Write the formulas for the following compounds:a. mercury(II) nitrateb. ammonium phosphatec. calcium silicated. lead(II) chromate

The energy radiated by the black hole would be approximately 3.4 x 1031 Joules per second.This means the black hole radiates 3.4 x 10^31 watts of energy.

The energy radiated by a non-spinning black hole that accretes 10-7 Msun per year can be computed using the formula L=ηMc2, where M is the accretion rate. Putting in the numbers, we find L=0.06(10-7 x 6.3 kg/s)(3 x 108 m/s)2 = 3.4 x 1031. Therefore, the energy radiated by the black hole would be approximately 3.4 x 1031 Joules per second.

To calculate the energy radiated by a non-spinning black hole that accretes 10^-7 Msun per year, you can use the formula L=ηMc^2, where L is the luminosity, η is the efficiency, M is the accretion rate, and c is the speed of light. Plugging in the numbers, L=0.06(10^-7 x 6.3 kg/s)(3 x 10^8 m/s)^2 = 3.4 x 10^31. This means the black hole radiates 3.4 x 10^31 watts of energy.

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Explain to the kids that since there is essentially no —which is required for fire—if the bag contains only pure carbon dioxide, the splint would burn out right away.

H2 - Hydrogen Pure hydrogen gas will burst into flames when a burning splint is added to it, making a popping sound. Oxygen (O2) A smouldering splint will rekindle when exposed to a sample of pure oxygen gas.

The flame goes out as a result of carbon dioxide replacing the oxygen it requires to burn (the effect). A popping sound is produced when a flame is near hydrogen because of how the gas burns.

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The atomic mass of gallium is approximately 69.8009 amu.

To find the atomic mass of gallium, we need to calculate the weighted average of the masses of its two isotopes, taking into account their relative abundances.

First, we calculate the contribution of each isotope to the atomic mass:

Contribution of gallium-69 = 68.9256 amu x 0.60108 = 41.4024 amu

Contribution of gallium-71 = 70.9247 amu x 0.39892 = 28.3985 amu

Then, we add these contributions to obtain the atomic mass of gallium:

Atomic mass of gallium = 41.4024 amu + 28.3985 amu = 69.8009 amu

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100.0 g of oxygen gas at a pressure of 1.5 atm and a temperature of 25°C would occupy a volume of 49.2 L.

To solve this problem, we can use the Ideal Gas Law, which states:

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.

We need to rearrange this equation to solve for the volume V:

V = (nRT) / P

where n is the number of moles of the gas, which we can calculate using the molar mass of oxygen gas:

n = m / M

where m is the mass of the gas and M is the molar mass of oxygen gas (32 g/mol).

n = 100.0 g / 32 g/mol = 3.125 mol

Now we can substitute the given values into the equation to find the volume:

V = (nRT) / P

V = (3.125 mol)(0.0821 L·atm/mol·K)(298 K) / 1.5 atm

V = 49.2 L

Therefore, 100.0 g of oxygen gas at a pressure of 1.5 atm and a temperature of 25°C would occupy a volume of 49.2 L.

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Assuming that the capacitors are distinct and not identical, there are eight possible combinations of capacitance that can be produced using all three capacitors in a circuit.

This is because each capacitor can either be included or excluded from the circuit, resulting in two possibilities for each capacitor. With three capacitors, there are 2x2x2 = 8 possible combinations.

For example, if C1 = 1μF, C2 = 2μF, and C3 = 3μF, the eight possible combinations would be 1μF, 2μF, 3μF, 1+2=3μF, 1+3=4μF, 2+3=5μF, 1+2+3=6μF, and no capacitor connected.

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a small amount of Tollens' reagent (ammoniacal silver nitrate) is added to the sugar solution and the mixture is heated. If a reducing sugar is present, it will reduce the silver ions in the Tollens' reagent to metallic silver, which will form a silver mirror on the inside of the test tube.

Based on the structures of D-gulose and D-psicose, it can be predicted that both sugars will give a positive result in the Tollens' test because they both have an aldehyde group that can act as a reducing agent. However, the intensity of the reaction may differ for each sugar.

D-gulose has an aldehyde group at carbon 1, which is in the linear form of the sugar, while D-psicose has an aldehyde group at carbon 2. Since D-gulose can easily convert to its linear form, it is expected to give a stronger positive result in the Tollens' test compared to D-psicose, which may show a weaker positive result due to the steric hindrance of the bulky ketone group at carbon 3.

In summary, the Tollens' test can be used to distinguish between solutions of D-gulose and D-psicose by observing the intensity of the silver mirror formed. D-gulose is expected to give a stronger positive result due to its ability to convert to the linear form, while D-psicose may show a weaker positive result due to steric hindrance.

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There are 10.196 grams of ethane present in the sample. there are 10.201 grams of ethane in the sample.

To calculate the grams of ethane present in a sample containing 0.3393 moles, you can use the formula:

grams = moles x molar mass

Plugging in the given values, we get:

grams = 0.3393 moles x 30.067 g/mol
grams = 10.196 g

Therefore, there are 10.196 grams of ethane present in the sample.

To calculate the grams of ethane present in a sample containing 0.3393 moles, you need to multiply the moles by the molar mass of ethane (30.067 g/mol):

Grams of ethane = moles × molar mass
Grams of ethane = 0.3393 moles × 30.067 g/mol

Grams of ethane = 10.201 g

So, there are 10.201 grams of ethane in the sample.

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The reading on a barometer would change if you were to take one on a trip from Los Angeles to Lake Tahoe, which is at a much higher altitude.

1. A barometer measures atmospheric pressure.
2. Atmospheric pressure decreases with an increase in altitude.
3. Los Angeles is at a lower altitude (approximately 305 feet or 93 meters above sea level) compared to Lake Tahoe (about 6,225 feet or 1,897 meters above sea level).
4. As you travel from Los Angeles to Lake Tahoe, the altitude increases.
5. Due to the increase in altitude, the atmospheric pressure decreases.
6. The barometer reading in Lake Tahoe will be lower than the reading in Los Angeles.

In conclusion, the barometer reading would be lower in Lake Tahoe compared to Los Angeles due to the higher altitude and the resulting decrease in atmospheric pressure.

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If you were to take a barometer on a trip from Los Angeles to Lake Tahoe, which is at a much higher altitude, the reading on the barometer would decrease as you ascend to higher altitudes.

A barometer is a device used to measure atmospheric pressure. The pressure at sea level is approximately 1013 hPa (hectopascals) or 29.92 inches of mercury (inHg). As you increase in altitude, the pressure decreases due to the decreased weight of the atmosphere above. For every 1000 feet increase in altitude, there is an approximate decrease of 1 inch of mercury (inHg) or 33 hPa in pressure.

Lake Tahoe has an elevation of approximately 6,225 feet, which is significantly higher than Los Angeles, which is only 233 feet above sea level. As a result, the atmospheric pressure at Lake Tahoe would be lower than the pressure in Los Angeles. Therefore, if you were to take a barometer from Los Angeles to Lake Tahoe, the reading on the barometer would decrease as you ascend to higher altitudes.

In summary, the reading on a barometer would decrease as you ascend to higher altitudes such as Lake Tahoe from Los Angeles.

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The moles of water that were removed from the sample by the heating process is 0.436 mol.

The Mass of evaporating dish = 6.251 g

The mass of dish + hydrate = 16.864 g

The mass of hydrate =  (mass of the dish + hydrate) - mass of the dish

The hydrate = 16.864 - 6.251 = 10.613 g

The Mass of dehydrated substance = (mass of the dish + dehydrate ) - mass of the dish

The dehydrated material = 11.13 - 8.365 = 2.765 g

The Mass of water evaporated = hydrate - dehydrate

The mass of the water = 10.613 - 2.765 = 7.848 g

The Molar mass of the water  = 18 g/mol

The number of moles of the water evaporated = mass / molar mass

The moles = 7.848 / 18 = 0.436 moles

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Therefore, 1.546 grams of SmO are equal to 9.3 x 10-3 moles of SmO.

Under humid conditions or at temperatures above 150 °C in dry air, the chemical compound Samarium oxide (Sm2O3) rapidly develops on the surface of samarium metal1. It is typically white to off yellow in color and is frequently seen as a very fine powder that resembles dust1.

Among the many applications for samarium oxide are:

It is used to absorb infrared light in optical and infrared absorbing glass1.

It serves as a neutron absorber in nuclear power reactor control rods1.

To convert moles to grams, multiply the mass of the material by the molecular weight or formula weight. The oxide catalyzes the dehydration and dehydrogenation of primary and secondary alcohols. In other words, it is a result of mass production.. To put it another way, it is the result of the substance's mass and molecular weight. Samarium oxide has a molecular weight of 166.3594 g/mol5.

Thus, we can use the following formula to convert 9.3 x 10-3 moles of SmO to grams:

grams are equal to the substance's mass (in moles) times its molecular weight.

Inputting the values, we obtain:

Grams are 9.3 x 10-3 moles, and 166.3594 g/mol equals (about) 1.546 grams.

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If a ball rolls straight into an object with a greater mass than a bowling pin, the ball will experience a force upon impact.

If the object is stationary, the ball will transfer its momentum to the object upon impact. The object will then move in the direction of the ball's momentum, and the ball will come to a stop. This is an example of an inelastic collision.

If the object is moving in the same direction as the ball, the collision will result in a transfer of momentum between the two objects. The ball will slow down, while the object will speed up. The final velocities of the ball and object will depend on their respective masses and velocities before the collision.

Therefore, the result of the collision will depend on the specific circumstances of the collision and the properties of the objects involved.

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The enthalpy of the given reactions are as follows:

A. -1415.15 kJ/mol

B. -774.5 kJ/mol

C. -27.7 kJ/mol

D. -411.5 kJ/mol

Using the given heat of formation values, we can calculate the enthalpy of the reactions using the formula: ΔH°rxn = ΣΔH°f (products) - ΣΔH°f (reactants)

a) SiF₄(g) -> Si(s) + 2F₂(g)

Products:

Si(s): ΔH°f = 0

2F₂(g): ΔH°f = 0

Reactants:

SiF₄(g): ΔH°f = 1415.15 kJ/mol

ΔH°rxn = (0 + 0) - 1415.15 = -1415.15 kJ/mol

Answer: -1415.15 kJ/mol

b) SO₃(g) + H₂O(g) ->H₂SO₄(aq)

Products:

H₂SO₄(aq): ΔH°f = -909.3 kJ/mol

Reactants:

SO₃(g): ΔH°f = -395 kJ/mol

H₂O(g): ΔH°f = 241.8 kJ/mol

ΔH°rxn = (-909.3) - ((-395) + 241.8) = -774.5 kJ/mol

Answer: -774.5 kJ/mol

c) 3K₂O₂(s) + 3H₂O(l) -> 6KOH(aq) + O₃(g)

Products:

6KOH(aq): ΔH°f = -2517.6 kJ/mol

O₃(g): ΔH°f = 142.7 kJ/mol

Reactants:

3K₂O₂(s): ΔH°f = -857.4 kJ/mol

3H₂O(l): ΔH°f = -241.8 kJ/mol

ΔH°rxn = (-2517.6 + 142.7) - ((-857.4) + (-241.8)) = -27.7 kJ/mol

Answer: -27.7 kJ/mol

d) Fe₃O₄(s) + 8HCl(g) -> 2FeCl₃(s) + FeCl₂(s) + 4H₂O(g)

Products:

2FeCl₃(s): ΔH°f = -1317.6 kJ/mol

FeCl₂(s): ΔH°f = -341.8 kJ/mol

4H₂O(g): ΔH°f = -483.6 kJ/mol

Reactants:

Fe₃O₄(s): ΔH°f = -1117.5 kJ/mol

8HCl(g): ΔH°f = -436.8 kJ/mol (note: this is the ΔH°f for 8HCl gas molecules, not 1 mole of HCl)

ΔH°rxn = ((-1317.6) + (-341.8) + (-483.6)) - ((-1117.5) + (-436.8)) = -411.5 kJ/mol

Answer: -411.5 kJ/mol

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The balanced chemical equation for the reaction between aluminum and silver nitrate is:

2 Al + 3 AgNO3 → 3 Ag + 2 Al(NO3)3

From the equation, we can see that 3 moles of aluminum nitrate (Al(NO3)3) are produced for every 3 moles of silver nitrate (AgNO3) consumed.

Therefore, if 0.75 moles of silver nitrate react, we can calculate the number of moles of aluminum nitrate produced as follows:

0.75 mol AgNO3 x (2 mol Al(NO3)3 / 3 mol AgNO3) = 0.50 mol Al(NO3)3

So, 0.50 moles of aluminum nitrate (Al(NO3)3) are obtained from the reaction of 0.75 mol of silver nitrate with a sufficient amount of aluminum.

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Coinage metals, which typically include copper, silver, and gold, are chosen over more active metals for use in coins because they are less reactive and more resistant to corrosion.

This ensures durability and preserves the appearance of the coins. Some more active metals like zinc or nickel are sometimes used in coins due to their lower cost and availability, while still maintaining adequate resistance to corrosion and wear for everyday use.

The reason why the last 4 miles in the activity series of metals, which are gold, silver, platinum, and palladium, are commonly referred to as the "coinage medals" is because they are highly resistant to corrosion and have a low reactivity towards other chemicals, making them ideal for use in coins. These metals are also very rare and valuable, which adds to their appeal as a currency.

More active metals such as zinc or nickel are sometimes used in coins because they are more abundant and less expensive than the "coinage metals". However, these metals tend to be more reactive and therefore more prone to corrosion and other chemical reactions, which can affect the appearance and value of the coins over time. Additionally, the use of these metals in coins is often limited to lower denominations or commemorative coins, rather than as a standard currency.

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The "coinage metals" are typically gold, silver, copper, and platinum, which are the last 4 metals in the activity series. These metals are chosen over more active metals for use in coins because they are relatively unreactive and do not corrode easily, making them ideal for coins that need to be durable and long-lasting. Additionally, these metals have been historically valued and used as currency, making them culturally significant as well.

However, some more active metals such as zinc or nickel are sometimes used in coins because they are cheaper and more readily available than the coinage metals. These metals may be used as an alloy with the coinage metals to make coins more affordable, or they may be used as a substitute for the more expensive metals in lower denomination coins. However, these metals are not as durable as the coinage metals and may corrode more easily, leading to shorter lifespans for the coins.

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Point source pollution, like industrial waste flowing directly into a river from a factory drain, is one example. Plastic grocery bags and parking areas were contaminated by pollutants including motor oil leaks.

Two instances of how humans directly affect the water in watersheds are the construction of dams and the rerouting of rivers. Mankind have used water as just a resource, obtaining our drinking water from watersheds. As water use may be controlled to be sustainable, this need not have a bad effect.

The effects of modern livestock farming, development, deforestation, and Dioxide emissions, among several other things, are accelerating the biodiversity decline because to desertification, ocean and river pollution, and global warming.

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The correct option is

In an endothermic response(reaction), the whole vitality(total energy) at the beginning of the response is more noteworthy than the full vitality at the conclusion of the response 

because endothermic responses retain warmth from the environment, which implies that the vitality of the framework increments.

An endothermic response may be a chemical response that retains warmth from the environment, which implies that the vitality of the framework increments.

This increment in vitality is utilized to break the bonds between the particles or atoms within the reactants, and the items are shaped from the modification of these iotas or atoms into unused bonds.

As a result, the whole vitality of the framework at the conclusion of the response is more noteworthy than the full vitality at the start of the response. This increment in vitality is ordinarily watched as an increment within the temperature of the framework or its environment. 

In an endothermic response, the whole vitality at the beginning of the response is less than the overall vitality at the end of the response

.

Usually, endothermic responses retain warmth from the environment, which implies that the vitality of the framework increments.

As a result, the entire vitality of the framework at the conclusion of the response is greater than the full energy at the start of the response. Subsequently,

The proper reply is "In an endothermic response(reaction), the whole vitality(total energy) at the beginning of the response is more noteworthy than the full vitality at the conclusion of the response ".

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The classification of the given reaction is "Redox" which is the correct answer based on the chemical equation of the reaction given in the question.

The classification for the given reaction is a double displacement reaction.
The given reaction is:

2I¯(aq) + Cl2(aq) → I2(aq) + 2Cl¯(aq)

This reaction involves the exchange of ions between the reactants, and it can be classified as a "Redox" reaction (which stands for reduction-oxidation). In this reaction, the iodide ions (I¯) are oxidized, as they lose electrons to form I2, while the chlorine (Cl2) is reduced, as it gains electrons to form chloride ions (Cl¯).
To summarize, the classification of the given reaction is "Redox."

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The classification for the reaction 2I¯(aq) + [tex]Cl_{2}[/tex](aq) → [tex]I_{2}[/tex](aq) + 2Cl¯(aq) is a redox reaction.

Redox reactions involve the transfer of electrons from one species to another, resulting in changes in oxidation states (the charge or electron density on an atom).  The reaction can be classified as a redox (reduction-oxidation) reaction. In this reaction, iodine (I¯) is being oxidized, and chlorine ( [tex]Cl_{2}[/tex]) is being reduced. To identify such reactions, we can

1. Identify the initial oxidation states: I¯ has an oxidation state of -1, and [tex]Cl_{2}[/tex] has an oxidation state of 0.
2. Identify the final oxidation states:  [tex]I_{2}[/tex] has an oxidation state of 0, and Cl¯ has an oxidation state of -1.
3. Observe the change in oxidation states: I¯ is oxidized from -1 to 0, while [tex]Cl_{2}[/tex] is reduced from 0 to -1.

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Therefore, the net ionic equation for the reaction is Copper(2+) (aq) + 2 chlorine- (aq) → Copper(II) chloride (aq).

Ammonium sulphate and Copper hydroxide precipitate are the first products of the reaction between copper sulphate and ammonium hydroxide.

Mixing copper(II) sulphate and ammonium chloride results in the following observations:

Conventional: When copper ions (Copper2+) from Copper(II) sulfate are present, a blue solution develops. The colour of Ammonium Chloride doesn't seem to have changed at all.

Ionic total: While Ammonium Chloride dissociates into Ammonium and Chlorine- ions in solution, Copper(II) sulfate dissociates into Copper2+ and Sulfate 2- ions.

Copper(II) sulfate (aq) + 2 Ammonium Chloride (aq) → Copper(II) Chloride (aq) + 2 Ammonium (aq) + Sulfate 2- (aq)

Net Ionic: The net ionic equation shows only the species involved in the reaction. In this case, the Copper2+ and the Cl- ions combine to form Copper(II) chloride.

Copper2+ (aq) + 2 Chlorine- (aq) → Copper(II) chloride (aq)

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Expected grams of aluminum oxide product from the given masses of reactants are 18.93 g.

Aluminum is chemical element with symbol Al and atomic number is 13.

4Al + 3O₂ → 2Al₂O₃

10 g Al × 1 mol Al / 26.98 g Al = 0.371 mol Al

4 g O₂ × 1 mol O₂ / 32.00 g O₂ = 0.125 mol O₂

We determine the limiting reactant by comparing the mole ratios of aluminum and oxygen in the balanced equation and reactant that produces  smaller amount of product is limiting reactant. In this case, aluminum is the limiting reactant because it produces only 0.1855 moles of aluminum oxide, which is less than the 0.25 moles of aluminum oxide produced by the oxygen:

0.371 mol Al × 2 mol Al₂O₃ / 4 mol Al = 0.1855 mol Al₂O₃

0.125 mol O₂ × 2 mol Al₂O₃ / 3 mol O2 = 0.2083 mol Al₂O₃

0.1855 mol Al₂O₃ × 101.96 g/mol = 18.93 g Al₂O₃

Therefore, expected grams of aluminum oxide product from the given masses of reactants are 18.93 g.

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The initial rate of decomposition of X is 0.0013 M/h.

The first-order rate law is given as:

Rate = k [X]

Where, k = rate constant

[X] = concentration of X

Since the partial pressure of X is given in the problem, we need to convert it to concentration using the ideal gas law:

PV = nRT

where:

P = partial pressure of X = 0.473 atm

V = volume of the flask = 20.0 L

n = number of moles of X

R = ideal gas constant = 0.08206 L atm K^-1 mol^-1

T = temperature of the flask (assumed constant) = 298 K

Solving for n,

n = PV/RT = (0.473 atm)(20.0 L)/(0.08206 L atm K^-1 mol^-1)(298 K) = 0.952 mol X

At t = 0, the concentration of X is [X]_0 = n/V = 0.952 mol/20.0 L = 0.0476 M.

Using the given data, we can calculate the rate constant (k) as follows:

ln([X]_0/[X]) = kt

where:

t = time = 5.6 hours

Substituting the given values,

ln(0.0476/0.0376) = k(5.6 hours)

Solving for k, we get:

k = (ln(0.0476/0.0376))/5.6 hours = 0.0263 h^-1

The initial rate of decomposition of X is given by:

Rate = k[X]_0 = (0.0263 h^-1)(0.0476 M) = 0.00125 M/h

Rounding off to 2 significant digits,

Initial rate of decomposition of X = 0.0013 M/h.

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The grams of N₂ would have to react to produce 31.5 kJ of the energy is 84 g.

The chemical equation is as :

N₂ + 3H₂   --->     2NH₃       ΔH = -96 kJ

The energy produces = 31.5 kJ

We have multiplied the factor so that the value of the enthalpy has also been multiplied.

The factor = 96 / 31.5 = 3

Thus, the balanced chemical equation is :

3N₂ + 9H₂   --->   6NH₃  

The moles of N₂ = 3 mol

The mass of the N₂ = moles × molar mass

The mass of the N₂ = 3 mol × 28 g/mol

The mass of the N₂ = 84 g

The amount of the N₂ would have to react to produce the 31.5 kJ of the energy is 84 g.

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This question is incomplete, the complete question is :

The following thermochemical equation is for the reaction of n2 with h2 to form nh3. how many grams of n2 would have to react to produce 31.5 kj of energy?

N₂ + 3H₂   --->     2NH₃      ΔH = -96 kJ

The forecast indicates a 20% chance of rain, which means there is a low probability of precipitation occurring.

Rain is a type of precipitation that occurs when water droplets fall from the atmosphere and reach the Earth's surface. It is a vital part of the water cycle, which involves the continuous movement of water from the Earth's surface to the atmosphere and back again.

Rain is formed when water vapor in the atmosphere condenses into tiny water droplets or ice crystals, which combine to form clouds. When the clouds become heavy enough, the water droplets or ice crystals fall to the ground as precipitation, which can take the form of rain, snow, sleet, or hail.

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To find the percent ionization of the acid, we need to first calculate the initial concentration of the acid (HA) before it dissociates.

Since the solution is originally 0.532 M in acid (HA), we can assume that the initial concentration of HA is also 0.532 M.

Next, we need to calculate the concentration of the conjugate base (A-) at equilibrium. We can use the equation for the dissociation of an acid:
HA + H2O ⇌ H3O+ + A-

We know that the hydronium concentration at equilibrium is 0.112 M, so the concentration of the conjugate base is also 0.112 M.

To calculate the percent ionization of the acid, we use the equation:
% ionization = (concentration of dissociated acid / initial concentration of acid) x 100

We can find the concentration of dissociated acid (H3O+) by subtracting the concentration of the conjugate base (A-) from the hydronium concentration:

[H3O+] = 0.112 M - 0 M = 0.112 M

Plugging in the values, we get:
% ionization = (0.112 M / 0.532 M) x 100 = 21.05%

Therefore, the percent ionization of the acid is 21.05%.

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The percent of ionization of an acid in solution of 0.532 M in acid HA i and have a hydronium concentration of 0.112 M is equals to the 21.1%.

The ionization of acids results hydrogen ions, thus, that's why compounds act as proton donors.

Molarity of solution = 0.532 M

At Equilibrium, hydronium concentration = 0.112 M

As we know, concentration is defined as the number of moles of substance in a litre of solution, that most of time concentration is replaced by molarity. So, concentration of acid solution, [ H A] = 0.532 M

Chemical reaction, [tex]HA (aq) + H_2O -> H_3O^{ +}+A^{-}[/tex]

percent of ionization of the acid =

[tex] \frac{ [ H_3O^{+}] }{ [ HA]} × 100 [/tex]

= (0.112/0.532) × 100

= 21.1%

Hence, required value is 21.1%.

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The mass percentage of sodium nitrate in a solution consisting of 73 g of sodium nitrate and 135 g of water is 35.1% and therefore we can say option (a) is correct.

The mass percentage is a concentration term used to describe concentration in solution. It is defined as the percentage of the mass of the solute per mass of the solution.

It can be written as the mass of solute/ mass of solution * 100

According to the question,

mass of the solute = 73 g

mass of the solvent = 135 g

mass of the solution = mass of solute + mass of solvent

= 73 + 105

= 208 g

mass percentage = [tex]\frac{73}{208} *100[/tex]

= 0.351 * 100

= 35.1 %

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The correct answer is c. 73.6%. The mass percentage of sodium nitrate in the solution is 73.6%.

To find the mass percentage of sodium nitrate in the solution, we need to first calculate the total mass of the solution.

Total mass of solution = mass of sodium nitrate + mass of water
Total mass of solution = 73 g + 135 g
Total mass of solution = 208 g

Next, we need to calculate the mass of sodium nitrate as a percentage of the total mass of the solution.

Mass percentage of sodium nitrate = (mass of sodium nitrate / total mass of solution) x 100%
Mass percentage of sodium nitrate = (73 g / 208 g) x 100%
Mass percentage of sodium nitrate = 35.1%

Therefore, the mass percentage of sodium nitrate in the solution is 73.6%.

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One should be careful with diethyl ether and tert-butanol when using a hot plate as they have low flash points and can easily ignite.

It is important to take proper precautions such as using a well-ventilated area and avoiding any sources of ignition. Sulfuric acid and glacial acetic acid are also potentially dangerous as they are corrosive and can cause severe burns if they come into contact with skin. Propanol and butanol have higher flash points and are generally safer to use on a hot plate.

When using a hot plate in an experiment, one should be particularly careful with diethyl ether and tert-butanol. Diethyl ether is highly flammable and volatile, while tert-butanol (2-methyl-2-propanol) can generate flammable vapors when heated. These compounds pose a risk of fire or explosion if not handled properly.

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Each mole of the Al(NO₃)₃ contains number of the moles of oxygen atoms 1.3 mol of the oxygen atoms.

The one mole of the Al(NO₃)₃  is as :

In the one mole of the Al(NO₃)₃ contain the 1 atom of the Aluminum,

The 3 atoms of the Nitrogen,

The 9 atoms of the oxygen

The 1 mole of the atoms = 6.022 × 10²³ atoms

The number of the oxygen atoms = 9 atoms of the oxygen

The number of the moles of the oxygen in the 1 mole of the Al(NO₃)₃ :

The 9 atoms of the oxygen = 1.3 mol

Thus, in the 1 mole of the Aluminum nitrate, there will be the 1.3 mol of the oxygen atom.

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Answer:The pH of a buffer prepared by adding 1.00 L of 1.0 M HCl to 750 ml of 1.5 M NaHCOO is 2.84

What is pH?

pH is a measure of the acidity of a solution.

pH is calculated from the negative logarithm to base ten of the hydrogen ions concentration of the solution.

For weak acids such as those used in the preparation of buffers, the acid dissociation constant, Ka are used to determine the pH of the solution.

Therefore, from the Ka of acetic acid, the pH of a buffer prepared by adding 1.00 L of 1.0 M HCl to 750 ml of 1.5 M NaHCOO is 2.84

The grams of N2 are required to completely react with 3.03 grams of H2 for the following balanced chemical equation is 14 g.

We may calculate the number of moles of H2 that will be used by dividing the amount of H2 that will be utilised by its molar mass. We may multiply that number by the molar mass of N2 to get how many grammes we should use. We can divide that mole quantity by 3 to determine how many moles of N2 the reaction will consume.

In the reaction 1 mole of N2 react with  3 mole of H2 and give 2 mole of NH3

mass of H2 = 3.03g

No of moles of H2 = 3.03g/2 gmol-1

         = 1.51 mole

1.51 mole of H2 require N2 = (1/3)× 1.51 moles  

        = 0.50 mole N2

molar mass of N2  =28g/mol

Mass of N2 require   = 0.50mole ×28g/mol

    = 14g

Mass of N2 require = 14g.

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The answer is C. 14.0 grams of N2 are required to completely react with 3.03 grams of H2.

The balanced chemical equation is:

N2 + 3H2 -> 2NH3

From the equation, we can see that 1 mole of N2 reacts with 3 moles of H2 to produce 2 moles of NH3.

To find out how many grams of N2 are required to react with 3.03 grams of H2, we first need to convert 3.03 grams of H2 to moles:

moles of H2 = mass of H2 / molar mass of H2
moles of H2 = 3.03 / 2.016
moles of H2 = 1.505

Now, we can use the mole ratio from the balanced equation to find out how many moles of N2 are required to react with 1.505 moles of H2:

moles of N2 = (1.505 mol H2) / (3 mol H2/1 mol N2)
moles of N2 = 0.5017

Finally, we can convert moles of N2 to grams of N2:

mass of N2 = moles of N2 x molar mass of N2
mass of N2 = 0.5017 x 28.02
mass of N2 = 14.04

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