In the periodic table, Francium has the largest atomic size, Chlorine has the highest electron affinity, Barium has the lowest ionization enthalpy in Group II, and gaseous elements at room temperature include noble gases and some non-metals like Fluorine and Oxygen.
The periodic table organizes elements in such a way that we can predict their properties based on their position. These properties, including atomic size, electron affinity, and ionization enthalpy, show clear patterns or trends within the table. As a tutor from the Brainly platform, let's address the specific properties asked in the question.
The element with the largest atomic size is typically found at the bottom left of the periodic table. Francium (Fr) is generally considered to have the largest atomic radius because it belongs to Group 1 and is the heaviest alkali metal. The atomic size increases down a group and decreases across a period from left to right.Highest electron affinity is generally observed in halogens. Chlorine (Cl), being very eager to gain an electron to achieve a noble gas electron configuration, has one of the highest electron affinities.The element with the lowest ionization enthalpy in Group II is expected to be at the bottom of the group. Barium (Ba), being at the lower part of Group II, has lower ionization energy compared to its counterparts up the group.Gases at room temperature from Group II are notable as they are non-existent; however, elements that are in gaseous form at room temperature from the periodic table include the noble gases such as Radon (Rn), Xenon (Xe), Krypton (Kr), Argon (Ar), Neon (Ne), Helium (He), and some non-metals like Fluorine (F), Chlorine (Cl), Oxygen (O), Nitrogen (N), and Hydrogen (H).A 115.0-g sample of oxygen was produced by heating 400.0 g of potassium chlorate.2KClO3 Right arrow. 2KCI + 3O2What is the percent yield of oxygen in this chemical reaction?Use Percent yield equals StartFraction actual yield over theoretical yield EndFraction times 100..
Answer:
73.4% is the percent yield
Explanation:
2KClO₃ → 2KCl + 3O₂
This is a decomposition reaction, where 2 moles of potassium chlorate decompose to 2 moles of potassium chloride and 3 moles of oxygen.
We determine the moles of salt: 400 g . 1. mol /122.5g= 3.26 moles of KClO₃
In the theoretical yield of the reaction we say:
2 moles of potassium chlorate can produce 3 moles of oxygen
Therefore, 3.26 moles of salt, may produce (3.26 . 3) /2 = 4.89 moles of O₂
The mass of produced oxygen is: 4.89 mol . 32 g /1mol = 156.6g
But, we have produced 115 g. Let's determine the percent yield of reaction
Percent yield = (Produced yield/Theoretical yield) . 100
(115g / 156.6g) . 100 = 73.4 %
Stoichiometry!
Please note:
- Use 6.022x1023 for avogadro’s number
- Ignore sig figs and do not round the final answer.
- Keep it to 1 decimal place.
Answer:
a) 13.2 moles [tex]2H_{2}O[/tex]
b) 79.33 grams of [tex]2H_{2}O[/tex]
Explanation:
First, we'll need to balance the equation
[tex]H_{2(g)} + O_{2(g)}[/tex] → [tex]H_{2}O_{(g)}[/tex]
There are 2 (O) on the left and only one on the right, so we'll add a 2 coefficient to the right.
[tex]H_{2(g)} + O_{2(g)}[/tex] → [tex]2H_{2}O_{(g)}[/tex]
Now there are 4 (H) on the right and only 2 on the left, so we'll add a 2 coefficient to the ([tex]H_{2}[/tex]) on the left.
[tex]2H_{2(g)} + O_{2(g)}[/tex] → [tex]2H_{2}O_{(g)}[/tex]
The equation is now balanced.
a) This can be solved with a simple mole ratio.
4.6 moles [tex]O_{2}[/tex] × [tex]\frac{2 moles H_{2}O}{1 mole O_{2}}[/tex] = 13.2 moles [tex]2H_{2}O[/tex]
b) This problem is solved the same way!
2.2 moles [tex]H_{2}[/tex] × [tex]\frac{2 moles H_{2}O}{2 moles H_{2}}[/tex] = 2.2 moles [tex]2H_{2}O[/tex]
However, this problem wants the mass of [tex]2H_{2}O[/tex], not the moles.
The molecular weight of [tex]2H_{2}O[/tex] is the weight of 4 (H) molecules and 2 (O) molecules (found on the periodic table). So,
4(1.008) + 2(15.999) = 36.03 g/mol
2.2 moles [tex]2H_{2}O[/tex] × [tex]\frac{36.03 g}{1 mol}[/tex] = 79.33 grams of [tex]2H_{2}O[/tex]
Given the half‑reactions and their respective standard reduction potentials 1. Cr 3 + + e − ⟶ Cr 2 + E ∘ 1 = − 0.407 V 2. Cr 2 + + 2 e − ⟶ Cr ( s ) E ∘ 2 = − 0.913 V calculate the standard reduction potential for the reduction half‑reaction of Cr(III) to Cr(s).
Answer:
The standard reduction potential for the reduction half‑reaction of Cr(III) to Cr(s) is -0.744 V
Explanation:
Here we have
1. Cr³⁺ + e − ⟶ Cr²⁺ E⁰₁ = − 0.407 V
2. Cr²⁺ + 2 e − ⟶ Cr ( s ) E⁰₂ = − 0.913 V
To solve the question, we convert, the E⁰ values to ΔG as follows
ΔG₁ = n·F·E⁰₁ and ΔG₂ = n·F·E⁰₂
Where:
F = Faraday's constant in calories
n = Number of e⁻
ΔG₁ = Gibbs free energy for the first reaction
ΔG₂ = Gibbs free energy for the second half reaction
E⁰₁ = Reduction potential for the first half reaction
E⁰₂ = Reduction potential for the second half reaction
∴ ΔG₁ = 1 × F × − 0.407 V
ΔG₂ = 2 × F × − 0.913 V
ΔG₁ + ΔG₂ = F × -2.233 V which gives
ΔG = n × F × ΔE⁰ = F × -2.233 V
Where n = total number of electrons ⇒ 1·e⁻ + 2·e⁻ = 3·e⁻ = 3 electrons
We have, 3 × F × ΔE⁰ = F × -2.233 V
Which gives ΔE⁰ = -2.233 V /3 = -0.744 V.
The standard reduction potential for the reduction half‑reaction of Cr(III) to Cr(s) is found by the summation of the standard reduction potentials of the half-reactions involved, giving -1.32V.
Explanation:To calculate the standard reduction potential for the reduction half-reaction of Cr(III) to Cr(s), we first need to understand that a redox reaction is a sum of an oxidation half-reaction and a reduction half-reaction. In the given half-reactions, the first is for Cr3+ being reduced to Cr2+ and the second one is for Cr2+ being reduced to Cr(s).
From the given standard reduction potentials, the 1st reaction has E°1 = -0.407 V and the 2nd reaction has E°2 = -0.913 V. Therefore, to get from Cr3+ to Cr(s), we add both these half-reactions together, which also means we add their potentials together. The sum gives us the potential for the entire reaction, which is E°total = E°1 + E°2 = (-0.407V) + (-0.913V) = -1.32V.
Thus, the standard reduction potential for the reduction half-reaction of Cr(III) to Cr(s) is -1.32V.
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Which is the formula mass of Na₂SO₄? *
1. 119 amu
2. 125 amu
3. 142 amu
4. 174 amu
Answer:
the answer is option 3.
3. What is a major disadvantage of
environmental science?
A. Lack of government funding
B. Inability to control variables
C. Outdated technology
Answer:
I think b if not then c hopefully this helps
What conditions are required for heat transfer between states of matter to occur? In what
direction will heat transfer occur?
Consider the conversion of succinate to fumarate in the Citric Acid Cycle (reaction below). This reaction is endergonic under standard conditions (ΔGo’≈ 6 kJ/mol). How might this reaction be made favorable under equilibrium conditions? Your answer should include the relationship of this reaction to a canonical electron transport chain (i.e. an electron transport chain that uses oxygen as a terminal electron acceptor).
Answer:
Succinate oxidation to fumarate The following reactions transform succinate to regenerate oxalacetate. The first of these reactions is carried out by an oxidation catalyzed by succinate dehydrogenase. The hydrogen acceptor is FAD, since the free energy change is insufficient to allow NAD to interact. The final product is fumarate.
Explanation:
The condensation reaction of GDP + Pi and the hydrolysis of Succinyl-CoA involve the H2O necessary to balance the equation.
C. How many grams of Cu react with 5.65 grams of AgNO3?
Answer: 1.1g of copper
Explanation: We begin by writing the balanced reaction equation.
So we have : Cu(s) + 2AgNO3 (aq) ------> Cu(NO3)2 (aq) + 2Ag (s)
Molar mass of AgNO3 = 107.87 + 14.01 + 3(16.0) = 169.88g/mol, for 2 moles we then have 2 x 169.88 =339.76g.
Atomic mass of Copper = 64g approximately
From the equation the following deductions can be made:
339.76g of AgNO3 reacts with 64g of copper
5.65g of AgNO3 would react with 64/ 339.76 x 5.65 =1.062 approx 1.1g of copper.
Answer:
we need 1.06g of Cu to react with 5.65g of AgNO₃
Explanation:
C u +2 AgNO₃ = Cu(NO₃ )₂ + 2 Ag
the mole ratio between Copper and silver nitrate is 1:2 meaning we need 2 moles of silver nitrate for every mole of copper that takes part in the reaction.
the molar mass of Cu is 63.55
the molar mass of AgNO₃ is 107.8682+ 14 + (16x3) = 107.8682+ 14 + 48= 169.87682≈169.88g
for every 63.55 g of Cu reaction with 2(169.88g of AgNO₃)= 339.736g of AgNO₃
if x grams of Cu react with 5.65 grams of AgNO₃
63.55g= 339.736g
x = 5.65g
cross multiply
339.736x = 5.65 x 63.55 = 359.0575
x = 359.0575/339.736 = 1.05687210069≈1.06
we need 1.06g of Cu to react with 5.65g of AgNO₃
Which part of an atom is most directlly involved in chemical bonding?
Calculate the number of moles of sodium hydroxide in 8g of sodium hydroxide (NaOH). Na=23 O=16 H=1
Answer:
0.2 mol
Explanation:
Use the given values to find the molar mass. The molar mass is 40 g/mol.
23 + 16 + 1 = 40
Using the molar mass, convert grams to moles.
(8 g)/(40 g/mol) = 0.2 mol
You have 0.2 mol of sodium hydroxide.
What is the representative particle for copper metal, Cu
The representative particle for copper metal, Cu, refers to the smallest unit which retains the properties of copper – in this case, a single copper atom.
Explanation:In chemistry, a representative particle refers to the smallest unit of a substance that still retains the properties of that substance. For a pure metal like copper (Cu), the representative particle is an atom. Therefore, the representative particle for copper metal, Cu, is a copper atom. Copper atoms join together in a crystal lattice structure forming the bulk material we see and use in everyday life like wires, coins etc. But the smallest unit of this structure, the unit which represents its basic physical and chemical properties, is a single copper atom.
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What percentage of radioactive substance remains after two half-lives
Answer:
After 2 half-lives there will be 25% of the original isotope, and 75% of the decay product. After 3 half-lives there will be 12.5% of the original isotope, and 87.5% of the decay product. After 4 half-lives there will be 6.25% of the original isotope, and 93.75% of the decay product.
Explanation:
A sample of chlorine gas occupies a volume of 775 mL at a pressure of 545 mmHg. Calculate the pressure of the gas (in mmHg) if the volume is reduced at constant temperature to 171 mL. Enter your answer in scientific notation.
Answer : The final pressure of the gas is, [tex]2.47\times 10^3mmHg[/tex]
Explanation :
Boyle's Law : It is defined as the pressure of the gas is inversely proportional to the volume of the gas at constant temperature and number of moles.
[tex]P\propto \frac{1}{V}[/tex]
or,
[tex]P_1V_1=P_2V_2[/tex]
where,
[tex]P_1[/tex] = initial pressure = 545 mmHg
[tex]P_2[/tex] = final pressure = ?
[tex]V_1[/tex] = initial volume = 775 mL
[tex]V_2[/tex] = second volume = 171 mL
Now put all the given values in the above equation, we get:
[tex]545mmHg\times 775mL=P_2\times 171mL[/tex]
[tex]P_2=2470.03mmHg=2.47\times 10^3mmHg[/tex]
Therefore, the final pressure of the gas is, [tex]2.47\times 10^3mmHg[/tex]
How many moles are in 65 g of carbon dioxide (CO2)?
Answer:
1.477mole
Explanation:
First, we'll begin by calculating the molar mass of CO2. This is illustrated below:
Molar Mass of CO2 = 12 + (2x16) = 12 + 32 = 44g/mol
Mass of CO2 from the question = 65g
Mole of CO2 =?
Number of mole = Mass/Molar Mass
Mole of CO2 = 65/44
Mole of CO2 = 1.477mole
Therefore, 65g of carbon dioxide (CO2) contains 1.477mole
What is meant by enthalpy change?
Enthalpy change is a measurement of heat energy absorbed or released during a chemical reaction at constant pressure. It reflects the net energy required to break the chemical bonds of reactants versus the energy released when new bonds form in products, determining if the reaction is endothermic or exothermic.
It represents the difference between the enthalpy (heat content) of the reactants and the products. When chemical bonds are broken in reactants, energy is absorbed, and when new bonds form in products, energy is released. If the energy required to break bonds is greater than the energy released when new bonds are formed, the reaction is endothermic and the enthalpy change is positive; conversely, if more energy is released during bond formation than is absorbed during bond breaking, the reaction is exothermic and exhibits a negative enthalpy change.
The enthalpy change can be calculated using the formula ΔH = Σ (bond dissociation energies of reactants) minus Σ (bond dissociation energies of products), which simplifies the overall energy balance of bond breaking and bond formation processes. This understanding allows scientists to predict whether a reaction will absorb or release energy and is crucial in fields such as thermodynamics and chemical engineering.
Calculate the mass, in grams, of the solvent present in a 32.2% solution that contains 14.7 g of NaBr.
Answer:
32g
Explanation:
We have to remember that for percent (w/w) concentration we usually write;
Percent concentration= mass of solute/mass of solution ×100
Since mass of solute= 14.7 g and percent concentration = 32.2%
Then
Mass of solution= mass of solute × 100/ percent concentration
Mass of solution= 14.7 ×100/32.2
Mass of solution= 46.7 g
Since mass of solution = mass of solute + mass of solvent
Mass of solute= 14.7 g
Mass of solution = 46.7g
Mass of solvent = 46.7g -14.7g = 32g
A 1.00 L flask was filled with 2.00 mol gaseous SO2 and 2.00 mol gaseous NO2 and heated. After equilibrium was reached, it was found that 1.30 mol gaseous NO was present. Assume that the reaction occurs under these conditions. Calculate the value of the equilibrium constant, K, for the following reaction. SO2(g) NO2(g) equilibrium reaction arrow SO3(g) NO(g)
Answer:
Explanation:
SO3 (g) + NO (g) U SO2 (g) + NO2 (g)
occurs under these conditions. Calculate the value of the equilibrium constant, Kc, for the above reaction.
SO3 (g) + NO (g) U SO2 (g) + NO2 (g)
Initial (M) 2.00 2.00 0 0
Change (M) −x −x +x +x
Equil (M) 2.00 − x 2.00 − x x x
2 2 c 3
2
c 2
[SO ][NO ]
[SO ][NO]
(2.00 )
=
= −
K
x K
x
Since the problem asks you to solve for Kc, it must indicate in the problem what the value of x is. The concentration of
NO at equilibrium is given to be 1.30 M. In the table above, we have the concentration of NO set equal to 2.00 − x.
2.00 − x = 1.30
x = 0.70
Substituting back into the equilibrium constant expression:
2c 2 2c 2
(2.00 )
(0.70)
(2.00 0.70)
= − = −
x KxK
Kc = 0.290
help me please ...online claases got me crazy
Answer:
1,000 - 2,000
Explanation:
Just look at the bar graph. The bar ends at about 1,500 which is between 1,000 - 2,000. That is the only valid answer for this problem.
The frequency of individuals with doctorates who work in the field of physical science is 0-1,000.
Physical science is a branch of natural science that studies non-living systems and the fundamental principles governing the physical world. It encompasses physics and chemistry, exploring the properties and behavior of matter and energy. Physics delves into the nature of space, time, motion, and force, addressing phenomena from the smallest particles to the vastness of the cosmos.
Chemistry, on the other hand, focuses on the composition, structure, properties, and changes of matter. Together, these disciplines provide a comprehensive understanding of the physical universe, forming the basis for technological advancements and contributing to our comprehension of the fundamental laws that govern the natural world.
Which situation will most likely cause the sustainability of an ecosystem to weaken?
A zebra is killed and eaten by a lion.
A disease wipes out several plant species.
A frog lays hundreds of eggs that hatch into tadpoles.
A dry season causes a few small trees to die.
Answer:
B. A disease wipes out several plant species.
Explanation:
This will be the most detrimental to the environment as it will decrease oxygen levels and many animals will lose their shelter.
- everything else happens every year and is common and normal.
A disease wiping out several plant species will most likely cause the sustainability of an ecosystem to weaken.
What is an ecosystem?Ecosystem is defined as a system which consists of all living organisms and the physical components with which the living beings interact. The abiotic and biotic components are linked to each other through nutrient cycles and flow of energy.
Energy enters the system through the process of photosynthesis .Animals play an important role in transfer of energy as they feed on each other.As a result of this transfer of matter and energy takes place through the system .Living organisms also influence the quantity of biomass present.By decomposition of dead plants and animals by microbes nutrients are released back in to the soil.
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A gas is sealed in a rigid canister at a temperature of –5.0°C and a pressure of 713 mmHg. Which of the following actions would most likely bring the gas to STP?
heating the canister
removing some of the gas from the canister
transfer some of the gas to a larger canister
Answer: heating the canister
Answer:
heating the canister
Explanation:
NaOH and bleach have several characteristics in common. They include all BUT one of the characteristics listed.
Answer:
Their pH is less than 7
Explanation:
Both of these compounds pHs are greater than 7.
Answer:
A) Their pH is less than 7
Explanation:
Their pH is less than 7 is not correct. Their pH is greater than seven, but not equal to seven.
Consider the reaction between 15.0 mL of a 1.00 M aqueous solution of AgNO3 and 10.0 mL of a 1.00 M aqueous solution of K2CrO4. When these react, a precipitate is observed. What is present in solution after the reaction is complete
The two solutions AgNO3 and K2CrO4 react to give a precipitate along with an aqueous solution of KNO3. Hence, after the reaction is complete, KNO3 remains in the solution.
Explanation:When the two aqueous solutions AgNO3 and K2CrO4 react, they produce solid silver chromate Ag2CrO4 as a precipitate and leave potassium nitrate KNO3 in the solution. The reaction is balanced as 2AgNO3 (aqueous) + K2CrO4 (aqueous) -> Ag2CrO4 (s) + 2KNO3 (aqueous). As the reaction proceeds, silver ions (Ag+) and chromate ions (CrO42-) combine to form the precipitate, leaving potassium ions (K+) and nitrate ions (NO3-) in the solution. Hence, after the reaction, the solution consists of KNO3 because it remains aqueous, with 1.00 M K+ and 1.00 M NO3-.
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The pressure in a car tire is 198 kPa at 27°C. After a long drive, the pressure
is 225 kPa. What is the temperature of the air in the tire? Assume that the
volume is constant. Make sure your answer is rounded to nearest whole
number and your final answer has the units of K.*
Answer : The temperature of the air in the tire is, 341 K
Explanation :
Gay-Lussac's Law : It is defined as the pressure of the gas is directly proportional to the temperature of the gas at constant volume and number of moles.
[tex]P\propto T[/tex]
or,
[tex]\frac{P_1}{T_1}=\frac{P_2}{T_2}[/tex]
where,
[tex]P_1[/tex] = initial pressure = 198 kPa
[tex]P_2[/tex] = final pressure = 225 kPa
[tex]T_1[/tex] = initial temperature = [tex]27^oC=273+27=300K[/tex]
[tex]T_2[/tex] = final temperature = ?
Now put all the given values in the above equation, we get:
[tex]\frac{198kPa}{300K}=\frac{225kPa}{T_2}[/tex]
[tex]T_2=340.9K\approx 341K[/tex]
Therefore, the temperature of the air in the tire is, 341 K
PLEASE HELP ME OUT!!
Acids are made with compounds made of a nonmetal and _______.
a. oxygen b. carbon c. sulfur d. nitrogen
Answer:
hydrogen
Explanation:
A bullet of mass 10 g is fired from a gun of mass 490 g. The bullet leaves the gun with a speed of 120 ms-1. Find the speed of recoil of the gun.
A. 2.4 ms-1 B. 4.8 ms-1
C. 6.4 ms-1 D. 7.4 ms-1
Final answer:
The recoil speed of the gun is found using the conservation of momentum. The calculation shows that the gun's recoil speed is approximately 2.4 ms-1. So the correct option is A.
Explanation:
The question posed is related to the concept of conservation of momentum in physics. When a bullet is fired from a gun, the bullet moves forward and the gun is pushed back due to recoil, both objects conserving the system's total momentum. To find the speed of recoil of the gun, you can use the formula:
[tex]m_{bullet}v_{bullet} = m_{gun}v_{gun}[/tex]
Where:
[tex]m_{bullet}[/tex] mass of the bullet (10 g or 0.01 kg)[tex]v_{bullet}[/tex] speed of the bullet (120 m/s)[tex]m_{gun}[/tex]e mass of the gun (490 g or 0.49 kg)[tex]v_{gun}[/tex]the recoil speed of the gunThus, solving for [tex]v_{gun}[/tex] :
[tex]v_{gun} = m_{bullet}v_{bullet} / m_{gun}[/tex]
[tex]v_{gun}[/tex]= (0.01 kg * 120 m/s) / 0.49 kg = 2.45 m/s
The speed of the recoil of the gun is therefore approximately 2.4 ms-1, which matches answer choice A.
all helium atoms have 2 protons. what is the atomic number of helium?
Answer:
All helium atoms have two protons, and no other elements have atoms with two protons. In the case of helium, the atomic number is 2. The atomic number of an element is usually written in front of and slightly below the element's symbol, like in the Figure below for helium.
How many liters of o2 at 298 k and 1.00 bar are produced in 1.25 hr in an electrolytic cell operating at a current of 0.0500 a?
Answer:
0.0144 L
Explanation:
Step 1:
Data obtained from the question. This includes:
Temperature (T) = 298k
Pressure (P) = 1 bar
Time (t) = 1.25 hr
Current (I) = 0.0500 A
Step 2:
Determination of the quantity of electricity (Q) used. This is illustrated below:
Q = it
Time (t) = 1.25 hr = 1.25 x 3600 = 4500 secs
Current (I) = 0.0500 A
Quantity of electricity (Q) =?
Q = it
Q = 0.05 x 4500
Q = 225C
Step 3:
Determination of the number of mole of O2 liberated in the process.
In the electrolytic process, O2 will be liberated according to the equation:
2O^2- + 4e- —> O2
From the above illustration, 4 faraday are needed to liberate 1 mole of O2.
1 faraday = 96500C
Therefore of 4 faraday = 4x96500C = 386000C
From the above equation,
386000C of electricity liberated 1 mole of O2.
Therefore, 225C will liberate = 225/386000 = 5.83x10^-4 mole of O2.
Step 4:
Determination of the volume of the O2 liberated.
Number of mole (n) = 5.83x10^-4 mole
Temperature (T) = 298k
Pressure (P) = 1 bar = 0.987 atm
Gas constant (R) = 0.082atm.L/Kmol
Volume (V) =?
Applying the ideal gas equation:
PV = nRT
The volume of O2 can be obtained as follow:
PV = nRT
0.987 x V = 5.83x10^-4 x298x0.082
Divide both side by 0.987
V = (5.83x10^-4 x298x0.082)/0.987
V = 0.0144 L
The volume of [tex]\(O_2\)[/tex] produced is approximately [tex]0.0145 \ liters[/tex] at [tex]298 \ K[/tex] and [tex]1.00[/tex] bar.
To determine the volume of [tex]\(O_2\)[/tex] gas produced in an electrolytic cell, we can follow these steps:
Calculate the total charge
The total charge [tex]\(Q\)[/tex] passed through the cell is given by the product of current [tex]\(I\)[/tex] and time [tex]\(t\)[/tex]:
[tex]\[ Q = I \cdot t \][/tex]
Given:
[tex]Current, \(I = 0.0500 \, \text{A}\)[/tex]
[tex]Time, \(t = 1.25 \, \text{hr} = 1.25 \times 3600 \, \text{s} = 4500 \, \text{s}\)[/tex]
[tex]\[ Q = 0.0500 \, \text{A} \times 4500 \, \text{s} = 225 \, \text{C} \][/tex]
Determine the amount of [tex]\(O_2\)[/tex] produced
The half-reaction for the production of [tex]\(O_2\)[/tex] in water electrolysis is:
[tex]\[ 2 \, \text{H}_2\text{O} \rightarrow \text{O}_2 + 4 \, \text{H}^+ + 4 \, \text{e}^- \][/tex]
From this reaction, we see that [tex]4[/tex] moles of electrons produce [tex]1[/tex] mole of [tex]\(O_2\)[/tex]
The Faraday constant [tex](\(F\))[/tex] is [tex]\(96485 \, \text{C/mol}\)[/tex].
The moles of electrons [tex](\(n_{\text{e}^-}\))[/tex] can be calculated as:
[tex]\[ n_{\text{e}^-} = \frac{Q}{F} = \frac{225 \, \text{C}}{96485 \, \text{C/mol}} = 0.00233 \, \text{mol} \][/tex]
From the stoichiometry of the half-reaction, [tex]4[/tex] moles of electrons produce [tex]1[/tex] mole of [tex]\(O_2\)[/tex]:
[tex]\[ n_{\text{O}_2} = \frac{n_{\text{e}^-}}{4} = \frac{0.00233 \, \text{mol}}{4} = 0.000583 \, \text{mol} \][/tex]
Calculate the volume of [tex]\(O_2\)[/tex] gas
Use the ideal gas law to find the volume of [tex]\(O_2\)[/tex] gas at [tex]298\ K[/tex] and [tex]1.00[/tex] bar:
[tex]\[ PV = nRT \][/tex]
Given:
Pressure, \[tex](P = 1.00 \, \text{bar} = 1.00 \times 10^5 \, \text{Pa}\)[/tex]
Temperature, [tex]\(T = 298 \, \text{K}\)[/tex]
Gas constant, [tex]\(R = 8.314 \, \text{J/(mol K)}\)[/tex]
Moles of [tex]\(O_2\), \(n = 0.000583 \, \text{mol}\)[/tex]
[tex]\[ V = \frac{nRT}{P} = \frac{0.000583 \, \text{mol} \times 8.314 \, \text{J/(mol K)} \times 298 \, \text{K}}{1.00 \times 10^5 \, \text{Pa}} \][/tex]
[tex]\[ V = \frac{1.447 \, \text{J}}{1.00 \times 10^5 \, \text{Pa}} = 1.447 \times 10^{-5} \, \text{m}^3 \][/tex]
[tex]\[ V = 0.01447 \, \text{L} \][/tex]
86.1 g of nitrogen reacts with lithium, how many grams of lithium will react?
Answer:
128g of Li, will react in this reaction
Explanation:
Before to start working, we need the reaction:
N₂ and Li react, in order to produce Li₃N (lithium nitride)
N₂ + 6Li → 2Li₃N
1 mol of nitrogen reacts with 6 moles of lithium
We convert the mass of N₂ to moles → 86.1 g . 1 mol/ 28g = 3.075 moles
1 mol of N₂ reacts with 6 mol of Li
Therefore, 3.075 moles of N₂ will react with 18.4 moles of Li
We conver the moles to mass → 18.4 mol . 6.94g / 1mol = 128 g
O3(g)+NO(g)<—>O2(g)+NO2(g) Write the equalibrium expression
Answer: [NO2][O2]/[NO] [O3]
Explanation: Kc = [NO2][O2]/[NO] [O3]
what happens when you mix MORE vinegar than baking soda?
Answer:
Excess Vinegar
Explanation:
You will have excess vinegar