Answer : The volume of the sample after the reaction takes place is, 15.93 liters.
Explanation : Given,
Moles of [tex]N_2[/tex] = 0.13 mole
Moles of [tex]O_2[/tex] = 0.26 mole
Initial volume of gas = 23.9 L
First we have to calculate the moles of [tex]NO_2[/tex] gas.
The balanced chemical reaction is :
[tex]N_2(g)+2O_2(g)\rightarrow 2NO_2(g)[/tex]
From the balanced reaction, we conclude that
As, 1 mole of [tex]N_2[/tex] react with 2 moles of [tex]O_2[/tex] to give 2 moles of [tex]NO_2[/tex].
So, 0.13 mole of [tex]N_2[/tex] react with [tex]2\times 0.13=0.26[/tex] moles of [tex]O_2[/tex] to give [tex]2\times 0.13=0.26[/tex] moles of [tex]NO_2[/tex].
According to the Avogadro's Law, the volume of the gas is directly proportional to the number of moles of the gas at constant pressure and temperature.
[tex]V\propto n[/tex]
or,
[tex]\frac{V_1}{V_2}=\frac{n_1}{n_2}[/tex]
where,
[tex]V_1[/tex] = initial volume of gas = 23.9 L
[tex]V_2[/tex] = final volume of gas = ?
[tex]n_1[/tex] = initial moles of gas = 0.13 + 0.26 = 0.39 mole
[tex]n_2[/tex] = final moles of gas = 0.26 mole
Now put all the given values in the above formula, we get the final temperature of the gas.
[tex]\frac{23.9L}{V_2}=\frac{0.39mole}{0.26mole}[/tex]
[tex]V_2=15.93L[/tex]
Therefore, the volume of the sample after the reaction takes place is, 15.93 liters.
If the partial molar volumes of the two components of a binary solution are VA 30 m2/mol and Vg 20 m3/mol and the mole fraction of component A is 0.4, find the molar volume of the solution.
Answer : The molar volume of the solution is, [tex]24m^3/mole[/tex]
Explanation : Given,
Partial molar volumes of component A = [tex]30m^3/mole[/tex]
Partial molar volumes of component B = [tex]20m^3/mole[/tex]
Mole fraction of component A = 0.4
First we have to calculate the mole fraction of component B.
As we know that,
[tex]\text{Mole fraction of component A}+\text{Mole fraction of component B}=1[/tex]
[tex]\text{Mole fraction of component B}=1-0.4=0.6[/tex]
Now we have to calculate the molar volume of the solution.
Expression used :
[tex]V_s=X_A\times \bar V_A+X_B\times \bar V_B[/tex]
where,
[tex]\bar V_A[/tex] = partial molar volumes of component A
[tex]\bar V_B[/tex] = partial molar volumes of component B
[tex]V_s[/tex] = molar volume of the solution
[tex]X_A[/tex] = mole fraction of of component A
[tex]X_B[/tex] = mole fraction of of component B
Now put all the give values in the above expression, we get:
[tex]V_s=0.4\times 30m^3/mole+0.6\times 20m^3/mole[/tex]
[tex]V_s=24m^3/mole[/tex]
Therefore, the molar volume of the solution is, [tex]24m^3/mole[/tex]
A pressure gauge on a welder's tank gives a reading of barometric pressure is 28.6 in Hg. Calculate the absolute pressure in the tank in 1. Ib/ft 2. in Hg 8
Hey there!:
1 in. Hg = 0.4911 psi
=> Absolute pressure in psi = 22.4 + 28.6 x 0.4911 = 36.445 psi
a) 1 psi = 144 lb/ft²
36.445 psi = 5248.15 lb/ft²
b) 1 psi = 2.036 in. Hg
36.445 psi = 74.21 in. Hg
Hope this helps!
1 in. Hg = 0.4911 psi
Absolute pressure in psi = 22.4 + 28.6 x 0.4911 = 36.445 psi
a) 1 psi = 144 lb/ft²
36.445 psi = 5248.15 lb/ft²
b) 1 psi = 2.036 in. Hg
36.445 psi = 74.21 in. Hg
What is effect of pressure on rate of reaction?Pressure plays a important role in rate of reaction to increase the pressure it increases the rate of reaction by increasing the collision.Pressure increase the concentration of gases it means that it increases the number of molecules per unit volume due to this collision of gases increases this will increase the temperature.
This increase of temperature will increase the rate of reaction. So we can say that the increases of pressure will increase the rate of reaction. Density is defined as the mass per unit volume it means that mass present in 1 meter cube is called density. The S.I unit of density is kg/m^3 and in C.G.S it is gram/cm^3
So In above statement we can understand that density, mass, and volume all are convert to each other it means that if we know any two variable then third one will be calculated easily.
Therefore, a) 1 psi = 144 lb/ft²
36.445 psi = 5248.15 lb/ft²
b) 1 psi = 2.036 in. Hg
36.445 psi = 74.21 in. Hg
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The following initial rate data apply to the reaction F2(g) + 2Cl2O(g) ® 2ClO2(g) + Cl2(g).Which of the following is the rate law (rate equation) for this reaction?rate = k[F2][Cl2O]rate = k[F2]2[Cl2O]2rate = k[F2][Cl2O]2rate = k[F2]2[Cl2O]4rate = k[F2]2[Cl2O]
Answer: k [[tex]F_{2}[/tex]] [tex][Cl_{2}O]^{2}[/tex]
Explanation: The given reaction is -
[tex]F_{2}(g) + 2Cl_{2}O(g) \rightarrow 2ClO_{2}(g) + Cl_{2}[/tex]
The rate law of the reaction is the written as the concentration of the reactant species rest to the power of its stiochiomeric coefficient.
Thus the rate law of the given reaction can be written as -
Rate = k [[tex]F_{2}[/tex]] [tex][Cl_{2}O]^{2}[/tex]
Rate law usually is determined from the slowest step of the reaction.
Given the following compounds and their vapor pressures (VP). Arrange them in increasing (lowest to highest) order of boiling points.ethane (C2H6); VP= 841 mm Hgpentane (C5H12); VP=420 mm Hgisopropyl alcohol (C3H5OH); VP=31.5 mm Hgacetone (C3H6O); VP=185 mm HgIsopropyl alcohol < acetone < pentane < ethaneEthane < pentane < isopropyl alcohol < acetoneEthane < pentane < acetone < isopropyl alcoholAcetone < isopropyl alcohol < pentane < ethaneThe strength of the intermolecular forces is also needed to determine the order.
Answer:
Increasing order of boiling point
isopropyl alcohol > acetone > pentane > ethane
Explanation:
Vapor pressure - the pressure exerted the gaseous molecules , on the walls of the container is called the vapor pressure.
Boiling point - the temperature at which the the vapor pressure of the liquid equals the external atmospheric pressure.
Both boiling point and vapor pressure are linked by the inter molecular forces between the atoms.
The compound with stronger inter molecular forces are tightly held , hence more amount of energy is required to vaporize. Therefore, higher boiling point , and in turn the vapor pressure will be lower .
And the compound with weaker inter molecular forces are loosely held , hence less amount of energy is required to vaporize. Therefore, lower boiling point , and in turn the vapor pressure will be higher .
Therefore,
The vapor pressure and boiling point will have an inverse relation.
Given , The vapor pressure of the compound in increasing order -
isopropyl alcohol < acetone < pentane < ethane .
Hence, the order of boiling point will be exactly reverse.
So , the Increasing order of boiling point,
isopropyl alcohol > acetone > pentane > ethane
How many grams of CuSO4 5 H20 are needed to make 1 liter of a 2% (w/v) CuSO4. REPORT TO THE NEAREST 1/10 G.
Answer:
19.9 grams of [tex]CuSO_4.5H_2O[/tex] will be needed.
Explanation:
Required strength of the solution = 2% (w/v)
This means that 2 gram of solute in 100 ml of solution.
Mass of [tex]CuSO_4.5H_2O[/tex] = 2 g
Moles of [tex]CuSO_4.5H_2O[/tex] =[tex]\frac{2 g}{249.68 g/mol}=0.008010 mol[/tex]
Volume of the solution = 100 mL = 0.1 L
Molarity of the solution:
[tex]M=\frac{0.008010 mol}{0.1 L}=0.08010 mol/L[/tex]
0.08010 moles of [tex]CuSO_4.5H_2O[/tex] are present 1 l of the solution.
Then mass of 0.08010 moles of [tex]CuSO_4.H_2O[/tex] will be:
0.08010 mol × 249.68 g/mol = 19.9993 g≈ 19.9 g
19.9 grams of [tex]CuSO_4.5H_2O[/tex] will be needed.
7.50 Draw all constitutional isomers with the molecular formula C4H9Br, and then arrange them in order of: (a) increasing reactivity toward an SN2 reaction. (b) increasing reactivity toward an E2 reaction
The isomers of C4H9Br are ranked in increasing reactivity towards SN2 and E2 reactions, taking into account steric hindrance and alkyl substitutions.
Explanation:The constitutional isomers for the molecule C4H9Br are: 1-bromobutane, 2-bromobutane, 1-bromo-2-methylpropane, and 2-bromo-2-methylpropane. (a) For SN2 reactions, reactivity increases with fewer alkyl substitutions (steric hindrance), thus the order is: 1-bromobutane > 2-bromobutane > 1-bromo-2-methylpropane > 2-bromo-2-methylpropane. (b) For E2 reactions, reactivity is greatest for bulkier bases and more alkyl substitutions, hence the order is: 2-bromo-2-methylpropane > 1-bromo-2-methylpropane > 2-bromobutane > 1-bromobutane.
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Pre-laboratory Questions 1. Using LeChâtelier’s principle, determine whether the reactants or products are favored and the direction of the shift in equilibrium (left or right) in the following examples. 2CO(g) + O2(g) 2CO2(g) + heat a. Remove O2 b. Lower the temperature c. Add CO d. Remove CO2 e. Decrease pressure
Which of the following is an acceptable structure for 2,5,5-trimethylhept-3-yne (CH3CH2)CH(CH3)C≡CCH2CH(CH3)2 CH3CH2C(CH3)2C≡CC(CH3)3 (CH3CH2)2C(CH3)C≡CCH2CH3 (CH3CH2)C(CH3)2C≡CCH(CH3)2 (CH3CH2CH2)CH(CH3)C≡CC(CH3)3
Answer:
D. CH₃CH₂C(CH₃)₂C≡CCH(CH₃)₂
Explanation:
You start numbering from the end closest to the triple bond (on the right). The triple bond is between C3 and C4, and there is one methyl group on C3 and two on C5.
A. CH₃CH₂CH(CH₃)C≡CCH₂CH(CH₃)₂ is wrong. The longest chain has eight C atoms, so the compound is an octyne.
B. CH₃CH₂CH(CH₃)C≡CC(CH₃)₃ is wrong. This is a molecule of 2,2,5-trimethylhept-3-yne.
C. (CH₃CH₂)₂C≡CCH₂CH₃ is wrong. This is a molecule of 6-ethyl-5-methylhept-3-yne.
E. CH₃CH₂CH₂CH(CH₃)C≡CC(CH₃)₃ is wrong. The longest chain has eight C atoms, so the compound is an octyne.
Final answer:
The correct structure for 2,5,5-trimethylhept-3-yne, according to IUPAC nomenclature, is the one that features a seven-carbon backbone with a triple bond beginning at the third carbon and contains three methyl groups precisely at the second, fifth, and fifth carbon positions. The only structure that fits these requirements is (CH₃CH₂)C(CH₃)₂C≡CCH(CH₃)₂.
Explanation:
To determine the acceptable structure for 2,5,5-trimethylhept-3-yne, it is essential to understand the basics of organic chemistry nomenclature. This name indicates a seven-carbon chain (hept-) with a triple bond starting from the third carbon (3-yne) and three methyl groups (CH3) attached to the second (2-), fifth (5-), and fifth (5-) carbons. Thus, the structure should broadly fit this description to be correct.
Reviewing the options provided, we look for the structure that accurately aligns with the IUPAC naming rules which prioritize the length of the carbon chain, the position of the triple bond, and the placement of the methyl groups.
(CH₃CH₂)CH(CH₃)C≡CCH₂CH(CH₃)₂: This structure incorrectly places two methyl groups at the end of the chain rather than along its length as indicated by the name.CH₃CH₂C(CH₃)₂C≡CC(CH₃)₃: This structure does not provide a heptane backbone and places too many methyl groups at the end.(CH₃CH₂)₂C(CH₃)C≡CCH₂CH₃: This option does not match the naming structure due to the incorrect placement of methyl groups and the structure appearing to have more than seven carbons in the main chain.(CH₃CH₂)C(CH₃)₂C≡CCH(CH₃)₂: This is the correct structure, as it fits the description of having a seven-carbon chain with a triple bond starting from the third carbon and three methyl groups at the second, fifth, and fifth positions. Correctly, the two methyl groups on the fifth carbon fulfill the naming requirements.(CH₃CH₂CH₂)CH(CH₃)C≡CC(CH₃)₃: This structure also misplaces methyl groups and misunderstands the main chain requirement.Give the percentage and whether w/w, w/v, or v/v of the following: 10g of NaoH + 90g solvent.
Answer:
The w/w 5 of the solution is 10 %.
Explanation:
w/w % : The percentage mass or fraction of mass of the of solute present in total mass of the solution.
[tex]w/w\%=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 100[/tex]
w/v %: The percentage of mass of the of solute present in total volume of the solution.
[tex]w/v\%=\frac{\text{Mass of solute}}{\text{Volume of solution}}\times 100[/tex]
v/v % : The percentage volume of the of solute present in total volumeof the solution.
[tex]v/v\%=\frac{\text{Volume of solute}}{\text{Volume of solution}}\times 100[/tex]
Mass of solution = Mass of solute + Mass of solvent
Mass of solute= 10 g of NaOH
Mass of solvent = 90 g
Mass of solution = 10 g + 90 g = 100 g
[tex]w/w\%=\frac{10 g}{100 g}\times 100 =10\%[/tex]
The w/w 5 of the solution is 10 %.
The reaction converting glycerol to glycerol-3-phosphate (energetically unfavorable) can be coupled with the conversion of ATP to ADP (energetically favorable): glycerol+HPO42−ATP+H2O⟶⟶glycerol-3-phosphate+H2OADP+HPO42−+H+ Part A What are the net products of the coupled reactions above?
Answer:
glycerol-3-phosphate, ADP, H⁺
Explanation:
The reaction of converting glycerol to glycerol-3-phosphate which makes is unfavorable and is coupled with the second reaction which involves conversion of ATP to ADP which is high energetically favorable.
Reaction 1: Glycerol + HPO₄²⁻ ⇒ Glycerol-3-phosphate + water
Reaction 2: ATP + H₂O ⇒ ADP + HPO₄²⁻ + H⁺
The coupled reaction of both the reactions become favorable. Thus, the overall coupled reaction is:
Glycerol + ATP ⇒ Glycerol-3-phosphate + ADP + H⁺
The net products are = glycerol-3-phosphate, ADP, H⁺
The net products of the coupled reactions as described are; glycerol-3-phosphate, ADP and H+.
According to the reaction;
According to the reaction;We are required to determine the net products of the coupled reactions above.The reactions given is as follows;
glycerol + (HPO4)²− + ATP+H2O ⟶⟶ glycerol-3-phosphate+H2O ADP+(HPO4)²- + H+
Since, the (HPO4)²- cancels out on both sides of the equation; the net products of the coupled reactions above are; glycerol-3-phosphate, ADP and H+.
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Be sure to answer all parts. When baking soda (sodium bicarbonate or sodium hydrogen carbonate, NaHCO3) is heated, it releases carbon dioxide gas, which is responsible for the rising of cookies, donuts, and bread. Write a balanced equation for the decomposition of the compound (one of the three products is Na2CO3). Do not include states of matter in your balanced equation. Calculate the mass of NaHCO3 required to produce 27.0 g of CO2. g NaHCO3
Answer: 1. [tex]2NaHCO_3\rightarrow Na_2CO_3+CO_2+H_2O[/tex]
2. 100.8 grams
Explanation:
According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.
Balanced equation for the decomposition of the compound sodium hydrogen carbonate is:
[tex]2NaHCO_3\rightarrow Na_2CO_3+CO_2+H_2O[/tex]
According to avogadro's law, 1 mole of every substance weighs equal to its molecular mass and contains avogadro's number [tex]6.023\times 10^{23}[/tex] of particles.
moles of [tex]CO_2=\frac{\text {given mass}}{\text {molar mass}}=\frac{27.0g}{44g/mol}=0.6moles[/tex]
According to stoichiometry:
1 mole of [tex]CO_2[/tex] is produced from 2 moles of [tex]NaHCO_3[/tex]
Thus 0.6 moles of [tex]CO_2[/tex] is produced from =[tex]\frac{2}{1}\times 0.6=1.2[/tex] moles of [tex]NaHCO_3[/tex]
Mass of [tex]NaHCO_3=moles\times {\text {Molar mass}}=1.2\times 84=100.8g[/tex]
Thus the mass of [tex]NaHCO_3[/tex] required to produce 27.0 g of [tex]CO_2[/tex] is 100.78 grams.
The balanced equation for the decomposition of sodium bicarbonate is 2 NaHCO3 → Na2CO3 + CO2 + H2O. To produce 27.0 g of CO2, you would need approximately 51.0 g of NaHCO3.
Explanation:The balanced chemical equation for the decomposition of sodium bicarbonate or baking soda is 2 NaHCO3 → Na2CO3 + CO2 + H2O. When heated, sodium bicarbonate decomposes into sodium carbonate, carbon dioxide, and water. This CO2 is what makes baked goods rise.
To calculate the mass of NaHCO3 required to produce 27.0 g of CO2, we use molar mass and stoichiometry. The molar mass of NaHCO3 is about 84 g/mol, and that of CO2 is about 44 g/mol. Since the reaction produces one mole of CO2 for every two moles of NaHCO3, we can set up the equation: (27g CO2 / 44 g/mol CO2) * (2 mol NaHCO3 / 1 mol CO2) * (84 g NaHCO3 / 1 mol NaHCO3) = 51.0 g NaHCO3
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When silver nitrate reacts with barium chloride, silver chloride and barium nitrate are formed. How many grams of silver chloride are formed when 10.0 g of silver nitrate reacts with 15.0 g of barium chloride? When silver nitrate reacts with barium chloride, silver chloride and barium nitrate are formed. How many grams of silver chloride are formed when 10.0 g of silver nitrate reacts with 15.0 g of barium chloride? 18.8 g 8.44 g 11.9 g 9.40 g
Answer : The mass of silver chloride formed are, 8.44 grams.
Explanation : Given,
Mass of [tex]AgNO_3[/tex] = 10.0 g
Mass of [tex]BaCl_2[/tex] = 15.0 g
Molar mass of [tex]AgNO_3[/tex] = 169.87 g/mole
Molar mass of [tex]BaCl_2[/tex] = 208.23 g/mole
Molar mass of [tex]AgCl[/tex] = 143.32 g/mole
First we have to calculate the moles of [tex]AgNO_3[/tex] and [tex]BaCl_2[/tex].
[tex]\text{Moles of }AgNO_3=\frac{\text{Mass of }AgNO_3}{\text{Molar mass of }AgNO_3}=\frac{10g}{169.87g/mole}=0.0589moles[/tex]
[tex]\text{Moles of }BaCl_2=\frac{\text{Mass of }BaCl_2}{\text{Molar mass of }BaCl_2}=\frac{15g}{208.23g/mole}=0.072moles[/tex]
Now we have to calculate the limiting and excess reagent.
The balanced chemical reaction is,
[tex]2AgNO_3+BaCl_2\rightarrow 2AgCl+Ba(NO_3)2[/tex]
From the balanced reaction we conclude that
As, 2 moles of [tex]AgNO_3[/tex] react with 1 mole of [tex]BaCl_2[/tex]
So, 0.0589 moles of [tex]AgNO_3[/tex] react with [tex]\frac{0.0859}{2}=0.0295[/tex] moles of [tex]BaCl_2[/tex]
From this we conclude that, [tex]BaCl_2[/tex] is an excess reagent because the given moles are greater than the required moles and [tex]AgNO_3[/tex] is a limiting reagent and it limits the formation of product.
Now we have to calculate the moles of [tex]AgCl[/tex].
As, 2 moles of [tex]AgNO_3[/tex] react to give 2 moles of [tex]AgCl[/tex]
So, 0.059 moles of [tex]AgNO_3[/tex] react to give 0.059 moles of [tex]AgCl[/tex]
Now we have to calculate the mass of [tex]AgCl[/tex].
[tex]\text{Mass of }AgCl=\text{Moles of }AgCl\times \text{Molar mass of }AgCl[/tex]
[tex]\text{Mass of }AgCl=(0.0589mole)\times (143.32g/mole)=8.44g[/tex]
Therefore, the mass of silver chloride formed are, 8.44 grams.
The mass of silver chloride (AgCl) formed when 10.0 g of silver nitrate reacts with 15.0 g of barium chloride is 8.44 g. The correct option is the second option 8.44 g
First, we will write a balanced chemical equation for the reaction between silver nitrate and barium chloride
2AgNO₃ + BaCl₂ → 2AgCl + Ba(NO₃)₂
This means, 2 moles of AgNO₃ will react with 1 mole of BaCl₂ to produce 2 moles of AgCl and 1 mole of Ba(NO₃)₂
From the question,
10.0 g of silver nitrate reacts with 15.0 g of barium chloride
First, we will determine the number of moles present in each reactant
For silver nitrate (AgNO₃)Mass = 10.0 g
Molar mass = 169.87 g/mol
Using the formula
[tex]Number \ of \ moles = \frac{Mass}{Molar \ mass}[/tex]
Then,
[tex]Number \ of\ moles\ of\ silver\ nitrate = \frac{10.0}{169.87}[/tex]
Number of moles of AgNO₃ = 0.05887 moles
For Barium chloride (BaCl₂)Mass = 15.0 g
Molar mass of BaCl₂ = 208.23 g/mol
∴ [tex]Number \ of\ moles\ of\ barium \ chloride= \frac{15.0}{208.23}[/tex]
Number of moles of BaCl₂ = 0.072036 moles
From the balanced chemical equation
2 moles of AgNO₃ will react with 1 mole of BaCl₂
∴ 0.05887 moles AgNO₃ will react with [tex]\frac{0.05887}{2}[/tex] mole of BaCl₂
[tex]\frac{0.05887}{2} = 0.029435[/tex]
This means only 0.029435 moles of BaCl₂ will react
(NOTE: AgNO₃ is the limiting reagent while BaCl₂ is the excess reagent)
Now,
Since 2 moles of AgNO₃ will react with 1 mole of BaCl₂ to produce 2 moles of AgCl
That means,
0.05887 moles AgNO₃ will react with 0.029435 moles of BaCl₂ to produce 0.05887 moles of AgCl
∴ The number of moles of silver chloride (AgCl) produced is 0.05887 moles
Now, to determine the mass (in grams) of silver chloride that are formed
From the formula,
Mass = Number of moles × Molar mass
Molar mass of AgCl = 143.32 g/mol
∴ Mass of silver chloride, AgCl, formed = 0.05887 moles × 143.32 g/mol
Mass of silver chloride formed = 8.4372484 g
Mass of silver chloride formed ≅ 8.44 g
Hence, the mass of silver chloride (AgCl) formed when 10.0 g of silver nitrate reacts with 15.0 g of barium chloride is 8.44 g. The correct option is the second option 8.44 g
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The titration of 10.00 mL of an HCl solution of unknown concentration requires 120.54 mL of a 0.100 M NaOH solution to reach the equivalence point. What is the concentration of the unknown HCl solution?
Answer:
Concentration of HCl is 1.2054 M
Explanation:
Acid-base titration is a type of qualitative analysis in which the concentration of the unknown acid/base is determined by treating it with a known concentration of base/acid. Indicators are used to determine the end point or the equivalence point (when all the acid/base is neutralized by base/acid).
At the end point,
Moles of acid = Moles of base
Since,
Molarity of a solution is given by the moles of solute per liter of the solution
i.e.,
Molarity = moles of solute / volume of solution (in L)
moles = Molarity * Volume
moles = M * V
M₁V₁ ( acid )= M₂V₂( base )
Where,
M₁ is the molarity of the acid
V₁ is the volume of the acid
M₂ is the molarity of the base
V₂ is the total volume of the base
From the question,
M₁ = ?
M₂ = 0.100 M
V₁ = 10.00 mL
V₂ = 120.54 mL
Using the above mentioned formula
M₁V₁ ( acid )= M₂V₂( base )
M₁ * 10.00mL = 0.100 M * 120.54 mL
M₁ = 1.2054 M
Hence, the concentration of of HCl is 1.2054 M
The concentration of the HCl solution is calculated by the formula c1v1 = c2v2, deriving from the one-to-one reaction between NaOH and HCl. The resulting concentration of the HCl solution is determined to be 1.21 M.
Explanation:The subject of this question is a chemistry concept related to acid-base titration. In this context, you're given that 10.00 mL of an HCl solution is titrated with 120.54 mL of a 0.100 M NaOH solution to reach the equivalence point. From this, we can figure out the concentration of the unknown HCl solution using the formula c1v1 = c2v2, where c represents concentration and v represents volume. The reaction between NaOH and HCl is one-to-one, so the number of moles of NaOH that react will be equal to the number of moles of HCl initially present in the solution.
To calculate the number of moles of NaOH used in the reaction, you multiply the volume of NaOH solution used (in liters) by its molarity. Here, that is (120.54 mL / 1000 mL/L) * 0.100 mol/L = 0.012054 mol NaOH. Since the moles of NaOH must equal the moles of HCl, the molarity (M) of the HCl solution is the moles of HCl divided by the volume of HCl solution in liters. Therefore, the concentration of HCl is 0.012054 mol / (10.00 mL / 1000 mL/L) = 1.21 M.
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Automobile air bags are inflated with nitrogen gas, which is formed by the decomposition of solid sodium azide (NaN3). The other product is sodium metal. Calculate the volume of nitrogen gas at 27 °C and 756 torr formed by the decomposition of 125 g of sodium azide.
Answer: 71.65 L
Explanation:
Decomposition of sodium azide is shown by equation below:
[tex]2NaN_3\rightarrow 2Na+3N_2[/tex]
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
[tex]\text{Number of moles of sodium azide}=\frac{125g}{65g/mol}=1.92moles[/tex]
According to stoichiometry:
2 moles of [tex]NaN_3[/tex] produce 3 moles of [tex]N_2[/tex]
Thus 1.92 moles of [tex]NaN_3[/tex] will produce=[tex]\frac{3}{2}\times 1.92=2.88[/tex] moles of [tex]N_2[/tex]
According to the ideal gas equation:
[tex]PV=nRT[/tex]
P = Pressure of the gas = 756 torr = 0.99 atm (1 torr= 0.0013 atm)
V= Volume of the gas = ?
T= Temperature of the gas = 27°C = 300 K (0°C = 273 K)
R= Gas constant = 0.0821 atmL/K mol
n= moles of gas= 2.88
[tex]V=\frac{nRT}{P}=\frac{2.88\times 0.0821\times 300}{0.99}=71.65L[/tex]
Thus the volume of nitrogen gas at 27 °C and 756 torr formed by the decomposition of 125 g of sodium azide is 71.65 L
The efficiency of a spray chamber scrubber is 95% for 20 μm particles. Everything else being constant, what is the efficiency for 10 μm particles?
a. 48%
b. 53%
c. 78%
d. 95%
Answer : The correct option is, (a) 48 %
Explanation :
As we are given the efficiency of a spray chamber is, 95 %
Using unitary method :
If the efficiency of [tex]20\mu m[/tex] particles is 95 %
Then, the efficiency of [tex]10\mu m[/tex] particles will be = [tex]\frac{10\mu m}{20\mu m}\times 95\%=47.5\%\approx 48\%[/tex]
Therefore, the efficiency for [tex]10\mu m[/tex] particles is, 48 %
Blood clotting factors are regulated by allosterie control.(T/F)
Answer:
False
Explanation:
Regulation of blood coagulation by anticoagulant pathways Regulation of coagulation is exerted at each level of the pathway, either by enzyme inhibition or by modulation of the activity of the cofactors.
The zero order reaction 2N2O→2N2+O2 has the reaction constant k is 6.28×10−3 molL s. If the initial concentration of N2O is 0.962 mol/L, what is the concentration of N2O after 10.0 seconds? Your answer should have three significant figures (three decimal places).
The concentration will be equal to 0.899 mol/L.
How can we arrive at this result?We will use the equation [tex]k= \frac{1}{t} ([A_0]-[A])[/tex]In this equation, we will replace the "[tex]t[/tex]" with 10 and the "[tex]A_0[/tex]" with 0.962.The value of "A" will be found as follows:
[tex]6.28*10^-^3=\frac{1}{10} (0.962-[A])\\A= 0.899 mol/L[/tex]
It is important to remember that the concentration of a chemical solution refers to the amount of solute that exists within the solvent.
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The concentration of N2O after 10 seconds in a zero order reaction with initial concentration of 0.962 mol/L and rate constant of 6.28×10^-3 mol/L/s is 0.899 mol/L.
Explanation:The question involves calculating the concentration of N2O after 10 seconds in a zero order reaction. The rate equation for a zero order reaction is given by rate = k, where the reaction rate is constant and does not depend on the concentration of the reactant. The formula to calculate the remaining concentration after a certain time in a zero order reaction is: [A] = [A]0 - kt, where [A] is the concentration at time t, [A]0 is the initial concentration, k is the rate constant, and t is the time.
In this case, the initial concentration of N2O is 0.962 mol/L, the rate constant k is 6.28×10−3 mol/L/s, and the time t is 10.0 seconds. Using the formula, we get [N2O] = 0.962 mol/L - (6.28×10−3 mol/L/s)(10.0 s) = 0.962 mol/L - 0.0628 mol/L = 0.899 mol/L.
In reaction 3, 4.4 mL of 5.6 M NaOH is reacted with an excess HCl. From extrapolation of the temperature curve, the change in temperature is 14°C. Calculate the Cp of the the calorimete
Answer : The [tex]C_p[/tex] of the the calorimeter is, [tex]98.35J/^oC[/tex]
Explanation :
First we have to calculate the moles of NaOH.
[tex]\text{Moles of }NaOH=\text{Concentration of }NaOH\times \text{Volume of solution}[/tex]
[tex]\text{Moles of }NaOH=5.6mole/L\times 0.0044L=0.02464mole[/tex]
conversion used for volume : (1 L = 1000 ml)
As we know that, HCl and NaOH are the strong acid and bases. So, the heat of neutralisation is -55.9 kJ/mole.
Now we have to calculate the heat released by neutralization.
As, 1 mole of heat released by neutralization = 55.9 kJ
So, 0.02464 mole of heat released by neutralization = [tex]0.02464\times 55.9kJ=1.377kJ[/tex]
Now we have to calculate the [tex]C_p[/tex] of the the calorimeter.
[tex]Q=C_p\times \Delta T[/tex]
where,
Q = heat of reaction = 1.377 kJ
[tex]C_p[/tex] = specific heat capacity of calorimeter =?
[tex]\Delta T[/tex] = change in temperature = [tex]14^oC[/tex]
Now put all the given values in this expression, we get:
[tex]1.377kJ=C_p\times 14^oC[/tex]
[tex]C_p=0.0983kJ/^oC=98.35J/^oC[/tex]
Therefore, the [tex]C_p[/tex] of the the calorimeter is, [tex]98.35J/^oC[/tex]
The Cp of the calorimeter is equal to 98.35J/ºC
Steps to get to this answer:First, it is necessary to know the molarity of NaOH. This will be done with the following equations:
[tex]Mole_N_a_O_H= 5.6 mol/L*0.0044L= 0.02464 mole[/tex]
We must identify the heat released during the neutralization reaction. For this, we must consider that the neutralization of 1 mol of HCl or NaOH releases -55.9 kJ/mole. In this case, we will use the following equation:[tex]\left \ {{1mole=-55.9 kJ} \atop {0.02464 mole=x}} \right.\\x= 55.9*0.02464 mole\\x= 1.377kJ[/tex]
From this, we can calculate the Cp of the calorimeter and this will be done with the following equation:[tex]Q=C_p*\Delta T\\1,377= C_p*14 \textdegree C\\C_p= \frac{1.377}{14} = 98.35J\textdegree C[/tex]
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If enzymatic reactions could actually be carried out at high temperatures, what would you expect for the rate of the catalyzed reaction at 100 ∘C as compared to that at 21 ∘C?
Answer:
Temperature of 100°C increase the reaction rate, due to the rise of the energetic collision between molecules that increase the collision, compared to the same reaction at 21 °C, who has less energy and for that reason will be more slow to react.
Explanation:
The enzymatic reactions are reactions between organic molecules named as substrate and some proteic biological structures called enzymes. These reactions can be simplified as a regular chemistry process between reagents to obtain a product, that in this case is the transformations of the substrate.
So we can use the following process:
Enzyme (E) + substrate (S) = product (P)
Of course, this an uncomplex way to see this process, just to understand this example. In reality, the product or these reactions involves a transformation of the substrate and the enzyme. But for now, let's just use this equation
Using just the letters:
E + S = P
Now, we can use the concept of rate or velocity on chemical equations to analyze the effect of temperature in the enzymatic reactions:
Remember that for any chemistry reactions, the rate depends on the capacity of the molecules to collide, these collisions will be major when the reagents have enough kinetic energy to move around and interact between them. The frequency of these collisions is affected by different variables such as temperature.
Temperature is equal to energy, so if to reactions are supplied by external energy like thermic energy, the molecules, in this case, enzyme and substrate can move faster, and the collides can be more frequent when the temperature increases.
In conclusion, the increases in temperature to 100°C, increase the reaction rate, due to the rise of the energetic collision between molecules that increase the collision, and are these who result in the product of a enzymatic reactions, compared to the same reaction at 21 °C, who has less energy and for that reason will be more slow to react.
A piston-cylinder device contains 257 grams of R134a at 60 kPa and -20°C. The R134a is heated until its temperature reaches 100°C. Determine the total change of volume.
Answer : The total change of volume is, 41.883 liters.
Explanation :
R134a is a 1,1,1,2-tetrafluoroethane. It is a hydro-fluorocarbon and haloalkane gaseous refrigerant.
First we have to calculate the volume at [tex]-20^oC[/tex].
Using ideal gas equation:
[tex]PV=nRT\\\\PV=\frac{w}{M}\times RT[/tex]
where,
n = number of moles
w = mass of R134a = 257 g
P = pressure of the gas = 60 Kpa
T = temperature of the gas = [tex]-20^oC=273+(-20)=253K[/tex]
M = molar mass of R134a = 102.03 g/mole
R = gas constant = 8.314 Kpa.L/mole.K
V = initial volume of gas
Now put all the given values in the above equation, we get :
[tex](60Kpa)\times V=\frac{257g}{102.03g/mole}\times (8.314Kpa.L/mole.K)\times (253K)[/tex]
[tex]V=88.305L[/tex]
Now we have to calculate the volume at [tex]100^oC[/tex] by using Charles's law.
Charles' Law : It is defined as the volume of gas is directly proportional to the temperature of the gas at constant pressure and number of moles.
[tex]V\propto T[/tex]
or,
[tex]\frac{V_1}{V_2}=\frac{T_1}{T_2}[/tex]
where,
[tex]V_1[/tex] = initial volume of gas = 88.305 L
[tex]V_2[/tex] = final volume of gas = ?
[tex]T_1[/tex] = initial temperature of gas = 253 K
[tex]T_2[/tex] = final temperature of gas = [tex]100^oC=273+100=373K[/tex]
Now put all the given values in the above formula, we get the final volume of the gas.
[tex]\frac{88.305L}{V_2}=\frac{253K}{373K}[/tex]
[tex]V_2=130.188L[/tex]
Now we have to calculate the total change of volume.
[tex]V_2-V_1=130.188-88.305=41.883L[/tex]
Therefore, the total change of volume is, 41.883 liters.
Mathematically define heat capacity with equations.
Answer: The amount of energy needed to heat the body by 1 ° C (Q)
Explanation: The heat capacity or as it is called the thermal mass is the amount of energy needed to heat the body by 1 ° C. When the body is said, it refers to any object in any aggregate state. The energy required for the body to warm up by 1 ° C is expressed in joules, and is obtained when the specific heat of the body multiplies with the body mass and with the change of temperature:
Q = m·c·ΔT
Q - heat capacity (J),
m - mass of the body (g),
c - specific heat of the body (J/g-°C),
ΔT - change in temperature (°C)
Benzoyl chloride, C6H5COCl, reacts with water to form benzoic acid, C6H5COOH, and hydrochloric acid. This first-order reaction is 25% complete after 26 s. How much longer would one have to wait in order to obtain 99% conversion of benzoyl chloride to benzoic acid?180 s210 s290 s390 s420 s
Answer: The time required for the conversion of benzoyl chloride to benzoic acid is 390 seconds.
Explanation:
The expression for the rate law of first order kinetics is given as:
[tex]\ln \frac{N}{N_o}=e^{-kt}[/tex] ......(1)
where,
[tex]N_o[/tex] = initial mass of isotope = 100 g
N = mass of the parent isotope left after the time = 100 - 25 = 75 g
t = time taken = 26 s
k = rate constant = ?
Putting values in above equation, we get:
[tex]\ln \frac{75}{100}=e^{-k\times 26}\\\\k=0.011065s^{-1}[/tex]
Now, we need to find the time when 99 % of the reaction is complete. Using equation 1 again, we get:
[tex]N_o[/tex] = initial mass of isotope = 100 g
N = mass of the parent isotope left after the time = 100 - 99 = 1 g
t = time taken = ? s
k = rate constant = [tex]0.011065s^{-1}[/tex]
Putting values in above equation, we get:
[tex]\ln \frac{1}{100}=e^{-0.011065\times t}\\\\t=416s[/tex]
Time required for the conversion of benzoyl chloride to benzoic acid = 416 - 26 = 390 s
To obtain 99% conversion of benzoyl chloride to benzoic acid, it would take approximately 420 seconds.
Explanation:This is a first-order reaction, and the question asks for the time needed to obtain 99% conversion of benzoyl chloride to benzoic acid. The reaction is 25% complete after 26 seconds, so we can use the half-life formula to calculate the time required for complete conversion. In a first-order reaction, the half-life is given by t1/2 = (0.693/k), where k is the rate constant. Since the reaction is 25% complete after 26 seconds, we can find the rate constant by rearranging the formula to k = 0.693 / t1/2. The rate constant is approximately 0.0267 s-1. Now, we can find the time required for 99% conversion using the formula t = (ln(1-0.99)/-k), where k is the rate constant. Solving this equation gives approximately 420 seconds.
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Complete and balance the molecular equation for the reaction of aqueous ammonium bromide, NH4BrNH4Br , and aqueous lead(II) acetate, Pb(C2H3O2)2Pb(C2H3O2)2 . Include physical states.
Answer: The molecular equation is given below.
Explanation:
Every balanced chemical equation follows law of conservation of mass.
This law states that mass can neither be created nor be destroyed but it can only be transformed from one form to another form. This means that total mass on the reactant side is equal to the total mass on the product side.
This also means that the total number of individual atoms on the reactant side will be equal to the total number of individual atoms on the product side.
The balanced chemical equation for the reaction of ammonium bromide and lead (II) acetate follows:
[tex]2NH_4Br(aq.)+(CH_3COO)_2Pb(aq.)\rightarrow 2CH_3COONH_4(aq.)+PbBr_2(s)[/tex]
This is a type of double displacement reaction because here, exchange of ions takes place.
Hence, the molecular equation is given below.
Compound A and compound B are constitutional isomers with molecular formula C4H9Cl. Treatment of compound A with sodium methoxide gives trans-2-butene as the major product, while treatment of compound B with sodium methoxide gives a different disubstituted alkene as the major product.Draw the structure of compound A.
Compound A will be 2-chloro-butane which in the presence of sodium methoxide will undergo the elimination of the halogen with a hydrogen from the neighboring atom carbon (not the one with which the halogen in bound), which have less hydrogen atoms, forming the trans-2-butene.
Answer:
2-chlorobutane.
Explanation:
Hello,
In this case, the treatment of the substance A with sodium methoxide to yield trans-2-butene stands for a dehydrohalogenation in the presence of a strong base (the sodium methoxide), such chemical reaction is undergone by alkyl halides, in this case with chlorine, based on its molecular formula. In such a way, on the attached picture, you will find both the structure of the 2-chlorobutane and its chemical reaction to yield the trans-2-butene throughout the usage of sodium methoxide.
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An analytical balance was used to weigh 200.01 mg of dry solid sodium acetate {NaC2H3O2 ~ NaAc MW = 82.0343, pKa (HC2H3O2 ~ HAc) = 4.76} which was transferred to a 100.00 mL volumetric flask that was then filled to the line with carbonate-free water and mixed well. Determine the pH of each of the solutions below.
A.) The original sodium acetate solution
B.) A 25.00 mL aliquot of the NaAc solution plus 25.00 mL of 0.100 M aqueous acetic acid (HAc) solution
C.) A 25.00 mL aliquot of the NaAc solution plus 15.00 mL of 0.100 M aqueous hydrogen chloride solution
Liquid octane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 97. g of octane is mixed with 150. g of oxygen. Calculate the maximum mass of water that could be produced by the chemical reaction. Round your answer to significant digits.
Answer: 61 grams
Explanation:
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles of octane}=\frac{\text{Given mass}}{\text{Molar mass}}=\frac{97g}{114g/mol}=0.85moles[/tex]
[tex]\text{Number of moles of oxygen}=\frac{\text{Given mass}}{\text{Molar mass}}=\frac{150g}{32g/mol}=4.69moles[/tex]
The chemical equation for the combustion of octane in oxygen follows the equation:
[tex]2C_8H_{18}+25O_2\rightarrow 16CO_2+18H_2O[/tex]
By stoichiometry of the reaction;
25 moles of oxygen react with 2 moles of octane
4.69 moles of oxygen react with=[tex]\frac{2}{25}\times 4.69=0.37[/tex] moles of octane
Thus, oxygen is the limiting reagent as it limits the formation of product and octane is the excess reagent.
25 moles of oxygen produce 18 moles of water
4.69 moles of oxygen produce=[tex]\frac{18}{25}\times 4.69=3.38[/tex] moles of water.
Mass of water produced=[tex]moles\times {\text{Molar mass}}=3.38\times 18g/mol=61g[/tex]
The maximum mass of water that could be produced by the chemical reaction is 61 grams.
To determine the maximum mass of water produced from the combustion of octane, one must first calculate the moles of octane and water using the balanced chemical equation. The stoichiometry shows that for every 2 moles of octane, 18 moles of water are produced. The final mass of water is calculated by considering the limiting reagent, which could be either octane or oxygen.
Explanation:To calculate the maximum mass of water that could be produced from the combustion of octane, we should start with the balanced chemical equation for octane (C8H18):
2 C8H18(l) + 25 O2(g) → 16 CO2(g) + 18 H2O(g)
First, we determine the moles of octane based on the given mass:
Mass of octane = 97.0 gMolar mass of octane = 114.23 g/molMoles of octane = Mass / Molar mass = 97.0 g / 114.23 g/molNext, we calculate the moles of water produced per mole of octane using the stoichiometry of the reaction:
From the equation, 2 moles of octane produce 18 moles of water.Thus, 1 mole of octane will produce 18 / 2 = 9 moles of water.Now, we find the mass of water produced using the moles of octane we calculated:
Molar mass of water = 18.02 g/molMaximum moles of water = Moles of octane × 9Maximum mass of water = Maximum moles of water × Molar mass of waterFinally, we check if oxygen is the limiting reagent by comparing moles of oxygen provided with the moles required based on the moles of octane we have. To calculate this, use the molar mass of oxygen (O2) which is 32.00 g/mol and the stoichiometry of the equation. If the available moles of oxygen are less than required, we must recalculate the maximum mass of water based on the moles of the limiting reagent (oxygen).
Note that we haven't provided actual numerical results, and you will need to perform the calculations to find the maximum mass of water that can be produced. Remember to round your final answer to the correct number of significant digits based on the given data in the question.
How many moles of oxygen are formed when 58.6 g of KNO3 decomposes according to the following reaction? 4 KNO3(s) → 2 K2O(s) + 2 N2(g) + 5 O2(g)
Answer:
0.725 mol
Explanation:
Moles are calculated as the given mass divided by the molecular mass.
i.e. ,
moles = ( mass / molecular mass )
since,
mass of KNO₃ = 58.6 g ( given )
Molecular mass of KNO₃ = 101 g / mol
Therefore,
moles of KNO₃ = 58.6 g / 101 g / mol
moles of KNO₃ = 0.58 mol
From the balanced reaction ,
4 KNO₃ (s) ---> 2K₂O (s) + 2N₂ (g) + 5O₂ (g)
By the decomposition of 4 mol of KNO₃ , 5 mol of O₂ are formed ,
hence, unitary method is used as,
1 mol of KNO₃ gives 5 / 4 mol O₂
Therefore,
0.58 mol of KNO₃ , gives , 5 / 4 * 0.58 mol of O₂
Solving,
0.58 mol of KNO₃ , gives , 0.725 mol of O₂
Therefore,
58.6g of KNO₃ gives 0.725 mol of O₂.
What is the freezing point of water made by dissolving 22.78 g of ethylene glycol (CH2(OH)CH2(OH)) in 87.95 g of water? The freezing-point depression constant of water is 1.86 oC/m.
Explanation:
When a non-volatile solute is added in a solvent then decrease in its freezing point is known as freezing point depression.
Mathematically, [tex]\Delta T = K_{f}m[/tex]
where [tex]\Delta T[/tex] = change in freezing point
[tex]K_{f}[/tex] = freezing point depression constant [tex]\text({in ^{o}C/mol/kg})[/tex]
m = molality
First, calculate the number of moles as follows.
No. of moles = [tex]\frac{\text{mass of solute}}{\text{molar mass of solute}}[/tex]
= [tex]\frac{22.78g}{62.07 g/mol}[/tex]
= 0.367 mol
Now, it is given that mass of solvent is 87.95 g. As there are 1000 grams in 1 kg.
So, [tex]87.95g \times \frac{1 kg}{1000 g}[/tex] = 0.08795 kg
Hence, molality of the given solution is as follows.
Molality = [tex]\frac{\text{no. of moles}}{\text{mass in kg}}[/tex]
= [tex]\frac{0.367 mol}{0.08795}[/tex]
= 4.172 mol/kg
Therefore, depression in freezing point will be as follows.
[tex]\Delta T = K_{f}m[/tex]
[tex]T_{solvent} - T_{mixture}[/tex] = [tex]K_{f}m[/tex]
Since, freezing point of pure water is [tex]0^{o}C[/tex]. Now, putting the given values as follows.
[tex]0^{o}C - T_{mixture}[/tex] = [tex]1.86^{o}C/m \times 4.172mol/kg[/tex]
= [tex]-7.759 ^{o}C[/tex]
Thus, we can conclude that the freezing point of water in the given mixture is [tex]-7.759 ^{o}C[/tex].
A light microscope would be most advantageous for viewing which of the following? a. small molecule such as water b. a DNA strand c. virus d. eukaryotic cells such as green algae
Answer:
D
Explanation:
The resolution of a light microscope is approximately 200 nanometers due to the wavelength of light. A eukaryotic cell is approximately 1 – 5 micrometers. This means that light can be used to view a eukaryotic cell because the wavelength of light is smaller than the size of the cell. The other options are of a smaller size than the light wavelength and hence lower wavelength beams (such as those of electron microscope) or laser can be used to view them.
Final answer:
The light microscope is best suited for viewing eukaryotic cells such as green algae, leveraging staining techniques to provide necessary contrast. It cannot resolve smaller entities like water molecules, DNA strands, or viruses, which are below the limit of light microscopy resolution.
Explanation:
A light microscope would be most advantageous for viewing eukaryotic cells such as green algae. Light microscopes can magnify cells up to approximately 400 times, which is suitable for viewing relatively large cellular structures. However, they are not powerful enough to resolve smaller structures such as water molecules, DNA strands, or viruses, which are below their limit of resolution. For improved observation, staining techniques are often used, as these colored chemicals make cellular components visible by providing the necessary contrast.
Other Microscope Types
A phase-contrast microscope is especially useful for viewing thick structures such as biofilms.
For very small surface structures of a cell, a scanning electron microscope (SEM) would be the best choice, as it provides a detailed three-dimensional image of the specimen's surface.
To see individual components of cells under a light microscope, scientists commonly use special stains.
TYPE OUT EXTREMELY NEATLY PLEASE AND SHOW ALL YOUR WORK. Use significant figures where appropriate
A sample of nitrogen gas occupies 120.0 L at 340 K and 2.30 atm. What are the number of moles of nitrogen in the sample?
Answer: The number of moles of nitrogen gas is 9.9 moles.
Explanation:
To calculate the mass of bromine gas, we use the ideal gas equation, which is:
PV = nRT
where,
P = Pressure of nitrogen gas = 2.30 atm
V = Volume of nitrogen gas = 120.0 L
n = Number of moles of nitrogen gas = ? mol
R = Gas constant = [tex]0.0821\text{ L atm }mol^{-1}K^{-1}[/tex]
T = Temperature of nitrogen gas = 340 K
Putting values in above equation, we get:
[tex]2.30atm\times 120.0L=n\times 0.0821\text{L atm }mol^{-1}K^{-1}\times 340K\\\\n=9.88mol\approx 9.9mol[/tex]
Rule of significant figures in case of multiplication and division:
The least number of significant figures in any number of the problem will determine the number of significant figures in the solution.
Here, the least precise number of significant figures are 2. Thus, the number of moles of nitrogen gas is 9.9 moles.