Answer:
Molar concentration of [tex]CuSO_4[/tex] solution = 0.0056 moles / liter
Explanation:
Looking at the chemical reaction we realize that the precipitate, formed by adding the iron powder, is cooper.
Then for finding the number of moles of precipitated copper:
[tex] number.of.moles= \frac{mass (g)}{molecular.mass (g/mole)}[/tex]
[tex]number.of.moles.of.solid.copper = \frac{0.089}{63.5} =1.4*10^{3}[/tex]
From the chemical reaction we deduce that [tex]1.4*10^{3}[/tex] moles of Cu equals to [tex]1.4*10^{3}[/tex] moles of [tex]CuSO_4[/tex] in the initial solution.
So molar concentration is defined as:
[tex]molar.concentration = \frac{number.of.moles}{solution.volume (liters)}[/tex]
[tex]molar.concentration.of.CuSO_4.solution= \frac{1.4*10^{3}}{0.250}= 5.6*10^{3} moles/liter = 0.0056 moles/liter[/tex]
Answer:
[tex]M=5.6x10^{-3}M[/tex]
Explanation:
Hello,
In this case, we first must consider the given already-balanced chemical reaction to realize that 89 mg of copper were recovered, moreover we can relate such mass with the employed moles of copper (II) sulfate via this reaction's stoichiometry as follows:
[tex]n_{CuSO_4}=89mg Cu*\frac{1gCu}{1000mgCu} *\frac{1molCu}{63.546gCu} *\frac{1molCuSO_4}{1molCu} \\n_{CuSO_4}=1.4x10^{-3}molCuSO_4[/tex]
Now, if we state the molarity (mol/L) as the required concentration, we apply its mathematical definition as shown below:
[tex]M=\frac{n_{CuSO_4}}{V_{sln}} =\frac{1.4x10^{-3}molCuSO_4}{0.250L} \\M=5.6x10^{-3}M[/tex]
Best regards.
The reaction 2NO(g)+Cl2(g)→2NOCl(g) is carried out in a closed vessel. If the partial pressure of NO is decreasing at the rate of 21 torr/min , what is the rate of change of the total pressure of the vessel
Answer : The rate of change of the total pressure of the vessel is, 10.5 torr/min.
Explanation : Given,
[tex]\frac{d[NO]}{dt}[/tex] =21 torr/min
The balanced chemical reaction is,
[tex]2NO(g)+Cl_2(g)\rightarrow 2NOCl(g)[/tex]
The rate of disappearance of [tex]NO[/tex] = [tex]-\frac{1}{2}\frac{d[NO]}{dt}[/tex]
The rate of disappearance of [tex]Cl_2[/tex] = [tex]-\frac{d[Cl_2]}{dt}[/tex]
The rate of formation of [tex]NOCl[/tex] = [tex]\frac{1}{2}\frac{d[NOCl]}{dt}[/tex]
As we know that,
[tex]\frac{d[NO]}{dt}[/tex] =21 torr/min
So,
[tex]-\frac{d[Cl_2]}{dt}=-\frac{1}{2}\frac{d[NO]}{dt}[/tex]
[tex]\frac{d[Cl_2]}{dt}=\frac{1}{2}\times 21torr/min=10.5torr/min[/tex]
And,
[tex]\frac{1}{2}\frac{d[NOCl]}{dt}=\frac{1}{2}\frac{d[NO]}[/tex]
[tex]\frac{d[NOCl]}{dt}=\frac{d[NO]}=21torr/min[/tex]
Now we have to calculate the rate change.
Rate change = Reactant rate - Product rate
Rate change = (21 + 10.5) - 21 = 10.5 torr/min
Therefore, the rate of change of the total pressure of the vessel is, 10.5 torr/min.
The rate of change of the total pressure of the vessel is 10.5 torr/min
The given reaction is expressed as:
[tex]\mathbf {2O_{(g)} + Cl_{2(g)} \to 2NOCl_{(g)}}}[/tex]
From chemical kinetics, the average rate (r) can be expressed as:
[tex]\mathbf{r = -\dfrac{1}{2}\dfrac{d[NO]}{dt}= -\dfrac{d[Cl_2]}{dt}=\dfrac{1}{2}\dfrac{d[NOCl]}{dt} }[/tex]
where;
the negative sign (-) indicates the rate of disappearance of the substances.∴
rate of disappearance of NO [tex]\mathbf{= -\dfrac{1}{2} \dfrac{d[NO]}{dt}}[/tex] rate of disappearance of Cl₂ = [tex]\mathbf{\dfrac{-d[Cl_2]}{dt}}[/tex] rate of appearance of NOCl = [tex]\mathbf{\dfrac{1}{2} \dfrac{d[NOCl]}{dt}}[/tex]We are being told that the partial pressure of NO is decreasing at 21 torr/min
i.e.
[tex]\mathbf{\dfrac{d[NO]}{dt}}[/tex] = 21 torr/minand we know that:
[tex]\mathbf{\dfrac{-d[Cl_2]}{dt}= -\dfrac{1}{2} \dfrac{d[NO]}{dt}}}[/tex]
∴
[tex]\mathbf{\dfrac{-d[Cl_2]}{dt}= -\dfrac{1}{2}(21 \ torr/min) }}[/tex] [tex]\mathbf{\dfrac{d[Cl_2]}{dt}= 10.5 \ torr/min }}[/tex]Similarly;
[tex]\mathbf{-\dfrac{1}{2} \dfrac{d[NOCl]}{dt} = \mathbf{-\dfrac{1}{2} \dfrac{d[NO]}{dt}}}[/tex] [tex]\mathbf{\dfrac{d[NOCl]}{dt} = \mathbf{ \dfrac{d[NO]}{dt}}}[/tex] [tex]\mathbf{\dfrac{d[NOCl]}{dt} =21 \ torr/min}}[/tex]Now, we need to determine the rate of change of the total pressure at which these substances are decreasing;
Rate change = rate of reactant - rate of product.
[tex]\mathbf{Rate \ change =} \mathbf{\mathbf{ \dfrac{d[NO]}{dt}} +\dfrac{d[Cl_2]}{dt} - \dfrac{d[NOCl]}{dt} }[/tex]
[tex]\mathbf{Rate \ change =} \mathbf{(21 \ torr/min) +(10.5 \ torr/min) -( 21 \ torr/min})[/tex]
Rate change = 10.5 torr/min
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Write the symbol for each of the following ions: (a) the ion with a 3+ charge, 28 electrons, and a mass number of 71 (b) the ion with 36 electrons, 35 protons, and 45 neutrons (c) the ion with 86 electrons, 142 neutrons, and a 4+ charge (d) the ion with a 2+ charge, atomic number 38, and mass number 87
Answer:
[tex]\boxed{\math{_{31}^{71}\text{Ga}^{3+}}}\boxed{\math{_{35}^{80}\text{Br}^{-}}} \boxed{\math{_{90}^{232}\text{Th}^{4+}}} \boxed{\math{_{38}^{87}\text{Sr}^{2+}}}[/tex]
Explanation:
(a)
If the ion has 28 electrons and a 3+ charge, it normally has 31 electrons. It must also have 31 protons. Atom 31 is gallium.
[tex]\text{The symbol for the ion is }\boxed{\mathbf{_{31}^{71}\textbf{Ga}^{3+}}}[/tex]
(b)
If the ion has 35 protons, it normally has 35 electrons. It has 36 electrons, so it has a negative charge. Atom 35 is bromine.
A = p + n = 35 + 45 = 80, so the isotopic mass number is 80
\text{The symbol for the ion is }\boxed{\mathbf{_{35}^{80}\textbf{Br}^{-}}}
(c)
If the ion has 86 electrons and a charge of 4+, it normally has 90 electrons. It must also have 90 protons. Atom 90 is thorium.
A = p + n = 90 + 142 = 232, so the isotopic mass number is 232.
[tex]\text{The symbol for the ion is }\boxed{\mathbf{_{90}^{232}\textbf{Th}^{4+}}}[/tex]
(d)
If the ion has 38 protons, it is strontium.
[tex]\text{The symbol for the ion is }\boxed{\mathbf{_{38}^{87}\textbf{Sr}^{2+}}}[/tex]
The symbol of ion will be:[tex]^{71}_{31}Ga^{+3}[/tex]
The symbol of ion will be:[tex]^{80}_{35}Br^{-1}[/tex]
The symbol of ion will be:[tex]^{232}_{90}Th^{+4}[/tex]
The symbol of ion will be: [tex]^{87}_{38}Sr^{+2}[/tex]
Explanation:
The symbol of an ion of an element is written as:
[tex]^A_ZX^C[/tex]
Where:
A = Mass number of the atom
Z = Atomic number of the atom
X = Symbol the atom
C = Charge on the atom
Given:
a) The ion with a 3+ charge, 28 electrons, and a mass number of 71.
b) The ion with 36 electrons, 35 protons, and 45 neutrons
c) The ion with 86 electrons, 142 neutrons, and a 4+ charge
d) The ion with a 2+ charge, atomic number 38, and mass number 87
To find:
The symbols of given ions
Solution:
a)
Charge on the ion = C = 3+
Number of electrons in the ion = 28
Number of electrons in ion's parent atom = 28 +3 = 31
Number of electrons in parent atom = Number of proton in ion/ paarent atom
= 31 = 31
Atomic number = Number of protons
Atomic number of the ion = Z = 31
Mass number of ion = A = 71
The element with atomic number 31 is gallium with the chemical symbol Ga, so the symbol of ion will be:
[tex]^{71}_{31}Ga^{+3}[/tex]
b)
Charge on the ion = C = ?
Number of electrons in the ion = 36
Number of protons in the ion = 35
Number of neutrons = 45
Number of electrons in parent atom = Number of protons in ion/ parent atom
= 35 = 35
Atomic number = Number of protons
Atomic number of the ion = Z = 35
Number of electrons in ion's parent atom = Number of protons in the ion = 35
Charge on the ion = C = 36 - 35 = 1-
Mass number of ion = Number of protons + Mass of neutrons
A = 35 + 45 = 80
The element with atomic number 35 is bromine with the chemical symbol Br, so the symbol of ion will be:
[tex]^{80}_{35}Br^{-1}[/tex]
c)
Charge on the ion = C = 4+
Number of electrons in the ion = 86
Number of electrons in ion's parent atom = 86+4 = 90
Number of electrons in parent atom = Number of protons in ion/ parent atom
= 90= 90
Atomic number = Number of protons
Atomic number of the ion = Z = 90
Number of neutrons = 142
Mass number of ion = Number of protons + Mass of neutrons
A = 90+ 142 = 232
The element with atomic number 90 is thorium with the chemical symbol Th, so the symbol of ion will be:
[tex]^{232}_{90}Th^{+4}[/tex]
d)
Charge on the ion = C = 2+
Atomic number of the ion = Z = 38
Mass number of ion = A = 87
The element with atomic number 38 is strontium with chemical symbol Sr, so the symbol of ion will be:
[tex]^{87}_{38}Sr^{+2}[/tex]
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At 20°C, an aqueous solution that is 24.0% by mass in ammonium chloride has a density of 1.0674 g/mL. What is the molarity of ammonium chloride in the solution? The formula weight of NH4Cl is 53.50 g/mol.
Answer:
Molarity = 4.79 M
Explanation:
Mass percentage -
Mass percentage of A is given as , the mass of the substance A by mass of the total solution multiplied by 100.
i.e.
mass % A = mass of A / mass of solution * 100
Given,
24% by mass of ammonium chloride,
so,
100 g solution contains , 24 g of ammonium chloride,
mass of solution = 100g
and mass of the solute , i.e. , ammonium chloride = 24 g .
Hence,
Moles -
Moles are calculated as the given mass divided by the molecular mass.
i.e. ,
moles = ( mass / molecular mass )
Given,
The molecular mass of ammonium chloride is 53.50 g /mol
moles of ammonium chloride = 24 g / 53.50 g /mol
moles of ammonium chloride = 0.449 mol
Density -
Density of a substance is given as the mass divided by the volume ,
Density = mass / volume ,
Volume = mass / density
Given ,
Density of ammonium chloride = 1.0764 g /mL
Calculated above , mass of solution = 100 g
volume of solution = 100 g / 1.0764 g/ mL
volume of solution = 93.69 mL
Since , 1 ml = 1/1000 L
volume of solution = 93.69 /1000 L
volume of solution = 0.09369 L
Molarity -
Molarity of a solution is given by the moles of solute per liter of the solution
Hence,
M = moles of solute / volume of solution (in L)
As calculated above,
moles of ammonium chloride = 0.449 mol
volume of solution = 0.09368 L
Putting in the above formula
Molarity = 0.449 mol / 0/09368 L
Molarity = 4.79 M
The molarity of ammonium chloride in the solution is 4.79 M
What is the molarity of a solution?The number of moles of solute dissolved in one liter of solution is the molarity (M) of the solution.
Calculation of molarity:
Given,
The mass of the solution is 24.0%
Density is 1.0674 g/ml
The weight of NH4Cl is 53.50 g/mol
Step 1: Convert Mass % into gram
Considering the mass of solution = 100g
24% of 100g
Mass is 24 gram
Step 2: Calculate the mole compound
Moles = mass divided from molecular mass
The molecular mass of ammonium chloride is 53.50 g /mol
Thus,
[tex]\bold{Moles = \dfrac{24}{53.50} = 0.449 mol}[/tex]
Step 3: Calculating the volume
[tex]\bold{Volume = \dfrac{mass}{density} }[/tex]
[tex]\bold{Volume = \dfrac{ 100}{1.0674 g/m} = 93.69 ml}[/tex]
[tex]\bold{Volume\; of \;solution= \dfrac{ 93.69}{1000 L} = 0.09369 L}[/tex]
Step 4: Now, Molarity of the compound is
[tex]\bold{Molarity = \dfrac{moles\; of\; solute}{volume\; of \;solution (in L)}}[/tex]
Mole of ammonium chloride = 0.449 mol
Volume of solution = 0.09368 L
By formula,
[tex]\bold{Molarity = \dfrac{0.449 mol}{0.09368 L} = 4.79 m}[/tex]
Thus, the Molarity of ammonium chloride is = 4.79 m.
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True or False Aluminum does not perform well at elevated temperatures
Answer:
False
Explanation:
Calculate the number of vacancies per cubic meter for some metal, M, at 773°C. The energy for vacancy formation is 0.97 eV/atom, while the density and atomic weight for this metal are 7.81 g/cm3 (at 773°C) and 67.81 g/mol, respectively.
Answer:
0.112eV/atom
Explanation:
since
p=m/v
then
pv=pv
7.81*0.97=67.81*V
V=7.58/67.81
V= 0.112eV/atom
A mixture of two compounds, A and B, was separated by extraction. After the compounds were dried, their masses were found to be: 119 mg of compound A and 97 mg of compound B. Both compounds were recrystallized and weighed again. After recrystallization, the mass of compound A was 83 mg and the mass of compound B was 79 mg. Calculate the percent recovery from recrystallization for both compounds.
Answer:
The percent recovery from re crystallization for both compounds A and B is 69.745 and 81.44 % respectively.
Explanation:
Mass of compound A in a mixture = 119 mg
Mass of compound A after re-crystallization = 83 mg
Percent recovery from re-crystallization :
[tex]\frac{\text{Mass after re-crystallization}}{\text{Mass before re-crystallization}}\times 100[/tex]
Percent recovery of compound A:
[tex]\frac{83 mg}{119 mg}\times 100=69.74\%[/tex]
Mass of compound B in a mixture = 97 mg
Mass of compound B after re-crystallization = 79 mg
Percent recovery of compound B:
[tex]\frac{79 mg}{97 mg}\times 100=81.44\%[/tex]
The density of a 3.37M MgCl2 (FW = 95.21) is 1.25 g/mL. Calulate the molality, mass/mass percent, and mass/volume percent. So far this is what I have, but I cannot get the correct values for the mass of the solution to even begin to figure out the volume.
Answer : The molality, mass/mass percent, and mass/volume percent are, 0.0381 mole/Kg, 25.67 % and 32.086 % respectively.
Solution : Given,
Density of solution = 1.25 g/ml
Molar mass of [tex]MgCl_2[/tex] (solute) = 95.21 g/mole
3.37 M magnesium chloride means that 3.37 gram of magnesium chloride is present in 1 liter of solution.
The volume of solution = 1 L = 1000 ml
Mass of [tex]MgCl_2[/tex] (solute) = 3.37 g
First we have to calculate the mass of solute.
[tex]\text{Mass of }MgCl_2=\text{Moles of }MgCl_2\times \text{Molar mass of }MgCl_2[/tex]
[tex]\text{Mass of }MgCl_2=3.37mole\times 95.21g/mole=320.86g[/tex]
Now we have to calculate the mass of solution.
[tex]\text{Mass of solution}=\text{Density of solution}\times \text{Volume of solution}=1.25g/ml\times 1000ml=1250g[/tex]
Mass of solvent = Mass of solution - Mass of solute = 1250 - 320.86 = 929.14 g
Now we have to calculate the molality of the solution.
[tex]Molality=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Mass of solvent}}=\frac{3.37g\times 1000}{95.21g/mole\times 929.14g}=0.0381mole/Kg[/tex]
The molality of the solution is, 0.0381 mole/Kg.
Now we have to calculate the mass/mass percent.
[tex]\text{Mass by mass percent}=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 100=\frac{320.86}{1250}\times 100=25.67\%[/tex]
The mass/mass percent is, 25.67 %
Now we have to calculate the mass/volume percent.
[tex]\text{Mass by volume percent}=\frac{\text{Mass of solute}}{\text{Volume of solution}}\times 100=\frac{320.86}{1000}\times 100=32.086\%[/tex]
The mass/volume percent is, 32.086 %
Therefore, the molality, mass/mass percent, and mass/volume percent are, 0.0381 mole/Kg, 25.67 % and 32.086 % respectively.
If 42.7 mL of a 0.208 M HCl solution is needed to neutralize a solution of Ca(OH)2, how many grams of Ca(OH)2 must be in the solution? Ca(OH)2(aq) + 2HCl CaCl2(aq) + 2H2O(l) g
Answer: The mass of calcium hydroxide that are present in the solution is 0.3256 grams.
Explanation:
To calculate the moles of a solute, we use the equation:
[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}[/tex]
We are given:
Volume of hydrochloric acid = 42.7mL = 0.0427 L (Conversion factor: 1 L = 1000 mL)
Molarity of the solution = 0.208 moles/ L
Putting values in above equation, we get:
[tex]0.208mol/L=\frac{\text{Moles of hydrochloric acid}}{0.0427L}\\\\\text{Moles of hydrochloric acid}=0.0088mol[/tex]
For the given chemical reaction:
[tex]Ca(OH)_2(aq.)+2HCl\rightarrow CaCl_2(aq.)+2H_2O(l)[/tex]
By Stoichiometry of the reaction:
2 moles of hydrochloric acid reacts with 1 mole of calcium hydroxide.
So, 0.0088 moles of hydrochloric acid will react with = [tex]\frac{1}{2}\times 0.0088=0.0044mol[/tex] of calcium hydroxide.
To calculate the mass of calcium hydroxide, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
Moles of calcium hydroxide = 0.0044 moles
Molar mass of calcium hydroxide = 74 g/mol
Putting values in above equation, we get:
[tex]0.0044mol=\frac{\text{Mass of calcium hydroxide}}{74g/mol}\\\\\text{Mass of calcium hydroxide}=0.3256g[/tex]
Hence, the mass of calcium hydroxide that are present in the solution is 0.3256 grams.
0.325 grams of Ca(OH)₂ must be in the solution of HCl to neutralize a solution.
How we calculate moles from molarity?Molarity of any solution is defined as:
M = n/V, where
n = no. of moles
V = volume
Given chemical reaction is:
Ca(OH)₂(aq) + 2HCl → CaCl₂(aq) + 2H₂O(l)
From the stoichiometry of the reaction, it is clear that:
2 moles of HCl = react with 1 mole of Ca(OH)₂
1 mole of HCl = react with 1/2 mole of Ca(OH)₂
Moles of HCl will be calculated by using the molarity formula and given data as:
Given concentration of HCl = 0.208M
Given volume of HCl = 42.7mL = 0.0427L
Moles of HCl = 0.208 × 0.0427 = 0.0088 moles
0.0088 moles of HCl = react with 0.0088 × 1/2 = 0.0044moles of Ca(OH)₂
Mass of Ca(OH)₂ will be calculated by using the formula:
n = W/M, where
W = required mass
M = molar mass = 72g/mole
W = 0.0044mole × 72g/mole = 0.325 grams
Hence, 0.325 grams of Ca(OH)₂ will be in the solution.
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The combination of coke and steam produces a mixture called coal gas, which can be used as a fuel or as a starting material for other reactions. If we assume coke can be represented by graphite, the equation for the production of coal gas is
2C(s)+2H2O(g)--->CH4(g)+CO2(g)
Determine the standard enthalpy change for this reactionf rom the following standard enthalpies of reactions:
C(s)+H2O(g)--->CO(g)+H2(g) delta H=131.3 kJ
CO(g)+H2O(g)--->CO2(g)+H2(g) delta H=-41.2 kJ
CH4(g)+H2O(g)--->3H2(g)+CO(g) delta H=206.1 kJ
Answer: The [tex]\Delta H^o_{rxn}[/tex] for the reaction is 15.3 kJ.
Explanation:
Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.
According to this law, the chemical equation is treated as ordinary algebraic expressions and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the intermediate reactions.
The chemical equation for the reaction of carbon and water follows:
[tex]2C(s)+2H_2O(g)\rightarrow CH_4(g)+CO_2(g)[/tex] [tex]\Delta H^o_{rxn}=?[/tex]
The intermediate balanced chemical reaction are:
(1) [tex]C(s)+H_2O(g)\rightarrow CO(g)+H_2(g)[/tex] [tex]\Delta H_1=131.3kJ[/tex] ( × 2)
(2) [tex]CO(g)+H_2O(g)\rightarrow CO_2(g)+H_2(g)[/tex] [tex]\Delta H_2=-41.2kJ[/tex]
(3) [tex]CH_4(g)+H_2O(g)\rightarrow 3H_2(g)+CO(g)[/tex] [tex]\Delta H_3=206.1kJ[/tex]
The expression for enthalpy of the reaction follows:
[tex]\Delta H^o_{rxn}=[2\times \Delta H_1]+[1\times \Delta H_2]+[1\times (-\Delta H_3)][/tex]
Putting values in above equation, we get:
[tex]\Delta H^o_{rxn}=[(2\times (131.3))+(1\times (-41.2))+(1\times (-206.1))]=15.3kJ[/tex]
Hence, the [tex]\Delta H^o_{rxn}[/tex] for the reaction is 15.3 kJ.
A certain heat engine operates between 800 K and 300 K. (a) What is the maximum efficiency of the engine? (b) Calculate the maximum work that can be done by for each 1.0 k) of hea a reversible process for each 1.0 kJ supplied by the hot source? t supplied by the hot source. (c) How much heat is discharged into the cold sink in
Answer :
(a) The maximum efficiency of the engine is, 62.5 %
(b) The maximum work done is, 0.625 KJ.
(c) The heat discharge into the cold sink is, 0.375 KJ.
Explanation : Given,
Temperature of hot body [tex]T_h[/tex] = 800 K
Temperature of cold body [tex]T_c[/tex] = 300 K
(a) First we have to calculate the maximum efficiency of the engine.
Formula used for efficiency of the engine.
[tex]\eta =1-\frac{T_c}{T_h}[/tex]
Now put all the given values in this formula, we get :
[tex]\eta =1-\frac{300K}{800K}[/tex]
[tex]\eta =0.625\times 100=62.5\%[/tex]
(b) Now we have to calculate the maximum work done.
Formula used :
[tex]\eta =\frac{Q_h-Q_c}{Q_h}=\frac{w}{Q_h}[/tex]
where,
[tex]Q_h[/tex] = heat supplied by hot source = 1 KJ
[tex]Q_c[/tex] = heat supplied by hot source
w = work done = ?
Now put all the given values in this formula, we get :
[tex]\eta =\frac{w}{Q_h}[/tex]
[tex]0.625=\frac{w}{1KJ}[/tex]
[tex]w=0.625KJ[/tex]
(c) Now we have to calculate the heat discharge into the cold sink.
Formula used :
[tex]w=Q_h-Q_c[/tex]
[tex]Q_c=Q_h-w[/tex]
[tex]Q_c=1-0.625[/tex]
[tex]Q_c=0.375KJ[/tex]
Therefore, (a) The maximum efficiency of the engine is, 62.5 %
(b) The maximum work done is, 0.625 KJ.
(c) The heat discharge into the cold sink is, 0.375 KJ.
Determine the molar solubility ( ???? ) of Zn(CN)2 in a solution with a pH=1.33 . Ignore activities. The ????sp for Zn(CN)2 is 3.0×10−16 . The ????a for HCN is 6.2×10−10 .
The molar solubility of Zn(CN)2 in a solution with a given pH and Ksp can be calculated using principles of acid-base equilibria and solubility. However, without exact initial concentrations, a complete calculation can't be provided. Similar applications of these principles are seen in the examples of Cadmium Sulfide (CdS) or mercury chloride mentioned previously.
Explanation:The question involves the determination of the molar solubility of Zinc Cyanide (Zn(CN)2) in a solution with a given pH. The main principle involved is the understanding of acid-base equilibria and solubility. The pH value provided implies an [H3O+] = 10^-1.33. The Ka for Hydrogen Cyanide (HCN) is given which can be used to calculate [CN-]. After these concentrations are calculated, they can be utilized to find the molar solubility of Zn(CN)2 using the given Ksp.
Based on the information provided, some calculations similar to those mentioned but applied to Zn(CN)2 will have to be performed. However, we do not have certain required values like the exact initial concentration of the solution. Therefore, a complete solution can't be provided
However, similar stoichiometry-based calculations are applied to other salts' equilibria for determining molar solubility like Cadmium Sulfide (CdS) and Dissolution stoichiometry of mercury chloride in the provided data. These applications demonstrate how such problems are typically approached and solved.
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To determine the concentration of X in an unknown solution, 1.00 mL of 8.48 mM S was added to 3.00 mL of the unknown X solution and the mixture was diluted to 10.0 mL. After chromatographic separation, this solution gave peak areas of 5473 and 4851 for X and S, respectively. Determine the concentration of S in the 10.0 mL solution.
The concentration of S in the 10.0 mL solution is calculated by considering the initial 8.48 mM concentration and the dilution factor due to the increase in volume to 10.0 mL. The final concentration of S is found to be 0.848 mM.
Explanation:To determine the concentration of S in the 10.0 mL solution, you must take into account the initial concentration of S and how it changes with dilution. Initially, 1.00 mL of 8.48 mM (millimolar) S was added to the solution. The dilution can be calculated using the formula C1V1 = C2V2, where C1 is the initial concentration, V1 is the initial volume, C2 is the final concentration, and V2 is the final volume after dilution.
The initial volume of S (V1) is 1.00 mL and the final volume of the diluted solution (V2) is 10.0 mL. Therefore, using the initial concentration of S (C1) as 8.48 mM, we can solve for the final concentration (C2) as follows:
C1V1 = C2V2
(8.48 mM)(1.00 mL) = (C2)(10.0 mL)
Solving for C2 gives us:
C2 = (8.48 mM)(1.00 mL) / (10.0 mL)
C2 = 0.848 mM
The final concentration of S in the 10.0 mL solution is 0.848 mM (millimolar).
If two protons and two neutrons are removed from the nucleus of an oxygen-16 atom, a nucleus of which element remains? Express your answer as an isotope (e.g., as 31H).
Answer:
[tex]_{6}^{12}\text{C}[/tex]
Explanation:
A particle with two protons and two neutrons is a helium nucleus.
Your unbalanced nuclear equation is:
[tex]_{8}^{16}\text{O} \longrightarrow \, _{x}^{y}\text{Z} + \, _{2}^{4}\text{He}[/tex]
The main point to remember in balancing nuclear equations is that the sums of the superscripts and of the subscripts must be the same on each side of the equation.
Then
8 = x + 2, so x = 8 - 2 = 6
16 = y + 4, so y = 16 - 4 =12
Element 6 is carbon, so the nuclear equation becomes
[tex]_{\ 8}^{16}\text{O} \longrightarrow \, _{\ 6}^{12}\text{C} + \, _{2}^{4}\text{He}[/tex]
Which one of the following is not true concerning Diels-Alder reactions?
a.The reaction is stereospecific.
b.The reaction mechanism has only one step.
c.The reaction mechanism involves a resonance stabilized carbocation.
d.The diene must be a conjugated diene.
Answer:Option c is incorrect
Explanation:
Diels-Alder reaction is a 4+2 cycloaddition reaction .
The reaction occurs between a diene and a dieneophile .
Generally In the Diels-alder reaction HOMO of the diene and LUMO of the dienophile react with proper orbital symmetry. But vice-versa can also be used.
HOMO-Highest occupied molecular orbital
LUMO-Lowest unoccupied molecular orbital
The primary driving force for the diels alder reaction is is the conversion of 2pi bonds into 2 sigma bonds. The sigma bonds are energetically more favorable than pi bonds.
The diels alder reaction happens through orbital interaction and hence the substituents on either the diene or dienophile do not change there stereochemistry in the product so it is a stereospecific reaction.Since predominantly only isomer would be produced. so a is correct.
The Diels alder reaction is a concerted( more bonds form at a time) reaction which means it is just a one step reaction. so statement b is correct.
The option c is incorrect as diels alder reaction occurs through orbital interaction in a pericyclic manner. Since Diels alder reactions are pericyclic in nature and occur through orbital symmetry they do not involve polar intermediates like carbocation or radicals.
The dienes must be conjugated as on account of conjugation stability of a diene also increases. Also since we know that due to conjugation the energy of LUMO decreases and that of HOMO increases and so HOMO is more reactive and generally we involve HOMO of the diene and LUMO of the dienophile. So conjugated dienes are important.
So for a bond formation to take place in a Diels-Alder reaction HOMO and LUMO with proper symmetry must overlap.
Match the correct definition with the correct term from questions 10-13: A. Internal energy B. Latent heat C. Chemical (bond) energy D. Nuclear energy 10.The internal energy associated with the atomic bonds in a molecule. 11. May be viewed as the sum of the kinetic and potential energies of the molecules 12. The internal energy associated with the bonds within the nucleus of the atom itself 13. The internal energy associated with the phase of a system.
Answer: A. Internal energy : May be viewed as the sum of the kinetic and potential energies of the molecules
B. Latent heat: The internal energy associated with the phase of a system.
C. Chemical (bond) energy : The internal energy associated with the atomic bonds in a molecule
D. Nuclear energy : The internal energy associated with the bonds within the nucleus of the atom itself
Explanation:
Internal energy is defined as the total energy of a closed system. Internal energy is the sum of potential energy of the system and the kinetic energy of the system. It is represented by symbol U.
Latent heat is the thermal energy released or absorbed by a thermodynamic system when the temperature of the system does not change. It is thus also called as hidden heat.
Chemical energy is the energy stored in the bonds of molecules.
Nuclear energy is the energy which is stored in the nucleus of an atom called as binding energy within protons and neutrons.
Be sure to answer all parts. Propane (C3H8) is a minor component of natural gas and is used in domestic cooking and heating. (a) Balance the following equation representing the combustion of propane in air. Include states of matter in your answer. C3H8(g) + O2(g) → CO2(g) + H2O(g) (b) How many grams of carbon dioxide can be produced by burning 8.11 moles of propane? Assume that oxygen is the excess reactant in this reaction. × 10 g Enter your answer in scientific notation.
Answer:
For a: The balanced chemical equation is given below.
For b: The mass of carbon dioxide produced will be [tex]1.07\times 10^3g[/tex]
Explanation:
For a:Every balanced chemical equation follows law of conservation of mass.
This law states that mass can neither be created nor can be destroyed but it can only be transformed from one form to another form.
This law also states that the total number of individual atoms on the reactant side must be equal to the total number of individual atoms on the product side.
For the given reaction, the balance chemical equation follows:
[tex]C_3H_8(g)+5O_2(g)\rightarrow 3CO_2(g)+4H_2O(g)[/tex]
All the substances are present in gaseous state.
For b:By Stoichiometry of the reaction:
1 mole of propane gas produces 3 moles of carbon dioxide gas.
So, 8.11 moles of propane gas will produce = [tex]\frac{3}{1}\times 8.11=24.33mol[/tex] of carbon dioxide gas.
Now, calculating the mass of carbon dioxide using equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
Molar mass of carbon dioxide = 44 g/mol
Moles of carbon dioxide = 24.33 mol
Putting values in above equation, we get:
[tex]24.33mol=\frac{\text{Mass of carbon dioxide}}{44g/mol}\\\\\text{Mass of carbon dioxide}=1070.52g[/tex]
Hence, the amount of [tex]CO_2[/tex] produced in the given reaction and expressed in scientific notation is [tex]1.07\times 10^3g[/tex]
The melting point of phenol is 40.5∘C and that of toluene is −95∘C. What is the best explanation for this difference? Select the correct answer below: a. The (−OH) group on phenol can form hydrogen bonds, and the −CH3 group on toluene cannot. b. Phenol has only one hydrogen on the −OH group available to form hydrogen bonds, so the hydrogen bond is stronger. c. In toluene, the hydrogen bond is spread over all three hydrogens on the methyl group, so the interaction is weaker overall. d. Phenol has a higher molecular mass than toluene. e. None of the above.
Answer:
None of the above
Explanation:
The (−OH) group on phenol can form hydrogen bonds, and the −CH3 group on toluene cannot.
Phenol has only one hydrogen on the −OH group available to form hydrogen bonds, so the hydrogen bond is stronger. In toluene, the hydrogen bond is spread over all three hydrogens on the methyl group, so the interaction is weaker overall.
Phenol has a higher molecular mass than toluene.
Answer:
a. The (−OH) group on phenol can form hydrogen bonds, and the −CH3 group on toluene cannot.
Explanation:
Hello,
We firs must consider that the hydroxyl functional group is present in phenol as a highly polar section into its structure. Thus, phenol molecules are strongly associated by the presence of hydrogen bonds which toluene does not have due to its apolarity.
Consequently, since associating interactions are present in the phenol but absent in the toluene, more energy must be supplied to the phenol to melt it down, that is why phenol's melting point is higher than toluene's that one.
Best energy
For the aqueous solution containing 75 mg of compound C in 1.0 mL of water, what will be the total amount of the solute C that will be extracted after seven (7) extractions using a certain solvent D if 0.90 mL of a certain organic solvent D was used for each extraction. Given Ksolvent D/water = 1.8.
Answer:
75 mg
Explanation:
We can write the extraction formula as
x = m/[1 + (1/K)(Vaq/Vo)], where
x = mass extracted
m = total mass of solute
K = distribution coefficient
Vo = volume of organic layer
Vaq = volume of aqueous layer
Data:
m = 75 mg
K = 1.8
Vo = 0.90 mL
Vaq = 1.00 mL
Calculations:
For each extraction,
1 + (1/K)(Vaq/Vo) = 1 + (1/1.8)(1.00/0.90) = 1 + 0.62 = 1.62
x = m/1.62 = 0.618m
So, 61.8 % of the solute is extracted in each step.
In other words, 38.2 % of the solute remains.
Let r = the amount remaining after n extractions. Then
r = m(0.382)^n.
If n = 7,
r = 75(0.382)^7 = 75 × 0.001 18 = 0.088 mg
m = 75 - 0.088 = 75 mg
After seven extractions, 75 mg (99.999 %) of the solute will be extracted.
What is the conjugate acid in this reaction? HC2H3O2(aq)+H2O(l)⇌H3O++C2H3O−2(aq) View Available Hint(s) What is the conjugate acid in this reaction? HC2H3O2 H2O H3O+ C2H3O−2
Hey there!
HC2H3O2(aq) + H2O(l) → H3O⁺(aq) + C2H3O2⁻(aq)
↓ ↓ ↓ ↓
acid base acid base
If we consider the only forward reaction H3O⁺is the conjugate acid of the base H2O . For reversse reaction CH3COOH is the conjugate acid of the base CH3COO⁻.
Hope this helps!
In the given reaction, after water (H2O) accepts a proton (H+) from acetic acid (HC2H3O2), it forms H3O+ (hydronium ion), which is the conjugate acid.
Explanation:In this reaction, the conjugate acid is the species that forms after a base has accepted a proton. So, here, the base is H2O and it accepts a proton, H+, from HC2H3O2 to become H3O+. Thus, in the reaction HC2H3O2(aq) + H2O(l) ⇌ H3O+ + C2H3O−2(aq), the conjugate acid that forms is H3O+ (hydronium ion).
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Ammonia gas is dissolved in a 1.00 ×10-4M solution of CuSO4to give an equilibrium concentration of [NH3] = 1.60 ×10-3M. Calculate the equilibrium concentration of Cu2+(aq) ions in the solution
Answer:
Explanation:
Mixing Ammonia gas into a solution of Copper(II) Sulfate will give Ammonium Sulfate and a precipitate of Copper(II) Hydroxide (Cu(OH)₂). The Ksp of Cu(OH)₂ is published => 2.2 x 10⁻²². Such gives a solubility* of the Cu(OH)₂ to be ~1.77 x 10⁻⁷M => [Cu⁺²] ~1.77 x 10⁻⁷M and [OH⁻] = 2(1.77 x 10⁻⁷)M = 3.53 x 10⁻⁷M. The reaction of Ammonium Hydroxide and Copper(II) Sulfate will generate 1 x 10⁻⁴ mole Cu(OH)₂ as a precipitate but only 1.77 x 10⁻⁷ mole of the hydroxide will remain in 1 Liter of solution b/c of extreme limited solubility.
*Solubility of 1:2 ionization ratio salts = CubeRt(Ksp/4).
Be sure to answer all parts. Nitric oxide (NO) reacts with oxygen gas to form nitrogen dioxide (NO2), a dark brown gas: 2NO(g) + O2(g) → 2NO2(g)
In one experiment, 0.857 mol of NO is mixed with 0.498 mol of O2.
Determine which of the two reactants is the limiting reactant. Calculate also the number of moles of NO2 produced. Limiting reactant: Moles of NO2 produced: moles
Answer: NO is the limiting reagent in the given reaction and 0.857 moles of [tex]NO_2[/tex] will be produced.
Explanation:
Limiting reagent is defined as the reagent which is present in less amount and it limits the formation of products.
Excess reagent is defined as the reagent which is present in large amount.
For the given chemical reaction:
[tex]2NO(g)+O_2(g)\rightarrow 2NO_2(g)[/tex]
We are given:
Moles of NO = 0.857 mol
Moles of oxygen = 0.498 mol
By stoichiometry of the reaction:
If 2 moles of NO reacts with 1 mole of oxygen gas.
So, 0.857 moles of NO will react with = [tex]\frac{1}{2}\times 0.857=0.4285mol[/tex] of [tex]O_2[/tex]
As, the given amount of oxygen gas is more than the required amount. So, it is considered as an excess reagent.
Thus, NO is considered as the limiting reagent because it limits the formation of products.
By Stoichiometry of the reaction:
If 2 moles of NO produces 2 moles of nitrogen dioxide gas.
So, 0.857 moles of NO will produce = [tex]\frac{2}{2}\times 0.857=0.857mol[/tex] of [tex]NO_2[/tex]
Hence, NO is the limiting reagent in the given reaction and 0.857 moles of [tex]NO_2[/tex] will be produced.
A chemist carefully measures the amount of heat needed to raise the temperature of a 894.0g sample of a pure substance from −5.8°C to 17.5°C . The experiment shows that 4.90kJ of heat are needed. What can the chemist report for the specific heat capacity of the substance? Round your answer to 3 significant digits.
An 894.0 g sample with a specific heat capacity of 2.35 × 10⁻⁴ kJ/g.° C, increases its temperature from -5.8 °C to 17.5 °C when absorbing 4.90 kJ of heat.
A chemist has a sample with a mass of 894.0 g. When it absorbs 4.90 kJ of heat its temperature increases from -5.8 °C to 17.5 °C. The chemist can calculate the specific heat capacity of the substance using the following expression.
[tex]Q = c \times m \times \Delta T[/tex]
where,
c: specific heat capacity of the substancem: mass of the sampleΔT: change in the temperature of the sample[tex]Q = c \times m \times \Delta T\\\\c = \frac{Q}{m \times \Delta T} = \frac{4.90 kJ}{894.0 g \times (17.5 \° C - (-5.8 \° C))} = 2.35 \times 10^{-4} kJ/g\° C[/tex]
An 894.0 g sample with a specific heat capacity of 2.35 × 10⁻⁴ kJ/g.° C, increases its temperature from -5.8 °C to 17.5 °C when absorbing 4.90 kJ of heat.
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The specific heat capacity of the substance is approximately 0.234 J/g°C. This was found by applying the formula for specific heat capacity (q = mcΔT) and doing the necessary calculations with the given values.
Explanation:To determine the specific heat capacity of the substance, we can use the formula q = mcΔT, where q is the heat energy absorbed, m is the mass of the substance, c is the specific heat capacity, and ΔT is the change in temperature. Given that q equals 4.90 kJ (or 4900 J to match the unit of specific heat capacity), m equals 894.0g and ΔT is 17.5°C - (-5.8°C) = 23.3°C, we can rearrange the formula to solve for c: c = q / (mΔT).
Substituting the given values, we find c = 4900 J / (894.0g * 23.3°C), yielding a specific heat capacity of approximately 0.234 J/g°C to three significant figures.
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An experiment shows that a 111 mL gas sample has a mass of 0.168 g at a pressure of 734 mmHg and a temperature of 34 ∘C. What is the molar mass of the gas?
hey there!:
Volume in liters ( V ) = 111 mL / 1000 => 0.111 L
Pressure in atm ( P ) = 734 / 760 => 0.965789 atm
temperature in Kelvin ( K ) = 34+273.15 => 307.15 K
Molar gas constant ( R ) = 0.0826 atm*L/mol*K
ideal gas equation :
p*V = n*R*T
moles of gas:
n = p*v / R*T
n = 0.965789 * 0.111 / 0.0826 * 307.15
n = 0.107202 / 25.37059
n = 0.004225 moles
Therefore:
Molar mass = mass / moles of gas
molar mass = 0.168 / 0.004225
molar mass = 39.76 g/mol
Hope this helps!
The molar mass of the gas was calculated by applying the ideal gas law, rearranging it for molar mass, and using the given quantities in the problem. Conversions to appropriate units were made, and the molar mass was found to be 31.1 g/mol.
Explanation:
The molar mass of a gas can be calculated by applying the ideal gas law, PV = nRT. By rearranging this to solve for the molar mass, using known quantities from the problem, and converting units appropriately, we find:
First, you need to convert the given volume from mL to L by dividing by 1000. So, 111 mL = 0.111 L.The pressure should also be in atmospheres, so we convert 734 mmHg to atm by dividing by 760, getting approximately 0.965 atm.The temperature must be converted to Kelvin, the SI unit of temperature. The formula to convert Celsius to Kelvin is K = C + 273.15, which gives us 307.15 K.From the definition of molar mass (mass/moles), we can express the number of moles as mass/molar mass. Substituting P, V, and T into the ideal gas law then allows us to solve for the molar mass.Applying these steps, we find that the molar mass of the gas is approximately 31.1 g/mol.
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Tooth enamel ( Ca5(PO4)3(OH) ) reacts with acid according to the reaction below: Ca_5(PO_4)_3(OH) (s) + H^+(aq) \leftrightarrow 5Ca^{2+}(aq) + 3HPO_4^{2-}(aq) + OH^-(aq) C a 5 ( P O 4 ) 3 ( O H ) ( s ) + H + ( a q ) ↔ 5 C a 2 + ( a q ) + 3 H P O 4 2 − ( a q ) + O H − ( a q ) What happens if we add more acid?
Final answer:
Adding more acid to tooth enamel causes increased solubility of enamel and may lead to cavities. The added acid reacts with hydroxide ions in the enamel, driving the reaction forward. Fluoride in dental products can help protect enamel by reducing solubility in acid.
Explanation:
When more acid, which in chemical terms is a source of H+ ions, is added to the solution where tooth enamel, chemically known as calcium hydroxyapatite Ca5(PO4)3(OH), is present, it results in the reaction shifting to the right according to Le Chatelier's principle. This process increases the solubility of the tooth enamel in acid, leading to enamel dissolution and potentially to the formation of dental cavities. The reaction with acid produces 5Ca2+(aq), 3HPO42-(aq), and OH-(aq). Hydroxide ions (OH-) react with added H+ ions to form water (2H2O), driving the reaction forward and increasing enamel solubility. To mitigate this effect, toothpastes and mouth rinses may contain fluoride compounds, replacing the strong base hydroxide in the enamel with the weaker base fluoride. This renders the enamel more resistant to acid attack by reducing the extent to which the equilibrium shifts upon acid addition.
Where can membrane proteins be located? a. entirely within the membrane b. on the surface of the membrane c. in either position d. The position of membrane proteins has never been determined
Answer:
C) in either position
Explanation:
There are two kinds of membrane protein:
a) Integral proteins: They have a fixed or permanent association with the membrane. They are majorly embedded in the middle layer of membrane. A kind of integral proteins are transmembrane proteins, they can cross the membrane and are path for transport of ions or molecule in and out cell.
b) Peripheral proteins: The are confined to the surface of membrane and are boned with ionic interactions. they are more in number as compared to integral proteins.
Compare suspension and emulsion polymerizations. Describe the microscopic environment of each process shows where the monomer, initiator and additives are located in the initial stages of the polymerization and after 25% conversion is achieved. (Draw clear pictures and label them) B. What are the advantages of producing polyethene using Ziegler Natta catalyst? C. Why is step-growth polymerization not applicable to ethene molecules?
Answer:
The main difference between suspension and emulsion polymerization is that suspension polymerization requires a dispersing medium, monomer(s), stabilizing agents and initiators whereas emulsion polymerization requires water, monomer and a surfactant.
Explanation:
The average atomic weight of copper, which has two naturally occurring isotopes, is 63.5. One of the isotopes has an atomic weight of 62.9 amu and constitutes 69.1% of the copper isotopes. The other isotope has an abundance of 30.9%. The atomic weight (amu) of the second isotope is ________ amu.
Answer: The atomic weight of the second isotope is 64.81 amu.
Explanation:
Average atomic mass of an element is defined as the sum of atomic masses of each isotope each multiplied by their natural fractional abundance
Formula used to calculate average atomic mass follows:
[tex]\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i[/tex] .....(1)
We are given:
Let the mass of isotope 2 be 'x'
Mass of isotope 1 = 62.9 amu
Percentage abundance of isotope 1 = 69.1 %
Fractional abundance of isotope 1 = 0.691
Mass of isotope 2 = 'x'
Percentage abundance of isotope 2 = 30.9%
Fractional abundance of isotope 2 = 0.309
Average atomic mass of copper = 63.5 amu
Putting values in equation 1, we get:
[tex]\text{Average atomic mass of copper}=[(62.9\times 0.691)+(x\times 0.309)][/tex]
[tex]x=64.81amu[/tex]
Hence, the atomic weight of second isotope will be 64.81 amu.
By using the weighted average formula and the given details, we can determine the atomic weight of the second naturally occurring isotope of copper approximately to be 64.93 amu.
Explanation:The atomic weight of an isotope is calculated by adding the products of the abundance percentages and the atomic weights of each isotope. For copper (Cu), one of its isotopes is Copper-63, which makes up 69.1% of naturally occurring copper, and it has an atomic weight of 62.9 amu. Therefore, we can calculate the atomic weight of the second isotope using the given average atomic weight of copper (63.5 amu) as follows:
Average atomic weight = (abundance of Copper-63 * atomic weight of Copper-63) + (abundance of Copper-65 * atomic weight of Copper-65)
63.5 = (0.691*62.9) + (0.309 * x)
After we calculate the first part, we subtract it from 63.5 to solve for the atomic weight (x) of the second isotope, known as Copper-65.
Therefore, the atomic weight of the second isotope of copper is approximately 64.93 amu.
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Liquid A has a vapor pressure of 264 torr at 20∘C, and liquid B has a vapor pressure of 96.5 torr at the same temperature. If 5.50 moles of liquid A and 8.50 moles of liquid B are combined to form an ideal solution, what is the total vapor pressure (in torr) above the solution at 20.0∘C?
Answer: 161.8 torr
Explanation:
According to Raoult's law, the vapor pressure of a component at a given temperature is equal to the mole fraction of that component multiplied by the vapor pressure of that component in the pure state.
[tex]p_1=x_1p_1^0[/tex] and [tex]p_2=x_2P_2^0[/tex]
where, x = mole fraction
[tex]p^0[/tex] = pressure in the pure state
According to Dalton's law, the total pressure is the sum of individual pressures.
[tex]p_{total}=p_1+p_2[/tex][tex]p_{total}=x_Ap_A^0+x_BP_B^0[/tex]
[tex]x_{A}=\frac{\text {moles of A}}{\text {moles of A+moles of B}}=\frac{5.50}{5.50+8.50}=0.39[/tex],
[tex]x_{B}=\frac{\text {moles of B}}{\text {moles of A+moles of B}}=\frac{8.50}{5.50+8.50}=0.61[/tex],
[tex]p_{A}^0=264torr[/tex]
[tex]p_{B}^0=96.5torr[/tex]
[tex]p_{total}=0.39\times 264+0.61\times 96.5=161.8torr[/tex]
The total vapor pressure above the solution is 161.8 torr.
Calculate the boiling point of a solution of 500.0 g of ethylene glycol (C2H6O2) dissolved in 500.0 g of water. Kf = 1.86°C/m and Kb = 0.512°C/m. Use 100°C as the boiling point of water.
Answer:
The boiling point of a solution of 500.0 g of ethylene glycol dissolved in 500.0 g of water is 108.258°C.
Explanation:
Elevation in boiling point : [tex]\Delta T_b[/tex]
[tex]\Delta T_b=T_b-T[/tex]
[tex]\Delta T_b=K_b\times m[/tex]
[tex]T_b[/tex] = Boiling point of the solution
T = Boiling point of pure solvent
[tex]K_b[/tex]= Molal elevation constant of solvent
m = molality of the solution
Molality of the ethylene glycol solution:
[tex]molality=\frac{\text{Moles of solute}}{\text{Mass of solvent in kg}}[/tex]
Moles of ethylene glycol = [tex]\frac{500.0 g}{62 g/mol}=8.0645 mol[/tex]
Mass of solvent that uis water = 500.0 g = 0.5000 kg
[tex]m=\frac{8.0645 mol}{0.5000 kg}=16.1290 m[/tex]
Molal elevation constant of water =[tex]K_b=0.512^oC/m[/tex]
[tex]\Delta T_b=0.512^oC/m\times 16.1290 m=8.258^oC[/tex]
Boiling point of the solution =[tex]T_b[/tex]
Boiling point of pure water = T = 100°C
[tex]T_b=T+\Delta T_b=100^oC+8.258^oC=108.258^oC[/tex]
The boiling point of a solution of 500.0 g of ethylene glycol dissolved in 500.0 g of water is 108.258°C.
Final answer:
To calculate the boiling point of the solution, use the equation ΔT = Kb * molality, where ΔT is the boiling point elevation, Kb is the boiling point elevation constant, and molality is the molal concentration of the solution. Calculate the molality by dividing the number of moles of ethylene glycol by the mass of water. Substitute the molality into the equation to calculate the boiling point elevation, and add this elevation to the boiling point of pure water (100°C) to find the boiling point of the solution.
Explanation:
To calculate the boiling point of the solution, we need to use the equation:
ΔT = Kb * molality
Where ΔT is the boiling point elevation, Kb is the boiling point elevation constant, and molality is the molal concentration of the solution.
First, we need to calculate the molality of the solution by dividing the number of moles of ethylene glycol by the mass of water. The number of moles of ethylene glycol can be found by dividing the mass of ethylene glycol by its molar mass, and the mass of water is given as 500.0 g.
Once we have the molality, we can substitute it into the equation to calculate the boiling point elevation. Finally, we add this elevation to the boiling point of pure water (100°C) to find the boiling point of the solution.
A 0.1375-g sample of solid magnesium is burned in a constant-volume bomb calorimeter that has a heat capacity of 3024 J/°C. The temperature increases by 1.126°C. Calculate the heat given off by the burning Mg, in kJ/g and in kJ/mol.
A 0.1375-g sample of solid magnesium is burned in a constant-volume bomb calorimeter (heat capacity of 3024 J/°C), causing a temperature increase of 1.126°C. The heat given off by the burning Mg is -24.76 kJ/g and -601.9 kJ/mol.
When a sample of magnesium is burned in a constant-volume bomb calorimeter that has a heat capacity (C) of 3024 J/°C, the temperature increases by 1.126°C (ΔT). We can calculate the heat absorbed by the calorimeter (Qc) using the following expression.
[tex]Qc = C \times \Delta T = \frac{3024J}{\° C} \times 1.126 \° C \times \frac{1kJ}{1000J} = 3.405 kJ[/tex]
According to the law of conservation of energy, the sum of the heat absorbed by the calorimeter and the heat released by the reaction (Qr) is zero.
[tex]Qc + Qr = 0\\\\Qr = -Qc = -3.405 kJ[/tex]
3.405 kJ are released by the combustion of 0.1375 g of Mg. The heat released per gram of Mg is:
[tex]\frac{-3.405kJ}{0.1375g} = -24.76 kJ/g[/tex]
Finally, we will convert -24.67 kJ/g to kJ/mol using the molar mass of Mg (24.31 g/mol).
[tex]\frac{-24.76kJ}{g} \times \frac{24.31g}{mol} = -601.9 kJ/mol[/tex]
A 0.1375-g sample of solid magnesium is burned in a constant-volume bomb calorimeter (heat capacity of 3024 J/°C), causing a temperature increase of 1.126°C. The heat given off by the burning Mg is -24.76 kJ/g and -601.9 kJ/mol.
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