balance the following equation in acidic solution using the lowest possible integers and give the coefficient of h . cr2o72- fe2 (aq) → cr3 (aq) fe3 (aq)

The chemiluminescence reaction should be conducted in a dark environment, at a specific temperature and pH, and in the presence of a suitable oxidizing agent in order to produce a successful reaction.

Chemiluminescence is a process that involves the emission of light from a chemical reaction. This type of reaction requires a specific environment in order to occur successfully.

First and foremost, the chemiluminescence reaction must be conducted in the absence of light. This is because the emission of light from the reaction can be easily masked by ambient light, making it difficult to detect. Therefore, a dark environment such as a darkroom or a light-tight box is typically used to perform chemiluminescence reactions.

Additionally, the chemiluminescence reaction requires specific temperature and pH conditions in order to occur. These conditions can vary depending on the specific reaction being performed, but generally, the reaction takes place at a low temperature (around room temperature) and at a slightly basic pH.

Finally, the reaction must be conducted in the presence of a suitable oxidizing agent, which is responsible for initiating the chemiluminescence process. Common oxidizing agents used in chemiluminescence reactions include hydrogen peroxide and luminol.

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The volume of 4.00 mol of methane gas at 18 °C and 1.40 atm is 68.295 litres.

To calculate the volume of 4.00 mol of methane gas, we can use the ideal gas law equation:

PV = nRT

where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature in Kelvin.

First, let's convert the temperature from Celsius to Kelvin:

kelvin = celsius + 273.15

So, T = 18 °C + 273.15 = 291.15 K.

Next, let's rearrange the ideal gas law equation to solve for volume:

V = (nRT)/P.

Plugging in the given values, we get:

V = (4.00 mol x 0.0821 L atm/mol K x 291.15 K)/1.40 atm
V = 68.295 litres

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The presence of CO2 must be maintained at a partial pressure of approximately 3.64 atm to keep the CO2 concentration at 0.12 M in soda at 25°C.

To maintain the presence of CO2 at a concentration of 0.12 M in soda at 25°C, you need to consider the solubility of CO2 in the soda and the partial pressure of CO2 above the liquid. Here's a step-by-step explanation:

1. Determine the solubility of CO2 in water at 25°C, which is approximately 3.3 x 10^-2 M (under 1 atm of pressure).
2. Since the desired concentration of CO2 (0.12 M) is higher than its solubility at 25°C and 1 atm, you need to increase the pressure of CO2 above the soda to achieve the desired concentration.
3. Use Henry's Law to calculate the partial pressure of CO2 required: P = C / KH, where P is the partial pressure, C is the concentration (0.12 M), and KH is the Henry's Law constant for CO2 in water at 25°C (3.3 x 10^-2 M/atm).
4. Plug in the values: P = 0.12 M / (3.3 x 10^-2 M/atm) = 3.64 atm (approximately).

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To calculate the volume of a gas, you can use the Ideal Gas Law equation: PV = nRT, where P is pressure, V is volume, n is moles, R is the gas constant, and T is temperature. Here, we are given n = 2.0 moles, T = 300 K, and P = 1.5 atm. First, we need to use the appropriate gas constant for the given pressure unit. For atm, R = 0.0821 L·atm/mol·K.

Now, plug in the values into the equation:

(1.5 atm)(V) = (2.0 moles)(0.0821 L·atm/mol·K)(300 K)

Solve for V:

V = (2.0 moles)(0.0821 L·atm/mol·K)(300 K) / (1.5 atm)

V ≈ 32.8 L

Therefore, the volume that 2.0 moles of the gas occupy at 300 Kelvins and 1.5 atmospheres is approximately 32.8 L.

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The balanced equation for the ionization of butanoic acid (CH3CH2CH2COOH) in water (H2O) is:

CH3CH2CH2COOH + H2O ⇌ CH3CH2CH2COO- + H3O+

This reaction involves the ionization of the carboxylic acid group (COOH) of butanoic acid, which releases a hydrogen ion (H+) into the solution. The resulting ion, CH3CH2CH2COO-, is the conjugate base of butanoic acid. The hydrogen ion combines with a water molecule to form the hydronium ion (H3O+). This reaction is an example of a proton transfer reaction, also known as acid-base reaction, in which an acid donates a proton (H+) to a base (water in this case) to form a conjugate acid-base pair.

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The element with the ground state electron configuration of [Ar] 4s2 3d3 is Vanadium (V)


1. Locate the noble gas element just before the given configuration, in this case, it is Argon (Ar).
2. Determine the atomic number of Argon (Ar), which is 18.
3. The configuration [Ar] 4s2 3d3 means that there are 2 electrons in the 4s subshell and 3 electrons in the 3d subshell.
4. Add these 2+3=5 electrons to the atomic number of Argon (18).
5. The resulting atomic number is 18+5=23.
6. Locate the element with atomic number 23 on the periodic table, which is Vanadium (V).

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The number of moles of the gas present in the container at 1195 K and pressure of 1.65 atm is 0.45 mole

From the question given above, the following data were obtained obtained:

Using the ideal gas equation, we can obtain the number of mole present in the container as follow:

PV = nRT

1.65 × 27 = n × 0.0821 × 1195

44.55 = n × 98.1095

Divide both sides by 98.1095

n = 44.55 / 98.1095

n = 0.45 mole

Thus, we can conclude that the number of mole of the gas present in the container is 0.45 mole

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The percent yield is 37.2%. The equation for the reaction between calcium carbonate and hydrochloric acid is: CaCO3 + 2HCl → CaCl2 + CO2 + H2O.

The molar mass of CaCO3 is 100.1 g/mol, and the molar mass of HCl is 36.5 g/mol. Using these values, we can calculate the theoretical yield of calcium chloride: 1 mol CaCO3 reacts with 2 mol HCl to produce 1 mol CaCl2, 155 g CaCO3 x (1 mol CaCO3 / 100.1 g) x (2 mol HCl / 1 mol CaCO3) x (36.5 g HCl / 1 mol HCl) x (1 mol CaCl2 / 1 mol HCl) x (110.98 g CaCl2 / 1 mol CaCl2) = 381.1 g CaCl2 (theoretical yield)

The actual yield of calcium chloride obtained is 142 g. To calculate the percent yield, we use the formula: Percent yield = (actual yield / theoretical yield) x 100, Percent yield = (142 g / 381.1 g) x 100 = 37.2%

Therefore, the percent yield is 37.2%.

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The effect of decreasing the concentration of Mg²⁺ is that it lowers the cell potential of the electrochemical cell because the cell potential depends on the concentrations of the reactants and products involved in the cell reaction.

In this case, the reduction of Mg²⁺ to Mg(s) is one of the half-reactions in the cell, and a decrease in the concentration of Mg²⁺ will shift the equilibrium of this half-reaction towards the reactants. This means that there will be less Mg(s) formed at the magnesium electrode, which will result in a lower concentration of electrons available to flow through the wire and voltmeter. As a result, the cell potential will decrease.
In summary, decreasing the concentration of the Mg2+ solution from 1.00 M to 0.75 M will decrease the cell potential of the spontaneous cell composed of Mg and Al electrodes.

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a. The reduction of half-reactions for each electrode is -0.763 V and +1.358 V from which electrode will electrons flow from the battery into a circuit if the electrode potentials are not too different from E values.

b. The battery will consume 5.30 grams or 0.0053 kilograms of Cl2 during 1 hour of constant current delivery.

(a) The reduction half-reaction for the zinc electrode is:

Zn²+ + 2 [tex]e^{-}[/tex] -> Zn(s) E° = -0.763 V

The reduction half-reaction for the chlorine electrode is:

2 [tex]Cl^{-}[/tex](aq) -> Cl2(l) + 2 [tex]e^{- E}[/tex]° = +1.358 V

The electrons will flow from the chlorine electrode to the zinc electrode because the chlorine electrode has a higher standard reduction potential and is therefore the cathode.

(b) First, we need to calculate the total charge transferred by the battery during 1 hour:

1 hour = 3600 seconds

Q = I × t = (1.00 × 10³ A) × (3600 s) = 3.60 × [tex]10^6[/tex] C

From the reduction half-reaction for the chlorine electrode, we can see that for every two electrons transferred, one mole of Cl2 is produced. Therefore, the number of moles of Cl2 produced is:

moles of Cl2 = (2/96500) × Q = (2/96500) × (3.60 × [tex]10^6[/tex] C) = 0.0746 mol

The molar mass of Cl2 is 70.90 g/mol, so the mass of Cl2 produced is:

mass of Cl2 = moles of Cl2 × molar mass of Cl2 = 0.0746 mol × 70.90 g/mol = 5.30 g

Therefore, the battery will consume 5.30 grams or 0.0053 kilograms of Cl2 during 1 hour of constant current delivery.

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The question is -

Consider the rechargeable battery:Zn(s) | ZnCl2(aq) || Cl-(aq) | Cl2(l) | C(s)

(a) Write reduction half-reactions for each electrode. From which electrode will electrons flow from the battery into a circuit if the electrode potentials are not too different from E values?

(b) If the battery delivers a constant current of 1.00 × 103 A for 1.00 h, how many kilograms of Cl2 will be consumed?

In certain minerals and is used in various applications, such as in the production of solar panels and semiconductors.

Tellurium, with an atomic number of 52, has six valence electrons. To obey the octet rule, it needs to gain two more electrons. Therefore, tellurium would be expected to form two covalent bonds.

If the molecule contains only one tellurium atom and only single bonds are formed, the formula of the compound that would form between tellurium and hydrogen is[tex]H_2Te[/tex]. This compound is known as hydrogen telluride or tellane.

It is true that tellurium is among the rarest of the elements in Earth's crust, with an abundance of only about 0.001 parts per million. However, it can be found in certain minerals and is used in various applications, such as in the production of solar panels and semiconductors.

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The hydrogen atom before the ionization happened was in its ground state. The energy required to remove an electron from a hydrogen atom is known as ionization energy.

In this case, the ionization energy was measured to be 0.850 electron volts (eV). When an electron is in the ground state of the hydrogen atom, it has the lowest energy possible.

The ground state electron is located in the lowest energy level or orbital, which is the 1s orbital. If an electron in the ground state absorbs energy equal to or greater than the ionization energy, it will become ionized and leave the atom.

Therefore, the hydrogen atom must have been in its ground state before the ionization occurred since ionization energy is the energy required to remove an electron from the ground state of an atom.

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The concentrations, we can't calculate the exact value of Ecell. However, we have found E°cell, which is 0.46 V.

Hi! To write the net cell equation for the given electrochemical cell, we need to consider the half-reactions occurring at the electrodes.

The half-reactions are:

Cu(s) → Cu²⁺(aq) + 2e⁻ (oxidation at anode)
Ag⁺(aq) + e⁻ → Ag(s) (reduction at cathode)

Now, we can combine the half-reactions to get the net cell equation:

Cu(s) + 2Ag⁺(aq) → Cu²⁺(aq) + 2Ag(s)

For calculating E°cell and Ecell at 25°C, we need the standard reduction potentials (E°) of both half-reactions:

E°(Cu²⁺/Cu) = +0.34 V
E°(Ag⁺/Ag) = +0.80 V

E°cell = E°(cathode) - E°(anode) = E°(Ag⁺/Ag) - E°(Cu²⁺/Cu) = 0.80 V - 0.34 V = 0.46 V

To calculate Ecell, we can use the Nernst equation:

Ecell = E°cell - (RT/nF) * lnQ

At 25°C, R = 8.314 J/(mol·K), T = 298 K, F = 96485 C/mol, and n = 2 (number of electrons transferred in the reaction). The reaction quotient (Q) is:

Q = [Cu²⁺]/[Ag⁺]²

Since we're not supposed to include the concentrations, we can't calculate the exact value of Ecell. However, we have found E°cell, which is 0.46 V.

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The type of intermolecular force that you would expect to see in all polar molecules, but not in non-polar ones, is dipole-dipole interaction.

This is because polar molecules have an uneven distribution of electron density, resulting in a partial positive and partial negative charge. The partial positive end of one polar molecule will be attracted to the partial negative end of another polar molecule, creating a dipole-dipole interaction. In non-polar molecules, there is no permanent dipole moment, so dipole-dipole interactions do not occur.
The type of intermolecular force you would expect to see in all polar molecules but not in non-polar ones is called "dipole-dipole" interactions. This force occurs due to the partial positive and partial negative charges in polar molecules, which attract the oppositely charged ends of neighboring polar molecules. Non-polar molecules, lacking these charges, do not experience dipole-dipole interactions.

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The poh of an aqueous solution of 0.460 m phenol (a weak acid), C6H5OH, cannot be determined without first knowing the value of its acid dissociation constant, Ka. Once Ka is known, the pKa can be calculated and from there the pH can be determined using the Henderson-Hasselbalch equation. Finally, the pOH can be calculated by subtracting the pH from 14.

To determine the pOH of a 0.460 M aqueous solution of phenol (C6H5OH), we first need to find the dissociation constant (Ka) for phenol. The Ka value for phenol is approximately 1.3 × 10⁻¹⁰.
Next, we'll use the Ka to calculate the hydrogen ion concentration, [H⁺]. Since phenol is a weak acid, we can use the approximation method for the reaction:

C6H5OH (aq) ⇌ C6H5O⁻ (aq) + H⁺ (aq)

Assuming x moles of C6H5OH dissociate, we get:

[C6H5O⁻] = [H⁺] = x
[C6H5OH] = 0.460 - x

As x is very small compared to 0.460 M, we can approximate [C6H5OH] ≈ 0.460 M. Now, we can write the expression for Ka:

Ka = [C6H5O⁻][H⁺] / [C6H5OH]
1.3 × 10⁻¹⁰ = x^2 / 0.460

Now, solve for x:
x ≈ 2.09 × 10⁻⁶ M
Since x represents the [H⁺] concentration, we can now calculate the pH:
pH = -log[H⁺] ≈ 5.68
Lastly, we can find the pOH using the relation between pH and pOH:
pH + pOH = 14
pOH = 14 - pH ≈ 14 - 5.68 ≈ 8.32
So, the pOH of the 0.460 M aqueous solution of phenol is approximately 8.32.

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The angular momentum of a particle is (a) a vector parallel to its momentum vector. This means that the direction of the angular momentum is perpendicular to the plane of motion and is dependent on the direction of the momentum vector.

The magnitude of the angular momentum is equal to the product of the particle's momentum and the perpendicular distance from the particle to the axis of rotation. It is important to note that the angular momentum is independent of the specific origin of coordinates, but it does depend on the position and momentum vectors of the particle.

Angular momentum is a vector quantity that represents the rotational analog of linear momentum. It is calculated using the cross product of the position vector (relative to the origin) and the linear momentum vector. When the position and momentum vectors are parallel, their cross-product is zero, resulting in zero angular momentum.

Angular momentum has both a direction and a magnitude, and both are conserved. Bicycles and motorcycles, flying discs, rifled bullets, and gyroscopes owe their useful properties to the conservation of angular momentum1. Conservation of angular momentum is also why hurricanes form spirals and neutron stars have high rotational rates

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To find which is a stronger acid, we need to compare their acid dissociation constants.

A stronger acid will have a higher acid dissociation constant.

A. Acid with a dissociation constant of 3.9×10^−4
B. Acid with a dissociation constant of 4.7×10^−3

Comparing these values, we can see that the acid dissociation constant for B (4.7×10^−3) is higher than that for A (3.9×10^−4).

Therefore, the stronger acid is the one with an acid dissociation constant of 4.7×10^−3.

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The compounds can be ranked in order of increasing reactivity as: h5ch18p20qab < h5ch18p20qac < h5ch18p20qaa.

To rank the compounds in order of increasing reactivity in electrophilic aromatic substitution, we need to consider the electron-withdrawing or electron-donating groups attached to the benzene ring. The more electron-withdrawing the group, the more reactive the compound will be in EAS.
a) h5ch18p20qaa - This compound has a nitro group attached to the benzene ring, which is a very strong electron-withdrawing group. Therefore, it will be the most reactive in EAS.
b) h5ch18p20qac - This compound has a chlorine atom attached to the benzene ring, which is a weaker electron-withdrawing group than nitro. Therefore, it will be less reactive than h5ch18p20qaa.
c) h5ch18p20qab - This compound has a methyl group attached to the benzene ring, which is an electron-donating group. It will be the least reactive in EAS.
Therefore, the compounds can be ranked in order of increasing reactivity as: h5ch18p20qab < h5ch18p20qac < h5ch18p20qaa.

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The new temperature of the gas is 400 â—¦C.

To solve this problem, we can use the Ideal Gas Law, which states that PV=nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature in Kelvin.

Assuming that the pressure is constant in this scenario, we can rewrite the Ideal Gas Law as V/T = constant. This means that the ratio of volume to temperature is constant for a given amount of gas at constant pressure.

We are given that the initial volume is 5 L and the final volume is 10 L. Therefore, the ratio of initial volume to initial temperature is V1/T1 = 5/T1, and the ratio of final volume to final temperature is V2/T2 = 10/T2.

Since the ratio of volume to temperature is constant, we can set these two ratios equal to each other and solve for T2:
5/T1 = 10/T2
T2 = (10/5) x T1
T2 = 2 x 200 â—¦C
T2 = 400 â—¦C

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The new volume of the balloon at a temperature of 57.0 degree C is 23.0 L.

we can use the combined gas law, which relates the initial and final conditions of a gas sample when one or more of the gas properties (temperature, pressure, volume) are changed while the others remain constant.

The combined gas law is expressed as:

(P₁V₁)/T₁ = (P₂V₂)/T₂

where P is the pressure, V is the volume, and T is the temperature of the gas.

We are given:

P₁ = P₂ (pressure remains constant)

V₁ = 20.9 L (initial volume)

T₁ = 300 K (initial temperature)

T₂ = 57.0 oC + 273.15 = 330.15 K (final temperature)

Plugging these values into the combined gas law, we get:

(P₁V₁)/T₁ = (P₂V₂)/T₂

(P₁)(20.9 L)/(300 K) = (P₂)(V₂)/(330.15 K)

Solving for V₂, we get:

V₂ = (P₁)(20.9 L)(330.15 K) / (300 K)(P₂)

Since P₁ = P₂, we can simplify:

V₂ = (20.9 L)(330.15 K) / 300 K = 23.0 L

Therefore, the new volume of the balloon at a temperature of 57.0 degree C is 23.0 L.

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Hi! Based on the provided information, it's difficult to determine the functional group without seeing the actual structure of the compound. However, I can give you a brief explanation of the three mentioned functional groups:

1. Ketone: A compound containing a carbonyl group (C=O) bonded to two other carbon atoms.
2. Aldehyde: A compound containing a carbonyl group (C=O) bonded to at least one hydrogen atom and one other carbon atom.
3. Carboxylic acid: A compound containing a carbonyl group (C=O) bonded to a hydroxyl group (OH).
Please provide the structure or description of the compound, so I can help you identify the specific functional group present.

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Using the example input 3,7,4,2,8,6, the array would look like this: [1, 2, 1, 1, 3, 2]. The longest monotonically increasing subsequence is 3,4,6 with a length of 3, which is the maximum value in the array.

To find the longest monotonically increasing subsequence of a sequence of n numbers, we can use a dynamic programming approach. We will create an array of length n, where each element in the array will represent the length of the longest monotonically increasing subsequence ending at that index.

We will start with a base case where the first element in the array will have a value of 1, as the longest monotonically increasing subsequence ending at that index is just the element itself. For every other index in the array, we will iterate through all the previous elements and check if they are smaller than the current element.

If so, we will add 1 to the length of the longest monotonically increasing subsequence ending at that index, which will be the maximum value of the longest monotonically increasing subsequence ending at any of the previous indices plus 1.

Finally, we will iterate through the entire array and find the maximum value, which will be the length of the longest monotonically increasing subsequence in the sequence. The time complexity of this algorithm is O(n²) because we need to iterate through all previous elements for every index in the array.

Using the example input 3,7,4,2,8,6, the array would look like this: [1, 2, 1, 1, 3, 2]. The longest monotonically increasing subsequence is 3,4,6 with a length of 3, which is the maximum value in the array.

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If a sample of gas is heated from 100◦C to 300◦C at constant pressure. The volume will increase by a factor of approximately 1.82.

If a sample of gas is heated from 100◦C to 300◦C at constant pressure, the volume will increase by a factor of approximately 1.82, assuming that the gas behaves ideally.

This can be calculated using Charles's Law, which states that at constant pressure, the volume of a fixed amount of gas is directly proportional to its absolute temperature.

The absolute temperature can be calculated by adding 273.15 to the Celsius temperature:

V1/T1 = V2/T2

where V1 and T1 are the initial volume and temperature, and V2 and T2 are the final volume and temperature. Converting 100◦C and 300◦C to absolute temperature gives:

V1/T1 = V2/T2

V1/(100 + 273.15) = V2/(300 + 273.15)

V1/373.15 = V2/573.15

V2 = V1 x (573.15/373.15) = V1 x 1.82

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According to Boyle's law, the pressure of a fixed amount of gas at a constant temperature is inversely proportional to its volume.

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The empirical formula of the substance is: [tex]CH^2O[/tex].

To determine the empirical formula, we need to find the simplest whole number ratio of atoms in the substance.
1. First, we need to find the moles of each element:
- Carbon: 0.200 mol
- Hydrogen: 0.400 mol
- Oxygen: 0.200 mol

2. Next, we need to divide each value by the smallest value to get the ratio of atoms:
- Carbon: 0.200 mol ÷ 0.200 mol = 1
- Hydrogen: 0.400 mol ÷ 0.200 mol = 2
- Oxygen: 0.200 mol ÷ 0.200 mol = 1

3. Finally, we write the chemical formula using the ratios as subscripts:
[tex]C^1H^2O^1[/tex]
However, we can simplify this formula by not writing the subscript 1:
[tex]CH^2O[/tex]

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Answer:

this are iconic compounds

NaCl: sodium chloride

NaBr: sodium bromide

NaF: sodium fluoride

NaI: sodium iodide

KF: potassium fluoride

KCl: potassium chloride

KI: potassium iodide

KBr: potassium bromide

LiI: lithium iodide

Li2O: lithium oxide

MgO: magnesium oxide

MgS: magnesium sulfide

MgSe: magnesium selenide

CaCl: calcium chloride

CaO: Calcium oxide

CaSe: Calcium selenide

The following ionic chemicals are present: c. strontium oxide; e. rubidium bromide; and f. ammonium chloride.

The ionic compounds among the given options. Ionic compounds are formed when a metal and a non-metal exchange electrons, resulting in a bond between positively charged cations and negatively charged anions.

Your answer: Among the given options, the ionic compounds are:
c. SrO (strontium oxide)
e. RbBr (rubidium bromide)
f. NH_4Cl (ammonium chloride)

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The mixture that would result in a buffered solution is mixing 100.0 mL of 0.100 M HCl with 100.0 mL of 0.100 M NH3 (Kb = 1.8 × 10–5).

This is because a buffered solution requires a weak acid and its conjugate base or a weak base and its conjugate acid. In this case, HCl is a strong acid and NH3 is a weak base. When they are mixed together, the NH3 will act as a base and react with the HCl to form NH4+ and Cl-. The NH4+ and NH3 form a conjugate acid-base pair, which makes the solution buffered.

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34.5 moles of water can be made when 23 moles of  NH₃ are consumed.

Chemical equation is a representation of a chemical reaction using symbols and formulas for the reactants and products involved. It shows the chemical species that are consumed and produced during a reaction, and it provides information about stoichiometry of the reaction.

By the given balanced chemical equation: 2 NH₃ + 3 CuO -> 3 Cu + N₂ + 3 H₂O

We observe that for every 2 moles of NH₃  consumed, 3 moles of water are produced. Therefore, we use a proportion to calculate the number of moles of water produced when 23 moles of NH₃  are consumed:

(3 moles  H₂O / 2 moles  NH₃ ) x 23 moles NH₃  = 34.5 moles H₂O

Therefore, 34.5 moles of water can be made when 23 moles of  NH₃  are consumed.

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The initial product is 2-pentyn-1-ol and the final products are 2-penten-1-ol and 3-penten-1-ol (cis/trans isomers).

The response of 2-pentyne (confined 2-yne) with H2O, H2SO4, and Hg2+ is a corrosive catalyzed hydration response. Within the sight of H2SO4 and Hg2+ impetus, the protonation of the triple bond happens first, trailed by the expansion of water to give an enol transitional. The enol middle of the road can then tautomerize to frame either a ketone or an aldehyde isomer.

The underlying item shaped will be the enol middle of the road, which can tautomerize to give the end results: 3-pentanone (a ketone) and 2-pentanone (an aldehyde). These two isomers are in harmony with one another.

The construction of 3-pentanone has a carbonyl gathering at the third carbon position, though 2-pentanone has a carbonyl gathering at the subsequent carbon position. The eventual outcomes are both established isomers, varying in the place of the carbonyl gathering. The component for the response includes the expansion of water across the triple bond, trailed by tautomerization of the enol halfway, which brings about the two isomers.

Generally, this response is an illustration of corrosive catalyzed hydration, which is a significant response in natural science, with various applications in the amalgamation of alcohols, ketones, and aldehydes.

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The given response is a corrosive catalyzed hydration of 2-pentyne or confined 2-yne.

The underlying items are the enol tautomer and the keto tautomer of 2-pentanone. The enol tautomer is shaped by protonation of the triple bond with H+ from H2SO4, trailed by assault of water atom on the carbocation. The keto tautomer is shaped from the enol tautomer by tautomerization.

The end results are the protected isomers of 3-pentanone and 2-pentanone. The keto tautomer of 2-pentanone modifies to shape 3-pentanone, which is more steady because of the stretching of the carbonyl gathering. The keto tautomer of 2-pentanone is the significant item since it is more steady than the enol tautomer.The response system is displayed underneath:

Introductory items:

Enol tautomer of 2-pentanone:

H2C=C(CH3)CH2C(=O)H

Keto tautomer of 2-pentanone:

CH3CH2C(=O)CH2CH3

End results:

3-pentanone:

CH3CH2CH2C(=O)CH2CH3

2-pentanone:

CH3CH2C(=O)CH(CH3)CH2

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