A barium hydroxide solution is prepared by dissolving 2.29 g of Ba(OH)2 in water to make 46.6 mL of solution. What is the concentration of the solution in units of molarity?
Concentration: 0.287 M

The barium hydroxide solution is used to titrate a perchloric acid solution of unknown concentration. Write a balanced chemical equation to represent the reaction between barium hydroxide and perchloric acid.
Chemical Equation:

If 25.5 mL of the barium hydroxide solution was needed to neutralize a 8.99 mL aliquot of the perchloric acid solution, what is the concentration of the acid?
Concentration:

Answer:

1.207 ××10⁻¹⁷ J

Explanation:

Step 1: Given and data

Step 2: Calculate the energy (E) of the photon

We will use the Planck-Einstein's relation.

E = h × ν

E = (6.626 × 10⁻³⁴ J.s) × ( 1.821 × 10¹⁶ s⁻¹) = 1.207 ××10⁻¹⁷ J

Answer:

33.8 g Solution

Explanation:

A chemistry student needs 15.0 g of methanol for an experiment. The concentration of ethanol in the solution is 44.4% w/w, that is, there are 44.4 g of methanol every 100 g of solution. The mass of solution that would contain 15.0 g of methanol is:

15.0 g Methanol × 100 g Solution/44.4 g Methanol = 33.8 g Solution

Since 33.8 g are required and 320. g are available, there is enough solution for the requirements.

Answer:

Aluminum tribromide: AlBr₃, however, it should be just aluminum bromide.

Sulfur dioxide: SO₂.

Beryllium hydride: BeH₂

Magnesium(II) oxide: MgO; however the roman numeral is not used in Mg as it just has one oxidation number.

Copper(II) oxide: CuO.

Calcium sulfate: CaSO₄

Nitric acid: HNO₃.

Explanation:

Hello there!

In this case, it seems that the formulas were not given, however, we can write the correct one for each given compound according to the widely used nomenclature rules as shown below:

Aluminum tribromide: AlBr₃, however, it should be just aluminum bromide.

Sulfur dioxide: SO₂.

Beryllium hydride: BeH₂

Magnesium(II) oxide: MgO; however the roman numeral is not used in Mg as it just has one oxidation number.

Copper(II) oxide: CuO.

Calcium sulfate: CaSO₄

Nitric acid: HNO₃.

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Answer:

H2O.....................

Answer:

Explanation:

[H+] = 0.1x0.03x2 + 0.2x0.03 = 0.012 mol

[OH-] = 0.001x0.05x2 = 0.0001 mol

=> [H+] dư = 0.012 - 0.0001 =0.0119 mol

pH = -log[H+] = 1.92

Answer: The molarity of NaCl solution is 0.0489 M

Explanation:

Molarity is defined as the amount of solute expressed in the number of moles present per liter of solution. The units of molarity are mol/L. The formula used to calculate molarity:

[tex]\text{Molarity of solution}=\frac{\text{Given mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (mL)}}[/tex] .....(1)

We are given:

Given mass of NaCl = 0.8 g

Molar mass of NaCl = 58.44 g/mol

Volume of the solution = 280 mL

Putting values in equation 1, we get:

[tex]\text{Molarity of solution}=\frac{0.8\times 1000}{58.44\times 280}\\\\\text{Molarity of solution}=0.0489M[/tex]

Hence, the molarity of NaCl solution is 0.0489 M

Answer:

C) isoleucine

Explanation:

Isoleucine is among nine necessary amino acids in humans (found in dietary proteins). It has a variety of physiological activities, including aiding tissue repair, nitrogenous waste detoxification, immunological stimulation, and hormonal production promotion. When attached at a binding site, they are capable of providing a side chain thereby increasing the hydrophobicity at the binding site.

Answer:

THE ANSWER IS: copper(I) sulfide.

hope this helped <3

Explanation:

Answer:

For the neutralization reaction between pyridine and propanoic acid, draw curved arrows to indicate the direction of electron flow.

Draw curved arrows to show the movement of electrons in this step of the mechanism.

Explanation:

According to Bronsted acid-base theory, an acid is a substance which is a proton donor.

Base is the proton acceptor.

In the given example, acid is propanoic acid and it loses the proton.

Pyridine is the base and it accepts the proton from propanoic acid.

The entire reaction is shown below:

Answer:

14.9 g

Explanation:

Step 1: Write the balanced equation

C₄H₁₀ + 6.5 O₂ ⇒ 4 CO₂ + 5 H₂O

Step 2: Calculate the moles corresponding to 9.57 g of C₄H₁₀

The molar mass of C₄H₁₀ is 58.12 g/mol.

9.57 g × 1 mol/58.12 g = 0.165 mol

Step 3: Calculate the moles of H₂O produced from 0.165 moles of C₄H₁₀

0.165 mol C₄H₁₀ × 5 mol H₂O/1 mol C₄H₁₀ = 0.825 mol H₂O

Step 4: Calculate the mass corresponding to 0.825 mol of H₂O

The molar mass of H₂O is 18.02 g/mol.

0.825 mol × 18.02 g/mol = 14.9 g

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Answer:

Identify the indicated protons in the following molecules as unrelated, homotopic, enantiotopic, or diastereotopic. a) Methyls a & b: _________ b) Ha & Hc: ________

Explanation:

Homotopic hydrogens:

Consider two hydrogens in the given molecule and replace one by one with a different atom say for example deuterium, then if the two molecules formed by replacing hydrogens are the same then the two hydrogens are called homotopic hydrogens.

After replacing the two hydrogens with a different atom then, enantiomers are formed then, the two hydrogens are called enantiotopic hydrogens.

After replacing the two hydrogens with a different atom then, diastereomers are formed then, the two hydrogens are called diastereotopic hydrogens.

In the methyl group, select two hydrogens and replace one hydrogen atom with a D-atom name the compound.

Again replace another hydrogen atom with D-atom.

Name it.

If both are the same then, the hydrogens are homotopic and they are shown below:

Hence, they are homotopic protons.

Answer:

INCREASE in the difference between the melting point measured and the true melting temperature.

Explanation:

Melting point of a compound is defined as the temperature at which the soils compound changes into liquid at the atmospheric pressure. There are different circumstances that can lead to inaccurate melting point. These include:

--> presence of impurities in the compound,

--> Molecular composition,

--> Force of attraction, and

--> Rapid heating of the compound.

Under the circumstances of rapid heating of the compound, there would be an increase in the melting point range when compared with the true melting point range of the compound.

The higher the heating rate, the more rapid the rise in oven temperature, increasing the difference between the melting point measured and the true melting temperature.

Answer: The molecule [tex]CH_{3}-CH_{3}[/tex] is expected to have a dipole moment of zero.

Explanation:

The product of magnitude of the charge calculated in electrostatic units is called dipole moment.

Formula for dipole moment is as follows.

Dipole moment = Charge (in esu) [tex]\times[/tex] distance (in cm)

Non-polar molecules have zero dipole moment.

For example, [tex]CH_{3}-CH_{3}[/tex] is a non-polar molecule so its dipole moment is zero.

[tex]H_{2}C=O[/tex] is a polar molecule so it will have dipole moment.

[tex]CH_{2}Cl_{2}[/tex] is a polar molecule so it will have dipole moment.

[tex]NH_{3}[/tex] has nitrogen atom as more electronegative than hydrogen atom. So, net dipole moment will be in the direction of nitrogen atom.

Thus, we can conclude that the molecule [tex]CH_{3}-CH_{3}[/tex] is expected to have a dipole moment of zero.

Answer:

0.2 M

Explanation:

Step 1: Given data

There are different ways to express the concentration of a solution. We will calculate molarity, which is one of the most used.

Step 2: Calculate the moles of sucrose

The molar mass of sucrose is 342.3 g/mol.

15 g × 1 mol/342.3 g = 0.044 mol

Step 3: Calculate the molarity of the solution

Molarity is equal to the moles of solute divided by the liters of solution.

M = 0.044 mol/0.2 L = 0.2 M

Answer:

it's B

Explanation:

Write the balanced equation: H2SO4 + 2KOH → K2SO4 +2H2O. So 2(moles KOH) = (moles H2SO4); 2(volume KOH)(concentration KOH) = (volume H2SO4)(concentration H2SO4); 2(40ml)(0.2M) = (30ml)(x); x = 0.53M

The concentration of  H₂SO₄ solution is equal to 0.133 M.

A neutralization reaction can be described as a chemical reaction in which an acid and base react together to form respective salt and water. When a strong acid such as HCl will react with a strong base such as NaOH the salt can be neither acidic nor basic.

When H₂SO₄ (a strong acid) reacts with KOH, the resulting salt will be  K₂SO₄ and water.

H₂SO₄    +  2KOH   → K₂SO₄  +  2H₂O

Given, the concentration of KOH solution = 0.2 M

The volume of the KOH solution = 40 ml = 0.040 ml

The number of moles of KOH, n = M × V = 0.2 × 0.04 = 0.008 mol

The volume of the H₂SO₄ = 30 ml = 0.03 L

The number of moles of H₂SO₄, n = 0.008/2 = 0.004 mol

The concentration of  H₂SO₄ solution = 0.004/0.03 = 0.133 M

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Answer:

LIBs have had a huge impact on our society. They enabled modern portable electronics such as laptops and mobile phones. And they are now enabling clean and low-carbon transport, be it via electric cars or even flying taxis, and grid-scale storage of renewable energy

Explanation:

The density (g/mL) for a 20.0-mL sample of a salt solution has a mass of 24.0 g is 1.2 g/ml.

Density is the mass per unit volume. Density is a scalar quantity. It is denoted by d and the symbol for density is given as rho, a Greek symbol. Density is calculated as mass divided by volume.

Mass is the quantity of matter in a physical body. The product of the molar mass of the compound and the moles of the substance are defined as mass.

Volume is the space occupied by a three-dimensional object. Volume is calculated by dividing mass by density.

Given, the 20.0-mL sample of a salt solution, which is the volume.

The mass of the solution is 24.0 g

To calculate the density

mass/volume

24.0 / 20.0 = 1.2 g/ml

Thus, the density of the given salt solution is 1.2 g/ml.

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Answer:

MgO

Explanation:

Answer: The empirical and molecular formula for the given organic compound is [tex]C_9H_{13}O_3N[/tex]

Explanation:

Let the mass of the compound be 100 g

Given values:

% of C = 59.0%

% of H = 7.15%

% of O = 26.20%

% of N = 7.65%

Mass of C = 59.0 g

Mass of H = 7.15 g

Mass of O = 26.20 g

Mass of N = 7.65 g

The number of moles is defined as the ratio of the mass of a substance to its molar mass. The equation used is:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] ......(1)

To formulate the empirical formula, we need to follow some steps:

Molar mass of C = 12 g/mol

Molar mass of H = 1 g/mol

Molar mass of O = 16 g/mol

Molar mass of N = 14 g/mol

Putting values in equation 1, we get:

[tex]\text{Moles of C}=\frac{59.0g}{12g/mol}=4.917 mol[/tex]

[tex]\text{Moles of H}=\frac{7.15g}{1g/mol}=7.15 mol[/tex]

[tex]\text{Moles of O}=\frac{26.20g}{16g/mol}=1.6375 mol[/tex]

[tex]\text{Moles of N}=\frac{7.65g}{14g/mol}=0.546 mol[/tex]

Calculating the mole fraction of each element by dividing the calculated moles by the least calculated number of moles that is 0.546 moles

[tex]\text{Mole fraction of C}=\frac{4.917}{0.546 }=9[/tex]

[tex]\text{Mole fraction of H}=\frac{7.15}{0.546 }=13[/tex]

[tex]\text{Mole fraction of O}=\frac{1.6375}{0.546 }=2.99\approx 3[/tex]

[tex]\text{Mole fraction of N}=\frac{0.546}{0.546 }=1[/tex]

The ratio of C : H : O : N = 9 : 13 : 3 : 1

The empirical formula of the compound becomes [tex]C_9H_{13}O_3N_1=C_9H_{13}O_3N[/tex]

To calculate the molecular formula, the number of atoms of the empirical formula is multiplied by a factor known as valency that is represented by the symbol, 'n'.

[tex]n =\frac{\text{Molecular mass}}{\text{Empirical mass}}[/tex] .....(2)

We are given:

Mass of molecular formula = 183 g/mol

Mass of empirical formula = 183 g/mol

Putting values in equation 2, we get:

[tex]n=\frac{183g/mol}{183g/mol}=1[/tex]

Multiplying this valency by the subscript of every element of empirical formula, we get:

[tex]C_{1\times 9}H_{1\times 13}O_{1\times 3}N_{1\times 1}=C_9H_{13}O_3N[/tex]

Hence, the empirical and molecular formula for the given organic compound is [tex]C_9H_{13}O_3N[/tex]

Answer:

Ionic compounds conduct electricity when molten (liquid) or in aqueous solution (dissolved in water), because their ions are free to move from place to place. Ionic compounds cannot conduct electricity when solid, as their ions are held in fixed positions and cannot move.

Explanation:

because their ions are free to move from place to place.

Answer: The ion that contribute to water hardness are:

--> a. Ca2+

--> b. (HCO)3^- and

--> c. Mg2+

While K+ DOES NOT contribute to water hardness.

Explanation:

WATER in chemistry is known as a universal solvent. This is so because it is polar in nature and dissolves most inorganic solutes and some polar organic solutes to form aqueous solutions. It is composed of elements such as hydrogen and oxygen in the combined ratio of 2:1.

Water is said to be HARD if it does not lather readily with soap. There are two types of water hardness:

--> Permanent hardness: This is mainly due to the presence of CALCIUM and MAGNESIUM ions in the form of soluble tetraoxosulphate(VI) and chlorides. These ions are removed by adding washing soda or caustic soda.

--> Temporary hardness: This is due to the presence of calcium HYDROGENTRIOXOCARBONATES. It can be removed by boiling and using slaked lime.

Therefore from the above given ions, Ca2+,(HCO)3^- and Mg2+ contributes to water hardness.

Answer: The millimoles of potassium iodide the chemist has added to the flask is 522 millimoles.

Explanation:

Given: Volume of KI = 370.0 mL (1 mL = 0.001 L) = 0.37 L

Molarity of KI solution = 1.41 mol/L

Now, moles of KI (potassium iodide) is calculated as follows.

[tex]Moles = Volume \times Molarity \\= 0.37 L \times 1.41 M\\= 0.5217 mol[/tex]

Convert moles into millimoles as follows.

1 mol = 1000 millimoles

0.5217 mol = [tex]0.5217 mol \times \frac{1000 millimoles}{1 mol} = 521.7 millimoles[/tex]

This can be rounded off to the value 522 millimoles.

Thus, we can conclude that the millimoles of potassium iodide the chemist has added to the flask is 522 millimoles.

Answer:

[tex]4.67\times 10^{-4}[/tex]

Explanation:

Given that,

The pH of a certain orange juice is 3.33.

We need to find the +ion concentration.

We know that,

[tex]pH=-log[H^+][/tex]

So,

[tex]3.33=-log[H^+]\\\\\[H^+=10^{-3.33}\\\\=4.67\times 10^{-4}[/tex]

So, the +ion concentraion is equal to [tex]4.67\times 10^{-4}[/tex].

Answer:

The mole is defined as a collection of 6.022 × 1023 particles.

The atomic mass given on a periodic table that is given in grams is the mass of

one mole (6.022 × 1023 particles) of that element

Explanation:

Answer: A volume of 20.49 milliliters of sodium formate do you need to measure to make this buffer (assuming the rest is formic acid).

Explanation:

Given: Total volume of the buffer = 21.0 mL

[tex]\frac{[HCOONa]}{[HCOOH]} = 4[/tex] ... (1)

It is assumed that the volume of HCOONa is x. Hence, volume of HCOOH is (21.0 - x) mL.

Hence,

[HCOONa] = Molarity [tex]\times[/tex] Volume

= 0.10 [tex]\times[/tex] x

= 0.1x mmol

Similarly, [HCOOH] = Molarity [tex]\times[/tex] Volume

= 0.10 [tex]\times[/tex] (21.0 - x) mmol

Using equation (1),

[tex]\frac{[HCOONa]}{[HCOOH]} = 4\\\frac{0.1x}{(21.0 - x)} = 4\\0.1x = 84.0 - 4x\\4.1x = 84.0\\x = 20.49 mL[/tex]

As x is the volume of sodium formate. Hence, 20.49 mL of sodium formate is required to make the buffer.

Thus, we can conclude that a volume of 20.49 milliliters of sodium formate do you need to measure to make this buffer (assuming the rest is formic acid).

Answer:

A

Explanation:

Answer:

Aldehydes and Ketones

Both aldehydes and ketones contain a carbonyl group, a functional group with a carbon-oxygen double bond. The names for aldehyde and ketone compounds are derived using similar nomenclature rules as for alkanes and alcohols, and include the class-identifying suffixes -al and -one, respectively:



In an aldehyde, the carbonyl group is bonded to at least one hydrogen atom. In a ketone, the carbonyl group is bonded to two carbon atoms:





As text, an aldehyde group is represented as –CHO; a ketone is represented as –C(O)– or –CO–.

In both aldehydes and ketones, the geometry around the carbon atom in the carbonyl group is trigonal planar; the carbon atom exhibits sp2 hybridization. Two of the sp2 orbitals on the carbon atom in the carbonyl group are used to form σ bonds to the other carbon or hydrogen atoms in a molecule. The remaining sp2 hybrid orbital forms a σ bond to the oxygen atom. The unhybridized p orbital on the carbon atom in the carbonyl group overlaps a p orbital on the oxygen atom to form the π bond in the double bond.

Like the C=OC=O bond in carbon dioxide, the C=OC=O bond of a carbonyl group is polar (recall that oxygen is significantly more electronegative than carbon, and the shared electrons are pulled toward the oxygen atom and away from the carbon atom). Many of the reactions of aldehydes and ketones start with the reaction between a Lewis base and the carbon atom at the positive end of the polar C=OC=O bond to yield an unstable intermediate that subsequently undergoes one or more structural rearrangements to form the final product (Figure 1).

Figure 1. The carbonyl group is polar, and the geometry of the bonds around the central carbon is trigonal planar.

The importance of molecular structure in the reactivity of organic compounds is illustrated by the reactions that produce aldehydes and ketones. We can prepare a carbonyl group by oxidation of an alcohol—for organic molecules, oxidation of a carbon atom is said to occur when a carbon-hydrogen bond is replaced by a carbon-oxygen bond. The reverse reaction—replacing a carbon-oxygen bond by a carbon-hydrogen bond—is a reduction of that carbon atom. Recall that oxygen is generally assigned a –2 oxidation number unless it is elemental or attached to a fluorine. Hydrogen is generally assigned an oxidation number of +1 unless it is attached to a metal. Since carbon does not have a specific rule, its oxidation number is determined algebraically by factoring the atoms it is attached to and the overall charge of the molecule or ion. In general, a carbon atom attached to an oxygen atom will have a more positive oxidation number and a carbon atom attached to a hydrogen atom will have a more negative oxidation number. This should fit nicely with your understanding of the polarity of C–O and C–H bonds. The other reagents and possible products of these reactions are beyond the scope of this chapter, so we will focus only on the changes to the carbon atoms: