list the 7 diatomic molecules

Answer:- pH = 3.37

Solution:- The balanced equation for the reaction of KOH with [tex]HNO_2[/tex] is written as:

[tex]HNO_2(aq)+KOH(aq)\rightleftharpoons KNO_2(aq)+H_2O(l)[/tex]

Nitrous acid is a weak acid and KOH is a strong base. So, at half-stoichiometric point half of the acid will be neutralized to form its conjugate base and half of the acid will still be remaining.

It means at half stoichiometric point the solution will have equal moles of weak acid(nitrous acid) and its conjugate base(nitrite ion). It will act as a buffer solution and the pH of the buffer solution is calculated by using Handerson equation:

[tex]pH=pKa+log(\frac{base}{acid}) [/tex]

Since, for the half stoichiometric point, [acid] = [base]

The ratio of their concentrations becomes 1 and the log of 1 is 0.

So, pH = pKa

pKa = - log Ka

[tex]pKa=-log(4.3*10^-^4)[/tex]

pKa = 3.37

So, pH = 3.37

Hence, the pH of the solution at half equivalence point will be 3.37.

The pH at the half-stoichiometric point for the titration of 0.22 M HNO₂(aq) with 0.1 M KOH(aq) is 3.4.

Let's consider the equation for the titration of 0.22 M HNO₂(aq) with 0.1 M KOH(aq).

HNO₂(aq) + KOH(aq) ⇒ KNO₂(aq) + H₂O(l)

At the half-stoichiometric point, we have a buffer system formed by equal concentrations of the weak acid (HNO₂) and its conjugate base (NO₂⁻).

We can calculate the pH using the Henderson-Hasselbach's equation.

[tex]pH = pKa+log(\frac{[NO_2^{-} ]}{[HNO_2]} )= pKa+log1=pKa=-log(4.3\times 10^{-4} )=3.4[/tex]

The pH at the half-stoichiometric point for the titration of 0.22 M HNO₂(aq) with 0.1 M KOH(aq) is 3.4.

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Answer:

Electrons exchange creating ions to form an ionic bond by electrostatic attraction.

Explanation: I just answered this on USA test prep.

The wavelength of a radiation beam with an energy of 2.39×10² kJ/mol can be calculated using the Planck's formula. Convert the energy to J/photon, then rearrange the formula to solve for wavelength. The wavelength for this energy is approximately 501 nm.

The energy of a beam of radiation can be used to determine its wavelength using Planck's formula, E=hv, where E is energy, h is Planck’s constant, and v is frequency. However, since we want to find the wavelength, we can modify the formula to make use of the speed of light: E=hc/λ. Here, c denotes the speed of light, and λ is wavelength.

First, convert the energy from kJ/mol to J/photon. As Avogadro's number (6.022 × 10²³ molecules/mol) is the number of photons in one mole, the energy per photon is (2.39×10⁵ J/mol) / (6.022×10²³ photon/mol) = 3.97×10⁻¹⁹ J/photon. Now, use Planck's formula to calculate the wavelength:

λ = hc/E = (6.63×10⁻³⁴ J.s × 3.0×10⁸ m/s) / 3.97×10⁻¹⁹ J = 5.01×10⁻⁷ m = 501 nm

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Answer:

The gas pressure is 5.15 atm

Explanation:

Given:

Volume of gas, V = 34.8 L

Moles of gas, n = 7.45 moles

Temperature, T = 19.9 C

To determine:

Pressure, P of the gas

Explanation:

Based on the ideal gas equation:

[tex]PV = nRT[/tex]

where P = pressure, V = volume, n = moles, T = temperature, R = gas constant

[tex]P = \frac{nRT}{V} \\\\P = \frac{7.45\ moles*0.0821\ Latm/mol-K*(19.9+273)\ K}{34.8\ L} =5.15\ atm[/tex]

This problem requires using the concept of pH and pOH. The “p” stands for the negative log so pH and pOH represent the negative log of the concentration of hydrogen or hydroxide ions.

Here is the solution:

pOH = -log [OH-]

pOH = -log [1.83x10^-7 M]

pOH = 6.74

pH + pOH = 14

pH = 14 - 6.74

pH = 7.26

pH = -log [H3O+]

7.26 = -log[H3O+]

[H3O+] = 5.46 x 10^-8 M

To calculate the mass percent, molality, and mole fraction of sulfuric acid in a lead storage battery, you need to perform calculations involving the solution's molarity, density, and the molar mass of H₂SO₄. You must find the mass of the solution, then determine the mass and moles of the solute and solvent.

To calculate the mass percent, molality, and mole fraction of sulfuric acid (H₂SO₄) in an automobile lead storage battery with 3.75 M sulfuric acid and a density of 1.230 g/mL, you need to perform several calculations.

Mass Percent.The mass percent of the solution is the mass of the solute (in this case, H₂SO₄) divided by the total mass of the solution multiplied by 100. Based on the volume of the solution and its molarity, you can find the number of moles of H₂SO₄, which helps to calculate mass, as you know the molar mass of H₂SO₄ is 98.079 g/mol.

Molality. Molality is the number of moles of solute per kilogram of solvent (water in this case). To find the molality, you would calculate the number of moles of sulfuric acid based on its molarity and then calculate the mass of the solvent in kilograms. Remember that the density and the volume of the solution can lead you to find the mass of the solution itself, from which you subtract the mass of the solute to get the mass of the solvent.

Mole Fraction.The mole fraction is the ratio of the number of moles of the solute to the total number of moles of all components in the solution. This requires you to know the moles of both the solute and the solvent.

Answer: Option (b) is the correct answer.

Explanation:

Atomic number is the sum of only total number of protons present in an element. Whereas mass number or atomic mass is the sum of total number of both protons and neutrons present in an element.

For example, given atom has atomic number is 8.  And, it is known that atomic mass of O is 16 g/mol.

Therefore, number of neutrons present in it will be calculated as follows.

                    Mass number = no. of protons + no. of neutrons

                         16 = 8 + no. of neutrons

                    no. of neutrons = 16 - 8  

                                               = 8

Hence, we can conclude that in the given isotope of oxygen-16 there are 8 protons and 8 neutrons.

Answer is: the percent ionizationof formic acid is 1,82%.
Chemical reaction: HCOOH(aq) ⇄ H⁺(aq) + HCOO⁻(aq).
pKa(HCOOH) = 3,77.

Ka(HCOOH) = 1,7·10⁻⁴.

c(HCOOH) = 0,5 M.

[H⁺] = [HCOO⁻] = x; equilibrium concentration.
[HA] = 0,1 M - x.
Ka = [H⁺] · [HCOO⁻] / [HCOOH].
0,00017 = x² / 0,5 M - x.
Solve quadratic equation: x = 0,0091 M.
α = 0,0091 M ÷ 0,5 M · 100% = 1,82%.

Answer : 1.91 %

Explanation : The steps to solve this problem are explained below;

1.  HCOOH ⇄ [tex]HCOO^{-} + H^{+} [/tex]      

Here Ka =([tex] [HCOO^{-}]_{eq} X  [H^{+}]_{eq} [/tex] )/  [tex][HCOOH]_{eq}[/tex]                                                        

As the equilibrium concentration of [tex] H^{+} [/tex] will be the pH of the solution.

∴ [tex] [H^{+}]_{eq}[/tex] = [tex]10^{(-2.02)} = 9.55 x [tex] 10^{-3} [/tex] M  

2. The initial concentration of HCOOH.  When it loses x moles from it as the acid undergoes dissociation to form [tex]HCOO^{-}[/tex] and [tex]H^{+}[/tex].  

3.  The moles present will be as

         [HCOOH] (M)                 [tex] [H^{+}] [/tex](M)           [tex][HCOO^{-}] [/tex](M)  

Initial       0.50                                0.00                                          0.00

After Change  -x                              +x                                            +x

Equilibrium      ( 0.50 -x)                     x                                              x

∴ Ka   =   (x) x (x)  / (0.50 - x)  

4.  Assuming that all of the [tex] H^{+}[/tex] comes from the acid, and none from water.  

As [tex] [H^{+}]_{eq}[/tex] = 9.55 x[tex]  10^{-3}[/tex] which is much higher than the 1.0 x[tex]  10^{-7 } [/tex] M [tex[H^{+}[/tex] from water.  

Also, the concentration of HCOOH will change very little, from 0.50  to 0.50 - 9.55 x [tex] 10^{-3}[/tex].  

The change in concentration can be ignored if it is less than 5% of the original concentration.  

∴ 0.50 M x 5% = 0.025, so the change in [HCOOH] in this problem can be ignored.    

Now,  Ka = (x)(x)/0.50 = (9.55 x [tex] 10^{-3})^{2}[/tex] /0.50= 1.82 x [tex] 10^{-4}[/tex]

Now, calculating the percent ionization for this problem.  

which will represent the relative number of acid molecules which dissociate. It is calculated as :

[tex] [H^{+}]_{eq} [/tex] x  100 /[tex] [HCOOH]_{i}[/tex]                                                        

∴ percent ionization = {(9.55 x [tex] 10^{-3})[/tex]/ (0.50)}x 100 = 1.91 %

This value of 1.91 % indicates that very little of this acid dissociates (ionizes) under these conditions.  

For strong acids and bases, the percent ionization is 100%.

Answer:

The correct answer is :'The equilibrium will shift to the right to favor the forward reaction'.

Explanation:

[tex]CH_4(g) + H_2O(g) \rightleftharpoons CO(g) + 3H2(g)[/tex]

According to Le-Chatlier's principle,When the pressure is increased the equilibrium shifts in the direction where number of moles of gas molecules are greater .

The equilibrium will shift towards the product side because there are more number of moles of gas re greater on product side. So, the equilibrium will shift in the right direction favoring the forward reaction.

The common name to the family member of phylum Echinodermata of marine family is echinoderm. They are usually characterized by a five-fold symmetry, and possess an internal skeleton of calcite plates.  They are found at every ocean depth.

The features of all adult echinoderm are:

- They have a five-fold symmetry.

- Body without segmentation.

- Spiny skin.

- Internal skeleton.

- found at every ocean depth.

0.3 liter of the 2.5 M NaOH solution is required to make 1 liter of a 0.75 M NaOH solution.

To find the volume of a NaOH solution needed, we can use the formula:

(M1)(V1) = (M2)(V2)

Where:

Plugging in the given values:

(2.5 M)(V1) = (0.75 M)(1 L)

Solving for V1:

V1 = (0.75 M)(1 L) ÷ (2.5 M) = 0.3 L

Therefore, 0.3 liter of the 2.5 M NaOH solution is required to make 1 liter of a 0.75 M NaOH solution.

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2 moles of hydrogen molecules forms 2 moles of water or 36 g of water. 2 molecule of hydrogen contains 1.2 × 10²⁴. Thus, 1.11× 10²⁴ hydrogen molecules will give 33.3 g of water.

One mole of any compound contains 6.022 × 10²³ molecules. This number is called Avogadro number. Thus,one mole water contains 6.022 × 10²³  water molecules.

2 moles of hydrogen molecules will give 2 moles of water. One mole of hydrogen molecule contains 6.022 × 10²³ molecules.

molar mass of water = 18 g/mol

2 moles of water = 36 g.

Then, 2 moles of H₂ = 2 × 6.022 × 10²³ =  1.2 × 10²⁴ molecules

These much hydrogen molecules produces 36 g of water. Thus, mass of water produced by 1.11× 10²⁴ hydrogen molecules is :

(1.11× 10²⁴ × 36 )/1.2 × 10²⁴  = 33.3 g.

Therefore, the mass of water produced from 1.11× 10²⁴ hydrogen molecules is 33.3 g.

To find more on Avogadro number, refer here:

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Answer: [tex]C_{3}H_{8} + 5O_{2} \rightarrow 3CO_{2} + 4H_{2}O[/tex] this is an example of a combustion reaction.

Explanation:

A reaction in which a compound reacts with oxygen and results in the formation of carbon dioxide and water is known as combustion reaction.

For example, [tex]C_{3}H_{8} + 5O_{2} \rightarrow 3CO_{2} + 4H_{2}O[/tex] is a combustion reaction.

Also, it is known that combustion reactions are exothermic in nature because heat is released during these reactions.

The hydrolysis of a protein results in the production of amino acids. This procedure involves the breaking of the peptide bonds within the protein using water, resulting in the formation of individual amino acids. Additionally, amino groups are converted into waste products such as urea.

Proteins are polymers largely composed of amino acids connected by peptide bonds. In a process known as hydrolysis, these proteins are broken down into smaller components through the introduction of water. As a result, the bonds within the protein molecule are broken, typically resulting in the formation of individual amino acids.

It's important to note that for an amino acid to participate in this breakdown procedure, its amino group must first be removed. Following this, the amino group is usually converted into ammonia. Mammalian organisms, however, convert this ammonia into a more efficient waste product known as urea.

Overall, in the hydrolysis of proteins, the primary compounds produced are amino acids, which may be further metabolized or reformed into other proteins.

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Answer:

See explanation

Explanation:

First, let's write the equation again.

CH4 + NH3 + O2 -----> HCN + H2O

In order to balance this equation, we need to see both sides of the reaction (Reactants and products), and count the elements and atoms there. it's usually begin with the metals, then non metals, and finally hydrogen and oxygen.

Let's begin with the Carbon and nitrogen.

In reactant we have 1 C and product the same, so theorically speaking, we don't need to balance, same thing happen with N (1 and 1).

However, when we look at H, we can see we have 7 in reactants (4 and 3) and only 3 in products (1 and 2), so we need to balance them. As we have those atoms in different compounds, we need to put a number in both compound so the sum gives an even number. In this case if we put a 2 in CH4, we'll get 8 hydrogen and the other 3, will be 11.

2CH4 + NH3 + O2 -----> HCN + H2O

To get an even number, let's put a 2 in NH3 too, the total is now 14.

2CH4 + 2NH3 + O2 -----> HCN + H2O

Let get 14 in the products, to do that we simply put a 2 in HCN and 6 in H2O.

2CH4 + 2NH3 + O2 -----> 2HCN + 2H2O

With that, we balance Hydrogen, and also Carbon and nitrogen were balanced too with this.

Now only the oxygen needs to be balanced. The 6 in H2O put the oxygen with 6, so we need to put a 3 in reactants, and with that, the equation is balanced:

2CH4 + 2NH3 + 3O2 -----> 2HCN + 6H2O

Now that we have the balanced equation, we can calculate the rest of the questions.

The limiting reagent is the reagent that it gets consumed first and completely in the reaction. To do that, we need to do stechiometry of the reagents, and the easier way to do that is with the number of moles.

moles are calculated:

n = m/MM

The molecular mass of CH4 is 16 g/mol and NH3 is 17 g/mol. According to the balanced equation, we have at least 2 moles of CH4 and 2 moles of NH3. The limiting reagent will be the reagent with the lower moles so:

nCH4 = 8/16 = 0.5 moles

nNH3 = 10/17 = 0.59 moles

We have more moles of NH3 than moles CH4, so the limiting reagent is CH4.

Now to get the mass of HCN, we need the moles. But we already know which is the limiting reagent and we know (according to the balanced reaction) that 2 moles of CH4 produces 2 moles HCN, so, the moles of CH4 are the same of HCN produced therefore:

moles CH4 = moles HCN = 0.5 moles

To get the mass we need the molecular mass of HCN which is 27 g/mol, so the mass:

m = 27 * 0.5

m = 13.5 g

This is the mass formed of HCN

To calculate the molarity of acetic acid in vinegar, convert the average mass percentage of acetic acid to grams, then convert grams to moles, and finally divide the moles by the volume in liters. The molarity of acetic acid in vinegar is 0.866 M.

To calculate the molarity of acetic acid in vinegar, we need to first convert the average mass percentage of acetic acid to grams. Assuming we have 100g of vinegar, the mass of acetic acid would be 5.2g (5.2% of 100g). Next, we need to convert grams of acetic acid to moles. The molar mass of acetic acid is 60.05 g/mol. Dividing the mass by the molar mass gives us the number of moles. Finally, we can calculate the molarity by dividing the moles by the volume in liters. Since the density of vinegar is given as 1.005 g/mL, we can assume the volume of 100g of vinegar is 100 mL, or 0.1 L.

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The molarity of the acetic acid in the vinegar is 0.871 M, calculated by finding the mass of acetic acid in a certain volume of vinegar and then using that to determine the number of moles of acetic acid and thus the molarity.

To get the molarity (M) of acetic acid in vinegar, we need to first find the mass of acetic acid in a certain volume of vinegar since molarity (M) is defined as moles of solute/liters of solution. The given data implies that 5.2% of the vinegar's mass is due to acetic acid. Hence, given vinegar's density (1.005 g/mL), in 1000 ml (or 1 liter) of vinegar, mass of vinegar = density * volume = 1.005 g/mL * 1000 mL = 1005 g. Given that 5.2% of this mass is acetic acid, mass of acetic acid is 5.2/100 * 1005 = 52.26g.

The molecular mass of acetic acid (CH3COOH) is approximately 60 g/mol. So, the number of moles of acetic acid = mass of acetic acid / molecular mass = 52.26 g / 60 g/mol = 0.871 moles. Therefore, the molarity (M) = moles of solute/liters of solution = 0.871 moles / 1 L = 0.871 M.

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Molecular formula, structural formula, and 3D drawing or model can be used to show differences between isomers.

The types of molecular representations that can be used to show differences between isomers are: