Explanation:
The given data is as follows.
n = 2 mol, P = 1 atm, T = 300 K
Q = +34166 J, W= -1216 J (work done against surrounding)
[tex]C_{v}[/tex] = [tex]\frac{3R}{2}[/tex]
Relation between internal energy, work and heat is as follows.
Change in internal energy ([tex]\Delta U[/tex]) = Q + W
= [34166 + (-1216)] J
= 32950 J
Also, [tex]\Delta U = n \times C_{v} \times \Delta T[/tex]
= [tex]3R \times (T_{2} - T_{1})[/tex]
32950 J = [tex]3 \times 8.314 J/mol K \times (T_{2} - 300 K)[/tex]
[tex]\frac{32950}{24.942} = T_{2} - 300 K[/tex]
1321.06 K + 300 K = [tex]T_{2}[/tex]
[tex]T_{2}[/tex] = 1621.06 K
Thus, we can conclude that the final temperature of the gas is 1621.06 K.
A chemical process requires 7.300 g-mol of calcium sulfate (CaSO.) for a reaction. How many grams are in 7.300 g-mol of calcium sulfate? Caso
Answer: The mass of copper sulfate is 1165 grams.
Explanation:
We are given:
Number of gram-moles of copper sulfate = 7.300 g-mol
We know that:
1 g-mol = 1 mol
So, number of moles of copper sulfate = 7.300 mol
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
Molar mass of copper sulfate = 159.6 g/mol
Moles of copper sulfate = 7.300 moles
Putting values in above equation, we get:
[tex]7.300mol=\frac{\text{Mass of copper sulfate}}{159.6g/mol}\\\\\text{Mass of copper sulfate}=1165g[/tex]
Hence, the mass of copper sulfate is 1165 grams.
How many moles of solute particles are present in 4.98 mL of 0.72 M (NH4)2SO4? When you have your answer, take the LOG (base 10) of your answer and enter it with 2 decimal places.
Answer: The number of moles of ammonium sulfate is 0.0036 moles and its logarithmic value is -2.44
Explanation:
To calculate the number of moles for given molarity, we use the equation:
[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}[/tex]
Molarity of ammonium sulfate solution = 0.72 M
Volume of solution = 4.98 mL
Putting values in above equation, we get:
[tex]0.72M=\frac{\text{Moles of ammonium sulfate}\times 1000}{4.98mL}\\\\\text{Moles of ammonium sulfate}=0.0036mol[/tex]
Taking the log (base 10) of the calculated moles of ammonium sulfate we get:
[tex]\log_{10}(0.0036)=-2.44[/tex]
Hence, the number of moles of ammonium sulfate is 0.0036 moles and its logarithmic value is -2.44
Final answer:
First, calculate the moles of (NH4)2SO4 using the molarity and volume, then find the base 10 logarithm of the result. The moles of (NH4)2SO4 are approximately 0.003586, and the logarithm is roughly -2.45.
Explanation:
To compute the moles of solute in a given volume of solution, one can use the formula moles of solute = Molarity (mol/L) × Volume of solution (L). For a 0.72 M solution of (NH4)2SO4, when given a volume of 4.98 mL (which is 0.00498 L), we calculate:
moles of solute = 0.72 mol/L × 0.00498 L = 0.0035856 mol
Now, to find the base 10 logarithm, we perform the following calculation:
LOG(moles of solute) = LOG(0.0035856) ≈ -2.45
What is the silver concentration in a solution prepared by mixing 250
mL of 0.310 M silver nitrate with 100 mL of 0.160 M of sodium chromate?
Answer:
0.221M
Explanation:
From the question ,
The Molarity of AgNO₂ = 0.310 M
Hence , the concentration of Ag⁺ = 0.301 M
The volume of AgNO₂ = 250 mL
and,
The Molarity of Sodium chromate = 0.160 M
The volume of Sodium chromate = 100 mL.
As the solution is mixed the final volume becomes ,
250mL +100mL = 350mL
Now, using the formula , to find the final molarity of the mixture ,
M₁V₁ ( initial ) = M₂V₂ ( final )
substituting the values , in the above equation ,
0.310M * 250ml = M₂ * 350ml
M₂ = 0.221M
Hence , the concentration of the silver in the final solution = 0.221M
Write a reaction to describe the behavior of Sr(C2H3O2)2 (s) in water. please be sure to include all phases.
Answer:
Sr(C₂H₃O₂)₂(s) + 2H₂O (l) ⇄ Sr(OH)₂(s) + 2C₂H₃OOH(aq)
Explanation:
Sr(C₂H₃O₂)₂ is a salt, formed by a metal cation (Sr⁺²) and an anion (C₂H₃O₂⁻). This ionic compound must ionize in water, making an equilibrium, which will react with the water equilibrium.
The equilibrium of the salt is:
Sr(C₂H₃O₂)₂(s) ⇄ Sr⁺²(aq) + C₂H₃O₂⁻(aq)
And the water equilibrium:
H₂O (l) ⇄ H⁺ (aq) + OH⁻(aq)
So, Sr⁺² must react with OH⁻ to form the hydroxide Sr(OH)₂, and C₂H₃O₂⁻ must react with H⁺ to form the acid C₂H₃OOH. Sr is a metal of group 2, so the base will be a little soluble in water, and the solid may precipitate. C₂H₃OOH is a weak acid, and soluble in water, so it will be in aqueous form. The reaction is:
Sr(C₂H₃O₂)₂(s) + 2H₂O (l) ⇄ Sr(OH)₂(s) + 2C₂H₃OOH(aq)
Final answer:
Strontium acetate (Sr(C2H3O2)2) dissociates in water to form Sr2+ and 2 C2H3O2- ions. The dissociation is represented by the equation Sr(C2H3O2)2 (s) → Sr2+ (aq) + 2 C2H3O2- (aq).
Explanation:
The behavior of Sr(C2H3O2)2 (s) in water can be described as a process where the solid salt dissociates into its constituent ions. When dissolved in water, strontium acetate separates into strontium ions and acetate ions according to the following equation:
Sr(C2H3O2)2 (s) → Sr2+ (aq) + 2 C2H3O2− (aq)
This representation is known as the dissociation equation for the ionic compound in water. The process demonstrates the compound breaking down from a solid to freely moving ions in an aqueous solution.
The value of the equilibrium constant for the following chemical equation is Kr 40 at 25°C. Calculate the solubility of Al(OH)s(s) in an aqueous solution buffered at pH 11.00 at 25°C. Al(OH)(s)OH (aq)Al(OH)l'(aq) + a) 4.0 x 1010 M b) 6.3 M d) 0.37 M c) 0.040 M
Answer:
0,040 M
Explanation:
The global reaction of the problem is:
Al(OH) (s) + OH⁻ ⇄ Al(OH)₂⁻(aq) K= 40
The equation of equilibrium is:
K = [tex]\frac{[Al(OH)_{2} ^-]}{[Al(OH)][OH^-]}[/tex]
The concentration of OH⁻ is:
pOH = 14 - pH = 3
pOH = -log [OH⁻]
[OH⁻] = 1x10⁻³
Thus:
40 = [tex]\frac{[Al(OH)_{2} ^-]}{[Al(OH)][1x10^{-3}]}[/tex]
0,04M = [tex]\frac{[Al(OH)_{2} ^-]}{[Al(OH)]}[/tex]
This means that 0,04 M are the number of moles that the solvent can dissolve in 1L, in other words, solubility.
I hope it helps!
What is the percent s character in an sp? hybridized orbital? 25% 33% 50% 67% 75%
Answer:
50%
Explanation:
The percent s - character , in a particular hybridization can be calculated from the following formula , i.e. ,
% s character = 1 / ( 1 + x ) * 100
Where , x = number of p orbitals .
Hence , from the question ,
The % s character in sp hybrid orbital is calculated as -
% s character = 1 / ( 1 + x ) * 100
x = 1 ( since , one p orbital is present in the sp hybridization )
% s character = 1 / ( 1 + 1 ) * 100
% s character = 1/ 2 * 100
% s character = 50 %.
Hence , the % s character in sp hybrid orbital = 50 % .
In sp hybridization, one s and one p atomic orbital combine to form two new hybrid orbitals. Each of these sp hybrid orbitals has 50% s character and 50% p character, so the percent s character in a sp hybridized orbital is 50%. This kind of hybridization is fundamental to understanding molecular geometry in chemistry.
Explanation:The term sp hybridization refers to the mathematical combination of one s and one p atomic orbital to form two new hybrid orbitals. Each hybrid orbital in a sp configuration contains 50% s character and 50% p character, therefore, the percent s character in a sp hybridized orbital is 50%. This orbital hybridization concept is a fundamental aspect of understanding molecular geometry in chemistry.
As an example, consider carbon. In its ground state, carbon has two unpaired electrons in separate 2p orbitals. If one of its 2s electrons is excited to a 2p orbital, the atom can then hybridize these three orbitals (one 2s orbital and two 2p orbitals) to form an sp hybridized state. The result of this is two equivalent sp orbitals arranged linearly at 180° to each other. This sp hybridization in carbon is seen in molecules like carbon dioxide (CO2).
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A sample of Ne pas is observed to effuse through a pour own barrier in 8.61 minutes. Under the same conditions, the same number of moles of an unknown as requires 13.0 minutes to effuse through the same bam The molar mass of the unknown as mol
Answer:
46.004 g/mol
Explanation:
Scottish physicist Thomas Graham formulated a law known as Graham's law of effusion in 1848. He conducted an experiment and found the relationship between the rate of effusion of a gas and its molar mass as:
[tex]r=\sqrt {\frac {1}{M}}[/tex]
where,
r is the rate of effusion of a gas
M is the molar mass of the gas.
Also, r = v/t
And for two gases taking different t₁ and t₂ to effuse, the formula is:
[tex]\frac {t_2}{t_1}=\sqrt {\frac {M_2}{M_1}}[/tex]
So,
For Neon :
[tex]t_1[/tex] = 8.61 minutes
[tex]M_1[/tex] = 20.1797 g/mol
For unkown gas:
[tex]t_2[/tex] = 13.0 minutes
[tex]M_2[/tex] = ? g/mol
[tex]\frac {13.0}{8.61}=\sqrt {\frac {M_2}{20.1797}}[/tex]
Molar mass of unknown gas = 46.004 g/mol
If a sodium chloride (NaCl) solution in water at 40°C has reached a concentration 50% calculate the quantity of NaCl crystals that will form once crystallization has been started in per 100 g water. The solubility of NaCl at 40°C is 36.6 g/100 g water.
Answer:
13.4g of NaCl will crystallize per 100g of water at 40°C.
Explanation:
Solubility stands for the maximum amount of solute that may stay dissolved per 100 g of water at a specific temperature. A solution with this concentration is known as a saturated solution. At 40°C, every 100 g of water, 36.6 g of NaCl can remain dissolved. If more solute is added to the solution the excess will crystallize and precipitate. Since prior to criystallization there are 50 g of NaCl dissolved, what will precipitate will be:
50g - 36.6g = 13.4g
Once crystallization is initiated in a 50% sodium chloride solution at 40°C, 13.4 g of NaCl crystals will form per 100 g of water. This is calculated by subtracting the solubility limit at 40°C (36.6 g) from the amount initially present in the solution (50 g).
If a sodium chloride (NaCl) solution in water at 40°C has a concentration of 50%, for every 100 g of water, there is 50 g of NaCl dissolved. Since the solubility of NaCl at 40°C is 36.6 g per 100 g of water, the solution is supersaturated because it contains more solute than can remain in solution at that temperature. When crystallization starts, the excess NaCl above its solubility limit will precipitate out.
To calculate the quantity of NaCl crystals that will form, we subtract the maximum solubility at 40°C from the total amount of NaCl currently dissolved:
Amount of NaCl present initially = 50 g
Maximum solubility at 40°C = 36.6 g
NaCl crystals that will form = Amount of NaCl present initially - Maximum solubility at 40°C
NaCl crystals that will form = 50 g - 36.6 g = 13.4 g
Therefore, 13.4 g of NaCl crystals will form per 100 g of water once crystallization has been started.
How many of the following statements about silver acetate, AgCH3COO, are true? i) More AgCH3CoO(S) will dissolve if the pH of the solution is reduced to 2.5 i) Less AgCH3CO0(s) will dissolve if AgNO; (s) is added to the aqueous solution iii) More AgCH3CoO(s) will dissolve if NaOH (aq) is added to the aqueous solution a) 0 b) c) 2 d) 3
Answer:
All three statements are true
Explanation:
Solubility equilibrium of silver acetate:
[tex]AgCH_{3}COO\rightleftharpoons Ag^{+}+CH_{3}COO^{-}[/tex]
If pH is increased then concentration of [tex]H^{+}[/tex] increases in solution resulting removal of [tex]CH_{3}COO^{-}[/tex] by forming [tex]CH_{3}COOH[/tex]. Hence, according to Le-chatelier principle, equilibrium will shift towards right. So more [tex]AgCH_{3}COO[/tex] will dissolveIf [tex]AgNO_{3}[/tex] is added then concentration of [tex]Ag^{+}[/tex] increases in solution resulting shifting of equilibrium towards left in accordance with Le-chatelier principle. So less [tex]AgCH_{3}COO[/tex] will dissolveInsoluble precipitate of AgOH is formed by adding NaOH in solution resulting removal of [tex]Ag^{+}[/tex]. So, more [tex]AgCH_{3}COO[/tex] will dissolveHence all three statements are true
Hellmann, Zelles, Palojarvi, and Bai published a paper in 1997 about extracting lipids from organic material. They reported using a mixture that was 85% n-hexane and 15 % dichloromethane (vol/vol). What is the mole fraction of each compound in the mixture? What is the mass fraction of each compound in the mixture? Skills: unit conversions, choosing a temporary basis
Answer:
Mass fraction: 73,6% n-hexane; 26,4% dichloromethane
Mole fraction: 73,0% n-hexane; 27,0% dichloromethane
Explanation:
With a basis of 100 mL:
Mass of n-hexane:
85 mL ×[tex]\frac{0,655g}{1mL}[/tex] = 55,7 g
Mass of dichloromethane
15 mL ×[tex]\frac{1,33g}{1mL}[/tex] = 20,0 g
Total mass = 20,0 g + 55,7 g = 75,7 g
Mass fraction of n-hexane:
[tex]\frac{55,7g}{75,7g}[/tex] =73,6%
Mass fraction of dichloromethane:
[tex]\frac{20,0g}{75,7g}[/tex] = 26,4%
Moles of n-hexane:
55,7 g ×[tex]\frac{1mol}{86,18 g}[/tex] = 0,65 moles
Mass of dichloromethane
20,0g ×[tex]\frac{1mol}{84,93 g}[/tex] = 0,24 moles
Total moles: 0,65 moles + 0,24 moles = 0,89 moles
Molar fraction of n-hexane:
[tex]\frac{0,65moles}{0,89moles}[/tex] =73,0%
Molar fraction of dichloromethane:
[tex]\frac{0,24moles}{0,89moles}[/tex] = 27,0%
I hope it helps!
Aluminum chloride, AICh is an inexpensive reagent used in many industrial processes. It is made by treating scrap aluminum with chlorine according to the following equation. 2 Aln + 3 Cla) → 2 AlCl3() a) If 13,49 g of Al and 35.45 g of Cl2 are allowed to react, how much AlCl is produced? b) How many grams of the excess reactant is left?!
Answer:
a) 44.442 grams of aluminium chloride is produced.
b) 8.996 grams of the excess reactant that id aluminum is left unreacted.
Explanation:
[tex]2Al(s) + 3Cl_2(g)\rightarrow 2AlCl_3(s)[/tex]
Moles of aluminum = [tex]\frac{13.49 g}{27 g/mol}=0.4996 mol[/tex]
Moles of chlorine gas = [tex]\frac{35.45}{35.5 g/mol}=0.4993 mol[/tex]
According to reaction, 1 mol aluminum reacts with 3 moles of chlorine gas.
Then 0.4996 moles of aluminum will react with:
[tex]\frac{3}{1}\times 0.4996 mol=1.4988 mol[/tex] of chlorine gas
Then 0.4993 moles of chlorine gas will react with:
[tex]\frac{1}{3}\times 0.4993 mol=0.1664 mol[/tex] of aluminum
As we can see that moles of an aluminium are in excess. Hence, aluminium is an excess reagent.
So moles of aluminum chloride will depend upon moles of chlorine gas.
According to reaction 3 moles of chlorine gas give 2 moles of aluminium chloride.
Then 0.4993 mole of chlorine gas will give:
[tex]\frac{2}{3}\times 0.4993 mol=0.3329 mol[/tex] of aluminium chloride .
Mass of 0.3329 moles of aluminium chloride :
=0.3329 mol × 133.5 g/mol= 44.442 g
44.442 grams of aluminium chloride is produced.
Moles of aluminium left unreacted = 0.4996 mol - 0.1664 mol = 0.3332 mol
Mass of 0.3332 moles of aluminum :
0.3332 mol × 27 g/mol = 8.996 g[/tex]
8.996 grams of the excess reactant that id aluminum is left unreacted.
Using stoichiometry, we calculate how much aluminum chloride is formed from given masses of aluminum and chlorine and determine the excess reactant remaining. This requires finding the limiting reactant, using the balanced chemical equation, and converting between moles and grams.
Explanation:To answer the question about how much aluminum chloride is produced when 13.49 g of Al and 35.45 g of Cl2 react, we need to use stoichiometry, which is the method of quantitatively relating the products and reactants in a chemical reaction. Here is the step-by-step explanation:
Determine the molar masses of the reactants and product (Al = 26.98 g/mol, Cl2 = 70.90 g/mol, AlCl3 = 133.34 g/mol).Calculate the number of moles of each reactant (n = mass/molar mass)Identify the limiting reactant by comparing the stoichiometric ratios from the balanced equation with the number of moles of each reactantUse the molar ratio from the balanced equation to find out how many moles of product can be formed by the limiting reactant.Convert the moles of AlCl3 back into grams using the molar mass of AlCl3 (AlCl3's mass = moles x molar mass).To find the excess reactant remaining:
Subtract the amount of limiting reactant that reacted, from the initial amount, to find the amount of excess reactant.
Using the steam tables, determine the enthalpy change for 1lb of water:
Condensing a saturated vapor at 212 Fahrenheit and 1 atm abs.
Answer:
change in enthalpy for 1 lb water is - 970.10872 Btu
Explanation:
Given data:
weight of water 1 lb
pressure 1 atm
temperature 212 Fahrenheit
From steam table
initial enthalpy [tex]h_1 = 1150.288 Btu/lb[/tex]
final enthalpy[tex] h_2 = 180.18 Btu/lb[/tex]
change in enthalpy is given as
[tex]\Delta h = m (h_2 -h_1)[/tex]
[tex]\Delta h = 1\times ( 180.18020 - 1150.288)[/tex]
[tex]\Delta h = - 970.10872 Btu[/tex]
change in enthalpy for 1 lb water is - 970.10872 Btu
Liquid n-pentane and liquid n-octane are mixed to form a stream flowing at a rate of 1,496.5 lbm/hr. An in-line density measurement determines the mixture density is 0.654 g/mL. Calculate the volumetric flow rate of the liquid n-pentane feed in cm3/sec. The specific gravities of n-pentane and n-octane are 0.631 and 0.708, respectively. Assume ideal mixing.
Answer:
202.20 cm^3/s
Explanation:
Hello,
I'm sending a photo showing the solution for this exercise.
The solution of the 2x2 system could be achieved by using a calculator or any available algebraic method.
Best regards.
How much does 269 mmol of magnesium sulfate heptahydrate weigh, in grams?
Answer:
66,3 g of MgSO₄· 7 H₂O
Explanation:
Some salts must be hydrated to increase solubility.
Magnesium sulfate heptahydrate has as molecular formula
MgSO₄· 7 H₂O
The molecular weight is the sum of atomic weights. Thus:
1 Mg × (24,305 g/mol) = 24,305 g/mol
1 S × (32,065 g/mol) = 32,065 g/mol
11 O × (15,999 g/mol) = 175,989 g/mol
14 H × (1,008 g/mol) = 14,112 g/mol
Molecular weight: 246,471 g/mol
269 mmol are 0,269 mol (m is mili: 1×10⁻³. It means 269×10⁻³ = 0,269)
0,269 mol of MgSO₄· 7 H₂O × (246,471 g / mol) = 66,3 g of MgSO₄· 7 H₂O
I hope it helps!
A turbine operates on superheated steam that enters at 80 bar absolute and 500˚C, and leaves the turbine at 7.5 bar absolute and 250˚C. The flow rate of the steam is 400 kg/min and the turbine operates adiabatically. The steam exiting the turbine goes to a heater, which is supplied with 2000
2kW of energy. The steam leaves the heater at 5 bar absolute. All of the pipes through which the
steam flows are 0.5 m (id) in diameter.
a. How much power is produced by the turbine? (4 marks)
b. What is the temperature of the steam leaving the heater? (4 marks)
c. What is the velocity (m/s) of the steam entering the turbine? (3 marks)
d. To what temperature can the steam at 80 bar be cooled before a liquid will appear?
The thallium Subscript 81 Superscript 208 Baseline Tl nucleus is radioactive, with a half-life of 3.053 min. At a given instant, the activity of a certain sample of thallium is 2400 Bq. Using the concept of a half-life, and without doing any written calculations, determine what the activity 9 minutes later is.
(A) 150 bq
(B) 2400 bq
(C) 1200 bq
(D) 600 bq
(E) 300 bq
Answer: (E) 300 bq
Explanation:
Half life is the amount of time taken by a radioactive material to decay to half of its original value.
Radioactive decay process is a type of process in which a less stable nuclei decomposes to a stable nuclei by releasing some radiations or particles like alpha, beta particles or gamma-radiations. The radioactive decay follows first order kinetics.
Half life is represented by [tex]t_{\frac{1}{2}[/tex]
Half life of Thallium-208 = 3.053 min
Thus after 9 minutes , three half lives will be passed, after ist half life, the activity would be reduced to half of original i.e. [tex]\frac{2400}{2}=1200[/tex], after second half life, the activity would be reduced to half of 1200 i.e. [tex]\frac{1200}{2}=600[/tex], and after third half life, the activity would be reduced to half of 600 i.e. [tex]\frac{600}{2}=300[/tex],
Thus the activity 9 minutes later is 300 bq.
How many grams of disulfur decafluoride are present in 2.15 moles of this compound? grams
Answer
Take account the molar mass of this compound (first of all u should know the formula, S2F10) so when u have the molar mass, you can get how many grams of it, are present in 2.15 moles. The right mass is 546,3 g.
Explanation:
Molar mass in the S2F10 should be, molar mass in S (32.06 x2) + molar mass of F (18.998 x 10) = 254,1 g/m
A chemist must prepare 400 mL of 1.00M aqueous aluminum sulfate working solution. He'll do this by pouring out 1.82 mol/L aqueous aluminum sulfate stock solution into a graduated cylinder and diluting it with distilled water. How many mL of the aluminum sulfate stock solution should the chemist pour out?
To prepare the 1.00M aqueous aluminum sulfate working solution, the chemist should pour out approximately 219.78 mL of the aluminum sulfate stock solution.
Explanation:To prepare a 1.00M aqueous aluminum sulfate working solution, the chemist needs to dilute the 1.82 mol/L aluminum sulfate stock solution. The final volume required is 400 mL.
Using the formula for dilution, (stock concentration) × (stock volume) = (final concentration) × (final volume),
we can rearrange the formula to solve for the stock volume:
Stock volume = (final concentration) × (final volume) / (stock concentration)
Substituting the given values,
Stock volume = (1.00 M) × (400 mL) / (1.82 mol/L) = 219.78 mL
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The chemist should pour out 220 mL of the aluminum sulfate stock solution.
To prepare the desired 400 mL of 1.00M aluminum sulfate solution, the chemist needs to use the concept of dilution, which is described by the equation:
[tex]\[ C_1V_1 = C_2V_2 \][/tex]
where:
-[tex]\( C_1 \)[/tex] is the concentration of the stock solution (1.82 mol/L),
- [tex]\( V_1 \)[/tex] is the volume of the stock solution needed (unknown),
- [tex]\( C_2 \)[/tex] is the concentration of the working solution (1.00M),
-[tex]\( V_2 \)[/tex] is the volume of the working solution (400 mL).
First, we need to ensure that all volumes are in the same unit, so we'll convert 400 mL to liters:
[tex]\[ 400 \text{ mL} = 0.400 \text{ L} \][/tex]
Now we can plug the known values into the dilution equation:
[tex]\[ (1.82 \text{ mol/L})V_1 = (1.00 \text{ mol/L})(0.400 \text{ L}) \][/tex]
Solving for [tex]\( V_1 \):[/tex]
[tex]\[ V_1 = \frac{(1.00 \text{ mol/L})(0.400 \text{ L})}{1.82 \text{ mol/L}} \][/tex]
[tex]\[ V_1 = \frac{0.400}{1.82} \text{ L} \][/tex]
[tex]\[ V_1 \approx 0.220 \text{ L} \][/tex]
Converting liters back to milliliters:
[tex]\[ 0.220 \text{ L} = 220 \text{ mL} \][/tex]
Therefore, the chemist should pour out 220 mL of the 1.82 mol/L aluminum sulfate stock solution to make 400 mL of a 1.00M solution.
The reaction is as follows: CH4 + 202 + CO2 + 2H2O The enthalpy of reaction between methane ( 2 ) and oxygen ( 2 ) is exothermic (-882 kJ/mol) with respect to the dissipation of methane. It is of interest to generate 126799 kJ of energy from the combustion. How many kg of methane will need to be combusted? Molecular weight: C-12 kg/kmol H-1 kg/kmol 0 - 16 kg/kmol Select one: O a. 2.300 O b. 2300.209 O c. 1.725 O d. 0.009
Answer:
2300j
Explanation:
How do the viscosities of liquids and low-density gases depend on the temperature and pres- sure?
Answer:
Effect of Temperature and Pressure on Viscosity
Case I Liquids
Viscosity of liquids decreases with the increase in temperature.
This is due to decrease in cohesive forces in case of liquids with the increase in temperature.
Viscosity of liquids increases with the increase in pressure.
This is due to the increase in compression due to rise in pressure.
Case II Gases
Viscosity of gases increases with the increase in temperature.
This is due to the increase in the number of collisions with the increase in temperature.
There is no considerable change in the viscosity of gases with the increase in pressure.
why is it that when you use an expression for work you don't use the -VdP but when you use an expression for dH you use PdV and VdP in it?
Answer:
Work is given by -PdV.
It clearly means that when volume of the system changes, work is done either on the system or by the system depending upon volume change.
So, pressure change term -VdP is not included because it is indicating a constant volume.
But in case of enthalpy,
H = U + PV
So, for enthalpy change we can write
dH = dU + PdV + VdP
Because enthalpy change is the sum of all energy changes in the system that includes internal energy, P-V work and effect of pressure change.
Final answer:
The expression for work in thermodynamics is typically shown as -pdV to denote work done by the system, and is path-dependent, while the enthalpy change includes PdV and VdP terms to represent heat and work interactions at constant pressure, since enthalpy is a state function.
Explanation:
The question relates to why the expression for work in thermodynamics is given as work done by the system (-pdV) rather than the work done on the system, and why in the expression for enthalpy change (dH) both PdV and VdP terms appear. In thermodynamics, work done by a system during expansion is often expressed as W = PextΔV, where Pext is the external pressure and ΔV is the change in volume. The negative sign in -pdV signifies that the system does work on its surroundings, thus losing energy. This convention is such that energy leaving the system is negative, aligning with the conventional definition of heat flow being negative when it leaves the system.
Enthalpy (H) is a state function, which unlike work or heat, is not path-dependent. The differential change in enthalpy, dH, for a small, reversible process is given by dH = dU + PdV + VdP, involving both pressure-volume work and the work done on the system due to pressure changes at constant volume. This dH formula accurately describes the total heat and work interactions of a system at constant pressure. The inclusion of VdP in the expression accounts for the work involved when there is a change in pressure at constant volume.
It is also important to note that the exact expression for work can vary depending on the specific thermodynamic process, such as isothermal (PV=nRT) or adiabatic processes and can become complex when accounting for non-ideal behaviors of gases, hence the significance of defining a when and b to zero for an ideal gas equation.
Ethyl chloride vapor decomposes by the first-order reaction C2H5Cl → C2H4 + HCl The activation energy is 249 kJ/mol and the frequency factor is 1.60 × 1014 s−1. Find the value of the specific rate constant at 896 K . Enter your answer numerically (to 4 decimal places) and in terms of the appropriate units for a first order reaction.
Answer : The value of rate constant is, [tex]0.3607s^{-1}[/tex]
Explanation :
The Arrhenius equation is written as:
[tex]K=A\times e^{\frac{-Ea}{RT}}[/tex]
Taking logarithm on both the sides, we get:
[tex]\ln k=-\frac{Ea}{RT}+\ln A[/tex] ............(1)
where,
k = rate constant
Ea = activation energy = 249 kJ/mol = 249000 kJ/mol
T = temperature = 896 K
R = gas constant = 8.314 J/K.mole
A = pre-exponential factor or frequency factor = [tex]1.60\times 10^{14}s^{-1}[/tex]
Now we have to calculate the value of rate constant by putting the given values in equation 1, we get:
[tex]\ln k=-\frac{249000J/mol}{8.314J/K.mol\times 896K}+\ln (1.60\times 10^{14}s^{-1})[/tex]
[tex]\ln k=-1.0198[/tex]
[tex]k=0.3607s^{-1}[/tex]
Therefore, the value of rate constant is, [tex]0.3607s^{-1}[/tex]
Ethyl chloride vapor decomposes by the first-order reaction. Given the activation energy is 249 kJ/mol and the frequency factor is 1.60 × 10¹⁴ s⁻¹, the rate constant at 896 K is 0.4870 s⁻¹.
What is a first-order reaction?A first-order reaction is a chemical reaction in which the rate of reaction is directly proportional to the concentration of the reacting substance.
Let's consider the first-order reaction for the decomposition of ethyl chloride.
C₂H₅Cl → C₂H₄ + HCl
The activation energy is 249 kJ/mol and the frequency factor is 1.60 × 10¹⁴ s⁻¹. We can find the value of the rate constant at 896 K using the Arrhenius equation.
[tex]k = A \times e^{-Ea/R \times T} \\\\k = (1.60 \times 10^{14}s^{-1} ) \times e^{-(249 \times 10^{3} J/mol)/(8.314 J/mol.K) \times 896K} = 0.4870 s^{-1}[/tex]
where,
k is the rate constant.A is the frequency factor.Ea is the activation energy.R is the ideal gas constant.T is the absolute temperature.Ethyl chloride vapor decomposes by the first-order reaction. Given the activation energy is 249 kJ/mol and the frequency factor is 1.60 × 10¹⁴ s⁻¹, the rate constant at 896 K is 0.4870 s⁻¹.
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A pharmacist received a prescription for 100 mL of a cefuroxime axetil suspension to contain 300 mg of drug in each 5 mL. the pharmacist has 100 mL of a suspension containing 250 mg/5 mL and also has 250-mg scored tablets of the drug. How many tablets should be pulverized and added to the suspension to achieve the desired strength? a. 1 tablet b. 2 tablets C. 3 tablets d. 4 tablets
Answer:
Option d. 4 tablets
Explanation:
Let's analyze what the pharmacist can offer: a 100 mL suspension with a concentration of 250 mg/5mL of the prescribed drug. We need to calculate the amount of drug that we have in 100 mL of suspension:
5 mL of suspension ----- 250 mg of drug
100 mL of suspension ---- x = 5000 mg of drug
Now, we take a look at what was really prescribed: a 100 mL suspension with a concentration of 300 mg/5mL of the prescribed drug. Again, we calculate the quantity of drug present in 100 mL of suspension:
5 mL of suspension ----- 300 mg of drug
100 mL of suspension ---- x = 6000 mg of drug
So, there's a difference of 1000 mg of drug per 100 mL of suspension, between what was prescribed by the doctor and what the pharmacist can offer. Therefore, considering that the tablets of the same drugs contain 250 mg of it, we would need to pulverize 4 tablets (4 × 250 mg) and add it to the 250mg/5mL of suspension to reach de prescribed concentration of cefuroxime axetil.
To achieve the desired strength of 300 mg in each 5 mL of cefuroxime axetil suspension, 4 tablets should be pulverized and added to the suspension.
Explanation:To achieve the desired strength of 300 mg of drug in each 5 mL of cefuroxime axetil suspension, the pharmacist can use the 250 mg/5 mL suspension and the 250-mg scored tablets of the drug. Here is how to calculate how many tablets should be added:
Find the concentration of the tablets by dividing 250 mg by 5 mL to get 50 mg/mL.Determine the amount of drug needed by multiplying 100 mL by 300 mg/5 mL to get 6000 mg.Calculate the volume of the tablets that will yield 6000 mg of drug by dividing 6000 mg by 50 mg/mL to get 120 mL.Since the pharmacist already has 100 mL of the 250 mg/5 mL suspension, 20 mL of the suspension needs to be substituted with the tablets.Convert 20 mL to amount of tablets by dividing 20 mL by 5 mL per tablet to get 4 tablets.Therefore, the pharmacist should pulverize and add 4 tablets to the suspension to achieve the desired strength.
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An air tight freezer measures 4 mx 5 m x 2.5 m high. With the door open, it fills with 22 °C air at 1 atm pressure.
a. Calculate the density of this air in kg/m3
b. After closing the door it is cooled down to 7 °C. How low will the pressure in the freezer
be in units of Pa and mmHg?
C. How many newtons of force will be needed to open the 1 m x 2 m door?
Answer:
(a) Density of the air = 1.204 kg/m3
(b) Pressure = 93772 Pa or 0.703 mmHg
(c) Force needed to open the door = 15106 N
Explanation:
(a) The density of the air at 22 deg C and 1 atm can be calculated using the Ideal Gas Law:
[tex]\rho_{air}=\frac{P}{R*T}[/tex]
First, we change the units of P to Pa and T to deg K:
[tex]P=1 atm * \frac{101,325Pa}{1atm}=101,325 Pa\\\\ T=22+273.15=293.15^{\circ}K[/tex]
Then we have
[tex]\rho_{air}=\frac{P}{R*T}=\frac{101325Pa}{287.05 J/(kg*K)*293.15K} =1.204 \frac{kg}{m3}[/tex]
(b) To calculate the change in pressure, we use again the Ideal Gas law, expressed in another way:
[tex]PV=nRT\\P/T=nR/V=constant\Rightarrow P_{1}/T_{1}=P_{2}/T_{2}\\\\P_{2}=P_{1}*\frac{T_{2}}{T_{1}}=101325Pa*\frac{7+273.15}{22+273.15}=101,325Pa*0.9254=93,772Pa\\\\P2=93,772 Pa*\frac{1mmHg}{133,322Pa}= 0.703 mmHg[/tex]
(c) To calculate the force needed to open we have to multiply the difference of pressure between the inside of the freezer and the outside and the surface of the door. We also take into account that Pa = N/m2.
[tex]F=S_{door}*\Delta P=2m^{2} *(101325Pa - 93772Pa)=2m^{2} *7553N/m2=15106N[/tex]
Deep sea divers use a mixture of helium and oxygen to breathe. Assume that a diver is going to a depth of 150 feet where the total pressure is 4.42 atm. The partial pressure of oxygen at this depth is to be maintained at 0.20 atm, the same as at sea level. What must be the percent by volume of oxygen in the gas mixture?
Answer:
4.525% is the percentage by volume of oxygen in the gas mixture.
Explanation:
Total pressure of the mixture = p = 4.42 atm
Partial pressure of the oxygen = [tex]p_1=0.20 atm[/tex]
Partial pressure of the helium = [tex]p_2[/tex]
[tex]p_1=p\times \chi_1[/tex] (Dalton law of partial pressure)
[tex]0.20 atm=4.42 atm\times \chi_1[/tex]
[tex]\chi_1=\frac{0.20 atm}{4.42 atm}=0.04525[/tex]
[tex]\chi_2=1-\chi_1=1-0.04525=0.95475[/tex]
[tex]chi_1+chi_2=1[/tex]
[tex]n_1=0.04525 mol,n_2=0.95475 mol[/tex]
According Avogadro law:
[tex]Moles\propto Volume[/tex] (At temperature and pressure)
Volume occupied by oxygen gas =[tex]V_1[/tex]
Total moles of gases = n = 1 mol
Total Volume of the gases = V
[tex]\frac{n_1}{V_1}=\frac{n}{V}[/tex]
[tex]\frac{V_1}{V}=\frac{n_1}{n}=\frac{0.04525 mol}{1 mol}[/tex]
Percent by volume of oxygen in the gas mixture:
[tex]\frac{V_1}{V}\times 100=\frac{0.04525 mol}{1 mol}\times 100=4.525\%[/tex]
How many milliliters of 2.19 M H2SO4 are required to react with 4.75 g of solid containing 21.6 wt% Ba(NO3)2 if the reaction is Ba2+ + SO42- → BaSO4(s)? x mL
Answer:
1.7927 mL
Explanation:
The mass of solid taken = 4.75 g
This solid contains 21.6 wt% [tex]Ba(NO_3)_2[/tex], thus,
Mass of [tex]Ba(NO_3)_2[/tex] = [tex]\frac {21.6}{100}\times 4.75\ g[/tex] = 1.026 g
Molar mass of [tex]Ba(NO_3)_2[/tex] = 261.337 g/mol
The formula for the calculation of moles is shown below:
[tex]moles = \frac{Mass\ taken}{Molar\ mass}[/tex]
Thus,
[tex]Moles= \frac{1.026\ g}{261.337\ g/mol}[/tex]
[tex]Moles= 0.003926\ mol[/tex]
Considering the reaction as:
[tex]Ba(NO_3)_2+H_2SO_4\rightarrow BaSO_4+2HNO_3[/tex]
1 moles of [tex]Ba(NO_3)_2[/tex] react with 1 mole of [tex]H_2SO_4[/tex]
Thus,
0.003926 mole of [tex]Ba(NO_3)_2[/tex] react with 0.003926 mole of [tex]H_2SO_4[/tex]
Moles of [tex]H_2SO_4[/tex] = 0.003926 mole
Also, considering:
[tex]Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}[/tex]
Molarity = 2.19 M
So,
[tex]2.19=\frac{0.003926}{Volume\ of\ the\ solution(L)}[/tex]
Volume = 0.0017927 L
Also, 1 L = 1000 mL
So, volume = 1.7927 mL
As a technician in a large pharmaceutical research firm, you need to produce 250. mL of a potassium dihydrogen phosphate buffer solution of pH = 7.05. The pKa of H2PO4− is 7.21. You have the following supplies: 2.00 L of 1.00 M KH2PO4 stock solution, 1.50 L of 1.00 M K2HPO4 stock solution, and a carboy of pure distilled H2O. How much 1.00 M KH2PO4 will you need to make this solution? (Assume additive volumes.)
Answer:
We will need 147.772 mL of KH2PO4 to make this solution
Explanation:
For this case we can give the following equation:
H2PO4 - ⇄ H+ + HPO42-
With following pH- equation:
pH = pKa + log [HPO42-]/[H2PO4-]
7.05 = 7.21 + log [HPO42-]/[H2PO4-]
-0.16 = log [HPO42-]/[H2PO4-]
10^-0.16 = [HPO42-]/[H2PO4-]
0.6918 = [HPO42-]/[H2PO4-]
Let's say the volume of HPO42-= x then the volume of H2PO4- will be 250 mL - x
Since both have a concentration of 1M = 1 mol /L
If we plug this in the equation 0.6918 = [HPO42-]/[H2PO4-]
0.6918 = x / (250 - x)
0.6918*250 - 0.6918x = x
172.95 = 1.6918x
x = 102.228 mL
The volume of HPO42- = 102.228 mL
Then the volume of H2PO4- = 250 - 102.228 = 147.772 mL
To control this we can plug this in the pH equation
7.05 = 7.21 + log [HPO42-]/[H2PO4-]
7.05= 7.21 + log (102.228 / 147.772) = 7.05
We will need 147.772 mL of KH2PO4 to make this solution
To produce the buffer solution, use the Henderson-Hasselbalch equation to determine the ratio of conjugate base to acid, ensuring that the concentration of KH2PO4 is higher than K2HPO4, and calculate the necessary volumes of 1.00 M stock solutions.
Explanation:To prepare a potassium dihydrogen phosphate buffer solution with a pH of 7.05, when the pKa of H2PO4− is 7.21, the Henderson-Hasselbalch equation can be used, which is pH = pKa + log([A−]/[HA]), where [A−] is the concentration of the conjugate base and [HA] is the concentration of the acid. Since the desired pH is slightly less than the pKa, you'll need a bit more acid (KH2PO4) than the base (K2HPO4). To find the exact amounts, set up the equation like this: 7.05 = 7.21 + log([K2HPO4]/[KH2PO4]). Solving for the ratio [K2HPO4]/[KH2PO4], we get a value less than 1, which indicates that the concentration of KH2PO4 must be higher than that of K2HPO4 in our buffer solution. Then, you can calculate the volume of each stock solution needed to reach the final total volume of 250 mL, remembering the molarities are both 1.00 M and that volumes are additive.
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According to the following reaction, how many grams of chloric acid (HClO3) are produced in the complete reaction of 31.6 grams of chlorine gas? 3 C12(g) + 3 H2O(1) ► 5 HCl(aq) + HCO3(aq) grams HC103
Answer:
[tex]m_{HClO_3}=12.7gHClO_3[/tex]
Explanation:
Hello,
Considering the reaction:
[tex]3Cl_2(g)+3H_2O(l)-->5HCl+HClO_3[/tex]
The molar masses of chlorine and chloric acid are:
[tex]M_{Cl_2}=35.45*2=70.9g/mol\\M_{HClO_3}=1+35.45+16*3=84.45g/mol[/tex]
Now, we develop the stoichiometric relationship to find the mass of chloric acid, considering the molar ratio 3:1 between chlorine and chloric acid, as follows:
[tex]m_{HClO_3}=31.6gCl_2*\frac{1molCl_2}{70.9gCl_2} *\frac{1molHClO_3}{3mol Cl_2} *\frac{85.45g HClO_3}{1mol HClO_3} \\m_{HClO_3}=12.7gHClO_3[/tex]
Best regards.
How many grams of NaCl (molecular weight = 58 Da) would you have to dissolve in 50ml to make a 2M stock solution? Answer:
Answer:
5.8 g
Explanation:
Molecular weight in Daltons is equivalent to the molecular weight in grams per mole.
The amount of NaCl required is calculated as follows:
(2 mol/L)(50 mL)(1 L/1000 mL) = 0.1 mol
This amount is converted to grams using the molar mass (58 g/mol).
(0.1 mol)(58 g/mol) = 5.8 g
Final answer:
To prepare a 2M stock solution of NaCl, dissolve 5.8 grams of NaCl in 50ml of water, using the molar mass of NaCl which is 58 g/mol.
Explanation:
To make a 2M stock solution of NaCl, you would need to dissolve the number of grams equivalent to 2 moles of NaCl in 50ml of water. Since the molecular weight of NaCl is 58 Da (or 58 g/mol), we calculate the mass as follows:
Calculate the number of moles needed: 2 moles/L × 0.050 L = 0.1 moles.Calculate the mass of NaCl required for 0.1 moles: 0.1 moles × 58 g/mol = 5.8 grams of NaCl.Therefore, to prepare a 2M solution, you would dissolve 5.8 grams of NaCl in 50ml of water.
A Chemist measured 184 mL using a graduated cylinder. Based on her experience, the volume did not seem to be correct. This volume was again measured using a calibrated standard device, from which it was determined that the volume of the liquid was actually 230.0 mL. What is the % error in this measurement?
Answer:
Percent error = 20%
Explanation:
The percent error is calculated using the following equation:
Percent error = |(approximate value - exact value)| / (exact value) x 100%
In this problem, the approximate value was 184 mL and the exact value was 230.0 mL
Percent error = |(184 mL - 230.0 mL)| / (230.0 mL) x 100% = 20%
Final answer:
The percent error of the chemist's measurement is 20%, which reflects the difference between the measured value using a graduated cylinder and the actual value obtained using a calibrated device.
Explanation:
The percent error in measurement can be determined using the formula: percent error = (|actual value - measured value| / actual value) × 100%. In this case, the actual volume is 230.0 mL, while the measured volume is 184 mL.
To calculate the percent error: percent error = (|230.0 mL - 184 mL| / 230.0 mL) × 100% = (46 mL / 230.0 mL) × 100% = 20%.
So, the percent error of the chemist's measurement is 20%. This value indicates how far the initial measurement was from the actual value, reflecting the importance of using calibrated instruments for accurate measurement. The example of the quality control chemist shows the significance of both precision and accuracy in chemical measurements.