ANSWER
[tex]undefined[/tex]EXPLANATION
Given that;
The mass of Al2(SO4)3 is 500 grams
The mass of Ca(OH)2 is 450 grams
The mass of CaSO4 is 596 grams
Follow the steps below to find the limiting reactant of the reaction
Step 1; Write the balanced equation of the reaction
[tex]\text{ Al}_2(SO_4)_{3(aq)}+\text{ 3Ca\lparen OH\rparen}_{2(aq)}\text{ }\rightarrow\text{ 2Al\lparen OH\rparen}_{3(s)}\text{ + 3CaSO}_{4(s)}[/tex]In the reaction above, 1 mole Al2(SO4)3 reacts with 3 moles Ca(OH)2 to give 2 moles Al(OH)3 and 3 moles CaSO4
Step 2; Determine the number of moles using the below formula
[tex]\text{ mole = }\frac{\text{ mass}}{\text{ molar mass}}[/tex]Recall, that the molar mass of Al2(SO4)3 is 342.15 g/mol and the molar mass of Ca(OH)2 is 74.093 g/mol
For Al2(SO4)3
[tex]\begin{gathered} \text{ Mole = }\frac{500}{\text{ 342.15}} \\ \text{ Mole = 1.461 moles} \end{gathered}[/tex]For Ca(OH)2
[tex]\begin{gathered} \text{ Mole = }\frac{\text{ 450}}{\text{ 74.093}} \\ \text{ Mole = 6.073 moles} \end{gathered}[/tex]Step 3; Find the limiting reactant of the reaction
To find the limiting reactant of the reaction, divide the moles of the reactant by the co-efficient of the compound
[tex]\begin{gathered} \text{ For Al}_2(SO_4)_3 \\ \text{ The mole ratio = }\frac{1.461}{1} \\ \text{ The mole ratio = 1.461 mol/wt} \\ \\ \text{ For Ca\lparen OH\rparen}_2 \\ \text{ The mole ratio = }\frac{\text{ 6.073}}{3} \\ \text{ The mole ratio = 2.024 mol/wt} \end{gathered}[/tex]Since the limiting reactant of the reaction is the reactant with the lowest number of mol/wt, then Al2(SO4)3 is the limiting reactant
[tex]\text{ The limiting reactant of the reaction is Al}_2(SO_4)_3[/tex]The excess reactant of the reaction is Ca(OH)2
Therefore, the no of moles of the excess reactant that is unreacted is
6.073 - 1.461 = 4.612 moles
Hence, the number of moles of the excess reactant that is unreacted is 4.621 moles
11.A substance that is corrosive and that conducts electricity in water can be classified as which of the following?Select one:a. An acid only.b. A base only.c. Either an acid or a base.d. Neither an acid or a base.
Answer
acids and bases are both corrosive furthermore both will conduct electricity depending on the strength of the acid or base
C is the correct option
What volume of Cl2 gas, measured at 688 torr and 36 °C, is required to form 22 g of NaCl?Express your answer using two significant figures.
ANSWER
Volume of the gas 5.3L
EXPLANATION;
Given that
The pressure of the chlorine gas is 688 torr
The temperature of the chlorine gas is 36 degrees Celcius
The mass of NaCl is 22 grams
Follow the steps below to find the volume of chlorine gas
Step 1; Write a balanced equation for the reaction
[tex]\text{ 2Na + Cl}_2\text{ }\rightarrow\text{ 2NaCl}[/tex]In the above reaction, 2 moles of Na react with 1 mole of chlorine to form 2 moles of NaCl
Step 2; Find the number of moles of NaCl using the below formula
[tex]\text{ mole = }\frac{\text{ mass}}{\text{ molar mass}}[/tex]Recall, that the molar mass of NaCl is 58.44 g/mol
[tex]\begin{gathered} \text{ mole = }\frac{22}{\text{ 58.44}} \\ \text{ moles = 0.376 mole} \end{gathered}[/tex]Step 3; Find the number of moles of Cl2 using a stoichiometry ratio
Let x represents the number of moles of Cl2
[tex]\begin{gathered} 1\text{ mole Cl}_2\text{ }\rightarrow\text{ 2 moles NaCl} \\ \text{ x moles Cl}_2\rightarrow\text{ 0.376 moles NaCl} \\ \text{ cross multiply} \\ \text{ x moles Cl}_2\text{ }\times2\text{ moles NaCl }=\text{ 1 mole Cl}_2\text{ }\times\text{ 0.376 mole NaCl} \\ \text{ Isolate x} \\ \text{ x = }\frac{\text{ 1 mole Cl}_2\times0.376moles\cancel{NaCl}}{2moles\cancel{NaCl}} \\ \\ \text{ x = }\frac{1\text{ }\times\text{ 0.376}}{2} \\ x\text{ = }\frac{0.376}{2} \\ \text{ x = 0.188 mol} \end{gathered}[/tex]The number of moles of Cl2 is 0.188 mol
Step 4; Find the volume of the gas using the ideal gas equation
[tex]\text{ PV = nRT}[/tex]Convert the temperature to degree kelvin
[tex]\begin{gathered} \text{ t = 36}\degree C \\ \text{ T = 36 + 273.15} \\ \text{ T = 309.15 K} \end{gathered}[/tex]Step 5; Find the volume of the gas
[tex]\begin{gathered} \text{ PV = nRT} \\ \text{ Recall, that R = 62.363 L. torr mol}^{-1}K^{-1} \\ \text{ 688}\times\text{ V =}^{\text{ }}0.188\text{ x 62.636 }\times\text{ 309.15} \\ \text{ 688 V = 3624. 500} \\ V\text{ = }\frac{\text{ 3645, 500}}{688} \\ \text{ v = 5.3 L} \end{gathered}[/tex]Therefore, volume of the gas 5.3L
Two different signs that indicate a chemical action has occurred. What time is showing in the picture on the left what sign is showing on the picture in the right?
The reaction that occurs on the left is called a single displacemt reaction
Read each question carefully and choose the letter of the best answer.
This element is nonmetal and appears in the periodic table.
The number that represents the position of the elements is called, the atomic number, it also represents the number of protons = numbers of electrons an atom has.
Answer: D. the atomic number
balanced chemical equation for .1 M barium nitrate and .1 m sodium sulfate
The balanced chemical equation for 0.1 M barium nitrate and 0.1 M sodium sulfate is -
2 Ba(NO₃)₂ + 2 Na₂SO₄ → 2 Ba(SO₄) + 2 NaNO₃
Balanced chemical equation are the equations having equal amount of reactants and the products formed. The reactants are placed on the left and products are placed on right forming a balanced chemical equation.
When 0.1 M barium nitrate reacts with the 0.1 M sodium sulfate, it gives an insoluble salt of barium sulfate and soluble salt of sodium nitrate as the product. This reaction is a type of precipitation reaction as precipitate is formed in the reaction. It is a double displacement reaction because the sulfate ions replace the nitrate ions to give the product.
Barium sulfate is known to be the most insoluble compound. It is formed as a precipitate.
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Calculate the volume of 0.875 mol/dm3 NaOH solution needed to make up a 100.0 ml of a diluted solution with a concentration of 0.235 mol/dm3 NaOH.
The volume of 0.875 mol/dm³ NaOH solution needed to make up a 100.0 ml of a diluted solution with a concentration of 0.235 mol/dm³ NaOH is 73.15 mL.
Given that :
molarity M1 = 0.875 mol/ dm³
M2 = 0.235 mol/ dm³
volume V2 = 100mL
using the formula , we get
M1 V1 = M2 V2
substituting the values in the formula we get :
0.875 × V1 = 0.235 × 100 mL
V1 = 26.857 mL
the volume added = V2 - V1 = 100 mL - 26.85 = 73.15
Thus, The volume of 0.875 mol/dm³ NaOH solution needed to make up a 100.0 ml of a diluted solution with a concentration of 0.235 mol/dm³ NaOH is 73.15 mL.
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Propane (C3H8) is a component of natural gas and is used in domestic cooking and heating. It burns according to the following reaction:
C3H8 + O2 → CO2 + H2O
a. Balance the equation representing the combustion of propane in air.
b. How many grams of carbon dioxide can be produced by burning 20.0 pounds of propane, the typical size of a BBG grill propane tank?
Assume that oxygen is the excess reagent in this reaction.
A. The balanced equation for the reaction is given as
C₃H₈ + 5O₂ -> 3CO₂ + 4H₂O
B. The mass (in grams) of carbon dioxide that can be produced by burning 20.0 pounds of propane is 27216.49 g
A. Balance equation
The balanced equation for the combustion of propane is:
C₃H₈ + 5O₂ -> 3CO₂ + 4H₂O
B. How to determin the mass (in grams) of CO₂ produced
The mass (in grams) of CO₂ produced from the reaction can be obtained as follow:
Molar mass of C₃H₈ = (12 × 3) +(8 × 1) = 44 g/molMass of C₃H₈ from the balanced equation = 44 × 1 = 44 g = 44 / 453.592 = 0.097 poundMolar mass of CO₂ = 12 + (16 × 2) = 44 g/molMass of CO₂ from the balanced equation = 3 × 44 = 132 gFrom the balanced equation above,
0.097 pound of C₃H₈ reacted to produce 132 g of CO₂
Therefore,
20 pound of C₃H₈ will react to produce = (20 × 132) / 0.097 = 27216.49 g of CO₂
Thus, the mass of CO₂ produced is 27216.49 g
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What is a molecular equation and what is this question asking please help?
In this question, we have a reaction between silver nitrate and potassium chloride, which will have the following chemical formula and also the following product:
AgNO3 (aq) + KCl (aq) --> AgCl (s) + KNO3 (aq)
Now we need to identify the ions that are components of the precipitate that was formed in this reaction
A precipitate is a product of a reaction that is not soluble in water in the conditions given in the reaction, therefore the precipitate will be a solid product of the reaction. The solid of the reaction is AgCl, therefore this is the precipitate
Now the ions that compose AgCl are Ag+ and Cl-, the answer will be the 3rd option
Howmany Mol MnO3areproducedwhen 4.30 kg of O2completelyreactsaccordingtothebalancedchemicalreaction:2Mn(s)+3O₂(g)→2MnO₃(s)Answer as equation
Answer:
89.58 moles of MnO3 are produced.
Explanation:
1st) From the balanced reaction we know that the relation between O2 and MnO3 is that 3 moles of O2 produces 2 moles of MnO3.
2nd) To build the equation, we must build a path from the given value of 4.30kg of O2 to moles of MnO3. So, it is important to figure out the relations that we need to build the equation.
The order to build the equation is:
• Start with, the given value of ,4.30kg of O2,.
,• We need to convert kg to grams, so we need the realtion between kg and g: ,1kg=1000g,.
,• To convert the grams of oxygen to moles, we need to use the molar mass of O2: ,32g/1mol,.
,• We need to use the relation between O2 and MnO3 from the balanced equation: ,3 moles of O2, produces, 2 moles of MnO3,.
[tex]4.30kg\text{ O}_2*\frac{1000g}{1kg}*\frac{1mol\text{ O}_2}{32g\text{ O}_2}*\frac{2moles\text{ MnO}_3}{3moles\text{ O}_2}=89.58moles\text{ MnO}_3[/tex]So, 89.58 moles of MnO3 are produced.
I dont understand how to find the one with smallest atomic radiusa. Calciumb. Cobaltc. potassiumd NickelI think its last one but why
Answer:
d. Nickel
Explanation:
The atomic radius is a periodic property, that is, we can see how it varies in the Periodic Table of Elements.
Atomic radius increases by a column (group) from top to bottom, and increases by a row (period) from right to left.
We can see the variation of atomic radius and the location of the given atoms in the following drawing:
Since K (potassium), Ca (calcium), Co (cobalt) and Ni (nickel) are located in the same row (period), to identify which one has the smallest atomic radius, we have to analyze the increase in horizontal atomic radius.
So, the Ni atom has the smallest atomic radius (because it is located more to the right compared to the other atoms in the same row) and K atom has the largest atomic radius.
What is the enthalpy change for the following reaction: C3H8(g) + 5O2 (g) --> 3CO2 (g) + 4H2O (l)ΔHof C3H8 (g) = -103.8 kJΔHof CO2 (g) = -393.5 kJΔHof H20 (l) = -285.8 kJ
The answer is -2219.9 kJ/mol
Match each colored row or group on the table to the family of elements it represents.Green (2nd column)Red (1st column)Purple (two rows below the rest of table)Blue (3rd to 12th columns)?Alkali metals?Alkaline earth metals?Transition metals?Lanthanides and actinides
Green column - Alkaline earth metals
Red column - Alkali metals
Purple column - Lanthanides and actinides
Blue column - Transition metals
Explanations:
The red block are made up of group 1 elements and these elements are known as Alkali metals.
The green column are the group 2 elements in the periodic table and are known as the Alkali earth metals.
For the blue columns, they are element from group 3 to 10 and they are known as the transition metals.
For the purple column, they are Lanthanides and actinides
Select the structure that correspondsto the molecule name:ethylmethylamineA.NHB. CH3CH₂NHCH3C. both
Ethylmethylamine is a secondary amine, that has one ethyl group attached to the amine group and one methyl group, which is a also attached to the amine group.
It is represented by:
Nevertheless it can be represented by CH3CH2NHCH3.
It means that the correct answer is c. Both.
0.650 moles of O₂(g) are added to 1.00 L flask and the internal pressure ismeasured at 20.00 atm. What is the temperature of the gas under theseconditions (in *C)? ________ °C?
1) List the known and unknown quantities.
Sample: O2
Amount of substance: 0.650 mol O2.
Volume: 1.00 L.
Pressure: 20.00 atm.
Ideal gas constant: 0.082057 L * atm * K^(-1) * mol^(-1)
Temperature: unknown.
2) Set the equation.
Ideal gas law.
[tex]PV=nRT[/tex]3) Plug in the known quantities and solve for T.
[tex](20.00\text{ }atm)(1.00\text{ }L)=(0.650\text{ }mol\text{ }O_2)(0.082057\text{ }L*atm*K^{-1}*mol^{-1})(T)[/tex][tex]T=\frac{(20.00atm)(1.00L)}{(0.650\text{ }mol\text{ }O_2)(0.082057\text{ }L*atm*K^{-1}*mol^{-1})}[/tex][tex]T=374.97\text{ }K[/tex]4) Convert K to ºC.
[tex]ºC=K-273.15[/tex][tex]ºC=374.94-273.15[/tex][tex]ºC=101.81\text{ }ºC[/tex]The temperature of the gas under these conditions is 101.81ºC.
.
find the natural logarithm of7.56
A student finds two green powders that look the same. He wants to figure out if the two powders are the same or different substances. Which of the following is
the best method to use?
I temperature of each powder
Mix the two green powders together and then test the properties.
A student finds two green powders that look the same he wants to figure out if the two powders are the same or different substances then mix the two green powders together and then test the properties
A powder is a dry bulk solid composed of many very fine particles that may flow freely when shaken or tilted and to figure out unknown two mixture powder are same or different then student want to mix the two green powders together and then test the properties because of properties student the two unknown powder are to be determined
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In the following equation, how many grams of CO2 are produced if 2 moles of O2 react completely with methane (CH4)? CH4 + 2O2 -> CO2 + 2H2OHELPFUL INFO: molar mass of CO2=44g/mol, molar mass of O2=32g/mol, molar mass of CH4=16g/mol, and molar mass of water = 18g/mol
Step 1 - "Reading" the equation
The given chemical equation is:
[tex]CH_{4(g)}+2O_{2(g)}\to CO_{2(g)}+2H_2O_{(l)}[/tex]The bigger numbers, those that come before the formulas of the substances, indicate the quantity in moles of each substance required in this rection.
We can "read" this reaction thus as :
one mole of CH4 reacts with 2 moles of O2 thus producing one mole of CO2 and two moles of H2O
As the exercise is specifically asking about the proportion between O2 and CO2, we can further simplify this statement to:
two moles of O2 produce one mole of CO2
Step 2 - Obtaining the "recipe" for the reaction
In order to obtain a relation in grams, we need to multiply the number of moles of each substance by its respective molar mass. Let's remember that:
two moles of O2 produce one mole of CO2
Therefore, converting to grams (molar masses: 32 g/mol for O2; 44 g/mol for CO2)
[tex]\begin{gathered} O_2\to2\text{ moles }\times32\text{ g/mole = 64g} \\ CO2\to1\text{ mole }\times44\text{ g/mole = 44g} \end{gathered}[/tex]We have now obtained the "recipe" for this reaction:
64g of O2 produce 44g of CO2
Pretty much as we would with a cake recipe, we can use this to predict how much CO2 is formed.
Since the reaction of two moles of O2 is exactly what we have already calculated, 44g of CO2 would be formed in this reaction.
Predict the shape of the molecule around the indicated atom.
Answer:
Linear (180 degrees)
Explanation:
Whenever you have a triple bond between two atoms, their shape is linear. This is because the density of the triple bond between them is heavy, causing the other things attached to the atom to be as far away as possible, creating a linear structure.
determine the mass (in grams) of c4h10 that is required to produce 14.7 g co2
According to the stoichiometry, 4.85 g of propane is required to produce 14.7 g carbon dioxide.
What is stoichiometry?
It is the determination of proportions of elements or compounds in a chemical reaction. The related relations are based on law of conservation of mass and law of combining weights and volumes.
Stoichiometry is used in quantitative analysis for measuring concentrations of substances present in the sample.
In the given problem as per balanced equation of combustion of propane 2 molecules of propane produce 8 molecules of carbon dioxide.That is, 116.24 g of propane produces 352.08 g of carbon dioxide.
∴14.7 g of carbon dioxide is produced by, 14.7×116.24/352.08=4.85 g of propane.
Thus, 4.85 g of propane is required to produce 14.7 g of carbon dioxide.
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If the volume is 15 and the mass of water is 14.9 what is the density
Answer:
0.9933
(Text Text Text)
What is the specific heat capacity of methanol (C) if it takes 7,490 joules of energy to heat 50.0 grams from 10.0 0C to 80.0 0C?
The formula for calculating the amount of heat energy absorbed is given as:
[tex]Q=mc\triangle t[/tex]Given the following
• mass m = 50.0g
,• Heat energy Q = 7490 Joules
,• change in temperature = 80 -10= 70 degrees celsius
Substitute
[tex]\begin{gathered} 7490=50\times c\times70 \\ c=\frac{7490}{3500} \\ c=2.14J\text{/}g^oC \end{gathered}[/tex]Hence the specific heat capacity of methanol is 2.14J/gC
Terrance found a small hair on the victim’s shirt. What type of evidence does this BEST represent?
A.
trace evidence
B.
biological evidence
C.
ammunitions evidence
D.
impression evidence
Answer:
Biological evidence, since it was a hair, which forms part of the human
Calculate the enthalpy of combustion (kJ/mol) of methyl linoleate (C₁₉H₃₄O₂), an important component of many biodiesel fuels made from corn oil and methanol, given the following standard enthalpies of formation: CO₂(g) = -393.5 kJ/mol H₂O(l) = -285.8 kJ/mol C₁₉H₃₄O₂(l) = -645.7 kJ/mol
Answer:
Explanation: 2 C19H34O2(l) + 53 O2(g) ==> 38 CO2(g) + 34 H2O(l) ... balanced equation
∆Hcombustion
∆H = ∑∆Hf products - ∑∆Hf reactants
∆H = [(38 x -393.5) + (34 x -285.8)] - (2 x -645.7) = (-14,953 + -9717) - (-1291)
∆H = -24,670 + 1291
∆H = 25,961 kJ
(be sure to check the math)
Find the molecular formula for a compound that contains 4.90 % N and 11.2 % O. The molecular mass of the compound is 92.0 g/mol.
The molecular formula for the compound that contains 4.90 g N and 11.2 g O is N₂O₄
How to determine molecular formulaWe'll begin by obtaining the empirical formula for the compound. This is illustrated below:
Nitrogen (N) = 4.90 gOxygen (O) = 11.2 gEmpirical formula =?Divide by their molar mass
N = 4.9 / 14 = 0.35
O = 11.2 / 16 = 0.7
Divide by the smallest
N = 0.35 / 0.35 = 1
O = 0.7 / 0.35 = 2
Thus, the empirical formula of the compound is NO₂
Finally, we shall determine the molecula formula. Details below
Molar mass of compound = 92 g/molEmpirical formula = NO₂Molecular formula =?Molecular formula = empirical × n = molar mass
[NO₂]ₙ = 92
[14 + (16×2)]ₙ = 92
46n = 92
Divide both side by 46
n = 92 / 46
n = 2
Molecular formula = [NO₂]ₙ
Molecular formula = [NO₂]₂
Molecular formula = N₂O₄
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Complete question:
Find the molecular formula for a compound that contains 4.90 g N and 11.2 g O. The molar mass of the compound is 92.0 g/mol.
Which of the listed species have the following electron configuration?
(See attached)
Answer:
Al 3+
Na+
Mg2+
F-
Explanation:
The element with the given electron that isn't an ion is Neon. It has 10 electrons in total.
Looking at the periodic table attached will help to see how the elements below are related to the one circled in red (Ne).
Al 3+ lacks 3 (because it has a 3+ charge) of its normal electrons so it has 10 and will have the same electron configuration as shown.
Na+ lacks one electron so it will also have 10 electrons instead of its normal 11.
Mg 2+ lacks two electrons so it will also have 10 total electrons.
F- has gained an extra electron so it now has 10 instead of 9.
Al3+, Na+, F-, and Mg2+ will have the given electronic configuration.
The electronic configuration given in the question is:
2p6, 2s2, 1s2
If it was a stable atom, the answer would have been Neon.
And if that would have been negative ion or anion (to be precise),
the answer would have been either F- or O2-.
But in the case of positive ions or we can say cations,
Al3+, Na+, and Mg2+ have the given electronic configurations.
Because the atom of Aluminum loses three electrons to achieve another state which is responsible for the formation of Al3+.
Similarly,
The atoms of Sodium and Magnesium also lose one and two electrons respectively to form Na+ and Mg2+.
So, we can conclude that Al3+, Na+, F- and Mg2+ will have the given electronic configuration.
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need help on chem homework
What is the relationship between n, ℓ, and the total number of nodes?
The number of planar nodes = ℓ.
The number of spherical nodes = n - ℓ - 1.
The relationship between the number nodes, the energy level n and the azimuthal quantum number l is (n-l -1) + l. Thus the total number of nodes is n-1.
What is node?Node is the region or point in the orbital where the probability of finding an electron is zero. Electrons are revolving around circular paths of fixed energy levels.
The principal quantum number n represents the number of main shells or the energy level of the electrons. Whereas, azimuthal quantum number represents the number of orbitals denoted as l.
The point where there is a probability of finding an electron is called antinode. The total node in an orbital is the sum of number of planar nodes and number of spherical nodes.
Thus total number of nodes = (n-l-1) + l
= n - 1.
Therefore, the total number of nodes is n-1. For example for n=3 level there will be two nodes.
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Given the following equation, if I make 18.5G of iron (III) sulphate, what is the percent yield? I picked B but I’m not sure if I’m correct
Answer:
B. The percent yield is 55.9%.
Explanation:
From the balanced reaction we know that 2 moles of FePO4 produce 1 mole of Fe2(SO4)3.
1st) It necessary to convert moles into grams, using the molar mass of FePO4 (151g/mol) and the molar mass of Fe2(SO4)3 (400g/mol):
- FePO4 conversion:
[tex]2moles*\frac{151g}{1mole}=302g[/tex]- Fe2(SO4)3 conversion:
In this case, it is not necessary the conversion, because there is 1 mole, so it is equal to 400g.
Now we know that 302g of FePO4 produce 400g of Fe2(SO4)3.
2nd) Now we have to calculate the Theoretical yield of Iron (III) sulfate, that is, the amount of Iron (III) sulfate that we must produce from 25g of Iron (III) phosphate:
[tex]\begin{gathered} 302gFePO_4-400gFe_2(SO_4)_3 \\ 25gFePO_4-x=\frac{25gFePO_4*400gFe_2(SO_4)_3}{302gFePO_4} \\ x=33.11gFe_2(SO_4)_3 \end{gathered}[/tex]So, the Theoretical yield is 33.11g.
3rd) Finally, we can calculate the Percent yield using the Theoretical yield (33.11g) and the Actual yield (18.5g):
[tex]\begin{gathered} PercentYield=\frac{ActualYield}{TheoreticalYield}*100\% \\ PercentYield=\frac{18.5g}{33.11g}*100\% \\ PercentYield=55.9\% \end{gathered}[/tex]So, the percent yield is 55.9%.
a sample of argon has a volume of 4.95 L and the pressure is 0.92 atm if the final temperature is 30 C the final volume is 5.7 L and the final pressure is .092 atm what was the initial temperature of the argon
The initial temperature of the argon is 2678 K
Given that,
Initial volume (V1) = 4.95 L
Initial pressure (P1) = 0.92 atm
Final temperature (T2) = 30 C = 303.15 K
Final volume (V2) = 5.7 L
Final pressure (P2) = 0.092 atm
We need to find the Initial temperature of the argon represented as T1
We Know that, In ideal gas law: PV = nRT
It can be written as, P1V1 / T1 = P2V2 / T2
(0.92 * 4.95) / T1 = (0.092 * 5.7) / 303.15
4.554 / T1 = 0.5244 / 303.15
4.554 / T1 = 0.0017
T1 = 4.554 / 0.0017
T1 = 2678 K
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Andrew wants to become a certified forensic pathologist. What is MOST likely to be true about his path to achieve this goal?
A.
It is fairly easy and doesn’t take a long time.
B.
There is no required residency or clinical fellowship.
C.
It requires a medical degree like a regular doctor.
D.
There are no schools that are currently offering this training.
Answer:
A
Explanation:
it is easy and doesn't take a long time
I keep getting 37.5 KPA but apparently that isn't the answer can anyone help?
ANSWER
The pressure of the gas is 281.25 mmHg
EXPLANATION
Given that;
The initial pressure of the gas is 75 mmHg
The intial temperature of the gas is 80K
The final temperature of the gas is 27 degrees Celcius
Follow the steps below to find the final pressure of the gas
Step 1; Convert the final temperature to degrees kelvin
[tex]\text{ T K = t}\degree C\text{ + 273.15}[/tex][tex]\begin{gathered} \text{ T K = 27 + 273.15} \\ \text{ T K = 300.15K} \end{gathered}[/tex]Step 2; Apply the Gay Lussac's law
[tex]\text{ }\frac{\text{ P1}}{\text{ T1}}\text{ = }\frac{\text{ P2}}{\text{ T2}}[/tex][tex]\begin{gathered} \text{ }\frac{\text{ 75}}{\text{ 80}}\text{ = }\frac{\text{ P2}}{\text{ 300.15}} \\ \text{ cross multiply} \\ \text{ 75 }\times\text{ 300.15 = P2 }\times\text{ 80} \\ 22511.25\text{ = 80 P2} \\ \text{ Divide both sides by 80} \\ \text{ P2 = }\frac{\text{ 22511.25}}{\text{ 80}} \\ \text{ P2 = 281.25mmHg} \end{gathered}[/tex]