What is the concentration of HNO3 if 5.00×10−2 mol are present in 905 mL of the solution?

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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:

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.

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.

To learn more about density, refer to the link:

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Answer:

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Answer:

Neutron has no charge while electron has a negative charge and proton has a positive charge

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

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:

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:

178.5g of NaCl is the maximum amount that can be dissolved

Explanation:

Solubility is defined as the maximum amount of solute that can be dissolved in aAt 20°C, the solubility of NaCl is 35.7g per 100g of water at 20°C. That means the maximum amount that can be dissolved of NaCl is:

500g water * (35.7g NaCl / 100g water) =

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 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.

A. Atoms can combine in many ways to form various compounds

and molecules.

Answer:

0050 hours

Explanation:

The size of the drops is regulated by types of tubing in microdrip tubing 1cc(1ml ) forms 60 gtt.

Gtt in 500 cc D5W is 500×60 gtt.= 30000 gtt.

Time required to infuse whole fluid with the rate of 60gtt /min = 30000gtt÷60gtt/min = 500minute.

= 8hours 20 minute.

Fluid infusion started at 1630 hours after and it ends after 8 hours 30 minutes .

Therefore, infusion will be completed at 1630 hours +0830 hours.= 0050 hours.

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: 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:

propanal

Explanation:

hope this helps :)

Answer:

C₂H₅OH + 3 O₂ → 2 CO₂ + 3 H₂O

2 NH₃ + 3.5 O₂ → 2 NO₂ + 3 H₂O

C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O

H₂SO₄ + 2 NaOH → Na₂SO₄ + 2 H₂O

3 NO₂ + H₂O → 2 HNO₃ + NO

Explanation:

We will balance the equation using the trial and error method.

C₂H₅OH + O₂ → CO₂ + H₂O

1) We balance C atoms by multiplying CO₂ by 2 and H atoms by multiplying H₂O by 3.

C₂H₅OH + O₂ → 2 CO₂ + 3 H₂O

2) We balance O atoms by multiplying O₂ by 3.

C₂H₅OH + 3 O₂ → 2 CO₂ + 3 H₂O

NH₃ + O₂ → NO₂ + H₂O

1) We balance H atoms by multiplying NH₃ by 2 and H₂O by 3.

2 NH₃ + O₂ → NO₂ + 3 H₂O

2) We balance N atoms by multiplying NO₂ by 2.

2 NH₃ + O₂ → 2 NO₂ + 3 H₂O

3) We balance O atoms by multiplying O₂ by 3.5

2 NH₃ + 3.5 O₂ → 2 NO₂ + 3 H₂O

C₃H₈ + O₂ → CO₂ + H₂O

1) We balance C atoms by multiplying CO₂ by 3 and H atoms by multiplying H₂O by 4.

C₃H₈ + O₂ → 3 CO₂ + 4 H₂O

2) We balance O atoms by multiplying O₂ by 5.

C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O

H₂SO₄ + NaOH → Na₂SO₄ + H₂O

1) We balance Na atoms by multiplying NaOH by 2.

H₂SO₄ + 2 NaOH → Na₂SO₄ + H₂O

2) We balance H and O atoms by multiplying H₂O by 2.

H₂SO₄ + 2 NaOH → Na₂SO₄ + 2 H₂O

NO₂ + H₂O → HNO₃ + NO

1) We balance H atoms by multiplying HNO₃ by 2.

NO₂ + H₂O → 2 HNO₃ + NO

2) We balance N atoms by multiplying NO₂ by 3.

3 NO₂ + H₂O → 2 HNO₃ + NO

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:

1 +-0.05cm

Explanation:

because this is more suitable

Answer:

AlCl3 + 3NaOH ———>Al(OH)3 + 3NaCl

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:

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:

MgO

Explanation:

Answer:

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

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: Pair 1 has Mg and Sr, Pair 2 has Kr and Ne, Pair 3 has As and P.

Explanation:

A periodic table is a group of elements presented in a tabular form where elements are arranged in a series of 7 rows and 18 columns.

The vertical columns are known as groups and horizontal rows are known as periods.

The elements having similar chemical properties are arranged in one group.

Magnesium (Mg) is the 12th element of periodic table placed at Group 2 and Period 3

Strontium (Sr) is the 38th element of periodic table placed at Group 2 and Period 5

Krypton (Kr) is the 36th element of periodic table placed at Group 18 and Period 4

Neon (Ne) is the 10th element of periodic table placed at Group 18 and Period 2

Arsenic (As) is the 33rd element of periodic table placed at Group 15 and Period 4

Phosphorus (P) is the 15th element of periodic table placed at Group 15 and Period 3

As magnesium and strontium are present in the same group, they will have similar chemical properties. Similarly, krypton and neon will form the second pair. Likewise, arsenic and phosphorus will form a pair.

Hence, Pair 1 has Mg and Sr, Pair 2 has Kr and Ne, Pair 3 has As and P.

Answer:

bleh

Explanation: