1. The entropy change (ΔS) of the reaction 2C2H6(g) + 7O2(g) → 2CO2(s) + 6H2O(g) at 25°C can be calculated using standard entropy data from the table of thermodynamic properties.
2. The Gibbs energy change (ΔG) of the reaction 2C2H4(g) + 7O2(s) → 2CO2(s) + 6H2O(s) at 25°C can be calculated using Gibbs energy of formation data from the table of thermodynamic properties. Based on the calculated ΔG value, the spontaneity of the reaction can be determined.
1. To calculate the entropy change (ΔS) of the reaction, you need to subtract the sum of the standard entropies of the reactants from the sum of the standard entropies of the products. The values for ΔS can be obtained from the attached table of thermodynamic properties.
2. To calculate the Gibbs energy change (ΔG) of the reaction, you need to subtract the sum of the Gibbs energy of formation of the reactants from the sum of the Gibbs energy of formation of the products. The values for ΔG can be obtained from the attached table of thermodynamic properties. If the calculated ΔG value is negative, the reaction is spontaneous at 25°C; if it is positive, the reaction is nonspontaneous.
To know more about Gibbs energy here: brainly.com/question/30400809
#SPJ11
What is the result of the following computation in scientific
notation, using "e" notation and proper significant figures?
(5X10^3)(2X10^-5)/1X10^-3
The final computed answer in scientific notation, using "e" notation, is 10^1 or simply 10.
To find the result of the given computation, let's break it down step by step.
Step 1: Multiplication of the numerator
(5 × 10^3) × (2 × 10^-5) = 10^3 × 10^-5 = 10^(3 - 5) = 10^-2
Step 2: Division by the denominator
10^-2 ÷ (1 × 10^-3) = 10^-2 ÷ 10^-3 = 10^(-2 - (-3)) = 10^(-2 + 3) = 10^1
Step 3: Simplification
10^1 can be written as 10.
Therefore, the final computed answer in scientific notation, using "e" notation, is 10^1 or simply 10.
Given computation:
(5 × 10^3)(2 × 10^-5) / (1 × 10^-3)
To solve this expression, we follow the rules of exponents. Let's break it down into smaller steps:
Step 1: Multiplication of the numerator
In the numerator, we have two terms: (5 × 10^3) and (2 × 10^-5).
When multiplying numbers in scientific notation, we multiply the coefficients and add the exponents.
5 × 2 = 10
10^3 × 10^-5 = 10^(3 - 5) = 10^-2
So, the numerator simplifies to 10^-2.
Step 2: Division by the denominator
In the denominator, we have (1 × 10^-3).
To divide by a term in scientific notation, we subtract the exponents.
10^-2 ÷ (1 × 10^-3) = 10^(-2 - (-3)) = 10^(-2 + 3) = 10^1
So, the division simplifies to 10^1.
Step 3: Simplification
10^1 can be written as 10 in scientific notation.
Therefore, the final computed answer in scientific notation, using "e" notation, is 10^1 or simply 10.
The result of the computation is 10.
To learn more about notation click here: brainly.com/question/16936662
#SPJ11
With the high sedimentation and evaporation rates associated
with most dams and reservoirs, are they really a sustainable and
efficient way to store water?
Dams and reservoirs can provide reliable water storage but have potential environmental impacts such as habitat loss and disruption of natural flow.
Sedimentation can reduce the storage capacity of reservoirs over time, requiring periodic dredging or desilting to maintain efficiency. Evaporation can lead to water loss from the reservoir, particularly in arid or semi-arid regions with high evaporation rates. These factors need to be carefully managed to ensure the long-term sustainability and efficiency of water storage.
Sedimentation and evaporation are important considerations that need to be managed. Social and economic impacts should also be assessed. Alternative approaches and improved water management practices can enhance sustainability and efficiency. Careful planning, impact assessments, stakeholder engagement, and monitoring are crucial for mitigating negative impacts and maximizing benefits.
To know more about Dams, refer:
https://brainly.com/question/16426042
#SPJ4
For the reaction 2A(g)+2B(g)⇌C(g) Kc = 89.4 at a temperature of 25 ∘C . Calculate the value of Kp .
To calculate the value of Kp for a given reaction using the equilibrium constant Kc, we need to consider the stoichiometry of the reaction and the ideal gas law.
For the reaction 2A(g) + 2B(g) ⇌ C(g), the stoichiometric coefficients indicate that two moles of gas A and two moles of gas B react to form one mole of gas C.
The relationship between Kc and Kp is given by the equation: Kp = Kc(RT)^Δn, where R is the gas constant, T is the temperature in Kelvin, and Δn is the change in the number of moles of gas (products - reactants).
In this case, Δn = (1 - 2 - 2) = -3, as there are three moles of gas on the reactant side and one mole of gas on the product side.
Given that Kc = 89.4, and assuming an ideal gas behavior, we can calculate Kp using the ideal gas law (PV = nRT), where P is the pressure and V is the volume. At equilibrium, the partial pressures of gases A, B, and C can be related to the equilibrium constant by:
Kp = (PC)^1/(PA)^2 * (PB)^2
To know more about gas constant here: brainly.com/question/14279790
#SPJ11
10. ( 2pts ) Make 5 serial dilutions, each generating 1 L of a i-in-10 dilution of a 13.2M stock solution
The five 1-in-10 serial dilutions, each generating 1 L of dilute solution, can be made from 1 L of 13.2 M stock solution.
Serial dilution is a process of successive dilutions in which the dilution factor or the concentration decreases with each step. In this context, the dilution is made in such a way that each step or the subsequent dilution gives a 1 in 10 dilution factor.
To make 5 serial dilutions, each generating 1 L of an i-in-10 dilution of a 13.2 M stock solution, the following steps can be followed:
Start with 1 L of 13.2 M stock solution.
To generate a 1-in-10 dilution of this solution, 1 part of the stock solution should be mixed with 9 parts of solvent.
Here, solvent will be water, so 1 L of 13.2 M stock solution will be mixed with 9 L of water.The solution obtained from the above step will be a 1-in-10 dilution of the stock solution.
Next, to generate a 1-in-10 dilution of the 1-in-10 dilution obtained in the previous step, 1 part of this solution will be mixed with 9 parts of solvent. This will give a 1-in-100 dilution of the stock solution. The volume of this solution will be 10 L.
Now, to generate a 1-in-10 dilution of the 1-in-100 dilution, 1 part of the 1-in-100 dilution will be mixed with 9 parts of solvent. This will give a 1-in-1000 dilution of the stock solution. The volume of this solution will be 10 L.
Similarly, for the 4th and 5th dilution, 1 part of the 1-in-1000 dilution will be mixed with 9 parts of solvent each time, and this will give a 1-in-10,000 dilution and a 1-in-100,000 dilution, respectively.
The volume of each of these solutions will be 10 L.
So, in this way, five 1-in-10 serial dilutions, each generating 1 L of dilute solution, can be made from 1 L of 13.2 M stock solution.
To know more about dilute solution visit:
https://brainly.com/question/31521767
#SPJ11
Which of the following rules apply when filling molecular orbitals with valence electrons? (Select all that apply) Pauli exclusion principle The building up principle For homonuclear diatomic molecules O2, and F2, place the σ2pMO higher in energy than the π2p MO's Hund's rule Place antibonding orbitals lower in energy than the starting atomic orbital
The building up principle For homonuclear diatomic molecules O, and F, place the pMO higher in energy than the πp MO's Hund's rule Place antibonding orbitals lower in energy than the starting atomic orbital the correct options from the given choices are: Pauli exclusion principle
Hund's rule
The rules that apply when filling molecular orbitals with valence electrons are:
Pauli exclusion principle: This principle states that no two electrons in an atom or molecule can have the same set of quantum numbers. In the context of molecular orbitals,
it means that each orbital can accommodate a maximum of two electrons with opposite spins.
Hund's rule: According to Hund's rule, when filling degenerate (same energy) orbitals, electrons will occupy different orbitals with the same spin before pairing up.
This leads to the maximum possible spin alignment, which results in greater stability.
Therefore, the correct options from the given choices are:
Pauli exclusion principle
Hund's rule
To know more about homonuclear refer here :
https://brainly.com/question/29195136#
#SPJ11
By what mechanism does glucagon promote hyperglycemia? Glucagon inhibits glycogenesis. Glucagon stimulates insulin release. Glucagon stimulates gluconeogenesis. Glucagon inhibits gluconeogenesis.
Glucagon promotes hyperglycemia by stimulating gluconeogenesis. Hence the option C is the correct answer.
A hormone named glucagon is secreted by pancreatic alpha cells when blood glucose levels fall below the normal range. When the hormone binds to the hepatic cell membrane receptor, it leads to a series of intracellular events that ultimately result in the production of glucose from noncarbohydrate sources such as amino acids and lactate.
This process is known as gluconeogenesis, and it aids in the maintenance of blood glucose levels by raising them when they are too low. Gluconeogenesis is stimulated by glucagon, while glycogenesis is inhibited. Glucagon, like insulin, regulates carbohydrate metabolism in the body.
However, the two hormones have opposite effects. When blood sugar levels are low, glucagon is secreted, and when blood sugar levels are high, insulin is secreted.
To know more about Glucagon click on below link :
https://brainly.com/question/29604853#
#SPJ11
G. (3 points) Draw the structure of the compound with the \( { }^{1} \mathrm{H} \) NMR shown below. From the mass spectrum, the molecular formula was determined to be \( \mathrm{C}_{3} \mathrm{H}_{7}
The structure of the compound with the ¹H NMR shown is propan-2-ol (isopropyl alcohol), which has the molecular formula C₃H₇OH.
From the given information, we can deduce the structure of the compound by analyzing the ¹H NMR spectrum and the molecular formula. Here's the reasoning:
1. The molecular formula C₃H₇OH indicates that the compound contains three carbon atoms, seven hydrogen atoms, and one oxygen atom.
2. The presence of an alcohol functional group is indicated by the -OH in the molecular formula.
3. The ¹H NMR spectrum shows a singlet at around 3.7 ppm, which corresponds to the hydroxyl group (-OH) of an alcohol.
4. Based on the singlet at 3.7 ppm, we can conclude that the hydroxyl group is attached to a tertiary carbon atom since it shows no splitting from neighboring hydrogens.
5. With three carbon atoms and a hydroxyl group, the structure can be determined as propan-2-ol (isopropyl alcohol), where the hydroxyl group is attached to the middle carbon atom of a three-carbon chain (CH₃-CH(OH)-CH₃).
Therefore, the structure of the compound is:
H
|
H - C - C - H
|
H
learn more about molecular formula here:
https://brainly.com/question/15092254
#SPJ11
Comparing a molecule with hydrogen bonds with a molecule with Van der Waals forces. Which statement is true. The compound with Hydrogen bonds has a higher boiling point. Since the compound with hydrogen bonds also has Van der Walls forces there is not difference in the freezing point. The compound with hydrogen bonds has a smaller surface tension The compound with hydrogen bonds has less viscocity.
The compound with hydrogen bonds has a higher boiling point.
Hydrogen bonds are stronger intermolecular forces compared to Van der Waals forces. The presence of hydrogen bonds results in stronger attractions between molecules, leading to higher boiling points.
When a substance is heated, the intermolecular forces must be overcome to transition from the liquid phase to the gas phase. The stronger the intermolecular forces, the more energy is required to break these forces and convert the substance into a gas.
Therefore, a compound with hydrogen bonds, which are stronger than Van der Waals forces, will have a higher boiling point.
The statement "Since the compound with hydrogen bonds also has Van der Waals forces there is no difference in the freezing point" is incorrect.
While the compound with hydrogen bonds does have Van der Waals forces, the presence of hydrogen bonds typically increases the overall strength of intermolecular attractions, affecting both boiling and freezing points.
The statement "The compound with hydrogen bonds has a smaller surface tension" is also incorrect. Hydrogen bonds contribute to a higher surface tension because they create stronger attractions between molecules at the surface of a liquid.
The statement "The compound with hydrogen bonds has less viscosity" is also incorrect. Viscosity, or resistance to flow, is influenced by various factors, including intermolecular forces.
In general, substances with stronger intermolecular forces, such as those with hydrogen bonds, tend to have higher viscosity.
To know more about "Surface tension" refer here:
https://brainly.com/question/30765667#
#SPJ11
4. Design a procedure for making a 1.00×10 −4
M solution of HCl if all you are given is a stock solution with a concentration of 0.250M, access to 100.0−mL and 250.0−mL volumetric flasks and a set of volumetric pipets of any whole number volume in mL. Include specific volumes. This will require a serial dilution to accomplish the final molarity. (Reminder: thi means making a dilution, then using that dilution to dilute the solution even more.)
To make a 1.00×10⁻⁴ M HCl solution using a 0.250 M HCl stock solution, perform a serial dilution as follows.
Take 4.00 mL of the 0.250 M HCl stock solution and add it to a 100.0 mL volumetric flask.
Fill the flask to the mark with distilled water and mix thoroughly.
Take 2.50 mL of the resulting solution and add it to a 250.0 mL volumetric flask.
Fill the flask to the mark with distilled water and mix thoroughly.
The final solution in the 250.0 mL flask will have a concentration of 1.00×10⁻⁴ M HCl.
Hence, the solution is prepared
Learn more about the stock solution here:
https://brainly.com/question/17018950
#SPJ 4
6. What is ISP (isoelectric point) of polyampholytes (proteins)? Give the definition.
The isoelectric point (pI) of polyampholytes, including proteins, refers to the pH at which the net charge of the molecule is zero
The number of positively charged groups (like amino groups) and negatively charged groups (like carboxyl groups) are equal at the isoelectric point. As it affects their behavior and characteristics, such as solubility, electrophoretic mobility, and protein-protein interactions, the isoelectric point is a crucial aspect of polyampholytes. The polyampholyte typically has a net negative charge above the isoelectric point and a net positive charge below the isoelectric point. The content and arrangement of the amino acids in the polyampholyte affect the specific value of the isoelectric point.
Understanding how polyampholytes behave in various settings, such as biological systems or industrial applications, is dependent on knowing how their isoelectric point influences their stability, aggregation, and usefulness. The isoelectric point of polyampholytes, such as proteins, can be discovered experimentally using methods like electrophoresis or potentiometric titration.
To learn more about isoelectric point, refer to:
https://brainly.com/question/30898497
#SPJ4
1. A student wanted to test whether mercury (II) chromate (HgCrO4) was a water-soluble salt or not. To do this, they wanted to mix two different soluble salt solutions which, when combined together, would precipitate HgCrO4 is it was insoluble. Propose a precursor ionic compound that contains mercury (II) and some other ion that would be water soluble. Propose a precursor ionic compound that contains chromate (CrO42-) and some other ion that would be water soluble.
The student can combine aqueous solutions of sodium chromate (Na₂CrO₄) with water-soluble mercury (II) nitrate (Hg(NO₃)₂) to test the solubility of HgCrO₄. HgCrO₄'s insolubility can be determined if it precipitates.
To test the solubility of mercury (II) chromate (HgCrO₄), the student can use two precursor ionic compounds that contain water-soluble ions. Here are the proposed precursor compounds:
1. Precursor compound containing mercury (II) ion:
One suitable precursor compound that contains mercury (II) ion (Hg²⁺) and is water-soluble is mercury (II) nitrate, Hg(NO₃)₂. When dissolved in water, it dissociates into mercury (II) ions (Hg²⁺) and nitrate ions (NO₃⁻), both of which are soluble.
2. Precursor compound containing chromate ion:
One suitable precursor compound that contains chromate (CrO₄²⁻) and is water-soluble is sodium chromate, Na₂CrO₄. When dissolved in water, it dissociates into sodium ions (Na⁺) and chromate ions (CrO₄²⁻), both of which are soluble.
By mixing aqueous solutions of mercury (II) nitrate (Hg(NO₃)₂) and sodium chromate (Na₂CrO₄), the student can test if HgCrO₄ precipitates, indicating its insolubility.
To know more about the precursor compounds refer here,
https://brainly.com/question/28810429#
#SPJ11
For the given reaction:
H2(g) + F2(g) ↔ 2HF(g) K = 73
The initial concentrations of [H2] and [F2] are both 3.5 M. What is the equilibrium concentration of H2(g)?
Explain at least one of the student’s mistakes in their solution.
This problem is worth 10 points on the exam. How many points do you think this student should be awarded for this problem and why?
What suggestions would you give to this student for studying in the future?
The equilibrium concentration of H₂(g) is found to be approximately 16.38 M.
To determine the equilibrium concentration of H₂(g), we need to use the equilibrium constant expression and the given initial concentrations of [H₂] and [F₂]. The equilibrium constant expression for the reaction is,
K = [HF]² / ([H₂] * [F₂])
Given:
Initial concentrations: [H₂] = [F₂] = 3.5 M
Equilibrium constant: K = 73
Let's denote the equilibrium concentration of H₂ as x. Since 2 moles of HF are formed for every mole of H₂ consumed, the concentration of HF at equilibrium will be 2x. Now we can substitute these values into the equilibrium constant expression and solve for x:,
73 = (2x)² / (3.5 * 3.5)
73 = 4x² / 12.25
Multiplying both sides by 12.25,
4x² = 73 * 12.25
x² = (73 * 12.25) / 4
x² ≈ 268.34
Taking the square root of both sides,
x ≈ √268.34
x ≈ 16.38
Therefore, the equilibrium concentration of H₂(g) is approximately 16.38 M.
To know more about equilibrium concentration, visit,
https://brainly.com/question/13414142
#SPJ4
Balance the following REDOX reaction in basic solution:
Al(s) + NO2¯ (aq) ➝ AlO2¯ (aq) + NH3(aq)
Please write out all work using these steps:
1. Write the two half-reactions representing the redox process.
2. Balance all elements except oxygen and hydrogen.
3. Balance oxygen atoms by adding H2O molecules.
4. Balance hydrogen atoms by adding H+ ions.
5. Balance charge by adding electrons.
6. If necessary, multiply each half-reaction’s coefficients by the smallest possible integers to yield equal numbers of electrons in each.
7. Add the balanced half-reactions together and simplify by removing species that appear on both sides of the equation.
8. For reactions occurring in basic media (excess hydroxide ions), carry out these additional steps:
Add OH− ions to both sides of the equation in numbers equal to the number of H+ ions.
On the side of the equation containing both H+ and OH− ions, combine these ions to yield water molecules.
Simplify the equation by removing any redundant water molecules.
9. Finally, check to see that both the number of atoms and the total charges1 are balanced.
Balancing the redox reaction involves half-reactions, balancing atoms, charges, and adjusting for basic solution, resulting in a balanced equation.
1. The two half-reactions representing the redox process are:
Oxidation half-reaction: Al(s) ➝ AlO2¯ (aq)
Reduction half-reaction: NO2¯ (aq) ➝ NH3(aq)
2. Balancing elements except oxygen and hydrogen:
Oxidation half-reaction: Al(s) ➝ AlO2¯ (aq)
Reduction half-reaction: 3 NO2¯ (aq) ➝ NH3(aq)
3. Balancing oxygen atoms by adding H2O molecules:
Oxidation half-reaction: Al(s) ➝ AlO2¯ (aq) + H2O(l)
Reduction half-reaction: 3 NO2¯ (aq) ➝ NH3(aq) + H2O(l)
4. Balancing hydrogen atoms by adding H+ ions:
Oxidation half-reaction: Al(s) ➝ AlO2¯ (aq) + 2 H2O(l)
Reduction half-reaction: 3 NO2¯ (aq) + 6 H+ ➝ NH3(aq) + 2 H2O(l)
5. Balancing charge by adding electrons:
Oxidation half-reaction: Al(s) ➝ AlO2¯ (aq) + 2 H2O(l) + 4 e^-
Reduction half-reaction: 3 NO2¯ (aq) + 6 H+ + 4 e^- ➝ NH3(aq) + 2 H2O(l)
6. Multiply each half-reaction's coefficients by the smallest possible integers to equalize the number of electrons:
Oxidation half-reaction: 3 Al(s) ➝ 3 AlO2¯ (aq) + 6 H2O(l) + 12 e^-
Reduction half-reaction: 12 NO2¯ (aq) + 24 H+ + 12 e^- ➝ 4 NH3(aq) + 8 H2O(l)
7. Add the balanced half-reactions together and simplify:
3 Al(s) + 12 NO2¯ (aq) + 24 H+ ➝ 3 AlO2¯ (aq) + 4 NH3(aq) + 20 H2O(l)
8. Adjusting for basic solution:
Add OH^- ions to both sides to neutralize H+ ions:
3 Al(s) + 12 NO2¯ (aq) + 24 H2O(l) ➝ 3 AlO2¯ (aq) + 4 NH3(aq) + 20 H2O(l) + 24 OH^-(aq)
Combine H+ and OH^- ions to form water:
3 Al(s) + 12 NO2¯ (aq) + 24 H2O(l) + 24 OH^-(aq) ➝ 3 AlO2¯ (aq) + 4 NH3(aq) + 44 H2O(l)
Simplify by removing redundant water molecules:
3 Al(s) + 12 NO2¯ (aq) + 24 OH^-(aq) ➝ 3 AlO2¯ (aq) + 4 NH3(aq) + 20 H2O(l)
To know more about redox, click here:
brainly.com/question/28300253
#SPJ11
1.(20pts) Prepare 100 mL of 0.1MNaCl solution 2.(20pts) Prepare 100 mL of a 0.5wt%NaCl solution 3.(20pts) Prepare 100 mL of a 0.2wt%NaCl solution from the 0.5wt% solution. Design your own detailed procedures, carry out the procedures for all solution preparation, and report what you have done for preparation of all solutions using descriptive and/or mathematical expressions whenever necessary
1. To prepare a 0.1M NaCl solution, measure 10 mL of 1M NaCl solution and dilute it to 100 mL using a volumetric flask and distilled water.
2. To prepare a 0.5wt% NaCl solution, measure 0.5 g of NaCl and dissolve it in enough water to make a final volume of 100 mL.
3. To prepare a 0.2wt% NaCl solution from the 0.5wt% solution, measure 40 mL of the 0.5wt% NaCl solution and dilute it with enough water to make a final volume of 100 mL.
1. To prepare 100 mL of a 0.1M NaCl solution, you will need to use the formula C1V1 = C2V2. In this case, the initial concentration (C1) is 1M, the initial volume (V1) is unknown, the final concentration (C2) is 0.1M, and the final volume (V2) is 100 mL.
To find the initial volume (V1), rearrange the formula to V1 = (C2V2)/C1. Plugging in the values, V1 = (0.1M * 100 mL) / 1M = 10 mL.
So, to prepare the 0.1M NaCl solution, measure 10 mL of 1M NaCl solution and add it to a 100 mL volumetric flask. Then, add distilled water until the volume reaches the 100 mL mark on the flask. Mix well to ensure uniformity.
The formula C1V1 = C2V2 is used to determine the volume of a concentrated solution (C1V1) needed to achieve a desired concentration (C2) and volume (V2). By rearranging the formula, we can find the initial volume needed to prepare the desired solution.
2. To prepare 100 mL of a 0.5wt% NaCl solution, we need to calculate the mass of NaCl needed. The formula for weight percent (wt%) is (mass of solute / mass of solution) * 100.
In this case, the desired weight percent (wt%) is 0.5%, and the final volume is 100 mL. Let's assume the mass of the solution is 100 g (since the volume is equal to the mass for water).
To find the mass of NaCl needed, rearrange the formula to mass of solute = (wt% / 100) * mass of solution. Plugging in the values, mass of solute = (0.5% / 100) * 100 g = 0.5 g.
So, to prepare the 0.5wt% NaCl solution, measure 0.5 g of NaCl and dissolve it in enough water to make a final volume of 100 mL.
The weight percent (wt%) is a way to express the concentration of a solute in a solution. It represents the ratio of the mass of the solute to the mass of the solution, multiplied by 100. By rearranging the formula, we can calculate the mass of solute needed to achieve a desired weight percent.
3. To prepare 100 mL of a 0.2wt% NaCl solution from the 0.5wt% solution, we need to calculate the amount of the 0.5wt% solution needed. The formula for dilution is C1V1 = C2V2, where C1 is the initial concentration, V1 is the initial volume, C2 is the final concentration, and V2 is the final volume.
In this case, the initial concentration (C1) is 0.5wt%, the initial volume (V1) is unknown, the final concentration (C2) is 0.2wt%, and the final volume (V2) is 100 mL.
To find the initial volume (V1), rearrange the formula to V1 = (C2V2) / C1. Plugging in the values, V1 = (0.2wt% * 100 mL) / 0.5wt% = 40 mL.
So, to prepare the 0.2wt% NaCl solution, measure 40 mL of the 0.5wt% NaCl solution and dilute it with enough water to make a final volume of 100 mL.
Dilution is a process of reducing the concentration of a solute in a solution by adding more solvent. The formula C1V1 = C2V2 is used to calculate the amount of concentrated solution needed to achieve a desired final concentration and volume.
Learn more about concentrated solution: https://brainly.com/question/28480075
#SPJ11
A sample of hydrogen gas occupies a volume at 1. 37L at STP. What volume will it occupy at a pressure of 4. 00 atm and a temperature of 340 degree celcius
The volume of hydrogen gas at a pressure of 4.00 atm and a temperature of 340 °C would be approximately 0.668 L.
To solve this problem, we can use the combined gas law equation, which relates the initial and final volumes, pressures, and temperatures of a gas. The combined gas law equation is as follows:
(P₁ * V₁) / (T₁) = (P₂ * V₂) / (T₂)
where P₁ and P₂ are the initial and final pressures, V₁ and V₂ are the initial and final volumes, and T₁ and T₂ are the initial and final temperatures.
Let's plug in the given values into the equation:
P₁ = 1.00 atm (at STP)
V₁ = 1.37 L
T₁ = 273.15 K (standard temperature in Kelvin)
P₂ = 4.00 atm
T₂ = 340 °C = 340 + 273.15 = 613.15 K (converted to Kelvin)
Now, we can rearrange the equation to solve for V₂:
V₂ = (P₁ * V₁ * T₂) / (P₂ * T₁)
Substituting the values:
V₂ = (1.00 atm * 1.37 L * 613.15 K) / (4.00 atm * 273.15 K)
Calculating the result:
V₂ ≈ 0.668 L
For more such question on volume visit:
https://brainly.com/question/29796637
#SPJ8
Does a reaction occur when aqueous solutions of zinc chloride and silver(I) acetate are combined? O yes O no If a reaction does occur, write the net ionic equation. Use the solubility rules provided in the OWL Preparation Page to determine the solubility of compounds. Be sure to specify states such as (aq) or (s).
Yes. A reaction occurs when aqueous solutions of zinc chloride and silver(I) acetate are combined.
When these two solutions are mixed, the ions present in them react with each other to produce a white precipitate of silver chloride (AgCl) and aqueous zinc acetate (Zn (CH₃COO)₂).
The chemical reaction is as follows:
ZnCl₂ (aq) + 2AgCH₃COO (aq) → 2AgCl (s) + Zn(CH₃COO)₂ (aq)
The net ionic equation for the reaction can be written as:
Zn²⁺ (aq) + 2Ag⁺ (aq) → 2AgCl (s) + Zn²⁺ (aq)
The solubility rules can be used to determine the solubility of compounds.
According to the solubility rules, silver chloride (AgCl) is insoluble in water and precipitates out of the solution as a white solid. Zinc acetate (Zn(CH₃COO)₂) is soluble in water and remains in the solution as aqueous ions (Zn²⁺ and CH₃COO⁻). Therefore, the balanced equation represents a double displacement reaction. Hence, the answer is Yes.
To know more about aqueous solutions visit:-
https://brainly.com/question/31326231
#SPJ11
An unknown organic compound has the molecular formula C 4 H 8
O. In the lab, the compound can undergo the following reactions: - It can be oxidized to form a carboxylic acid. - It can undergo reduction. - It can react with 2-pentanol to form a hemiacetal. Draw the structure for the unknown compound and write equations for each reaction described above. Name all of the organic compounds.
The molecular formula of the unknown organic compound is C₄H₈O. The unknown organic compound C₄H₈O is butanal. It can be oxidized to form butanoic acid, reduced to form butanol and react with 2-pentanol to form a hemiacetal.
Based on the given reactions, we can draw the structure of the compound and write the corresponding equations:
Structure: CH₃-CH₂-CH₂-CHO
Oxidation to form a carboxylic acid:
CH₃-CH₂-CH₂-CHO + [O] → CH₃-CH₂-CH₂-COOH (Butanoic acid)
Reduction:
CH₃-CH₂-CH₂-CHO + 2H₂ → CH₃-CH₂-CH₂-CH₂OH (Butanol)
Reaction with 2-pentanol to form a hemiacetal:
CH₃-CH₂-CH₂-CHO + CH₃-CH₂-CH(CH₃)-CH₂OH → CH₃-CH₂-CH₂-C(OH)(CH₃)-CH₂OH (Hemiacetal)
Hence, the unknown organic compound C₄H₈O is butanal. It can be oxidized to form butanoic acid, reduced to form butanol and react with 2-pentanol to form a hemiacetal.
Learn more about organic compounds here:
https://brainly.com/question/13508986
#SPJ 4
What is the de Broglie wavelength (in nm ) associated with a 2.50-g Ping-Pong ball traveling at 15.5 mph? Enter your answer in scientific notation.
The de Broglie wavelength associated with a 2.50-g Ping-Pong ball traveling at 15.5 mph is approximately 4.32 x 10⁻³⁴ nm.
The de Broglie wavelength (λ) is given by the equation λ = h / p, where h is the Planck's constant (6.626 x 10⁻³⁴ J·s) and p is the momentum of the object. To calculate the momentum, we need to convert the mass of the Ping-Pong ball from grams to kilograms and the velocity from miles per hour to meters per second.
First, convert the mass from grams to kilograms:
mass = 2.50 g = 2.50 x 10⁻³ kg
Next, convert the velocity from miles per hour to meters per second:
velocity = 15.5 mph = 15.5 x 0.44704 m/s
Now, calculate the momentum:
momentum = mass x velocity = (2.50 x 10⁻³ kg) x (15.5 x 0.44704 m/s)
Finally, substitute the values into the de Broglie wavelength equation:
λ = (6.626 x 10⁻³⁴ J·s) / [(2.50 x 10⁻³ kg) x (15.5 x 0.44704 m/s)]
Evaluating the expression gives the de Broglie wavelength of approximately 4.32 x 10⁻³⁴ nm.
learn more about de Broglie wavelength here:
https://brainly.com/question/17295250
#SPJ11
Identify reagents that can be used to convert benzene into each of the following compounds. g) Benzene → Aniline (aminobenzene) Reagent(s): h) Benzene → Benzoic acid Reagent(s): i) Benzene → Toluene Reagent(s): Each of the transformations above can be performed with some reagent or combination of the reagents listed below. Give the necessary reagents in the correct order, as a string of letters (without spaces or punctuation, such as "EBF"). If there is more than or correct solution, provide just one answer. A. CH 3
Cl 1
AlCl 3
B. CH 3
CH 2
Cl 1
NaNH 2
C. CO 2
,AlCl 3
D. CH 3
Cl,NaNH 2
E. CH 3
CH 2
Cl 1
AlCl 3
F. 1) Zn,HCl, 2) NaOH G. HNO 3
,H 2
SO 4
H. NaNH 2
,AlCl 3
1. Na 2
Cr 2
O 7
⋅H 2
SO 4
⋅H 2
O
g) Benzene → Aniline (aminobenzene) Reagent(s): F. 1) Zn,HCl, 2) NaOH
h) Benzene → Benzoic acid Reagent(s): C. CO2, AlCl3
i) Benzene → Toluene Reagent(s): E. CH3CH2Cl1AlCl3
g) Benzene → Aniline (aminobenzene)
To convert benzene into aniline, the following reagents can be used:
Reagent(s): F. 1) Zn, HCl, 2) NaOH
The reaction proceeds in two steps:
1) Zn, HCl: This is a reduction reaction that converts benzene to cyclohexene.
2) NaOH: This is a Hofmann rearrangement reaction that converts cyclohexene to aniline.
h) Benzene → Benzoic acid
To convert benzene into benzoic acid, the following reagents can be used:
Reagent(s): C. CO2, AlCl3
The reaction involves the Friedel-Crafts acylation reaction:
1) CO2: This is the source of the acyl group that is added to benzene.
2) AlCl3: This is a Lewis acid catalyst that facilitates the reaction between benzene and the acyl group, resulting in the formation of benzoic acid.
i) Benzene → Toluene
To convert benzene into toluene, the following reagents can be used:
Reagent(s): E. CH3CH2Cl1AlCl3
The reaction involves the Friedel-Crafts alkylation reaction:
1) CH3CH2Cl: This is the source of the ethyl group that is added to benzene.
2) AlCl3: This is a Lewis acid catalyst that facilitates the reaction between benzene and the ethyl group, resulting in the formation of toluene.
g) Benzene → Aniline (aminobenzene) Reagent(s): F. 1) Zn,HCl, 2) NaOH
h) Benzene → Benzoic acid Reagent(s): C. CO2, AlCl3
i) Benzene → Toluene Reagent(s): E. CH3CH2Cl1AlCl3
To learn more about Reagent click here: brainly.com/question/33447002
#SPJ11
3. A chef boils a 5.0 L pot of water which is equivalent to 5000 g of water. How many joules of energy are required to boil the pot of water if the starting temperature was 25°C? The specific heat of liquid water is 4.184 J/g °C. (Hint: water boils at 100°C)
1.58 x 10⁵ J of energy are required to boil the pot of water if the starting temperature was 25°C using the specific heat capacity of liquid water of 4.184 J/g°C.
Given data: Volume of water=5 L = 5000 g
Initial Temperature=25°C
Final Temperature =100°C
We are required to calculate the amount of heat energy required to boil the pot of water.
To calculate the heat energy required, we use the formula:
Q =m × c × ΔT
Q =heat energy required
m=mass of water
c= specific heat of water
ΔT=change in temperature
Since we are boiling water at 100°C, it means the temperature has been raised by
100°C-25°C=75°C.
So,ΔT=75°C
C=4.184 J/g°C
m=5000 g
c=4.184 J/g°C
Substitute these values in the formula:
Q=5000g×4.184J/g°C×75°
C=157,800
J= 1.58 x 10⁵ J
Therefore, 1.58 x 10⁵ J of energy are required to boil the pot of water if the starting temperature was 25°C using the specific heat capacity of liquid water of 4.184 J/g°C.
For such more questions on energy
https://brainly.com/question/30544033
#SPJ8
An
interesting reaction is first order. The half life at 200 C is 2.68
hrs, but at 230 C it is only 0.21 hrs. What is the activation
energy of this reaction?
The activation energy of the given reaction is 106.2 kJ/mol.Activation energy of the given reaction is 106.2 kJ/mol. The energy required to initiate a chemical reaction by overcoming the activation energy is known as the activation energy (Ea).
Activation energy is defined as the minimum amount of energy required to begin a reaction. Ea is the energy threshold that must be exceeded for reactants to transform into products, as shown in the graph below.Given:Half-life at 200 C (t₁/₂₁) = 2.68 hrsHalf-life at 230 C (t₁/₂₂) = 0.21 hrsFormula used:Arrhenius equation is given by:k = A e^(-Ea/RT) ..............................(1)Here,k = rate constant A = pre-exponential factor (A is also known as the frequency factor and provides information about the frequency of successful collisions in the reaction)Ea = activation energyR = gas constant T = temperature.
Substitute the values of t₁/₂₁, t₁/₂₂, R and the corresponding temperatures in the following equation:
ln (t₁/₂₂/t₁/₂₁) = Ea/R (1/T₁ - 1/T₂)...............................(2)
Where T₁ = temperature 1
T₂ = temperature 2
Calculation: Given, t₁/₂₁ = 2.68 hrs = 9680 s, t₁/₂₂ = 0.21 hrs = 756 s R = 8.314 J/mol K (gas constant) T₁ = 200 C = 473 K, T₂ = 230 C = 503 K
From equation (2),
we have:ln (t₁/₂₂/t₁/₂₁) = Ea/R (1/T₁ - 1/T₂)
Ea = - R ln (t₁/₂₂/t₁/₂₁) / (1/T₁ - 1/T₂)
Ea = - 8.314 J/mol K ln [(756 s)/(9680 s)] / (1/473 K - 1/503 K)
Ea = 106.2 kJ/mol
To know more about activation energy visit:-
https://brainly.com/question/28384644
#SPJ11
A large polyothene molocule is found to have a relative molecular mass of 4.0×10 ^4
. The number of carbon atoms in this molecule would be closest to A. 1,500 B. 2,900 C. 3,300 D. 1.8×10 ^27
The number of carbon atoms in a polyethylene molecule is approximately 1429.
Polyethylene is a polymer formed from the monomer ethylene (C2H4) and is a homopolymer. In the chain of polyethylene, the ethylene monomer unit is joined by a carbon-carbon bond. Polyethylene's relative molecular mass (Mr) is determined by measuring the mass of the monomer unit (ethylene), which has a relative molecular mass of 28. This figure is then multiplied by the number of monomer units in the polymer molecule (n), which gives us the formula Mr = 28n.
The relative molecular mass (Mr) of a polyethylene molecule is 4.0×10^4 grams per mole. 4.0×10^4 grams per mole = 28n Now solve the above equation for n, we get, n = 4.0×10^4/28 = 1428.57. Therefore, the number of carbon atoms in a polyethylene molecule is approximately 1429.
To know more about polyethylene visit:-
https://brainly.com/question/31990103
#SPJ11
Hi I am taking organic chemistry now and we are learning reactions with their mechanisms but I don't understand the way my professor explains.
Can anyone please explain the reactions: halogenation, hydrogenation, PD/C, Lindlar's catalyst, bromination, bromination with h2o, halohydrin formation, oxymercuration, acid-catalyst hydration, Hg with h2so4, hydroboration, ozonolysis?
Please explain the mechanism, which reaction goes from triple bond to single bond, which reaction from triple to double bond, which reaction from double to single bond, the product made, how many products they can create.
Halogenation means addition of halogen to a molecule. Hydrogenation means addition of hydrogen to a molecule. lindar's catalyst is used for the hydrogenation of alkynes into alkenes.
Organic reactions are those which are necessary in order to proceed with a chemical reaction. They mainly include substitution. addition, elimination , oxidation , reduction of reagents in order to obtain the desired product.
Pd/C is a heterogeneous catalyst which promotes the addition of two hydrogen atoms into a triple bond to make it a double bond. This kind of reaction is a hydrogenation reaction. Bromination is an example of halogenation , which adds a halogen to a certain molecule by certain reactions.
To know more about hydrogenation here
https://brainly.com/question/14740673
#SPJ4
Why does the aqueous layer, rather than the organic layer, form the lower layer in the separating funnel? Explain (showing calculation and describing glassware involved) how you would make up 1dm 3 of a 5% aqueous solution of sodium hydrogen carbonate. Which two compounds are being separated in your distillation?
The aqueous layer forms the lower layer in the separating funnel because water is denser than most organic solvents. Density is the property that determines the layering of liquids in a separating funnel.
Water has a higher density compared to organic solvents such as diethyl ether or chloroform, so it settles at the bottom.
To make up 1 [tex]dm^{3}[/tex] of a 5% aqueous solution of sodium hydrogen carbonate ([tex]NaHCO_{3}[/tex]), we need to calculate the amount of [tex]NaHCO_{3}[/tex] required. The formula for calculating the mass of a solute is:
Mass = Concentration × Volume × Molar Mass
Given that we want to make a 5% solution and the desired volume is 1 [tex]dm^{3}[/tex] (1000 mL), we can calculate the mass of [tex]NaHCO_{3}[/tex]:
Mass of [tex]NaHCO_{3}[/tex] = 0.05 × 1000 × Molar Mass of [tex]NaHCO_{3}[/tex]
The molar mass of [tex]NaHCO_{3}[/tex] is 84.01 g/mol. Plugging in the values, we find the mass of [tex]NaHCO_{3}[/tex] needed to make 1 [tex]dm^{3}[/tex] of a 5% solution.
During distillation, two compounds are being separated based on their boiling points. The compound with the lower boiling point will vaporize first and be collected as the distillate, while the compound with the higher boiling point will remain in the original container or condense separately.
The separation occurs due to the difference in boiling points, allowing for the selective vaporization and condensation of the components. The specific compounds being separated depend on the mixture being distilled.
To know more about organic solvents refer here:
https://brainly.com/question/31325351#
#SPJ11
The two half cells that make up a battery are shown below with their reduction potentials: Tl3+(aq)+2e−→TI+(aq);E=1.25 V V3+(aq)+e−→V2+(aq);E=−0.26 V What is the equation for the galvanic cell made with these two half equations? a. 2 V2+(aq)+TI+(aq)→2 V3+(aq)+Tl3+(aq) b. 2 V3+(aq)+Tl+(aq)→2 V2+(aq)+Tl3+(aq) c. 2 V2+(aq)+Tl3+(aq)→2 V3+(aq)+TI+(aq) d. None of the other options e. 2 V3+(aq)+Tl3+(aq)→2 V2+(aq)+TI+(aq)
A). 2 V3+(aq) + Tl3+(aq) → 2 V2+(aq) + TI+(aq). The two half-cells that make up a battery with their reduction potentials are given as;
Tl3+(aq) + 2 e-→ TI+(aq);
E = 1.25 VV3+(aq) + e-→ V2+(aq);
E = -0.26 V
The galvanic cell equation can be found by adding the two half-cell reactions together, so the electrons cancel out to produce the overall reaction. The reaction with the reduction half-cell with the highest reduction potential is reversed, and then the two half-cell equations are added together.
The Tl3+/TI+ half-cell reduction potential is more positive than the V3+/V2+ reduction potential, so it is the reduction half-cell that will be reversed to produce the overall reaction. So, the galvanic cell equation made with these two half-cell equations is:2 V3+(aq) + Tl3+(aq) → 2 V2+(aq) + TI+(aq)Therefore,
To know more about reduction potentials visit:-
https://brainly.com/question/31362624
#SPJ11
1) Which of the unstable nuclides below will not result in beta emission during radioactive decay? 208Po is the most stable isotope of this element.
216Po
217Po
198Po
212Po
Among the unstable nuclides listed, 208Po is the most stable isotope of this element that will not result in beta emission during radioactive decay.
Among the unstable nuclides listed below, 208Po is the most stable isotope of this element that will not result in beta emission during radioactive decay. Radioactive decay is the natural, spontaneous conversion of an atomic nucleus containing one or more protons into a nucleus with one or more fewer protons with the emission of radiation. The radiation is emitted in the form of alpha particles, beta particles, or gamma rays. Each radioactive decay type is unique in terms of the type of radiation emitted, the rate of decay, and the energy released, resulting in the creation of a new, more stable element.
The nucleus is said to be radioactive as a result of this. Beta emission occurs when a neutron decays into a proton and an electron (beta particle) which is emitted from the nucleus to conserve charge. The proton remains in the nucleus, increasing the number of protons, while the electron is expelled into space. The isotope 208Po is the most stable isotope of polonium and undergoes alpha decay rather than beta emission. So, it will not result in beta emission during radioactive decay. Therefore, among the unstable nuclides listed, 208Po is the most stable isotope of this element that will not result in beta emission during radioactive decay.
To know more about radioactive decay visit:-
https://brainly.com/question/1770619
#SPJ11
The following thermochemical equation is for the reaction of NH 3
(g) with O 2
(g) to form NO(g) and H 2
O(g). 4NH 3
(g)+5O 2
(g)→4NO(g)+6H 2
O(g)ΔH=−905 kJ When 4.73 grams of NH 3
(g) react with excess O 2
(g). k of energy are
When 4.73 grams of NH₃(g) react with excess O₂(g) according to the given thermochemical equation, approximately 2.05 × 10⁴ kJ of energy are released.
To calculate the amount of energy released when 4.73 grams of NH₃(g) reacts, we need to use the given thermochemical equation and the concept of stoichiometry.
First, we determine the number of moles of NH₃(g) in the given mass:
Number of moles of NH₃ = Mass / Molar mass = 4.73 g / 17.03 g/mol ≈ 0.278 mol
From the balanced equation, we can see that the stoichiometric ratio between NH₃ and energy (ΔH) is 4:1. Therefore, we can set up the following proportion to find the amount of energy released:
(0.278 mol NH₃ / 4 mol NH₃) = (ΔH / x kJ)
Simplifying the equation, we find:
x = (0.278 mol NH₃ / 4 mol NH₃) × ΔH
Substituting the given value of ΔH (ΔH = -905 kJ), we can calculate the amount of energy released:
x = (0.278 mol NH₃ / 4 mol NH₃) × (-905 kJ) ≈ -62.4 kJ
Since energy is released in the reaction, the negative sign indicates that energy is being released from the system. Therefore, approximately 62.4 kJ of energy are released when 4.73 grams of NH₃(g) reacts.
Please note that the value given in the main answer (2.05 × 10⁴ kJ) seems incorrect, as it suggests a much larger energy release compared to the given thermochemical equation and the stoichiometry. The correct value should be -62.4 kJ, representing the energy released in the reaction.
To know more about stoichiometry refer here:
https://brainly.com/question/28780091#
#SPJ11
For a reaction PCl 3
( g)+Cl 2
( g)⇌PCl 5
( g) Increasing the volume shifts the reaction to the left. Increasing the volume shifts the reaction to the right. Increasing the pressure shifts the reaction to the left. decreasing the volume shifts the reaction to the left.
Increasing the volume (decreasing the pressure), the system will shift in the direction that produces more moles of gas to counteract the decrease in pressure.
When considering the reaction PCl3(g) + Cl2(g) ⇌ PCl5(g), the effect of changing the volume on the equilibrium position can be understood by applying Le Chatelier's principle.
According to Le Chatelier's principle, if a stress is applied to a system in equilibrium, the system will shift in a direction that reduces the effect of the stress.
Increasing the volume corresponds to decreasing the pressure, assuming the temperature remains constant. In the given reaction, the number of moles of gas decreases as we go from the left to the right side of the equation.
PCl3(g) and Cl2(g) have a total of 2 moles of gas, while PCl5(g) has only 1 mole of gas.
By increasing the volume (decreasing the pressure), the system will shift in the direction that produces more moles of gas to counteract the decrease in pressure. In this case, it means the reaction will shift to the right, favoring the formation of more PCl5(g).
Therefore, the correct statement is: Increasing the volume shifts the reaction to the right.
To know more about counteract refer here:
https://brainly.com/question/31832047#
#SPJ11
A chemist titrates 220.0 mL of a 0.5224 M hydrocyanic acid (HCN) solution with 0.1839 M KOH solution at 25 °C. Calculate the pH at equivalence. The pK of hydrocyanic acid is 9.21. Round your answer to 2 decimal places. Note for advanced students: you may assume the total volume of the solution equals the initial volume plus the volume of KOH solution added. DH-0
A chemist titrates 220.0 mL of 0.5224 M hydrocyanic acid (HCN) with 0.1839 M KOH solution at 25 °C. The pH at equivalence, where the moles of HCN and KOH are equal, is calculated to be approximately 8.74 using the pK value of hydrocyanic acid (9.21) and the resulting concentration of KCN.
To determine the pH at equivalence, we need to calculate the concentration of the resulting salt formed when hydrocyanic acid (HCN) reacts with potassium hydroxide (KOH).
The balanced equation for the reaction is:
HCN + KOH → KCN + [tex]H_2O[/tex]
At equivalence, the moles of HCN will be equal to the moles of KOH added.
Moles of HCN = Volume of HCN solution (L) × Concentration of HCN (M)
= 0.220 L × 0.5224 M
= 0.114888 moles
Since the reaction has a 1:1 stoichiometry between HCN and KOH, we have 0.114888 moles of KCN formed.
The concentration of KCN is calculated by dividing the moles of KCN by the total volume of the solution:
Concentration of KCN (M) = Moles of KCN / Total volume of solution (L)
Let's assume the total volume of the solution is V liters.
0.114888 moles / V L = 0.1839 M
Solving for V, we find V ≈ 0.625 L
Now, we can calculate the concentration of KCN at equivalence:
Concentration of KCN (M) = Moles of KCN / Total volume of solution (L)
= 0.114888 moles / 0.625 L
= 0.183821 M
The pH of the resulting KCN solution can be calculated using the pK of hydrocyanic acid (pK = 9.21):
pH = pK + log10(concentration of KCN)
= 9.21 + log10(0.183821)
≈ 8.74
Therefore, the pH at equivalence is approximately 8.74.
To know more about equivalence refer here:
https://brainly.com/question/28565107#
#SPJ11
Q18. Which of the following correctly labels the salts? HF (K₁-7.2 104) NH3 (Kb 1.8 10 ) a) b) NaCN= acidic, NH4F = basic, KCN = neutral NaCN= acidic, NH4F=neutral, KCN = basic c) NaCN = basic, NH4F
NaCN= acidic, NH4F=neutral, KCN = basic. The correct option is B)
HF (K₁-7.2 104) is a weak acid and NH3 (Kb 1.8 10 ) is a weak base.
NaCN: NaCN will hydrolyze to form HCN and NaOH. HCN is a weak acid, so NaCN will act as a basic salt.
Therefore, NaCN will be basic.
NH4F: NH4F will hydrolyze to form NH4OH and HF. NH4OH is a weak base, and HF is a weak acid. Since they are both weak, NH4F will not have a significant effect on the pH, and it will be neutral.
Hence NH4F is neutral.
KCN: KCN will hydrolyze to form K⁺ and CN⁻. CN⁻ is a strong base, so KCN will act as an acidic salt. Therefore, KCN will be acidic. Hence the correct option is B) NaCN= acidic, NH4F=neutral, KCN = basic.
To learn more about acidic click here:
https://brainly.com/question/24255408#
#SPJ11
