Answer:
Partial pressure of CO₂ is 406.9 mmHg
Explanation:
To solve the question we should apply the concept of the mole fraction.
Mole fraction = Moles of gas / Total moles
We have the total moles of the mixture, if we have the moles for each gas inside. (3.63 moles of O₂, 1.49 moles of N₂ and 4.49 moles of CO₂)
Total moles = 3.63 mol O₂ + 1.49 mol N₂ + 4.49 mol CO₂ = 9.61 moles
To determiine the partial pressure of CO₂ we apply
Mole fraction of CO₂ → mol of CO₂ / Total moles = P. pressure CO₂ / Total P
Partial pressure of CO₂ = (mol of CO₂ / Total moles) . Total pressure
We replace values: (4.49 moles / 9.61 moles) . 871 mmHg = 406.9 mmHg
Answer:
Partial pressure O2 = 329 mmHg
Partial pressure N2 = 135 mmHg
Partial pressure CO2 = 407 mmHg
Explanation:
Step 1: Data given
Number of moles oxygen (O2) = 3.63 moles
Number of moles nitrogen (N2) = 1.49 moles
Number of moles carbon dioxide (CO2) = 4.49 moles
Total pressure = 871 mmHg
Step 2: Calculate total number of moles
Total moles = moles O2 + moles N2 + moles CO2
Total moles = 3.63 + 1.49 + 4.49
Total moles = 9.61 moles
Step 3: Calculate the mol ratio
Mol ratio number of moles compound / total moles
Mol ratio O2 = 3.63 moles / 9.61 moles
Mol ratio O2 = 0.378
Mol ratio N2 = 1.49 moles / 9.61 moles
Mol ratio N2 = 0.155
Mol ratio CO2 = 4.49 moles / 9.61 moles
Mol ratio CO2 = 0.467
Step 4: Calculate partial pressure
Partial pressure = mol ratio * total pressure
Partial pressure O2 = 0.378 * 871 mmHg
Partial pressure O2 = 329 mmHg
Partial pressure N2 = 0.155 * 871mmHg
Partial pressure N2 = 135 mmHg
Partial pressure CO2 = 0.467 * 871 mmHg
Partial pressure CO2 = 407 mmHg
Cycloalkanes are (saturated/unsaturated) compounds.
(which one)
Answer:
Saturated .
Explanation:
When toluene is used in free radical bromination, a very small amount of product is formed that contains only carbons and hydrogens and no bromine. Show the structure of that product and the arrow curved mechanism of how it is formed starting from the alkyl radical intermediate of the reaction.(9 pts)
Answer:
Explanation:
The product formed that contains only carbons and hydrogens after free radical bromination of toluene is 1,2-diphenylethane.....
Please go through the attached file for the diagrams .
Answer:
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Explanation:
write a balanced equation for the reaction between hydrogen peroxide H202 and Fe2+ to produce Fe3+ and H2O in acidic solution
The balanced chemical equation for the reaction between hydrogen peroxide (H2O2) and Fe2+ in acidic solution to produce Fe3+ and H2O is: 2 H2O2 + 2 Fe2+ + 2 H+ → 2 Fe3+ + 2 H2O
In this reaction, hydrogen peroxide (H2O2) acts as an oxidizing agent, while iron(II) ions (Fe2+) are being oxidized to iron(III) ions (Fe3+). The hydrogen peroxide molecules donate oxygen atoms to the iron(II) ions, causing them to undergo oxidation. The reaction takes place in an acidic solution, which provides the necessary protons (H+) to balance the reaction and ensure the overall charge neutrality.
Balancing the equation is crucial to ensure that the same number of atoms of each element are present on both sides of the reaction arrow, preserving the law of conservation of mass. In this balanced equation, there are two hydrogen (H) atoms, four oxygen (O) atoms, and two iron (Fe) atoms on both sides, demonstrating that mass is conserved in the chemical reaction.
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Final answer:
The balanced chemical equation in an acidic solution for the oxidation of Fe2+ with hydrogen peroxide to form Fe3+ and water is 2 Fe2+ (aq) + H2O2 (aq) + 2 H+ (aq) → 2 Fe3+ (aq) + 2 H2O (l).
Explanation:
The balanced equation for the reaction between hydrogen peroxide (H2O2) and Fe2+ to produce Fe3+ and H2O in an acidic solution is given by:
2 Fe2+ (aq) + H2O2 (aq) + 2 H+ (aq) → 2 Fe3+ (aq) + 2 H2O (l)
In this reaction, Fe2+ is oxidized to Fe3+, and H2O2 (hydrogen peroxide) acts as the oxidizing agent, being reduced to water (H2O). The presence of H+ indicates that the reaction occurs in an acidic solution. To balance the equation, take into account the transfer of electrons, the conservation of mass, and the charge balance on both sides of the equation.
The expected first intermediate formed during a halohydrin reaction is:
a. a cyclic oxonium ion.
b. the most stable carbocation with OH on the adjacent carbon.
c. ahalonium ion.
d. the most stable carbanion.
e. the most stable carbocation with X on the adjacent carbon.
Final answer:
The first intermediate formed during a halohydrin reaction is a halonium ion. This intermediate is key in the process of adding a halogen and an OH group to adjacent carbons in the formation of halohydrins.
Explanation:
The expected first intermediate formed during a halohydrin reaction is c. ahalonium ion. In the presence of a halogen and water, the reaction of an alkene leads to the formation of a halonium ion intermediate. Specifically, the alkene undergoes an addition reaction with a halogen such as bromine or chlorine to form a three-membered ring halonium intermediate. This is followed by the nucleophilic attack by water, which opens the ring and leads to the formation of the halohydrin, with the halogen and an OH group on two adjacent carbons. Halohydrins such as bromohydrins or chlorohydrins are important intermediates in organic synthesis.
Element X reacts with hydrogen gas at 200°C to form compound Y. When Y is heated to a higher temperature, it decomposes to the element X and hydrogen gas in the ratio of 559 mL of H2 (measured at standard temperature and pressure) for 1.00 g of X reacted. X also combines with chlorine to form a compound Z, which contains 63.89 percent by mass of chlorine. Deduce the identity of X. The symbol of element X is
Answer:
X is calcium with the symbol Ca
Explanation:
From the first statement:
X reacted with H2 to produce Z i.e
X + H2 —> Y
Y is heated to form X and H2 i.e
Y —> X + H2
Let us obtain moles of H2 produced. This is illustrated below:
Volume of H2 produced = 559 mL = 0.559 L.
I mole of H2 occupy 22.4L at stp
Therefore, b mol of H2 will occupy 0.559 L at stp i.e
b mol of H2 = 0.559/22.4
b mol of H2 = 0.025 mole
Next 0.025 mole of H2 to gram..
Molar Mass of H2 = 2x1 = 2g/mol
Mole of H2 = 0.025 mole
Mass = number of mole x molar Mass
Mass of H2 = 0.025 x 2
Mass of H2 = 0.05g
From the question given, we were told that 1g of X reacted. This means that 1g of X reacted with 0.05g of H2. Now let us determine the mass of X that will react with 1 mol ( i.e 2g) of H2. This is illustrated below:
From the reaction,
It was discovered that 1g of X reacted with 0.05g of H2.
Therefore, P g of X will react with 2g of H2 i.e
P g of X = 2/0.05 = 40g/mol
The molar mass of X is 40g/mol.
Now let us consider the second statement to see if we'll obtained the same result as 40g/mol of X
From the second statement:
X also combines with chlorine to form a compound Z, which contains 63.89 percent by mass of chlorine i.e
X + Cl2 —> Z
Molar Mass of Z (X + 2Cl) = X + (35.5 x 2) = X + 71.
Z contains 63.89% by mass of Cl2.
We can obtain the molar mass of X as follow:
Percentage by mass of Cl2 in the compound Z is given by:
Mass of Cl2/Molar Mass x100
63.89/100 = 71/(X + 71)
Cross multiply to express in linear form
63.89(X + 71) = 100 x 71
Clear the bracket
63.89X + 4536.19 = 7100
Collect like terms
63.89X = 7100 - 4536.19
63.89X = 2563.81
Divide both side by 63.89
X = 2563.81/63.89
X = 40g/mol
Now we can see that in both experiments, the molar mass of X is 40g/mol.
Comparing the value of X i.e 40g/mol with that from the periodic table, X is calcium with the symbol Ca
A chemist dissolves 751.mg of pure nitric acid in enough water to make up 290.mL of solution. Calculate the pH of the solution. Be sure your answer has the correct number of significant digits.
Answer:
1.4
Explanation:
Mass of pure nitric acid = 751mg
Volume of solution = 290mL
Unknown:
pH of the solution = ?
Solution:
To solve this problem, we need the concentration of the acid in the aqueous form.
This is given by molarity;
Molarity = [tex]\frac{number of moles }{volume}[/tex]
Since the number of moles of nitric acid is unknown, we can easily solve for it.
Number of moles of nitric acid = [tex]\frac{mass}{molar mass}[/tex]
molar mass of HNO₃ = 1 + 14 + 3(16) = 63g/mol
mass of nitric acid = 751mg = 0.751g
Number of moles = [tex]\frac{0.751}{63}[/tex] = 0.012mole
Volume of solution = 290mL = 0.29dm³
Now molarity of the solution = [tex]\frac{0.012}{0.29}[/tex] = 0.041moldm⁻³
Since:
pH = -log [H₃O⁺]
HNO₃ + H₂O → H₃O⁺ + NO₃⁻
1moldm⁻³ 1moldm⁻³ 1moldm⁻³
0.041moldm⁻³ 0.041moldm⁻³ 0.041moldm⁻³
pH = -log[0.041] = 1.4
II. Practice An ideal gas occupies 5 L at atmospheric pressure and 300 K (point A). It is warmed at constant volume to 3 atm (point B). Then it is allowed to expand isothermally to 1 atm (point C) and at last compressed isobarically to its original state. A. How many moles of gas are being used? B. Find the temperature at point C. C. Find the work done on the gas in each process. D. Find the amount of heat added to/removed from the gas in one cycle.
Answer:
The process can be represented as shown in the figure below; having got the diagram, we can solve for the questions.
A. the number of moles of gas used
n = PV/ RT = (1.013 *10^5 Pa) * (5.0 *10^-3 m^3) / (8.314 * 300)
n = 5.065 * 10^2 / 2494.2
n = 0.00203 *10^2
n = 0.203 moles
B. Temperature at point C (Tc)
Pa/Ta= Pb/Tb
Tb = Pb *Ta / Pa
Tb = 3 * 300 / 1
Tb = 900 K
Since Tb = Tc = 900 K
C. For process AB,
work done is zero
For process BC,
work done = -nRTbln (Vc/Vb)
W = -(0.203 * 8.314 * 900 ln (3)
W = -(1.518 kJ ln 3
W = -1.67 kJ
For process CA,
W = -P V =-nRT
W = -(0.203 * 8.314 * (-600))
W = 1.01 kJ
Explanation:
A sample of an unknown gas effuses in 12.5 min. An equal volume of H2 in the same apparatus under the same conditions effuses in 2.42 min. What is the molar mass of the unknown gas
Answer:
53.4 gMol-1
Explanation:
Let the mass of the unknown gas be M
Let the molar mass of hydrogen gas be 2×1=2gMol-1
Time for diffusion of unknown gas = 12.5 min
Time for diffusion of hydrogen= 2.42 min
From Graham's law:
t1/t2=√M1/M2
Hence:
2.42/12.5= √2/M
Hence M= 53.4 gMol-1
Answer:
The molar mass of the unknown gas is 40.06 g/mol
Explanation:
Graham's law of effusion states that the rate of effusion of a gaseous substance is inversely proportional to the square root of its molar mass.
[tex]\frac{R_{b} }{R_{a} } = \sqrt{\frac{M_{a} }{M_{b} } }[/tex] [tex]= \frac{t_{a} }{t_{b} }[/tex] where R = rate of effusion, M = molar mass and t= time of effusion
⇒ [tex]\sqrt{\frac{2 g/mol}{x g/mol} } = \frac{162secs}{725secs}[/tex]
x g/mol = [tex]\frac{2}{0.223448^{2} }[/tex]
= 40.06 g/mol
Ice is placed in a beaker of room temperature water, in which direction does heat flow?
Answer:
The heat will begin to flow from the room temperature water into the ice.
Explanation:
Hi there,
To get started, recall the principles of heat flow, and the laws of thermodynamics. Most importantly, remember that energy in the form of heat flows from a mass with higher heat towards a mass with lower heat.
In this case, ice has less heat than the room temperature water. Thus, when placed in the beaker, the heat will begin to flow from the room temperature water into the ice, warming up the ice, and melting it into water. The system will then reach a single final temperature, where it will be in equilibrium.
Final temperature of a solution with just two water masses (like ice and room temp water) can be calculated with the following formula:
[tex]T_F=\frac{m_1T_1_i+m_2T_2_i}{m_1+m_2}[/tex] where m is mass, and Ti is initial temperature. Though this is not needed for this problem.
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11:56 0220...
M
Due Tue
1. Write a balanced equation for the reaction between hydrogen
peroxide, H2O2 and Fe2+ to produce Fe3+ and H20. In acidic
solution.
ED
FEBIH
WEB
RE
RE
ON
Answer:
H2O2+ Fe2+ + 2H+ = 2H2O + Fe3+
Explanation:
H2O2 is reduced to H2O and Fe2+ is oxidized to Fe3+
The balanced equation for the reaction between hydrogen peroxide and Fe2+ in acidic solution is: H₂O₂(aq) + 2Fe²⁺(aq) + 2H⁺(aq) → 2Fe³⁺(aq) + 2H₂O(l). This balances both hydrogen and oxygen atoms by combining oxidation and reduction half-reactions.
To balance the chemical equation, we will follow these steps:
Identify the reactants and products: H₂O₂ and Fe²⁺ (reactants) form Fe³⁺ and H₂O (products) in the presence of H⁺ ions, as the solution is acidic.
Write the unbalanced equation: H₂O₂(aq) + Fe²⁺(aq) → Fe³⁺(aq) + H₂O(l).
Balance the atoms involved, particularly keeping hydrogen and oxygen in mind:
Separate the oxidation and reduction half-reactions:
Oxidation (H₂O₂ → O):
H₂O₂ → 2H₂O + 2e⁻
Reduction (Fe²⁺ → Fe³⁺):
Fe²⁺ → Fe³⁺ + e⁻
Combine the half-reactions, ensuring that the number of electrons is balanced:
This means multiplying the reduction half-reaction by two: 2(Fe²⁺ → Fe³⁺ + e⁻).
Thus, the combined balanced equation is:
H₂O₂(aq) + 2Fe²⁺(aq) + 2H⁺(aq) → 2Fe³⁺(aq) + 2H₂O(l).
This gives us the balanced equation for the reaction between hydrogen peroxide and Fe²⁺ to produce Fe³⁺ and water in an acidic solution.
What atomic or hybrid orbital on the central I atom makes up the sigma bond between this I and an outer Cl atom in iodine pentachloride, ICl5 ? orbital on I What are the approximate Cl-I-Cl bond angles ? (list all possible separated by a space)
The central iodine atom in ICl5 uses sp3d2 hybrid orbitals, formed from d, s, and p orbitals for bonding. It forms sigma bonds with chlorine atoms through their p orbitals. The bond angles in ICl5 are approximately 90 and 180 degrees.
Explanation:In iodine pentachloride (ICl5), the central iodine atom will utilize its d orbitals along with its s and p orbitals to form the necessary hybrid orbitals for bonding. Specifically, it adopts the sp3d2 hybridization to form a total of six hybrid orbitals; one for each of the five chlorine atoms and one for the lone pair. The sigma bond between the I and the outer Cl atoms is formed by the overlap of the sp3d2 hybrid orbital on I and a p orbital on Cl.
The geometry of ICl5 is square pyramidal, resulting from the six electron pairs (five bonding pairs from the chlorine atoms and one lone pair). The Cl-I-Cl bond angles in a square pyramidal structure are approximately 90 degrees and 180 degrees.
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The central iodine atom in iodine pentachloride, ICl5, forms a sigma bond with an outer chlorine atom using a sp³d hybrid orbital. The molecular structure of ICl5 is a square pyramid, with approximate Cl-I-Cl bond angles of 90° and 180°.
Explanation:In the molecule iodine pentachloride, ICl5, the central iodine (I) atom has 7 valence electrons. To allow for five bonds (five pairs of shared electrons), the atom needs to utilize five orbitals. It uses the 5s orbital, the three 5p orbitals, and one of the 5d orbitals, therefore forming a set of five sp³d hybrid orbitals. The sigma bond between I and an outer Cl atom is formed by the overlap of these hybrid orbitals, which concentrate the electron density along the internuclear axis.
The structure of ICl5 is a square pyramid, which is slightly distorted due to the presence of lone pairs. In this structure, there are four Cl-I-Cl bond angles of 90 degrees each and one bond angle of 180 degrees. Therefore, the approximate Cl-I-Cl bond angles are 90° and 180°.
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Use scientific reasoning to explain what happens if 0.87 mol NaCl is dissolved in pure water creating 3.00 L of solution, what is its molarity of the resulting solution?
Answer : The molarity of the resulting solution is, 0.29 M
Explanation :
When NaCl dissolved in water then it dissociates to give sodium ions and chloride ions.
Given,
Moles of [tex]NaCl[/tex] = 0.87 mol
Volume of solution = 3.00 L
Molarity : It is defined as the number of moles of solute present in one liter of volume of solution.
Formula used :
[tex]\text{Molarity}=\frac{\text{Moles of }NaCl}{\text{Volume of solution (in L)}}[/tex]
Now put all the given values in this formula, we get:
[tex]\text{Molarity}=\frac{0.87mol}{3.00L}=0.29mole/L=0.29M[/tex]
Therefore, the molarity of the resulting solution is, 0.29 M
How many moles of water are present in 15.00 ml of water
that has a density of 0.9956 g/ ml.
MW for Hydrogen 1.00 g/ 1 mole; MW for Oxygen 16.00 g/ 1 mole
Answer : The moles of water present in solution are, 0.8297 moles.
Explanation :
First we have to calculate the mass of water.
[tex]\text{Mass of water}=\text{Density of water}\times \text{Volume of water}[/tex]
[tex]\text{Mass of water}=0.9956g/mL\times 15.00mL=14.934g[/tex]
Now we have to calculate the moles of water.
[tex]\text{Moles of water}=\frac{\text{Mass of water}}{\text{Molar mass of water}}[/tex]
Molar mass of water = (2 × Molecular weight of hydrogen) + Molecular weight of oxygen
Molar mass of water = (2 × 1.00g/mol) + 16.00 g/mol
Molar mass of water = 18.00 g/mol
[tex]\text{Moles of water}=\frac{14.934g}{18.00g/mol}[/tex]
[tex]\text{Moles of water}=0.8297mol[/tex]
Therefore, the moles of water present in solution are, 0.8297 moles.