Electrolysis of nitric acid produces hydrogen gas at the cathode and nitrogen dioxide, oxygen gas, and hydrogen ions at the anode.
When an aqueous solution of 4mol/L nitric acid is electrolyzed in an electrolytic cell using graphite electrodes, the following chemical symbols for all the ions present in the electrolytic cell are obtained:A) The anode equation: 2HNO₃ → 2NO₂ + O₂ + 2H⁺ + 2e⁻The anions present in the electrolytic cell include nitrate ions (NO₃⁻) and chloride ions (Cl⁻). The nitrate ions, which are the conjugate base of nitric acid (HNO₃), are attracted to the anode where they lose electrons to produce nitrogen dioxide (NO₂), oxygen gas (O₂), and hydrogen ions (H⁺). Nitric oxide (NO) and nitrogen gas (N₂) can also be formed as by-products.B) The cathode equation: H⁺ + e⁻ → 1/2H₂The cations present in the electrolytic cell include hydrogen ions (H⁺) and nitrate ions (NO₃⁻). Hydrogen ions (H⁺) in the electrolytic cell are reduced by gaining electrons at the cathode to produce hydrogen gas (H₂). Nitrate ions (NO₃⁻) from nitric acid are not reduced at the cathode but migrate toward the anode. In summary, nitric acid, when electrolyzed in an electrolytic cell using graphite electrodes, produces hydrogen gas at the cathode and nitrogen dioxide, oxygen gas, and hydrogen ions at the anode. The conjugate base of nitric acid, nitrate ions, is present as anions in the electrolytic cell. The cations present in the electrolytic cell are hydrogen ions and nitrate ions.For more questions on Electrolysis
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Science Question!
Please order by correct order if Answer and please be Real!
Answer:
matter undergoes
chemical changes such as burning and rusting.
physical changes such as evaporating and melting.
matter has
chemical properties such as reacting with oxygen and changing when heated.
physical properties such as luster and volume.
4. Styrene (A) and Butadiene (B) are to be polymerized in a
series of mixed-flow reactors, each of volume 25 m3. The rate
equation is first order with respect to A and B:
−rA = kACACB
where kA = 10−5 m3·kmol−1·s−1
The initial concentration of styrene is 0.8 kmol·m−3 and
of butadiene is 3.6 kmol·m−3. The feed rate of reactants
is 20 t·h−1. Estimate the total number of reactors required
for polymerization of 85% of the limiting reactant. Assume
the density of reaction mixture to be 870 kg·m−3 and the
molar mass of styrene is 104 kg·kmol−1 and that of butadiene
54 kg·kmol−1
The total number of reactors required for polymerization of 85% of the limiting reactant is 4.
The calculation of the total number of reactors required for polymerization of 85% of the limiting reactant for Styrene (A) and Butadiene (B) is explained below.
Given data: Volume of each reactor, V = 25 m³.
The rate equation is, -rA = kACACB ,where kA = 10⁻⁵ m³·kmol⁻¹·s⁻¹
Initial concentration of Styrene = CA0 = 0.8 kmol·m⁻³ .Initial concentration of Butadiene = CB0 = 3.6 kmol·m⁻³
Feed rate of reactants = 20 t·h⁻¹Density of reaction mixture = ρ = 870 kg·m⁻³
Molar mass of Styrene = MStyrene = 104 kg·kmol⁻¹Molar mass of Butadiene = MButadiene = 54 kg·kmol⁻¹
The limiting reactant in the polymerization is the reactant that gets consumed first. Let's assume that Butadiene is the limiting reactant since it has the lowest initial concentration.
Mass balance equation for Butadiene,
FA0 = CA0.V.QFA = ρ.V.Q.CB
Where FA0 is the initial flow rate of Styrene, Q is the total volumetric flow rate of reactants.
Since the reaction is first-order with respect to both Styrene and Butadiene,-rA = -rB = kACACBVolume of reactant fed in 1 h = Q × 3600s = 20,000 kg
For a batch of 85% limiting reactant conversion,
Total moles of Butadiene fed in 1 h, nB = CB0.V.Q × 3600 × 0.85
Moles of Styrene required to react with 85% of Butadiene, n
Styrene = nB (MButadiene/MStyrene) = 15.08 V.Qkg
Number of moles of Styrene per reactor required to reach the above requirement in 1 h,
nStyrene/reactor = nStyrene/Total Number of Reactors Total Volume of all Reactors= nStyrene/ (Total Volume of Reactors/V)
Number of Reactors required = Total Volume of Reactors / V = nStyrene / (nStyrene/reactor) = 15.08 V.Qkg / (CA0 × V × kA × CB0) ≈ 3.36 → 4Reactors Hence, the total number of reactors required for polymerization of 85% of the limiting reactant is 4.
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If the pressure, volume, and the number of moles of a gas are known, which is needed to calculate the universal gas constant from the ideal gas law?the temperature of the gas the molar volume of the gasthe molar mass of the gasthe partial pressure of the gas
If the pressure, volume, and the number of moles of a gas are known, the temperature of the gas is needed to calculate the universal gas constant from the ideal gas law.
The synthesis of the following four rules led to the ideal gas law:
1) Boyle's Law: According to this rule, pressure is inversely related to a gas's volume and molecular weight at constant temperature.
P ∝ [tex]\frac{1}{V}[/tex] (At a certain temperature and molecular count)
2) Charles' Law: According to this rule, the volume of a gas with constant pressure and moles is precisely proportionate to its temperature.
V ∝ T (With the same pressure and mole count)
3) According to Gay-Lussac's third law, pressure is directly proportional to the gas's temperature for a gas with a fixed volume and number of moles.
P ∝ T (At constant volume and mole-count)
4) According to Avogadro's Law, at constant pressure and temperature, the volume of a gas is directly proportionate to its molecular weigh
V ∝ n (With respect to constant pressure and temperature)
Ideal gas Equation :
PV = nRT
where,
P stands for gas pressure.
Gas temperature is denoted by T.
The amount of gas molecules is N.
N is the number of gas moles.
R is the gas constant
So, in order to compute the gas constant, we must first know the gas's temperature.
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985.2 moles of nitrogen, how many moles of ammonia can produce?
Answer:
985.2 moles of nitrogen can produce 1970.4 moles of ammonia.
Explanation:
The balanced chemical equation for the production of ammonia from nitrogen is:
N2 + 3H2 → 2NH3
From the balanced equation, we can see that 1 mole of nitrogen reacts with 3 moles of hydrogen to produce 2 moles of ammonia.
So, to determine how many moles of ammonia can be produced from 985.2 moles of nitrogen, we need to use the mole ratio from the balanced chemical equation as follows:
985.2 moles N2 x (2 moles NH3 / 1 mole N2) = 1970.4 moles NH3
Therefore, 985.2 moles of nitrogen can produce 1970.4 moles of ammonia.
5. Which of the functional groups contain(s) nitrogen?
Explanation:
Functional groups containing nitrogen are amines and amides.
The general formula for amines is:
RNH₂, where R = longer hydrocarbon chain.
The general formula for amides is:
RCONH₂, where R = longer hydrocarbon chain.
See attached diagram for general structural formula.
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What is the density at STP of NOz gas (molar
mass = 46.01 g/mol) in grams per liter?
Answer:
We can use the ideal gas law, PV = nRT, to solve for the density at STP (standard temperature and pressure). At STP, the temperature is 273.15 K and the pressure is 1 atm. We know the molar mass of NO2 is 46.01 g/mol. We also know that 1 mole of any gas at STP occupies a volume of 22.4 L.
First, we can calculate the number of moles of NO2 at STP:
n = PV/RT = (1 atm)(22.4 L)/(0.08206 L·atm/mol·K)(273.15 K) = 1.00 mol
Next, we can calculate the mass of 1 mole of NO2:
46.01 g/mol
Finally, we can calculate the density of NO2 at STP:
density = mass/volume = (46.01 g/mol)/(22.4 L) = 2.054 g/L
Therefore, the density at STP of NO2 gas (molar mass = 46.01 g/mol) in grams per liter is 2.054 g/L.
Explanation:
Re-read the Topic 2 Learning Activities titled “Glycolysis” and “Overview of Photosynthesis”. What makes these necessary fundamental processes? Use an argument from the reading to support your answer. In what ways are these two processes similar? How are they different?
Glycolysis and photosynthesis are fundamental processes that are necessary for the survival of living organisms. They are similar in that they both involve the conversion of energy, but differ in the source of energy used, the location of the process, and the requirement for oxygen.
Glycolysis and photosynthesis are two necessary fundamental processes. Glycolysis is a metabolic pathway that occurs in the cytoplasm of cells. The glycolysis process is necessary because it produces ATP, which is the energy required for all cellular activities.
The energy is produced by breaking down glucose into two pyruvate molecules.Photosynthesis is the process by which green plants make their food. During photosynthesis, light energy is converted into chemical energy, which is stored in glucose molecules. This process is also necessary as it provides food and oxygen for most living organisms to survive.In terms of similarities, both glycolysis and photosynthesis are processes that involve the conversion of energy.
In glycolysis, glucose is converted into pyruvate and ATP, while in photosynthesis, light energy is converted into chemical energy. Both processes are also vital to the survival of living organisms.The primary difference between the two processes is the source of energy used. Glycolysis uses glucose as the primary energy source while photosynthesis uses light energy from the sun.
Glycolysis occurs in the cytoplasm of cells while photosynthesis takes place in the chloroplasts of plant cells. Glycolysis is an anaerobic process that does not require oxygen, while photosynthesis is an aerobic process that requires oxygen and releases it as a byproduct.
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In NH3+H2O > NH4OH which is being oxidized and which is being reduced?
Answer:
It doesn't look like there is any oxidation going on to me.
Explanation:
Oxidation: loss of electrons, Reduction: gain of electrons
in NH3, the charges are (-3 +3)=0. in NH4OH, the charge is (-3 +4 -2 +1)=0
Unless I'm wrong (which is def possible), N keeps a -3 charge, H is always +1, O is always -2, and both sides of the equation are neutral over all.
Given Kc = 2367 at 999 K, calculate Kp for the reaction at equilibrium: CS₂(g) + 3Cl₂(g) → S₂Cl3(g) + CCl4(8) R = 0.08206 L atm K-¹ mol-¹
The value of Kp for the given reaction at equilibrium is approximately 192,986.689 L atm mol⁻¹.
To calculate the equilibrium constant Kp for the given reaction, we can use the relationship between Kc and Kp, which is expressed as:
Kp = Kc * (RT)^Δn
Where:
- Kp is the equilibrium constant in terms of partial pressures.
- Kc is the equilibrium constant in terms of concentrations.
- R is the ideal gas constant (0.08206 L atm K⁻¹ mol⁻¹).
- T is the temperature in Kelvin.
- Δn is the change in the number of moles of gas (sum of products - sum of reactants).
In this case, the reaction involves four moles of gas on the left-hand side (reactants) and five moles of gas on the right-hand side (products). Therefore, Δn = 5 - 4 = 1.
Given that Kc = 2367 at 999 K, we can substitute these values into the equation:
Kp = 2367 * (0.08206 L atm K⁻¹ mol⁻¹ * 999 K)^1
Simplifying the expression:
Kp = 2367 * (81.367 L atm mol⁻¹)
Calculating the product:
Kp ≈ 192,986.689 L atm mol⁻¹
Therefore, the value of Kp for the given reaction at equilibrium is approximately 192,986.689 L atm mol⁻¹.
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Select the correct terms to complete this statement about charged particles.
Like charges attract | repel, and opposite charges attract repel. According to Coulomb's law, as the distance between two charged particles decreases, the force between the particles decreases I increases. As the magnitude of the charges decreases, the force decreases | increases.
Like charges repel each other, while opposite charges attract each other. This principle is one of the fundamental aspects of electrostatics. According to Coulomb's law, the force between two charged particles is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.
As the distance between two charged particles decreases, the force between them increases. This is because the closer the particles are, the stronger the electric field they create, leading to a stronger force of interaction.
On the other hand, as the magnitude of the charges decreases, the force between the particles also decreases. This is because the force is directly proportional to the product of the charges. If one or both of the charges are smaller, the force they exert on each other will be weaker.
In summary, according to Coulomb's law, decreasing the distance between charged particles increases the force between them, while decreasing the magnitude of the charges decreases the force. This understanding of the relationship between charge, distance, and force is crucial in explaining the behavior of charged particles and the interactions between them.
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