What is the concentration of h2so4 if 12. 3 ml of 0. 200 m naoh solution is needed to neutralize 10. 0 ml of h2so4 solution ?.

Answers

Answer 1

The concentration of H₂SO₄ in the original solution is 0.123 M.

Balanced chemical equation for the neutralization reaction between sulfuric acid (H₂SO₄) and sodium hydroxide (NaOH) is;

H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O

From the equation, we can see that one mole of sulfuric acid reacts with two moles of sodium hydroxide. Therefore, we can use the following equation to calculate the moles of sulfuric acid present in the 10.0 mL of H₂SO₄ solution;

moles of H₂SO₄ = moles of NaOH / 2

To calculate the moles of NaOH, we can use the following equation;

moles of NaOH = Molarity x Volume (in liters)

The volume of NaOH used is 12.3 mL, which is 0.0123 L.

Substituting the given values into the equation;

moles of NaOH = 0.200 mol/L x 0.0123 L = 0.00246 moles

Now we can calculate the moles of H₂SO₄;

moles of H₂SO₄ = 0.00246 moles / 2 = 0.00123 moles

Finally, we can calculate the concentration of the H₂SO₄ solution in units of moles per liter (M);

Molarity of H₂SO₄ = moles of H₂SO₄ / volume of H₂SO₄ (in liters)

The volume of H₂SO₄ used is 10.0 mL, which is 0.0100 L.

Substituting the values we know;

Molarity of H₂SO₄ = 0.00123 moles / 0.0100 L

= 0.123 M

Therefore, the concentration of H₂SO₄ is  0.123 M.

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Related Questions

what is the temperature of nitrogen molecules contained in an 8.1- m3 volume at 3.0 atm if the total amount of nitrogen is 1900 mol ?

Answers

The temperature of the nitrogen molecules in the given volume is 126.4 K.

To solve this problem, we can use the ideal gas law:

PV = nRT

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

We are given P, V, and n, so we can rearrange the ideal gas law to solve for T:

T = PV / nR

Substituting the given values, we get:

T = (3.0 atm) × ([tex]8.1 m^3[/tex]) / (1900 mol × 8.314 J/(mol*K))

T = 126.4 K

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what is the minimum integer value of the stoichiometric coefficient for substance needed to balance this chemical equation: ?

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The minimum integer value of the stoichiometric coefficient for substance HCl(aq) needed to balance this chemical equation is 6.

In the field of chemistry known as stoichiometry, needed quantitative data is ascertained by exploiting relationships between the reactants and/or products of a chemical process. Stoichiometry literally translates as the measure of elements because the Greek words stoikhein and metron both mean element and measure, respectively. To comprehend the links between products and reactants and why they exist, which necessitates understanding how to balance processes, is necessary before using stoichiometry to do calculations regarding chemical reactions.

Chemical equations using chemical symbols are widely used to represent chemical processes in chemistry. The reactants are shown on the left side of the equation, and the products are shown on the right, with a single or double arrow separating the two sides to indicate the reaction's direction. When discussing solubility constants, the significance of single and double arrows is significant, but we won't go into detail about it in this module. An equation must have an equal number of atoms on both its left and right sides in order to be balanced. By increasing the coefficients, one can achieve this.

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Which one of the following salts produces neutral solutions when it is dissolved in water?
a. NH4F
b. LiOCl
c. BaBr2
d. CaSO3
e. (NH4)2SO4

Answers

Out of the given salts, the one that produces a neutral solution when dissolved in water is NH4F. When an acid and a base react, they produce salt and water. Some salts can produce an acidic or basic solution depending on the nature of the acid and base used.

In the case of NH4F, it is a salt produced from the reaction of a weak acid (NH4OH) and a strong base (HF). Therefore, it has the ability to produce a neutral solution when dissolved in water. LiOCl, BaBr2, CaSO3, and (NH4)2SO4 are all produced from the reaction of a strong acid and a strong base or a weak acid and a weak base, making them acidic, basic or even amphoteric (able to produce both acidic and basic solutions) when dissolved in water.  For example, if a strong acid and a weak base react, the resulting salt will produce an acidic solution. On the other hand, if a weak acid and a strong base react, the resulting salt will produce a basic solution.

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when doing buffer region equation, what does an acid and base look like?

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The equation which represents an acid-base or a buffer solution is represented below-

pH = pKₐ + log([A⁻]/[HA])

One way to determine the pH of a buffer is by using the Henderson–Hasselbalch equation, which is

pH = pKₐ + log([A⁻]/[HA])

In the above equation, [HA] and [A⁻] refer to the equilibrium concentrations of the conjugate acid–base pair used to create the buffer solution. For the titration of a weak acid with a strong base, the pH curve is initially acidic and has a basic equivalence point (pH > 7). The section of curve between the initial point and the equivalence point is known as the buffer region.

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Water with density 1000kg/m3 is moving at 0.50m/s through a cylindrical tube with a diameter of 0.10m. The tube then narrows to a diameter of 0.05m. The mass flow rate in the narrow section of pipe is most nearly

Answers

The mass flow rate in the narrow section of the pipe is also 3.93 kg/s.

What is density?

Density is the measurement of how tightly a material is packed together. It is defined as the mass per unit volume.

To solve this problem, we can use the principle of conservation of mass, which states that the mass flow rate must be the same at any point along the pipe. Therefore, we can use the density and velocity of the water at the beginning of the pipe to calculate the mass flow rate, and then use the continuity equation to find the velocity of the water in the narrow section of the pipe.

The cross-sectional area of the tube at the beginning is A1 = (π/4)*(0.10m)² = 0.00785 m².

The volume flow rate of water at the beginning is Q₁ = A₁*v₁, where v₁ = 0.50 m/s is the velocity of the water. Therefore, Q₁ = 0.00785 m² * 0.50 m/s = 0.00393 m³/s.

The mass flow rate at the beginning is m1 = ρ*Q₁, where ρ = 1000 kg/m³ is the density of water. Therefore, m₁ = 1000 kg/m³ * 0.00393 m³/s = 3.93 kg/s.

According to the continuity equation, the mass flow rate at the narrow section of the pipe is the same as at the beginning, so m₂ = m₁ = 3.93 kg/s.

The cross-sectional area of the tube at the narrow section is A₂ = (π/4)*(0.05m)² = 0.00196 m².

The velocity of the water in the narrow section is v2 = Q2/A2, where Q2 is the volume flow rate of water in the narrow section. Therefore, Q2 = v2*A2.

Using the continuity equation, we have:

m₁ = m₂ = ρ*Q₂ = ρ*v₂*A₂

Substituting the values we found:

3.93 kg/s = 1000 kg/m³ * v₂ * 0.00196 m²

Solving for v₂, we get:

v₂ = 3.93 kg/s / (1000 kg/m³ * 0.00196 m²) = 2.01 m/s

Therefore, the mass flow rate in the narrow section of the pipe is also 3.93 kg/s.

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when we react a weak base with a strong acid, the component of the base that contributes to the ph of the solution is the:select the correct answer below:acidbaseconjugate acidconjugate base

Answers

When we react a weak base with a strong acid, the component of the base that contributes to the ph of the solution is the conjugate acid.

When a weak base is reacted with a strong acid, the weak base is protonated by the acid to form its conjugate acid. The conjugate acid of the weak base is the species that contributes to the pH of the solution since it is the one that can donate protons to water and increase the concentration of H₃O⁺ ions.

The conjugate base of the weak base, on the other hand, is typically very weakly basic and does not affect the pH significantly.

For example, when ammonia (NH₃), a weak base, reacts with hydrochloric acid (HCl), a strong acid, it forms ammonium chloride (NH₄Cl). The ammonium ion (NH₄⁺), which is the conjugate acid of NH₃, contributes to the pH of the solution by donating protons to water to form H₃O⁺ ions. |

The chloride ion (Cl⁻), which is the conjugate base of HCl, does not affect the pH significantly.

Hence when a strong acid reacts with a weak base, the part of the base that affects the solution's pH is called the conjugate acid.

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What color does blue litmus turn in the presence of an acid?.

Answers

Answer:

Blue litmus turn to red color in the presence of an acid

blue litmus turns red in the presence of acid

Calculate the pH of 0.15 M Co(NO3)2.
For [Co(OH2)6]2+, Ka = 5.0 × 10−10
a. 4.52
b. 4.74
c. 4.88
d. 5.06
e. 5.28

Answers

To calculate the pH of 0.15 M Co(NO3)2, we need to first determine if the solution is acidic, basic, or neutral. Since Co(NO3)2 is a salt, it will dissociate into Co2+ and NO3- ions in water. Neither of these ions is acidic or basic on their own, so the solution will be neutral.

However, the presence of the Co2+ ion can slightly hydrolyze water and create a small amount of H+ ions, making the solution slightly acidic. To calculate the pH, we need to use the equilibrium constant expression for this reaction:

Co2+ + H2O ⇌ CoOH+ + H+

The equilibrium constant expression for this reaction is:

K = [CoOH+][H+]/[Co2+]

Since the solution is neutral, we can assume that [H+] = [OH-] = 1.0 x 10^-7 M. We also know that [Co2+] = 0.15 M, and since CoOH+ is a weak acid, we can assume that it dissociates only slightly and [CoOH+] ≈ 0. Therefore, we can simplify the equilibrium constant expression to:

K = [H+]^2/[Co2+]

Plugging in the values we know:

1.0 x 10^-7 = (x)^2/(0.15)

Solving for x gives us:

x = 3.87 x 10^-4 M

Taking the negative log of this value gives us the pH:

pH = -log(3.87 x 10^-4) = 3.41

Therefore, the pH of 0.15 M Co(NO3)2 is approximately 3.41.

Note: It is important to check that the assumption made for [CoOH+] is valid. If it dissociates more than assumed, it will affect the pH calculation. However, in this case, the assumption is valid since CoOH+ is a weak acid and its dissociation is expected to be minimal.

The correct answer options were not provided, but the calculated pH value of 3.41 falls between d. 5.06 and e. 5.28, suggesting that neither of those options is correct.

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why is the splitting between the energy levels greater for higher lying energy levels than for lower lying energy levels

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Answer:

The splitting between energy levels is greater for higher lying energy levels than for lower lying energy levels because of the Coulomb force, which is the force of attraction or repulsion between charged particles.

In an atom, the positively charged nucleus exerts an attractive force on the negatively charged electrons, holding them in orbit around the nucleus. However, the electrons also repel each other due to their negative charges. The net result is that the energy levels of the electrons in an atom are determined by a balance between the attractive and repulsive forces acting on them.

The Coulomb force is proportional to the product of the charges of the interacting particles and inversely proportional to the square of the distance between them. As the distance between the nucleus and the electron increases, the Coulomb force becomes weaker, resulting in smaller energy differences between adjacent energy levels. Conversely, as the distance between the nucleus and the electron decreases, the Coulomb force becomes stronger, resulting in larger energy differences between adjacent energy levels.

Since higher lying energy levels are farther away from the nucleus than lower lying energy levels, the Coulomb force is weaker for the higher energy levels, resulting in larger energy differences between adjacent energy levels. This is why the splitting between energy levels is greater for higher lying energy levels than for lower lying energy levels.

What does the N stand for in geometric random variables?

Answers

In the context of geometric random variables, the letter "N" usually stands for the number of trials required to obtain the first success in a sequence of independent Bernoulli trials with a fixed probability of success, denoted by "p".

A geometric random variable models the probability distribution of the number of failures that occur before the first success in a sequence of independent trials, each with a probability of success "p". The number of trials required to obtain the first success is a random variable that follows a geometric distribution. The probability mass function of a geometric random variable N is given by P(N = k) = (1 - p)^(k-1) * p, where k is the number of trials required to obtain the first success. The expected value of a geometric random variable is E[N] = 1/p.

The geometric distribution is commonly used in various fields, such as reliability analysis, queueing theory, and statistical inference.

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"What is the molar solubility of AgCl in 0.10 M NaCN if the colorless complex ion Ag(CN) 2 - forms? K sp for AgCl is 1.8 × 10 ^-10 and K f for Ag(CN) 2 - is 1.0 × 10^ 21.
0.050 M
0.10 M
0.40 M
0.20 M"

Answers

The answer is 0.050 M. The molar solubility of a compound depends on its solubility product (Ksp), which is the equilibrium constant for the dissociation of the compound into its constituent ions in water.

What is Molar Solubility?

Molar solubility is defined as the number of moles of a solute that can dissolve in one liter of solvent (usually water) to form a saturated solution at a particular temperature and pressure. It is typically denoted by the symbol "s" and has units of mol/L.

Substituting Ksp and Kf into the expression for x:

Ksp = [Ag+][Cl-] =[tex]x^{2}[/tex]

Kf = [[tex]Ag(CN)^{2}[/tex]-]/[Ag+][tex][CN-]^{2}[/tex]

[Ag(CN)2-] = [Ag+]

x = [Ag+]

x^2 = Ksp = 1.8 × [tex]10^{-10[/tex]

x = √(1.8 × [tex]10^{-10}[/tex]) = 1.34 × [tex]10^{-5}[/tex]

[Ag(CN)2-] = 5.6 ×[tex]10^{-7}[/tex] M

Since [Ag(CN)2-] = [Ag+], the molar solubility of AgCl in the presence of 0.10 M NaCN is:

x = [Ag+] = [Ag(CN)2-] = 5.6 × [tex]10^{-7}[/tex] M

Therefore, the answer is 0.050 M.

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what happened to the ph of the solution as you came near to adding stoichiometrically equivalent amounts of base to your acids? why do you think that this occurred

Answers

As we add a base to an acidic solution, the pH of the solution increases.

The pH will continue to increase as we add more base until it reaches a certain point where the amount of base added is stoichiometrically equivalent to the amount of acid present in the solution. At this point, the pH of the solution will be at its highest point, known as the equivalence point.

At the equivalence point, all the acid has reacted with the base, and the solution contains only the salt and water formed by the reaction. The pH of the solution at the equivalence point will depend on the strength of the acid and base used.

The reason for the increase in pH as we add base is due to the neutralization reaction that takes place between the acid and the base. The acid donates a proton (H⁺) to the base, which accepts the proton and becomes a conjugate acid. This reaction reduces the concentration of H⁺ ions in the solution, which causes the pH to increase.

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an ammonia solution of unknown concentration is titrated with a solution of hydrochloric acid. the hcl solution is 1.25 m, and 5.19 ml are required to titrate 12.61 ml of the ammonia solution. what is the molarity of the ammonia solution?

Answers

The molarity of the NH₃ solution that is titrated with a solution of hydrochloric acidis 0.5147 mol/L.

To find the molarity of the NH₃ solution, we need to use the balanced chemical equation for the reaction between NH₃ and HCl:
NH₃ + HCl → NH₄Cl

From the equation, we can see that the stoichiometric ratio between NH₃ and HCl is 1:1. This means that the moles of HCl used in the titration are equal to the moles of NH₃ in the ammonia solution.

First, let's calculate the moles of HCl used:
moles HCl = molarity x volume in liters
moles HCl = 1.25 mol/L x (5.19/1000) L
moles HCl = 0.0064875 mol

Since the stoichiometric ratio between NH₃ and HCl is 1:1, the moles of NH₃ in the ammonia solution are also 0.0064875 mol.

Next, let's calculate the molarity of the NH₃ solution:
molarity NH₃ = moles NH₃ / volume in liters
molarity NH₃ = 0.0064875 mol / (12.61/1000) L
molarity NH₃ = 0.5147 mol/L

Therefore, the molarity of the NH₃ solution is 0.5147 mol/L.

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Draw the kinetic and the thermodynamic addition products formed when one equivalent of hbr reacts with the compound shown. Draw a single product for each. Ignore stereochemical or chiral isomers.

Answers

When one equivalent of HBr reacts with the given compound, two different products can be formed: kinetic and thermodynamic.The kinetic product is formed through the faster reaction pathway, which usually involves a lower activation energy.

In this case, the kinetic product is formed by adding the HBr molecule to the more substituted carbon of the double bond. This results in a more stable intermediate, which can then form the kinetic product through proton transfer. The kinetic product is shown below:
[Insert image of kinetic product].The thermodynamic product is formed through the slower reaction pathway, which usually involves a higher activation energy.


In this case, the thermodynamic product is formed by adding the HBr molecule to the less substituted carbon of the double bond.This results in a less stable intermediate, which can then form the thermodynamic product through proton transfer. The thermodynamic product is shown below:[Insert image of thermodynamic product]

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g the magnitude of the crystal field splitting affects the magnetic properties of a complex ion because it affects the pairing of electrons. group of answer choices true false'

Answers

True; the magnitude of the crystal field splitting affects the magnetic properties of a complex ion because it affects the pairing of electrons.

The magnitude of the crystal field splitting refers to the energy difference between the d orbitals in a complex ion, which determines the electronic configuration and the extent of electron pairing. If the splitting is small, the electrons may pair up in the lower energy orbitals, leading to a diamagnetic complex.

If the splitting is large, the electrons may occupy higher energy orbitals, resulting in an unpaired electron and a paramagnetic complex. Therefore, the magnetic properties of a complex ion are directly affected by the magnitude of the crystal field splitting. Hence, the statement "the magnitude of the crystal field splitting affects the magnetic properties of a complex ion because it affects the pairing of electrons" is true.

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the half-lives of different radioisotopes are given in the table. radioisotope half-life (min) argon-44 12 lead-196 37 potassium-44 22 indium-117 43 how long would it take, in minutes, for the amount of argon-44 to decrease from 80.0 mg to 10.0 mg?

Answers

It would take approximately 23.0 minutes for the amount of argon-44 to decrease from 80.0 mg to 10.0 mg.

Time it takes for  argon-44 to decrease from 80.0 mg to 10.0 mg

To solve this problem, we can use the formula for radioactive decay:

N(t) = N0 * (1/2)^(t/T)

where N(t) is the amount of the radioisotope at time t, N0 is the initial amount, T is the half-life, and t is the time elapsed.

We can rearrange this formula to solve for t:

t = T * log(N(t)/N0) / log(1/2)

Putting in the values for argon-44:

N0 = 80.0 mg

N(t) = 10.0 mg

T = 12 min

t = 12 min * log(10.0 mg / 80.0 mg) / log(1/2)

t = 12 min * (-1.32193) / (-0.69315)

t = 23.0 min (rounded to 2 decimal places)

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in which area of the periodic table would the elements with the highest density be found? group of answer choices

Answers

The elements with the highest density are found in the metals group of the periodic table, specifically in the late transition metals and the rare earth elements.

These elements have a high atomic number in periodic table, which means that they have a large number of protons in their nuclei, and therefore a high number of neutrons as well. As a result, they have a high atomic mass and a high density.

Some examples of elements with high density include:

Platinum (Pt)

Gold (Au)

Mercury (Hg)

Tantalum (Ta)

Tungsten (W)

Germanium (Ge)

Cerium (Ce)

It's worth noting that the density of an element can also be affected by its crystal structure, as different crystal structures can have different densities.  

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Correct Question:

in what area of the periodic table would the elements with the highest density be found?

I really need help here, please i beg

Answers

The specific heat of the metal is approximately 994.3 J/(kg°C). This means that it takes 994.3 J of energy to raise the temperature of one kilogram of this metal by one degree Celsius.

We can use the formula for heat energy,

Q = m x c x ΔT,

We know the values for Q, m, and ΔT from the problem:

Plugging in the given values, we get:

7690 J = 0.085 kg x c x (100°C - 11.2°C)

We can simplify this equation by subtracting the initial temperature from the final temperature:

7690 J = 0.085 kg x c x 88.8°C

Solving for c, we get:

c = 7690 J / (0.085 kg x 88.8°C)

c = 994.3 J/(kg°C)

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--The complete Question is, an 85 gm piece of metal is boiling and has an initial temp of 100 degrees after being removed from the water the metal cools down to 11.2 degrees celsius. 7690 j of energy is released what is the specific heat of the metal?--

At a particular temperature, a sample of pure water has a Kw of 5.1×10−11. What is the hydronium concentration of this sample?

Answers

The hydronium ion concentration, [H₃O⁺] = 7.14 x 10⁻⁶ M, which is calculated in the below section.

The value of Kw = 5.1 x 10⁻¹¹

In the autoionization of water, a proton is transferred from one water molecule to another to produce a hydronium ion (H₃O⁺) and a hydroxide ion (OH⁻). The equilibrium expression for this reaction is Kw = [H₃O⁺][OH⁻],

The concentration of hydronium ion and hydroxyl ion when a water molecule dissociates is the same which is 1 mol.

Kw = [H₃O] [OH⁻]

5.1 x 10⁻¹¹ = [H₃O⁺]²

[H₃O⁺] = √(5.1 x 10⁻¹¹)

[H₃O⁺] = 7.14 x 10⁻⁶ M

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A 21.8 g sample of ethanol (C2H5OH) is burned in a bomb calorimeter according to the following reaction. If the temperature rises from 25.0 °C to 62.3 °C, determine the heat capacity of the calorimeter. The molar mass of ethanol is 46.07 g mol-1. C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(g) ΔrU = -1235 kJ mol-1

Answers

Heat capacity of the calorimeter is 15.68 kJ/°C when a 21.8 g sample of ethanol is burned in a bomb calorimeter and the temperature rises from 25.0 °C to 62.3 °C, with a ΔrU of -1235 kJ/mol.

What is the heat capacity of a bomb calorimeter when a given mass of ethanol is burned?

To calculate the calorimeter's heat capacity, use the following formula:

q = CΔT

Where q is the amount of heat absorbed by the calorimeter, C is its heat capacity, and T is the temperature change.

First, let's calculate the amount of heat released by the combustion of ethanol. We can use the given value of ΔrU and the number of moles of ethanol burned to calculate the heat released:

n = m/M

n = 21.8 g / 46.07 g/mol = 0.473 mol

qrxn = ΔrU * n

qrxn = -1235 kJ/mol * 0.473 mol = -585.16 kJ

Since the heat released by the reaction is absorbed by the calorimeter, we have:

qcal = -qrxn

qcal = 585.16 kJ

Finally, we can utilise the values of qcal and T to compute the calorimeter's heat capacity:

C = qcal / ΔT

C = 585.16 kJ / (62.3 °C - 25.0 °C)

C = 585.16 kJ / 37.3 °C

C = 15.68 kJ/°C

Therefore, the heat capacity of the calorimeter is 15.68 kJ/°C.

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Determine the density of nh3 gas at 435 k and 1. 00 atm.

Answers

To determine the density of NH3 gas at 435 K and 1.00 atm, 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.

First, we need to find the number of moles of NH3 gas:

n = PV/RT

n = (1.00 atm) x V / [(0.08206 L atm/mol K) x (435 K)]

Assuming the volume of NH3 gas is 1 L:

n = (1.00 atm) x (1 L) / [(0.08206 L atm/mol K) x (435 K)]
n = 0.0276 mol

Next, we can use the formula for density:

density = mass/volume

To find the mass of NH3 gas, we can use its molar mass of 17.03 g/mol:

mass = n x molar mass
mass = 0.0276 mol x 17.03 g/mol
mass = 0.47 g

Therefore, the density of NH3 gas at 435 K and 1.00 atm is:

density = mass/volume
density = 0.47 g/1 L
density = 0.47 g/L
To determine the density of NH3 gas at 435 K and 1.00 atm, we can use the Ideal Gas Law formula, which is PV = nRT. Here's a step-by-step explanation:

1. Write down the given information:
  - Temperature (T) = 435 K
  - Pressure (P) = 1.00 atm
  - We also need the molar mass of NH3, which is 14.01 (N) + 3 * 1.01 (H) = 17.03 g/mol

2. Rearrange the Ideal Gas Law formula to solve for the number of moles (n) in one liter of NH3 gas:
  - n = PV/RT

3. Substitute the given values:
  - n = (1.00 atm) * (1 L) / (0.0821 L·atm/mol·K) * (435 K)
  - n ≈ 0.0279 mol/L

4. Calculate the mass of NH3 in one liter:
  - Mass = n * molar mass
  - Mass = 0.0279 mol/L * 17.03 g/mol ≈ 0.475 g/L

5. Determine the density of NH3 gas:
  - Density = Mass / Volume
  - Density ≈ 0.475 g/L / 1 L ≈ 0.475 g/L

So, the density of NH3 gas at 435 K and 1.00 atm is approximately 0.475 g/L.

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Describe the building of the simple filtration apparatus....

Answers

To build a simple filtration apparatus, you will need the following materials:

FunnelFilter paperBeaker or flaskMaterial to be filtered

Steps to build a simple filtration apparatus:

1. Choose a funnel that fits securely on top of the beaker or flask.

2. Fold a filter paper in half and then in half again to create a cone shape that fits inside the funnel.

3. Place the folded filter paper in the funnel, making sure that the paper touches the sides of the funnel to create a seal.

4. Wet the filter paper with a small amount of the solvent that will be used to filter the material.

5. Place the funnel on top of the beaker or flask, making sure that the funnel is centered and level.

6. Pour the material to be filtered onto the filter paper in the funnel.

7. Allow the solvent to pass through the filter paper, carrying the desired particles with it, and collecting in the beaker or flask below.

8. Dispose of the filter paper and collected material appropriately.

Note: The size and shape of the funnel, filter paper, and container may vary depending on the amount and nature of the material being filtered.

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A student studies about the planets in the solar system. The student makes the following list of characteristics of a planet to help identify it:


W: Planet is a gaseous planet.
X: Planet has more than two moons.
Y: Planet takes longer than one year to revolve around the sun.
Z: Planet was formed after the sun.


Which of the characteristics listed above cannot be used to help identify the planet?

W

X

Y

Z

Answers

Answer:

Z: Planet was formed after the sun.

Explanation:

Characteristic Z: The planet was formed after the sun cannot be used to help identify the planet since all the planets in the solar system, including Earth, were formed after the sun. Therefore, this characteristic applies to all planets in the solar system, and it cannot be used to distinguish one planet from another.

Good luck! >:))))

Which substance will not react with hydrochloric acid to form a salt?AgZnCuCO3Ca(OH)2

Answers

When an acid reacts with a base, it forms a salt and water, in a process known as neutralization.

However, not all substances can react with hydrochloric acid (HCl) to form a salt. Calcium hydroxide (Ca(OH)2) is a base that will not react with HCl to form a salt, but rather it will form calcium chloride and water. On the other hand, Ag, Zn, Cu, and CO3 are all metals and anions that can react with HCl to form their respective chlorides. Thus, the reaction between an acid and a base can vary depending on the specific substances involved, and the resulting product will be determined by their chemical properties.

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For a certain process at 27°C, ΔG = +210.6 kJ and ΔH = −168.2 kJ. What is the entropy change for this process at this temperature? Express your answer in the form, ΔS = ____ J/K.a. 1.26 × 103 J/Kb. −1.26 × 103 J/Kc. −141.3 J/Kd. +141.3 J/Ke. +628.3 J/K

Answers

We can use the equation ΔG = ΔH - TΔS, where ΔG is the change in Gibbs free energy, ΔH is the change in enthalpy, T is the temperature in Kelvin, and ΔS is the change in entropy.

First, we need to convert the temperature from Celsius to Kelvin:
27°C + 273.15 = 300.15 K

Now we can plug in the values we have:
ΔG = +210.6 kJ
ΔH = -168.2 kJ
T = 300.15 K
ΔS = ?

ΔG = ΔH - TΔS
ΔS = (ΔH - ΔG) / T
ΔS = (-168.2 kJ - (+210.6 kJ)) / 300.15 K
ΔS = -78.4 kJ / 300.15 K
ΔS = -261.15 J/K

Since the question asks for the entropy change in Joules per Kelvin, we can convert the answer from kJ/K to J/K:
ΔS = -261.15 J/K = -0.26115 kJ/K

Therefore, the answer is:
ΔS = -0.26115 kJ/K or approximately -261 J/K
The closest answer choice is (c) -141.3 J/K, but none of the choices are an exact match.

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Draw a Born-Haber cycle for BaBr₂ and calculate ∆Hf using the following values:
BaBr₂ lattice energy = +1950 kJmol-1
Ba atomization energy= 175
Ba 1st ionization energy = 503
Ba 2nd ionization energy = 965
Br₂ bond enthalpy = 193
Bromine 1st electron affinity= -325

Answers

∆Hf(BaBr₂) = -262 kJ/mol. Ba atomization energy (175) and 1st ionization energy (503) overcome the lattice energy (+1950), making the formation of BaBr₂ exothermic.

Explanation: A Born-Haber cycle shows the steps involved in the formation of an ionic compound. For BaBr₂, the cycle includes formation of Ba atoms from solid Ba, dissociation of Br₂, ionization of Ba to form Ba⁺, electron affinity of Br to form Br⁻, and formation of the solid BaBr₂ lattice. The lattice energy (+1950 kJ/mol) is overcome by the sum of the atomization energy (175 kJ/mol) and the first ionization energy of Ba (503 kJ/mol), as well as the exothermicity of adding electrons to Br to form Br⁻ (-325 kJ/mol). The overall enthalpy change (∆Hf) for the formation of BaBr₂ is -262 kJ/mol, indicating that the process is exothermic and spontaneous.

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What is the molarity of a 250. Ml h2so4 solution that was made from a 20. 0 ml of a 10. 0 m stock solution?.

Answers

The main answer to your question is that the molarity of the 250 mL H₂SO₄ solution is 0.8 M.



To find the molarity, we can use the dilution formula, M₁V₁ = M₂V₂, where M₁ and V₁ are the molarity and volume of the stock solution, and M₂ and V₂ are the molarity and volume of the diluted solution. Given M₁ = 10.0 M, V₁ = 20.0 mL, and V₂ = 250 mL, we can solve for M₂:
M₂ = (M₁V₁) / V₂
M₂ = (10.0 M × 20.0 mL) / 250 mL
M₂ = 200.0 mM / 250 mL
M₂ = 0.8 M

We can assume that the density of the stock solution and the diluted solution are the same (which is reasonable for aqueous solutions), and use the formula: Molarity = moles of solute / volume of solution (in liters).

We can calculate the moles of H2SO4 in the 20.0 mL of stock solution: moles = M x V = 10.0 M x 0.0200 L = 0.200 mol When we dilute this to 250. mL,


Summary: The molarity of the 250 mL H₂SO₄ solution that was made from a 20.0 mL of a 10.0 M stock solution is 0.8 M.

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3. Theoretically, how many moles of a ketone can reduce one mole of sodium borohydride?
. 1 c. 3 e. 5 b. 2 d. 4

Answers

The answer is c. 3. Sodium borohydride (NaBH4) is a strong reducing agent that can reduce aldehydes and ketones to their corresponding alcohols.

What is borohydride?

Borohydride is an inorganic compound that is composed of boron and hydrogen. It is a type of hydride, which is a compound that contains hydrogen and a metal or metalloid. Borohydride has a number of commercial and industrial uses, such as in the production of fuel cells, biodegradable detergents, and pharmaceuticals. It is also used as a reducing agent in the synthesis of organic compounds. Borohydride can be produced through a variety of methods, such as electrochemical reduction and thermal decomposition. Borohydride is a strong reducing agent, and its use as a reagent in organic synthesis has been increasing over the past few years due to its ease of use, low cost, and high reactivity.

For each mole of NaBH4, three moles of ketone can be reduced.

Therefore the correct option is C.

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comment on the validity of this statment to some extent every electron shields the nuclear charge from every other electron

Answers

To some extent, the statement that every electron shields the nuclear charge from every other electron is valid.


Electrons have a negative charge and are attracted to the positively charged nucleus of an atom. However, electrons also repel each other due to their negative charges. As a result, each electron in an atom experiences both attraction to the nucleus and repulsion from other electrons.

The shielding effect occurs when electrons in the inner energy levels of an atom shield the positive charge of the nucleus from the outer electrons. This shielding reduces the net attractive force that the outer electrons experience from the nucleus.

However, it is important to note that not all electrons in an atom shield the nuclear charge equally. Electrons in higher energy levels (further from the nucleus) are less effective at shielding the nuclear charge than electrons in lower energy levels (closer to the nucleus).

Additionally, electrons in the same energy level do not shield each other completely because they still experience some repulsion from each other.

Therefore, while the statement is valid to some extent, it is not a complete description of the complex interactions between electrons and the nucleus in an atom.

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How many milliliters of 0.0850 M NaOH are required to titrate each of the following solutions to the equivalence point:
40.0mL of 0.0900 M HNO3

Answers

To find the milliliters of 0.0850 M NaOH required to titrate 40.0 mL of 0.0900 M HNO3 to the equivalence point, you can use the formula: So, 42.4 milliliters of 0.0850 M NaOH are required to titrate 40.0 mL of 0.0900 M HNO3 to the equivalence point.

moles of acid = moles of base

First, find the moles of HNO3:
moles of HNO3 = (volume in L) * (molarity)
moles of HNO3 = (0.0400 L) * (0.0900 M) = 0.00360 moles

Since the ratio of acid to base is 1:1, the moles of NaOH required will also be 0.00360 moles. Now, find the volume of NaOH needed:

volume of NaOH = (moles of NaOH) / (molarity of NaOH)
volume of NaOH = (0.00360 moles) / (0.0850 M) = 0.0424 L

Finally, convert the volume to milliliters:

0.0424 L * 1000 = 42.4 mL

The moment in a chemical reaction when the quantity of one reactant has entirely interacted with the amount of another reactant is known as the equivalence point in chemistry. The reactants have now been devoured in the ideal ratios, causing one reactant to be completely consumed by the other.The equivalency point is achieved, for instance, in acid-base titration when the ratio of acid to base reaches unity. The solution is now neutral, not acidic or basic. Typically, an indicator whose colour changes when the reaction reaches the stoichiometric point is used to determine the equivalence point.

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