A sample of oxygen gas is compressed from 30.6 L to 1.8 L at constant temperature pressure of 1.8 atm. Calculate the amount of energy in joules when the system releases 1.5 KJ of heat?
Answer:
the change in the internal energy of the system is 3,752.67 J
Explanation:
Given;
initial volume of the gas, V₁ = 30.6 L
final volume of the gas, V₂ = 1.8 L
constant pressure of the gas, P = 1.8 atm
Energy released by the system, Q = 1.5 kJ = 1,500 J
Apply pressure-volume work equation, to determine the work done on the gas;
w = -PΔV
w = -P(V₂ - V₁)
w = - 1.8 atm(1.8 L - 30.6 L)
w = 51.84 L.atm
w = 51.84 L.atm x 101.325 J/L.atm
w = 5,252.67 J
The change in the internal energy of the system is calculated as;
ΔU = Q + w
Since the heat is given out, Q = - 1,500 J
ΔU = -1,500 J + 5,252.67 J
ΔU = 3,752.67 J
Therefore, the change in the internal energy of the system is 3,752.67 J
Given 0.60 mol CO2, 0.30 mol CO, and 0.10 mol H20, what is the partial pressure of the CO if the total pressure of the mixture was 0.80 atm?
Answer:
Explanation:
/ means divided by
* means multiply
1. formula is
partial pressure = no of moles(gas 1)/ no of moles(total)
0.30 mol CO/0.60 mol CO2 + 0.30 mol CO + 0.10 mol H20 ->
.3/(.6+.3+.1) =
.3/1 =
.3 =
partial pressure of CO
2.
.3 * .8 atm = .24
khanacademy
quizlet
The partial pressure of the CO is 0.24 atm if the total pressure of the mixture was 0.80 atm.
Dalton's Law of Partial pressureDalton's Law of partial pressure states that the total pressure exerted by non reacting gaseous mixture at a constant temperature and given volume is equal to the sum of partial pressure of all gases.
Dalton's Law of partial pressure using mole fraction of gas
Partial pressure of carbon monoxide (CO) = Mole fraction of carbon monoxide (CO) × Total pressure
Now, we have to find the first mole fraction of CO
Mole fraction of carbon monoxide (CO) = [tex]\frac{\text{moles of solute}}{\text{total moles of solute}}[/tex]
= [tex]\frac{\text{moles of CO}}{\text{moles of CO}_2 + \text{moles of CO} + \text{moles of H}_{2}O}[/tex]
= [tex]\frac{0.30}{0.60 + 0.30 + 0.10}[/tex]
= [tex]\frac{0.30}{1}[/tex]
= 0.3
Now, put the value in above equation, we get that
Partial pressure of carbon monoxide (CO)
= Mole fraction of carbon monoxide (CO) × Total pressure
= 0.3 × 0.8
= 0.24 atm
Thus, the partial pressure of the CO is 0.24 atm is the total pressure of the mixture was 0.80 atm.
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What would be the specific mathematical effect on the reaction rate if you carried out the sodium iodide-in-acetone reactions on the alkyl halides using an iodide solution half as concentrated? ("Slower" or "faster" is not specific enough.)
Answer:
Slower
Explanation:
The reaction between alkyl halides and sodium iodide-in-acetone is an SN2 reaction. The rate of reaction depends on the concentration of the alkyl halide as well as the concentration of the sodium iodide. It is a bimolecular reaction.
This means that if the concentration of any of the reactants is halved, the rate of reaction decreases accordingly.
Therefore, if the iodide solution is half as concentrated, the reaction is observed to be slower in accordance with the rate law;
Rate = k[alkyl halide] [iodide]
A gas mixture is made by combining 8.7 g each of Ar, Ne, and an unknown diatomic gas. At STP, the mixture occupies a volume of 17.28 L. What is the molar mass of the unknown gas
Answer: Molar mass of the unknown gas is 73.153 g/mol.
Explanation:
Given: Mass of each gas = 8.7 g
Volume = 17.28 L
Let us assume that the molar mass of gas is m g/mol.
Molar mass of Ar is 40 g/mol and Ne is 20 g/mol.
Hence, total moles of each gas are as follows.
[tex](\frac{8.7}{40} + \frac{8.7}{20} + \frac{8.7}{m}) mol[/tex]
At STP, the total volume of these gases is as follows.
[tex](\frac{8.7}{40} + \frac{8.7}{20} + \frac{8.7}{m}) mol \times 22.4 L = 17.28 L\\(\frac{8.7}{40} + \frac{8.7}{20})22.4 L + \frac{8.7}{m} \times 22.4 L = 17.28 L\\14.616 + \frac{8.7}{m} \times 22.4 L = 17.28 L\\\frac{8.7}{m} \times 22.4 L = (17.28 L - 14.616)\\\frac{8.7}{m} \times 22.4 L = 2.664 \\m = 73.153 g/mol[/tex]
Thus, we can conclude that molar mass of the unknown gas is 73.153 g/mol.
Liquid octane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 10.3 g of octane is mixed with 23. g of oxygen. Calculate the maximum mass of water that could be produced by the chemical reaction. Round your answer to significant digits.
Answer:
9.36 g
Explanation:
The equation of the reaction is;
C8H18(g) + 25/2 O2(g) ----> 8CO2(g) + 9H2O(g)
Number of moles of octane = 10.3g/ 114 g/mol = 0.09 moles
1 mole of octane yields 9 moles of water
0.09 moles of octane yields 0.09 × 9/1 = 0.81 moles of water
Number of moles of oxygen = 23g/32g/mol = 0.72 moles
12.5 moles of oxygen yields 9 moles of water
0.72 moles of oxygen yields 0.72 × 9/12.5 = 0.52 moles of water
Hence oxygen is the limiting reactant;
Maximum mass of water produced = 0.52 moles of water × 18 g/mol = 9.36 g
Me please answer as follows
Answer:
no reaction occurs .that is no product
Based upon the intermolecular forces present, rank the following substances according to the expected boiling point for the substance.
a. HCl
b. NaCl
c. N2
d. H2O
1.rain pours from the sky
2.leaves of the plant dried
3.fluffy clouds form in the sky
4.bathing suit dries after swim
5.water puddles disappear
A.Evaporation
B.Condensation
C.Precipitation
D.Transpiration
Yan po pag pipilian
Answer:
1.Precipitation
2.Transpiration
3.Condensation
4.Evaporation
5.Evaporation
3.Condensation
Explanation:
Rain pours from the sky occurs due to the process of precipitation, leaves of the plant dried due to the process of transpiration in which the water is evaporated from the body of plant, fluffy clouds form in the sky occurs in the process of condensation, bathing suit dries after swim is due to evaporation in which water is removed and goes into the atmosphere and water puddles disappear due to the process of evaporation. Evaporation is the removal of water from the any surface whereas transpiration is the removal of water from plant body parts.
State two conditions necessary for an esterification reaction to take place
Explanation:
Esterification occurs when a carboxylic acid reacts with an alcohol. This reaction can only occur in the presence of an acid catalyst and heat. It takes a lot of energy to remove the -OH from the carboxylic acid, so a catalyst and heat are needed to produce the necessary energy.
Answer:
The Esterification ProcessThe Esterification ProcessEsterification occurs when a carboxylic acid reacts with an alcohol. This reaction can only occur in the presence of an acid catalyst and heat. It takes a lot of energy to remove the -OH from the carboxylic acid, so a catalyst and heat are needed to produce the necessary energy.
The Esterification ProcessEsterification occurs when a carboxylic acid reacts with an alcohol. This reaction can only occur in the presence of an acid catalyst and heat. It takes a lot of energy to remove the -OH from the carboxylic acid, so a catalyst and heat are needed to produce the necessary energy.Once the -OH has been removed, the hydrogen on the alcohol can be removed and that oxygen can be connected to the carbon. Because the oxygen was already connected to a carbon, it is now connected to a carbon on both sides, and an ester is formed.
The Esterification ProcessEsterification occurs when a carboxylic acid reacts with an alcohol. This reaction can only occur in the presence of an acid catalyst and heat. It takes a lot of energy to remove the -OH from the carboxylic acid, so a catalyst and heat are needed to produce the necessary energy.Once the -OH has been removed, the hydrogen on the alcohol can be removed and that oxygen can be connected to the carbon. Because the oxygen was already connected to a carbon, it is now connected to a carbon on both sides, and an ester is formed.The methyl acetate that was formed is an ester. In this image, the green circle represents what was the carboxylic acid (in this case acetic acid), and the red circle represents what was the alcohol (in this case methanol):
This reaction lost an -OH from the carboxylic acid and a hydrogen from the alcohol. These two also combine to form water. So any esterification reaction will also form water as a side product.
One of the purposes of this lab is to determine the order of the reaction with respect to the Allura Red dye by creating first and second-order graphs for all four trials. The correct order of the reaction is the one where the slopes of the graphs for the four trials are roughly the same. Why is this important when choosing the order of the reaction
Answer: Hello the options related to your question are attached below
The slope is related to the rate constant so all four trials should have the same slope since the reactions are all the same ( Option C )
Explanation:
It is important when choosing the order of the reaction because the concentration of the bleaches used in the four trials are in excess hence their slopes have to be roughly the same and also because the reactions are similar and they where done at the same temperature, hence the slope of the first and second-order graphs will be the same.
If 12.3 g of Cu is deposited at the cathode of an electrolytic cell after 5.50 h, what was the current used?
Answer:
1.88 A
Explanation:
Let's consider the reduction of copper in an electrolytic cell.
Cu²⁺ + 2 e⁻ ⇒ Cu
We can calculate the charge used to deposit 12.3 g of Cu using the following relations.
The molar mass of Cu is 63.55 g/mol.1 mole of Cu is deposited when 2 moles of electrons circulate.1 mole of electrons has a charge of 96486 C (Faraday's constant).The charge used is:
[tex]12.3 g \times \frac{1 molCu}{63.55gCu} \times \frac{2molElectron}{1molCu} \times \frac{96486C}{1molElectron} = 3.73 \times 10^{4} C[/tex]
We can convert 5.50 h to seconds using the conversion factor 1 h = 3600 s.
5.50 h × 3600 s/1 h = 1.98 × 10⁴ s
The current used is:
I = q/t = 3.73 × 10⁴ C/1.98 × 10⁴ s = 1.88 A
4. What is the percent yield of a reaction that produces 12.5 g CF2Cl2 from 32.9 g of CCl4 and excess HF
Answer:
Percent yield = 48.3%
Explanation:
The reaction is:
CCl₄ + 2HF → CF₂Cl₂ + 2HCl
1 mol of CCl₄ reacts with 2 moles of hydrofluoric acid in order to produce 1 mol of CF₂Cl₂ and 2 moles of hydrogen chloride.
HF is in excess, so the limiting reagent is the CCl₄.
We convert mass to moles:
32.9 g . 1mol / 153.8g = 0.214 moles
Ratio is 1:1. In conclussion: 0.0813 moles of CCl₄ can produce 0.0813 moles of CF₂Cl₂. We convert moles to mass, to determine the theoretical yield:
0.214 mol . 120.91g /mol = 25.8 g
Percent yield = (Yield produced /Theoretical yield) . 100
Percent yield = (12.5 g/ 25.8g) . 100 = 48.3%
Which of the following is a physical change?
True or false: Boron contains 2s22p1 valence electrons, so only one p orbital is needed to form molecular orbitals.
Answer:
True
Explanation:
The valence orbitals of boron are 2s2 2p1. We have to recall that all the valence orbitals whether full or empty are involved in the formation of molecular orbitals.
The number of molecular orbitals formed is equal to the number of atomic orbitals that are combined.
Since there are two valence orbitals and there is only one p orbital among the valence orbitals, it is true that only one p orbital is needed to form molecular orbitals in boron.
What are the uses of Sulphuric acid?
Answer:
The major use of sulfuric acid is in the production of fertilizers, e.g., superphosphate of lime and ammonium sulfate. It is widely used in the manufacture of chemicals, e.g., in making hydrochloric acid, nitric acid, sulfate salts, synthetic detergents, dyes and pigments, explosives, and drugs.
The major use of sulfuric acid is in the production of fertilizers, e.g., superphosphate of lime and ammonium sulfate. It is widely used in the manufacture of chemicals, e.g., in making hydrochloric acid, nitric acid, sulfate salts, synthetic detergents, dyes and pigments, explosives, and drugs.
Which equation obeys the law of conservation of
mass?
Answer:2C4H10+2C12+12O2 4CO2+CC14+H20
100.0 mL of a 0.780 M solution of KBr is diluted to 500.0 mL. What is the new concentration of the solution?
o Calculate the pH of 500 ml of buffer solution containing .2M of ammonium sulphate and 0.3M ammonium hydroxide which 5 2.1% ionized in dilute solution. (kb of ammonium hydroxide is 1.8×10^-5)
Answer is 9
pKb=−logK
b=−log10^-5=5
A mixture of ammonium hydroxide and ammonium chloride forms a basic buffer solution.Henderson's equation for the basic buffer solution is as given below-pOH=pKb+log[acid]*[salt]
Substitute values in the above expression.
pOH=5+log0.1*0.1=5
Hence, the pH of the solution is pH=14−pOH=14−5=9
What do you mean by pH ?potential of hydrogena measure of the acidity or alkalinity of a solution equal to the common logarithm of the reciprocal of the concentration of hydrogen ions in moles per cubic decimetre of solution. Pure water has a pH of 7, acid solutions have a pH of less than 7, and alkaline solution with a pH greater than 7.
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A quantity of 1.435 g of naphthalene , was burned in a constant-volume bomb calorimeter. Consequently, the temperature of the water rose from 20.28oC to 25.95oC If the heat capacity of the bomb plus water was , calculate the heat of combustion of naphthalene on a molar basis; that is, find the molar heat of combustion.
Answer:
molar heat of combustion = -5156 *10³ kJ/mol
Explanation:
A quantity of 1.435 g of naphthalene , was burned in a constant-volume bomb calorimeter. Consequently, the temperature of the water rose from 20.28oC to 25.95oC If the heat capacity of the bomb plus water was 10.17 kJ/°C, calculate the heat of combustion of naphthalene on a molar basis; that is, find the molar heat of combustion.
Step 1: Data given
Mass of naphthalene = 1.435 grams
Initial temperature of water = 20.28 °C
Final temperature of water = 25.95 °C
heat capacity of the bomb plus water was 10.17 kJ/°C
Molar mass naphtalene = 128.2 g/mol
Step 2:
Qcal = Ccal * ΔT
⇒with Qcal =the heat of combustion
⇒with Ccal = heat capacity of the bomb plus water = 10.17 kJ/°C
⇒with ΔT = the difference in temperature = T2 - T1 = 25.95 - 20.28 = 5.67°C
Qcal = 10.17 kJ/°C * 5.67 °C
Qcal = 57.7 kJ
Step 3: Calculate moles
Moles naphthalene = 1.435 grams / 128.2 g/mol
Moles naphthalene = 0.01119 moles
Step 4: Calculate the molar heat of combustion
molar heat of combustion = Qcal/ moles
molar heat of combustion = -57.7 kJ/ 0.01119 moles
molar heat of combustion = -5156 *10³ kJ/mol
Write the number of sig. fig. in four numbers given in the sentence below. An (one) octopus has 8 legs. 13 octopi have 104 legs.
Give four answers.
A. Infinity, Infinity, Infinity, Infinity
B. 1, 1, 2, 3
C. Infinity, Infinity, 2, 3
D. No answer text provided.
Answer:
1, 1, 2, 3
Explanation:
The numbers 1 and 8 both have 1 sig. fig.
The number 13 has 2 sig. figs.
The number 104 has 3 sig. figs.
En la fermentación del alcohol, la levadura convierte la glucosa en etanol y dióxido de carbono:
C6H12O6(s) → 2C2H5OH(l) + 2CO2(g)
Si reaccionan 5.97 g de glucosa y se recolectan 1.44 L de CO2 gaseoso, a 293 K y 0.984 atm, ¿cuál
es el rendimiento porcentual de la reacción
Answer:
88.9%
Explanation:
Primero convertimos 5.97 g de glucosa a moles, usando su masa molar:
5.97 g ÷ 180 g/mol = 0.0332 molDespués calculamos la cantidad máxima de moles de CO₂ que se hubieran podido producir:
0.0332 mol C₆H₁₂O₆ * [tex]\frac{2molCO_2}{1molC_6H_{12}O_6}[/tex] = 0.0664 mol CO₂Ahora calculamos los moles de CO₂ producidos, usando los datos de recolección dados y la ecuación PV=nRT:
0.984 atm * 1.44 L = n * 0.082 atm·L·mol⁻¹·K⁻¹ * 293 Kn = 0.0590 molFinalmente calculamos el rendimiento porcentual:
0.0590 mol / 0.0664 mol * 100% = 88.9%Indicate how the concentration of each species in the chemical equation will change to reestablish equilibrium after reactant or product is added.
2CO(g) + O2(g) ⇌ 2CO2
Answer:
Indicate how the concentration of each species in the chemical equation will change to reestablish equilibrium after reactant or product is added.
[tex]2CO(g) + O2(g) <=> 2CO2[/tex]
Explanation:
When the reactants concentration increases, then the equilibrium will shift towards products and when the concentration of products increases, then equilibrium will shift towards reactants.
So, increases in concentration of carbon monoxide (CO) shifts the equilibrium to favor the formation of carbondioxide.
Similarly increase in concentration of oxygen also favor the formation of product carbon dioxide.
Increase in concentration of CO2 favors the formation of CO and O2.
Decrease in product concentration also favors the formation of product.
Decrease in reactant concentration favors the formation of reactants only.
You are asked to prepare a buffer solution with a pH of 3.50. The following solutions, all 0.100 M, are available to you: HCOOH, CH3COOH, H3PO4 , NaCHOO, NaCH3COO, and NaH2PO4. What would be the best combination to make the required buffer solution? Select one:
a. NaH2PO4 and NaCHOO
b. H3PO4 and NaH2PO4
c. NaH2PO4 and HCOOH
d. CH3COOH and NaCH3COO e. HCOOH and NaCHOO
can someone helo me with this
Answer:
e. HCOOH and NaCHOO
Explanation:
For a buffer solution, both an acid and its conjugate base are required.
With the information above in mind, we can discard options a) and c), as those combinations are not of an acid and its conjugate base.
Now it is a matter of comparing the pKa (found in literature tables) of the acids of the remaining three acids:
H₃PO₄ pKa = 2.12CH₃COOH pKa = 2.8HCOOH pKa = 3.74The acid with the pKa closest to the desired pH is HCOOH, so the correct answer is e. HCOOH and NaCHOO
Suppose you ran this reaction without triethylamine and simply used an excess of reactant 1. At the end of the reaction, your methylene chloride solution would contain mostly reactant 1 and the product. What would you do to remove reactant 1 from the solution
ummm is that chemistry?
Answer:
is this chem
Explanation:
4.005 X 74 X 0.007 = 2.10049
Answer:
2.07459
Explanation:
this is the correct answer.
discuss the benefits of observing good safety measures in relation to increase in productivity within a pharmaceutical laboratory?
Answer:
Pharmaceutical laboratory helps in devloping and conducting research, vaccines. Various kinds of drugs and chemical substances used and are produced at a Pharmaceutical laboratory.
The pharmaceutical laboratories performs with various hazardous substances that results in exposure to various chemicals, biological substances and radiation. To avoid any injury or infection labs need to maintain all safety measures.
Spillage and relaseing chemical substances can be lethal during transportaions by safety measures for heling in for manufacturing of such therapeutic agents spillage and avoid wastage.
Maintaining good safety standards in the pharmaceuticals laboratory will help promote the health of technicians and workers which in turn will increase productivity and attain positive outcomes.
4.106
Calculate the moles and the mass of solute in each of the following solutions.
(a) 150.0 mL of 0.245 M CaCl2
molarity = moles of solute / volume of solution
moles of solute = molarity × volume of solution
moles of solute = 0.245 mol/L × 0.1500 L
moles of solute = 0.03675 mol
moles of solute = 0.0368 mol
-----------------------------------------------------------
Solution: (mass of solute)Step 1: Calculate the molar mass of solute.
molar mass of solute = (40.08 g/mol × 1) + (35.45 g/mol × 2)
molar mass of solute = 110.98 g/mol
Step 2: Calculate the mass of solute.
mass of solute = moles of solute × molar mass of solute
mass of solute = 0.03675 mol × 110.98 g/mol
mass of solute = 4.08 g
Note: The volume of solution must be expressed in liters (L).
Answer:
[tex]\boxed {\sf \bold {0.0368 \ mol \ CaCl_2}}}}[/tex]
[tex]\boxed {\sf \bold {4.08 \ g \ CaCl_2}}}}}[/tex]
Explanation:
1. Moles of SoluteMolarity is a measure of concentration in moles per liter.
[tex]molarity= \frac {moles \ of \ solute}{liters \ of \ solution}[/tex]
In this solution, there are 150.0 milliliters of solution and the molarity is 0.245 M CaCl₂ or 0.245 mol CaCl₂ per liter.
First, convert the milliliters to liters. There are 1000 milliliters in 1 liter.
[tex]{150 \ mL * \frac{1 \ L}{1000 \ mL}= \frac{150}{1000} \ L = 0.150 \ L[/tex]Now, substitute the known values (molarity and liters of solution) into the formula. The moles of solution are unknown, so we can use x.
[tex]0.245 \ mol \ CaCl_2 /L= \frac{ x}{0.150 \ L}[/tex]
We are solving for x, so we must isolate this variable. It is being divided by 0.150 L. The inverse of divisions is multiplication, so we multiply both sides by 0.150 L.
[tex]0.150 \ L *0.245 \ mol \ CaCl_2 /L= \frac{ x}{0.150 \ L} * 0.150 L[/tex]
[tex]0.150 \ L *0.245 \ mol \ CaCl_2 /L=x[/tex]
The units of liters cancel.
[tex]0.150 *0.245 \ mol \ CaCl_2 =x[/tex]
[tex]0.03675 \ mol \ CaCl_2[/tex]
The original measurements have 3 significant figures, so our answer must have the same.
We should round to the ten thousandths place. The 5 to the right of this place tells us to round the 7 up to an 8.
[tex]\bold {0.0368 \ mol \ CaCl_2}[/tex]
2. Mass of the SoluteWe can convert mass to moles using the molar mass. These values are found on the Periodic Table. They are the same as the atomic masses, but the units are grams per mole (g/mol) instead of atomic mass units.
The solute is calcium chloride: CaCl₂. Look up the molar masses of the individual elements.
Ca: 40.08 g/mol Cl: 35.45 g/molNotice that chlorine has a subscript of 2. We must multiply the molar mass by 2.
Cl₂: 35.45 *2= 70.9 g/molAdd calcium's molar mass.
CaCl₂: 40.08 + 70.9 =110.98 g/molUse the molar mass as a ratio.
[tex]\frac {110.98 \ g\ CaCL_2}{ 1 \ mol \ CaCl_2}[/tex]
Multiply the moles of calcium chloride we calculated above.
[tex]0.0368 \ mol \ CaCl_2 *\frac {110.98 \ g\ CaCL_2}{ 1 \ mol \ CaCl_2}[/tex]
The units of moles of calcium chloride cancel.
[tex]0.0368 *\frac {110.98 \ g\ CaCL_2}{ 1 }[/tex]
[tex]4.084064 \ g\ CaCl_2[/tex]
Round to 3 significant figures again. For this number, it is the hundredths place. The 4 in the thousandths place tells us to leave the 8.
[tex]\bold {4.08 \ g \ CaCl_2}[/tex]
refer to pic plssss
Answer:
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Que es la actividad física y en qué mejora
A scientific hypothesis is
ANSWER:
predictive.
testable.
explanatory.
all of the above.
Answer:
All of the above.
Explanation:
For a scientific hypothesis to be considered a hypothesis, it has to be testable. When conducting a lab experiment, it also allows the tester to predict what might occur during and after the experimentation. They are also explanatory. For example, theories are hypotheses that have been verified and can explain why something in nature takes place.
