if a 250ml solution of 0.10M is made, calculate the mass of HCI contained in this solution

Answer:

24 g/mol.

Explanation:

To calculate the molar mass of the salt (z), we need to use the formula:

concentration = number of moles / volume of solution

We know that the concentration is 0.80 mol/dm^3, and the volume of the solution is 250 cm^3, which is equivalent to 0.25 dm^3. Therefore, we can rearrange the formula to solve for the number of moles of salt (z):

number of moles = concentration x volume of solution

number of moles = 0.80 mol/dm^3 x 0.25 dm^3

number of moles = 0.20 mol

Next, we can calculate the mass of salt (z) in the solution using the formula:

mass = number of moles x molar mass

We know that the mass of salt (z) is 4.8 g, and we just calculated that the number of moles is 0.20 mol. Therefore, we can rearrange the formula to solve for the molar mass:

molar mass = mass / number of moles

molar mass = 4.8 g / 0.20 mol

molar mass = 24 g/mol

Therefore, the molar mass of the salt (z) is 24 g/mol.

Answer:24 g/mo

Explanation:

Answer:  C

Explanation:

Nuclear reactions involve a change in an atom's nucleus, usually producing a different element. Chemical reactions, on the other hand, involve only a rearrangement of electrons and do not involve changes in the nuclei.

Magnesium oxide turns a white powder as a result. Magnesium creates by transferring two electrons oxygen atoms. This reaction is exothermic. Magnesium + oxygen → magnesium oxide. 2Mg + O2 → 2MgO.

An illustration of a combination reaction is the burning of magnesium ribbon to produce magnesium oxide. One chemical splits into two compounds, one with a high oxidation state and the other with a low oxidation state, in a disproportionation reaction.

The evolution of light and heat causes an exothermic chemical reaction that results in fire. Three essential components—oxygen, heat, and fuel—must all be present for such a fire to start. The kind of reaction that results in flames is referred to as a combustion reaction in chemistry.

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The specific heat capacity of the metal, given that the metal was heated to 97 °C and transferred to water at 20.5 °C, is 0.203 Cal/gºC

We'll begin by obtaining the heat absorbed by the water. Details below:

Q = MCΔT

Q = 86 × 1 × 3.6

Q = 309.6 Cal

Finally, we shall determine the specific heat capacity of the metal. Details below:

Q = MCΔT

-309.6 = 20.9 × C × -72.9

-309.6 = -1523.61 × C

Divide both sides by -1523.61

C = -309.6 / -1523.61

C =  0.203 Cal/gºC

Thus, we can conclude that the specific heat capacity of the metal is 0.203 Cal/gºC

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Several sizes of Cu particles' optical absorption spectra are also described; these spectra contain the plasmon band in the 560–580 nm range and a UV band at 222 nm, becoming flatter with increasing particle size.

From the calibration figure, calculate the copper (II) sulphate concentrations in your two unknown samples. Useful wavelength is 700 nm.

By testing an unknown CuSO4 solution's absorbance with a Colorimeter, you can ascertain its concentration. The corresponding concentration can be obtained on the horizontal axis of the graph by placing the absorbance of the unknown on the vertical axis.

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Answer:

use the prefix trans-

Explanation:

trans = opposite orientation across double bond its this one for apex

cis = same orientation across double bond.

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The ability to interact with web browser software is provided by the user interface. It offers a visual user interface that enables entry of commands and response from the programme.

Via a Web browser, a user can interact with data or software running on a remote server using a Web user interface or Web app. The user interacts with the content on a web browser, which functions as a client, after downloading it from the web server.

A web browser is a piece of software that enables a user to view and interact with content that may be found on a website, including text, photos, videos, music, and other types of media. Hyperlinks to other web pages at the same or different websites can be found in the text and images on a web page.

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Answer:

99585.48 Pa or approximately 0.981 atm.

Explanation:

To convert the barometric pressure from inches to SI units (Pascals), we can use the following conversions:

1 inch = 25.4 mm

1 mm = 0.1 cm

1 cm = 10 mm

1 mmHg = 133.322 Pa

1 atm = 760 mmHg

Therefore, we can calculate the pressure in SI units as follows:

Convert inches to mm: 29.4 inches x 25.4 mm/inch = 746.76 mm

Convert mmHg to Pa: 746.76 mm x 133.322 Pa/mmHg = 99585.48 Pa

Convert atm to Pa: 1 atm x 760 mmHg/atm x 133.322 Pa/mmHg = 101325 Pa

Calculate the pressure in SI units: 99585.48 Pa/101325 Pa/atm = 0.981 atm

Therefore, the barometric pressure in SI units is 99585.48 Pa or approximately 0.981 atm.

With a radius of 15,299.4 miles (24,622 kilometers), Neptune is about 4 times wider than Earth. If Earth have been the size of a nickel, Neptune would be about as big as a baseball.

From an common distance of 2.8 billion miles (4.5 billion kilometers), Neptune is 30 astronomical units away from the Sun.

Image end result for How is the outer layer of earth one-of-a-kind from the outer layer of Neptune

The outermost layer of Neptune is the atmosphere, forming about 5-10% of the planet's mass, and extending up to 20% of the way down to its core.

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The extent of Neptune is 6.3 x 1013 km3. You could healthy fifty seven Earths inner Neptune and nevertheless have room to spare. A day on Earth is 24 hours, but a day on Neptune is 16 hours and 6 minutes. A yr on Earth is, um, 1 12 months obviously, whilst a year on Neptune is 164.79 years.

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The mass percentage of KClO₃ in the initial mixture, given that the initial mixture has a mass of 7.25 g, is 62.2%

First, we shall determine the molar mass of KClO₃ and KCl. Details below:

For KClO₃

Molar mass of KClO₃ = 39 + 35.5 + (3 × 16)

Molar mass of KClO₃ = 39 + 35.5 + 48

Molar mass of KClO₃ = 122.5 g/mol

For KCl

Molar mass of KCl = 39 + 35.5

Molar mass of KCl = 74.5 g/mol

Next, we shall determine the mass of KClO₃ in the initial mixture. Details below:

Mass of KClO₃ = [molar mass of KClO₃ / molar mass of (KClO₃ + KCl)] × mass of mixture

Mass of KClO₃ = [122.5 / (122.5 + 74.5)] × 7.25

Mass of KClO₃ = 4.51 g

Finally, we shall determine the mass percentage of KClO₃. Details below:

Mass percentage of KClO₃ = (mass of of KClO₃ / mass of mixture) × 100

Mass percentage of KClO₃ = (4.51 / 7.25) × 100

Mass percentage of KClO₃ = 62.2%

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Answer:

Explanation:

The moles of sodium hydroxide (NaOH) from the experiment is 0.0518 mol.

Qsurroundings = -Qrxn, so Qsurroundings = -(m x C x ΔT) = - (50.0 g x 4.184 J/g°C x (28.5°C - 20.0°C)) = - 3464.96 J; Qrxn = -Qsurroundings = 3464.96 J

ΔHsol = Qrxn / moles of NaOH = 3464.96 J / 0.0518 mol = -66,871.12 J/mol or -66.87 kJ/mol

The thermochemical equation for the dissociation of sodium hydroxide:

2NaOH(s) → 2Na+(aq) + 2OH-(aq) + heat

or

NaOH(s) → Na+(aq) + OH-(aq) + heat

The enthalpy diagram shows an energy input (endothermic) for the dissociation of NaOH.

Some assumptions made to calculate #2 include assuming the specific heat capacity of water is the same as the specific heat capacity of the NaOH solution and that the density of the solution is the same as water.

The actual value of ΔHsol for NaOH is -44.51 kJ/mol, so the percent error is:

|(-44.51 kJ/mol - (-66.87 kJ/mol)) / -44.51 kJ/mol| x 100% = 50.21%

The breaking and forming of chemical bonds are responsible for the enthalpy change. In the case of the dissociation of NaOH, the bond between the Na and OH groups is broken, which requires an input of energy, making the process endothermic.

From the experiment, the moles of sodium hydroxide (NaOH) is found to be 0.0518 mol.

What is Moles?

In chemistry, a mole is a unit used to measure the amount of a substance, typically atoms or molecules. One mole of a substance is defined as the amount of that substance which contains the same number of particles (atoms, molecules, or ions) as there are in exactly 12 grams of carbon-12

In the experiment described, 2.00 grams of sodium hydroxide (NaOH) was added to 50.0 mL of water in a coffee cup. The temperature of the water was measured before adding the NaOH and recorded as the initial temperature (T initial). The NaOH was then added to the water and the mixture was stirred gently with a thermometer, while the temperature was recorded every 30 seconds for 3 minutes or until it peaked. The highest temperature reached during this time was recorded as the final temperature (T final).

Using the formula Q = mCΔT, where Q is the amount of heat transferred, m is the mass of the water, C is the specific heat of water, and ΔT is the change in temperature, the amount of heat released or absorbed by the NaOH and water mixture can be calculated.

Since the reaction between NaOH and water is an exothermic reaction, the heat released by the reaction is equal to the heat absorbed by the water.

From the moles of NaOH added and its molar mass, the mass of NaOH can be calculated.

Then, the heat of solution can be calculated using the formula:

Heat of solution = Q / moles of NaOH

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If the density of mercury is 13.6 g/mL, the volume of a 155-gram sample of mercury is 11.397 mL.

It follows that:

Mercury has a density of 13.6 g/mL.

155 grammes make to the mercury's weight.

The fact is,

A three-dimensional space enclosed by an object or thing is referred to as its volume.

Mass times volume equals density.

13.6 = Volume 155

quantity = 155/13.6

11.397 mL is the capacity.

As a result, assuming mercury has a density of 13.6 g/mL, a 155-gram sample of mercury has a volume of 11.397 mL.

The complete question is:

The density of mercury is 13.6 g/mL. What is the volume of a 155-gram sample of mercury?

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Because sodium is such a highly reactive metal, it interacts with water quickly to produce sodium hydroxide and hydrogen gas. The correct chemical formula is: H2O + Na(s) = NaOH (aq) + H2 (g)

Hydrochloric acid and lithium oxide react, neutralising the acid. Lithium chloride and water are the results of the process.

A larger surface area is exposed to the water as the molten metal flows across it. Moreover, among all alkali metals, lithium has the largest hydrated radius. This reduces the ionic mobility, which causes the molten metal to move more slowly.

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H atoms communicate attractively even when they are infinitely apart. Each electron is drawn to the other centre as two H atoms get close to one another. The electrons of two H atoms draw one another when they come close to one another.

When two atoms of H come close to one another, one electron of each atom repels one nucleus while the other nucleus attracts the other electron. H atoms communicate attractively even when they are infinitely apart.

They cannot make a covalent bond if they are too far apart because their individual 1s orbitals cannot overlap; instead, they remain as two separate objects.

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For each color of paper, the part of white light that is reflected depends on the color of the paper. When white light strikes an object, some of the light is absorbed by the object, some of it is transmitted through the object, and some of it is reflected.

The color of the object that we see is the color of the light that is reflected by the object.

For example, when white light strikes blue paper, the blue color of the paper absorbs all the other colors of the spectrum except blue, which is reflected back to our eyes. This is why we see the paper as blue. Similarly, when white light strikes red paper, the red color of the paper absorbs all the other colors except red, which is reflected back to our eyes. This is why we see the paper as red.

In summary, the color of an object is determined by the color of the light that is reflected by the object, and the color of the light that is reflected depends on the color of the object and the colors of the spectrum that are absorbed or transmitted by the object.

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Answer:

To determine the number of moles of precipitate formed in the reaction, we first need to find the limiting reagent.

From the balanced chemical equation, we know that 3 moles of Ca(NO₃)₂ react with 2 moles of K₃PO₄ to produce 1 mole of Ca₃(PO₄)₂.

Let's first calculate the number of moles of Ca(NO₃)₂ present in 52.9 mL of 0.400 M solution:

moles of Ca(NO₃)₂ = concentration × volume

= 0.400 mol/L × 0.0529 L

= 0.02116 mol

Since the stoichiometric ratio of Ca(NO₃)₂ to K₃PO₄ is 3:2, we would need 3/2 times the number of moles of K₃PO₄ to react completely with the given amount of Ca(NO₃)₂. However, the problem states that we have an excess of K₃PO₄, which means that all of the Ca(NO₃)₂ will react with the available K₃PO₄.

Therefore, the number of moles of Ca₃(PO₄)₂ that will be formed is equal to the number of moles of Ca(NO₃)₂ used in the reaction, which is 0.02116 mol.

Hence, 0.02116 moles of Ca₃(PO₄)₂ precipitate will be formed when 52.9 mL of 0.400 M Ca(NO₃)₂ is mixed with excess K₃PO₄ in the given chemical reaction.

Answer:

The balanced chemical equation for the reaction between H₂PO₂ and NaOH is:

H₂PO₂ + 3 NaOH → Na₃PO₄ + 3 H₂O

Explanation:

From the equation, we can see that 1 mole of H₂PO₂ reacts with 3 moles of NaOH to produce 1 mole of Na₃PO₄ and 3 moles of H₂O.

The volume of NaOH used in the titration is 150.0 mL, which is equivalent to 0.150 L. The molarity of the H₂PO₂ solution used is 0.16 M, which means that 0.16 moles of H₂PO₂ were used in the titration.

According to the balanced equation, 1 mole of H₂PO₂ reacts with 3 moles of NaOH. Therefore, the number of moles of NaOH used in the titration is three times the number of moles of H₂PO₂ used:

Number of moles of NaOH = 3 × 0.16 = 0.48 moles

The volume of NaOH used is 0.150 L, so the molarity of NaOH can be calculated as follows:

Molarity of NaOH = Number of moles of NaOH / Volume of NaOH used

Molarity of NaOH = 0.48 moles / 0.150 L

Molarity of NaOH = 3.2 M

Therefore, the molarity of NaOH is 3.2 M, which is represented by the sky blue color.

The car took 3 seconds to accelerate from 15 m/s to 30 m/s with an acceleration of 5 m/s^2.

We can use the following kinematic equation to solve this problem:

v = u + at

Where

Given:

u = 15 m/s (initial velocity)

v = 30 m/s (final velocity)

a = 5 m/s^2 (acceleration)

Substituting the given values into the equation, we get:

30 m/s = 15 m/s + 5 m/s^2 × t

Simplifying and solving for t, we get:

5 m/s^2 × t = 15 m/s

t = 15 m/s ÷ 5 m/s^2 = 3 seconds

Therefore, the car took 3 seconds to accelerate from 15 m/s to 30 m/s with an acceleration of 5 m/s^2.

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The two emissions required for oxygen-17 to become nitrogen-13 are a positron emission and a neutrino emission.

A neutrino is a subatomic particle that is similar to an electron in many ways but differs in that it lacks an electrical charge and has a very small mass that might potentially be zero.

Positron is a particle with the same mass as an electron but has a positive charge.

The transformation of oxygen-17 to nitrogen-13 requires the emission of two particles: a positron (also called a positive beta particle) and a neutrino.

The first emission is the positron, which is a particle with the same mass as an electron but with a positive charge. During the emission, a proton in the oxygen-17 nucleus converts into a neutron, and the positron and a neutrino are produced. The positron quickly annihilates with an electron, releasing two gamma rays.

The second emission is the neutrino, which is a subatomic particle with a very small mass and no electrical charge. The neutrino is emitted during the decay of the nitrogen-13 nucleus as it transitions to its ground state.

Therefore, the two emissions required for oxygen-17 to become nitrogen-13 are a positron emission and a neutrino emission.

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Potential energy is changed into kinetic energy throughout this process, which is the heat produced during reactions. On the contrary, this happens in an endothermic process.

Bonds break, new bonds form, and protons and electrons move from a structure with greater potential energy to one with lower potential energy during an exothermic reaction. Potential energy is changed into kinetic energy throughout this process.

The energy that is held inside the bonds and phases of the reactants and products is measured by potential energy. The internal energy includes this potential energy. The internal energy of chemical reactions, also known as enthalpy, stands in for the system's total energy.

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The elements are the simplest chemical forms and they cannot be broken down through chemical reactions. There are many elements in the given formulas.

What are elements?

The elements are defined as those substances whose atoms all have the same number of protons. The elements are considered as the building blocks of matter. Each element has an atomic number and a symbol.

Each atom is regarded as an element. The elements create bonds to form molecules. The isotopes are the elements with the same atomic number but different mass numbers.

NaC₂HO₄ - 'N' , 'C', 'H','O'

H₂F₅BLi - 'H','F','B','Li'

He₂PSO₄ - 'He', 'P','S','O'

He₂O₄PH - 'He', 'O','P','H'

The relative solubility is, % relative error = |(s - s') / s| × 100 by using concentrations instead of activities. The % Relative error is [tex]1.42*10^-9[/tex]

The % relative error in solubility, can determine by:
Step 1: Calculate the solubility using concentrations
- Determine the solubility product constant (Ksp) of each compound.
- Write the balanced chemical equation and expression for Ksp.
- Solve for the solubility (s) in terms of concentrations.

Step 2: Calculate the solubility using activities
- Replace concentrations in the Ksp expression with activities (use activity coefficients if needed).
- Solve for the solubility (s') using activities.

Step 3: Calculate the % relative error
- Use the formula: % relative error = |(s - s') / s| × 100

Assuming that the ionic strength is low, we can use the Debye-Hückel equation to calculate the activity coefficients at 25°C and 0.0500 M ionic strength:

log γ± = -0.5091 (z+ z- √(I)) / (1+√(I))

where z+ and z- are the charges of the cation and anion, respectively, and I is the ionic strength.

For KNO3, z+ = 1 and z- = 1, so I = [tex]1/2 (1^2 *0.0500 + 1^2*0.0500)[/tex] = 0.0250. Therefore, the activity coefficients are:

γ±(K+) = γ±(NO_3-) = 0.790

Using the activity coefficients, we can calculate the ion concentrations and then the solubility:

(a) CuCl:

[Cu+] = aCu+ × [CuCl] = [tex](0.3 nm) *√(Ksp/[CuCl]) = 3.17*10^-6 M[/tex]

[Cl-] = [Cu+]

Solubility = [CuCl] = [Cu+] = [Cl-] = [tex]3.17*10^-6 M[/tex]

Using concentrations instead of activities, we have:

[Cu+] = [Cl-] = √(Ksp/[CuCl]) = [tex]3.16*10^{-3}[/tex] M

Solubility = [CuCl] = [Cu+] = [Cl-] = [tex]3.16*10^{-3}[/tex] M

% Relative error = |[tex](3.17*10^{-6} - 3.16*10^{-3})/3.17×10^{-6}[/tex]| × 100% ≈ 99.7%

(b) Fe(OH)2:

[tex][Fe^2+] = aFe^2+ * [Fe(OH)_2] = (0.84 nm) * √(Ksp/[Fe(OH)2]) = M[OH-] = 2[Fe^2+][/tex]

Solubility =[tex][Fe(OH)_2] = [Fe^{2+}][/tex]= [tex]1.42*10^{-9} M[/tex]

Using concentrations instead of activities, we have:

[tex][Fe^2+] = [OH^-] = √(Ksp/[Fe(OH)_2])[/tex] = [tex]3.16*10^-8[/tex] M

Solubility = [tex][Fe(OH)_2] = [Fe^{2+}][/tex]= [tex]3.16*10^-8[/tex] M

% Relative error = [tex]1.42*10^-9[/tex]

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The complete question is:

Calculate the % relative error in solubility by using concentrations instead of activities for the following compounds in 0.0500 M KNO3 using the thermodynamic solubility products listed in Appendix 2.

(a) CuCl (aCu+ = 0.3 nm)

(b) Fe(OH)2

(c) Fe(OH)3

(d) La(IO3)3

(e) Ag3AsO4 (αAsO43֊= 0.4 nm)

The compound would have six hydrogen bonds.

It has 3 H bond acceptors. It can form six H bonds with an identical molecule. It can form three hydrogen bonds with water.

Hydrogen bonds are a type of intermolecular force that occurs between a hydrogen atom bonded to an electronegative atom (such as nitrogen, oxygen, or fluorine) and a nearby electronegative atom on another molecule.

The hydrogen bond is a weak electrostatic attraction between the partially positive hydrogen and the partially negative atom, which is typically a lone pair of electrons on the other molecule.

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Yes, the three astronauts can safely make the trip back to Earth with only the CO2 absorbers contained in the lunar module.

The Lunar Module (LM) was a spacecraft built by the United States and used in the Apollo program to land humans on the Moon. The LM was designed and built to be used only in the vacuum of space, and it had no capability to operate in the Earth's atmosphere or on the surface of the Moon.

To determine this, we can calculate the amount of glucose needed to sustain the astronauts for the 3-day trip.

We know that each astronaut needs 2500 kilocalories per day, and there are 4 kilocalories per gram of glucose.

Therefore, each astronaut needs 625 grams of glucose per day, or 1875 grams of glucose total for the 3-day trip.

We can then convert this to moles of glucose needed, using the molar mass of glucose (180.156 g/mol).

Therefore, 1875 grams of glucose is equal to 10.4 moles of glucose.

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Answer: 69.9 g  

Explanation:

To calculate the formula mass of an unknown acid using the volume of NaOH required to reach the endpoint of the titration, you will need to use the following formula: Formula mass = (molarity of NaOH) x (volume of NaOH) x (molar ratio of NaOH to acid) / (moles of acid).

Here are the steps to follow: Write the balanced equation for the reaction between the acid and NaOH.

Record the volume of NaOH which is used in the titration.

Determine the molarity of the NaOH solution. This can be done by dividing the number of moles of NaOH used in the titration by the volume of NaOH used.

Determine the molar ratio of NaOH to acid from the balanced equation.

Calculate the number of moles of acid that reacted with the NaOH by multiplying the volume of NaOH used in the titration by the molarity of NaOH.

Calculate the formula mass of the acid by plugging in the values for the molarity of NaOH, volume of NaOH, molar ratio of NaOH to acid, and moles of acid into the formula given above.

For example, suppose that you titrated an unknown acid with 0.100 M NaOH, and it took 25.0 mL of NaOH to reach the endpoint. The chemical equation for the reaction is:

Acid + NaOH → NaA + H₂O

The molar ratio of NaOH to acid is 1:1. Let's assume that you used 0.025 moles of NaOH in the titration. Then:

Molarity of NaOH = 0.100 M

Volume of NaOH =25.0 mL =0.0250 L

Moles of acid = (0.100 M) x (0.0250 L) x (1 mol acid / 1 mol NaOH) = 0.00250 mol acid

Now, let's assume that the formula mass of the acid is X. Then:

Formula mass = (0.100 M) x (0.0250 L) x (1 mol acid / 1 mol NaOH) / (0.00250 mol acid) =  X g/mol

Therefore, the formula mass of the unknown acid is X g/mol.

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--The given question is incomplete, the complete question is

"How to calculate the formula mass of an unknown acid using the manual titration volume of NaOH with which it was titrated, used to reach the endpoint?"--

Answer:

A 3.0 M solution of NaOH has 3.0 moles of NaOH per liter of solution. There are 0.25 L of solution (250mL⋅1L1000mL), so there are 0.25L⋅3.0mol/L=0.75mol of NaOH. The molar mass of NaOH is 40.0 g/mol, so there are 0.75mol⋅40.0g/mol=30g of NaOH, 30.

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