the point in a titration at which the indicator changes is called the

Considering the combined law equation, the new pressure of the sample is 723.81 mmHg.

Boyle's law states that the pressure of a gas in a closed container is inversely proportional to the volume of the container, when the temperature is constant.

This law is expressed mathematically as:

P×V=k

where

Charles' law states that the volume of a gas in a closed container is directly proportional to the temperature of the container, when the temperature is constant.

This law is expressed mathematically as:

V÷T= k

where

Charles' law states that the pressure of a gas in a closed container is directly proportional to the temperature of the container, when the temperature is constant.

This law is expressed mathematically as:

P÷T= k

where

Combined law equation is the combination of three gas laws called Boyle's, Charlie's and Gay-Lusac's law:

(V×P) ÷T= k

Considering an initial state 1 and a final state 2, it is fulfilled:

(V₁×P₁) ÷T₁= (V₂×P₂) ÷T₂

In this case, you know:

Replacing in the definition of combined law equation:

(5 L× 760 mmHg) ÷273 K= (6 L× P₂) ÷312 K

Solving:

[(5 L× 760 mmHg) ÷273 K]× (312 K÷ 6 L)=  P₂

723.81 mmHg= P₂

Finally, the new pressure is 723.81 mmHg.

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Answer:

2FeO + C ---> 2Fe +CO2

Explanation:

Answer:

17 dozens

Explanation:

204 divided by 12 equals 17 and to check your work multiply 17 by 12 and you get 204.

Answer:

yes

Explanation:

because hydrogen  can only form these bonds with highly electronegative atoms.

Answer:

No because hydrogen can only form these bonds with highly electronegative atoms.

The molar concentration of a 12% sodium chloride solution is approximately 2.05 M.

To determine the molar concentration of a 12% sodium chloride solution, we need to convert the given percentage concentration into molarity.

First, we need to understand that the percentage concentration refers to the mass of the solute (sodium chloride) relative to the total mass of the solution.

In this case, a 12% sodium chloride solution means that there are 12 grams of sodium chloride in 100 grams of the solution.

To convert this into molar concentration, we need to consider the molar mass of sodium chloride, which is 58.5 g/mol.

We can start by calculating the number of moles of sodium chloride in 12 grams:

Moles of sodium chloride = mass of sodium chloride / molar mass of sodium chloride

Moles of sodium chloride = 12 g / 58.5 g/mol = 0.205 moles

Next, we calculate the volume of the solution in liters using the density of the solution. Since the density is not provided, we assume a density of 1 g/mL for simplicity:

Volume of solution = mass of solution / density

Volume of solution = 100 g / 1 g/mL = 100 mL = 0.1 L

Finally, we calculate the molar concentration (Molarity) by dividing the number of moles by the volume in liters:

Molar concentration = moles of solute / volume of solution

Molar concentration = 0.205 moles / 0.1 L = 2.05 M

Therefore, the molar concentration of a 12% sodium chloride solution is approximately 2.05 M.


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We can use the formula M₁V₁ = M₂V₂ to determine the moles remaining after the volume is decreased due to evaporation..  

Since M₁V₁ = M₂V₂    (starting mole × starting volume = end mole × end volume)

⇒  (0.125 mol)(265 mL) =  M₂ (186 mL)

         33.125 mol/mL =  186 mL × M₂

⇒                      M₂ = 33.125 mol/mL ÷ 186 mL

                              = 0.178 mol

Answer:

0.02 M

Explanation:

Molarity = moles/volume (L)

125 mL =0.125L

M = 2.5/0.125 = 0.02 M

The rate of diffusion is defined as gas flowing through a space at a given rate at a unit time.

Graham's law of diffusion states that a gas's rate of diffusion is inversely proportionate to its density. Diffusion is the term used to describe the transfer of gas molecules along a concentration gradient. This phenomenon is known as effusion when it allows them to escape through a small hole while under pressure.

Gas diffusion rate is determined by time divided by the volume of the gas diffusion.

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Answer:

nothing

Explanation:

The correct increasing order of solubility for the given gases is: NH₃ < N₂ < CO₂

A chemical's solubility is the maximum amount of that substance that will dissolve in a specific amount of solvent at a specific temperature. Solubility has several practical applications, including water filtering, beverage manufacturing, and vitamin storage.

The ability of a medicine to dissolve is essential to its effectiveness. A drug substance could be absorbed without it, which results in limited bioavailability. Drugs with poor solubility can potentially cause problems with metabolism or permeability, interaction with other medications, or the requirement for prolonged drug release.

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The molarity of the solution will be 3.45 M

Dilution is the process of adding a solvent to get a more dilute solution.

The moles(n) before and after dilution are the same.

Can be formulated :

M₁V₁=M₂V₂  

M₁ = Molarity of the solution before dilution  

V₁ = volume of the solution before dilution  

M₂ = Molarity of the solution after dilution  

V₂ = Molarity volume of the solution after dilution

The initial solution

345 mL of a 2.5 M NaCl solution.

M₁=2.5 M

V₁=345 ml

The solution after boiled :

V₂=250 ml

\(\tt M_2=\dfrac{M_1.V_1}{V_2}\\\\M_2=\dfrac{2.5\times 345}{250}\\\\M_2=3.45~M\)

Answer:

Protons = 89

Neutrons = 142

Electrons = 89

Option C

Explanation:

Hello,

Were required to find the proton, neutron and electron of an element Y

The atomic mass of element Y is composed of the protons and neutrons while the atomic number is the exact amount of protons present in the element.

Element = ₈₉²³¹Y

Atomic number = 89

Number of protons = 89

Number of electrons = 89

Atomic mass = 231

Atomic mass = protons + neutrons

231 = 89 + neutrons

Neutrons = 231 - 89

Neutrons = 142

Therefore the number of protons is 89, number of neutrons is 142 and number of electrons is 89.

Note: for an electrically neutral atom, the number of protons must be equal to the number of electrons

When the experiment begins after 10 minutes, approximately 20 grams of barium-122 will be left from the initial 80-gram sample.

The half-life of a radioactive substance is the time it takes for half of the sample to decay. In this case, barium-122 has a half-life of 5 minutes. This means that after every 5 minutes, the amount of barium-122 is halved.

Given that it takes 10 minutes to set up the experiment, we need to determine how much barium-122 will remain at that point. Since the half-life of barium-122 is 5 minutes, we can calculate the number of half-lives that have passed in 10 minutes, which is 2 half-lives (10 minutes / 5 minutes per half-life).

Each half-life halves the amount of barium-122, so after 2 half-lives, we are left with 1/4 (1/2 * 1/2) of the original amount. Therefore, 1/4 of the 80-gram sample remains, which is 20 grams.

Hence, when the experiment begins after 10 minutes, approximately 20 grams of barium-122 will be left from the initial 80-gram sample.

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classified the oxides as acidic, basic, amphoteric or neutral: SrO - Basic oxide, SnO2 - Amphoteric oxide, Al2O3 -Amphoteric oxide, CO - Neutral oxide, NO - Neutral, oxide, As2O5 - Acidic oxide, K2O - Basic oxide, Cl2O7 - Acidic oxide

Basic oxide: A basic oxide is an oxide that reacts with water to form a base, which can neutralize an acid. Basic oxides typically contain a metal and oxygen, and they tend to have a high pH (greater than 7). Examples of basic oxides include Na2O, CaO, and MgO.

Amphoteric oxide: An amphoteric oxide is an oxide that can act as either an acid or a base, depending on the conditions. Amphoteric oxides usually contain a metalloid (an element with properties of both metals and nonmetals) or a metal that can exhibit multiple oxidation states. Examples of amphoteric oxides include Al2O3 and ZnO.

Neutral oxide: A neutral oxide is an oxide that neither reacts with water to form an acid nor reacts with water to form a base. Neutral oxides are typically covalent compounds (meaning they contain nonmetals bonded to oxygen) and have a pH of 7. Examples of neutral oxides include N2O and CO.

Acidic oxide: An acidic oxide is an oxide that reacts with water to form an acid. Acidic oxides typically contain a nonmetal and oxygen, and they tend to have a low pH (less than 7). Examples of acidic oxides include SO3, CO2, and NO2.

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Answer:

Answer :

The compounds are separated by using a suitable filtration technique.

Explanation:

Explanation:

The compounds are separated by using a suitable filtration technique. NaCl remains in the filtrate, but charcoal remains on the filter paper. Crystals of NaCl can be obtained by the method of evaporation.

The Fischer projections of the four aldotetroses, D- glucose, and L‑glucose are shown in the attached figures below.

In chemistry, the Fischer projection can be described as a two-dimensional representation of a 3-D organic molecule by projection. Fischer projections were proposed for carbohydrates and used by chemists, particularly in organic chemistry and biochemistry.

The main aim of Fischer projections is to show the chirality of a molecule and to differentiate between a pair of enantiomers. Some notable uses involve drawing sugars and depicting isomers.

All chemical bonds are represented as horizontal lines or vertical lines. The carbon chain is represented vertically and represented by the center of crossing lines. The orientation of the carbon chain is represented in such a way that the first carbon C₁ is at the top.

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Answer:

Neither

Explanation:

because they aren't designed to be charged

Answer:

37.84 Liters

Explanation:

(see picture)

Answer:

the type of liquid used

Explanation:

It is the variable that Rose is directly controlling

Answer:

Explanation:

Mitosis is crucial to this process. Mitosis is the reason we can grow, heal wounds, and replace damaged cells. Mitosis is also important in organisms which reproduce asexually: this is the only way that these cells can reproduce. This is the one key process that sustains populations of asexual organisms.Jul 22, 2020

Mitosis has an impact on life because it regulates the growth and regeneration of billions of cells within the body.

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The percent ionization of the base is 0.010%. Option C is correct.

First, we can use the pOH value to find the pH of the solution;

pH + pOH = 14

pH = 14 - 4.6 = 9.4

Next, we can use the pH and the concentration of the solution to find the pKa of the conjugate acid of the base.

pH = pKa + log([A⁻]/[HA])

9.4 = pKa + log([A⁻]/[HA])

pKa = 9.4 - log([A⁻]/[HA])

Assuming the base is weak, we can also approximate the percent ionization using the equation;

% ionization = [A⁻]/[HA] x 100

Since the solution is 0.25 M, we can assume that [HA] = 0.25 M - [A-]. Substituting these values into the equation above and using the pKa we calculated, we get;

% ionization = [A⁻]/[HA] x 100

% ionization = [A⁻]/(0.25 M - [A⁻]) x 100

% ionization = \(10^{(-pKa +pH)}\) x 100

% ionization = \(10^{(-9.4+4.6)}\) x 100

% ionization = 0.010%

Hence, C. is the correct option.

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

If a 0.25 m solution of a base is found to have a pOH of 4.6 at equilibrium, what is the percent ionization of the base? select the correct answer below: A) 0.00010% B) 0.0010% C) 0.010% D) 0.10%"--

Answer:

4Al + 3O2 ------> 2Al2O3

the mole ratio is 4:2

you divide 250 by 2

you get 125 and that's the mass produced

O-linked oligosaccharides are commonly attached to the —oh group of amino acids.

Glycosylation is a post-translational modification of proteins in which glyco-linked oligosaccharides are added to proteins. This process is found in eukaryotes, primarily in the endoplasmic reticulum and Golgi apparatus, and involves the enzymatic attachment of sugar units to specific sites on a protein molecule.

These short chains of saccharide sugar residues, called glycosyl sidechains, are examples of glycoconjugates which are covalently attached to hydroxyl groups on the protein.

The most common types of glycosylation sites are the N-linked glycosylation sites, which have an attached sugar group, often an N-acetylglucosamine or N-acetylmannosamine connected to an asparagine sidechain, and the O-linked glycosylation sites, which have a sugar group, usually a galactose or mannose sugar, attached to a hydroxyl group on serine, threonine or hydroxylysine residues.

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Answer:

Electron configuration can be carried out in two steps which can be:

Sublevel notation

Shell notations.

In the sublevel notation, the sequence of filling electrons into the orbitals of the sublevels are guided by some principles:

the maximum number of electrons in the orbital of sublevels are two for s-sublevel, six for p-sublevel, ten for d-sublevel and fourteen for f-sublevel. This indicates that the maximum number of electrons in an orbital is two.

Aufbau's prinicple states that sublevel with lower energies are filled up before those with higher energies.

Pauli exclusion principle states that no two electrons in an atom can have the same set of the four quantum numbers.

Hund's rule states that electrons go into degenerate orbitals singly first before pairing occurs.

Using the shell notation, numbers are used to denote the sum of electrons in all orbitals each energy level.

For an atom of Be:

Sublevel notation for the 4 electrons 1s²2s²

Shell notation 2,2

Other representation is using the electron dot structure.

Answer:

Their calculation was way too off to the actual one.

Explanation:

Answer:

They need a density calculation so they can get the right vaule. Their calculation was way off compare to the acutal vuale.

Explanation:

The value of b in the equation \(\( f(t) = 85 \cdot b^t \)\) represents the decay factor of the radioactive substance. To determine the value of \( b \), we can use the information that the half-life of the substance is 21 years.

The half-life of a radioactive substance is the time it takes for half of the substance to decay. In this case, the half-life is 21 years, which means that after 21 years, the amount of the substance remaining will be half of the initial amount.

We can use this information to set up an equation:

\(\(\frac{1}{2} = b^{21}\)\)

To solve for b, we need to take the 21st root of both sides of the equation:

\(\(b = \left(\frac{1}{2}\right)^{\frac{1}{21}}\)\)

Using a calculator, we can evaluate this expression:

\(\(b \approx 0.965\)\)

Therefore, the value of b in the equation \(\( f(t) = 85 \cdot b^t \)\) is approximately 0.965.

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