of ions as:
[tex]2NH_4^+(aq)+CO_3^{2-}(aq)+Ca^{2+}(aq)+2ClO_4^{-}(aq)\rightarrow CaCO_3(s)+2NH_4^{+}(aq)+2ClO_4^-(aq)[/tex]
Spectator ions are defined as the ions which does not get involved in a chemical equation or they are ions which are found on both the sides of the chemical reaction present in ionic form.
The ions which are present on both the sides of the equation are ammonium and chlorate ions and hence are not involved in net ionic equation.
Hence, the net ionic equation is :
[tex]CO_3^{2-}(aq)+Ca^{2+}(aq)\rightarrow CaCO_3(s)[/tex]
What is an example of a molecular compound
Answer:
Molecular compounds are inorganic compounds that take the form of discrete (covalent) molecules. Examples include such familiar substances as water (H2O) and carbon dioxide (CO2).
If the SN2 reaction of an aromatic alcohol with an alkyl halide, like the synthesis of nerolin, is successful, what changes would be seen in the IR spectrum for the product compared to the starting material
Answer:
O-H stretch signal at 3300 cm-1
Explanation:
In this question, we can start with the reaction mechanism for the synthesis of Nerolin. We have to start with naphthalen-2-ol adding NaOH we can produce the alkoxide. Then this alkoxide can react by an Sn2 reaction with bromomethane to produce Nerolin (see figure 1).
In the starting molecule (naphthalen-2-ol) we have an "OH" group. Therefore we will have an O-H stretch signal around 3300 cm^-1. The alcohol signals are very broad and very intense, so this will be the main signal for the initial molecule. In the final product, we dont have the "OH" therefore this signal will disappear (see figure 2).
I hope it helps!