Structures of compounds people use every day are shown. From which group of unsaturated hydrocarbons is each derived? 2 carbons are double bonded together, with H bonded to the left, and C H 2 bonded below left, above right, and below right. In a segment of an ongoing chain, 3 iterations of this structure are bonded together by single bonds from the bottom right C H 2 of one structure to the bottom left C H 2 of the next.

The pH of the solution at 25 degree celsius of 1.3 × 10⁻⁶ moles of a sample of Sr(OH)₂ is 10.02.

The pH of any solution gives an idea about the acidic and basic nature of the solution and the equation of pH will be represented as:

pH + pOH = 14

Given that,

Moles of Sr(OH)₂ = 1.3 × 10⁻⁶ mol

Volume of solution = 25mL = 0.025L

The concentration of Sr(OH)₂ in terms of molarity = 1.3×10⁻⁶/0.025

                                                                                    = 5.2×10¯⁵M

Dissociation of Sr(OH)₂ takes place as:

                               Sr(OH)₂ → Sr²⁺ + 2OH⁻

From the stoichiometry of the reaction 1 mole of Sr(OH)₂ produces 2 moles of OH⁻.

Given that the base is a strong base and that it entirely dissociates into its ions, the hydroxide ion concentration is 5.2×10¯⁵×2 = 1.04×10¯⁴ M.

pOH = -log[OH⁻]

pOH = -log(1.04×10¯⁴)

pOH = 3.98

Now we put this value on the first equation we get,

pH = 14 - 3.98 = 10.02

Therefore, the value of pOH is 10.02.

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The age of rock that is containing 0.628 g Uranium-238 is 2032631864.35 years.

Half-life can be given as the time required by the object to reduced to half of its initial concentration. The concentration remained can be given as:

Final concentration = Initial concentration * 1/2 ^ (t/t1/2)

The initial concentration of rock has been the remaining uranium-238 and lead-206 cumulative concentration. Thus, the initial concentration is given as:

Initial concentration = Uranium-238 + Lead-206

Initial concentration = 0.068 g + 0.025 g

Initial concentration = 0.093g

The final concentration of Uranium-238 remained = 0.68 g

The half life given = 4.5 × 10⁹ years

The age (t) of the rock can be given as:

[tex]0.068=0.093\left(\frac{1}{2}\right)^{\frac{t}{4.5\times \:10^9}}\\\\\frac{t\times \:10^{-9}}{4.5}\ln \left(\frac{1}{2}\right)=\ln \left(\frac{68}{93}\right)\\\\t=2032631864.35961[/tex]

Thus, the age of rock that has remaining 0.068 g Uranium-238 is 2032631864.35 years.

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436 atoms of Pb 207 are left in the rock.

Alpha and beta decays are the mechanism through which uranium transforms into the lead. While 235U and its daughter nuclides only experience seven alpha and four beta decays, 238U and its daughter nuclides suffer eight alpha and six beta decays overall.

When there are too few neutrons in relation to protons, alpha decay occurs. There is the discharge of an alpha particle, which consists of two protons and two neutrons.

When there are more neutrons than protons, beta decay can happen. A high-speed electron is ejected from a proton that transforms spontaneously from a neutron.

Since gamma radiation is emitted as a wave rather than a particle, the atomic structure of an atom remains unaltered. Atomic fission is the primary cause of neutron emission. At one time, many neutrons can be released.

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Answer:

3.97 x 10²² molecules

Explanation:

To find the moles, you need to use the Ideal Gas Law:

PV = nRT

In this equation,

-----> P = pressure (atm)

-----> V = volume (L)

-----> n = moles

-----> R = Ideal Gas constant (0.08206 atm*L/mol*K)

-----> T = temperature (K)

Before you can plug the values into the equation, you need to (1) convert the pressure from torr to atm (by dividing by 760), then (2) convert the volume from mL to L (by dividing by 1000), and then (3) convert the temperature from Celsius to Kelvin (by adding 273).

P = 800.0 torr / 760 = 1.05 atm               R = 0.08206 atm*L/mol*K

V = 1500 mL / 1000 = 1.5 L                       T = 19.0 °C + 273 = 292 K

n = ? moles

PV = nRT

(1.05 atm)(1.5 L) = n(0.08206 atm*L/mol*K)(292 K)

1.579 = n(23.96)

0.0659 = moles

Now, you need to convert moles to molecules using Avogadro's Number.

Avogadro's Number:

6.022 x 10²³ molecules = 1 mole

 0.0659 moles          6.022 x 10²³ molecules
------------------------  x  -------------------------------------  =  3.97 x 10²² molecules
                                               1 mole

The initial ph of the 0. 10M ch3cooh  (aq) solution is  -0.64.

The salt sodium acetate is present at the equivalency point. It is a salt of a strong base and a weak acid. The expression "its pH" provides the information.

pH = 7 + 0.5p + 0.5logC

where C is the salt concentration. Monobasic acid is Acetic acid here.

25 ml of 0.10 M NaOH solution are needed for the titration of 25 ml of 0.10 M acetic acid.

Total volume = 25 + 25 = 50ml

The salt concentration = C = 0.10×  = 0.05

The pH for weak acid is given by,

pH = [tex]\frac{1}{2}[/tex] (p[tex]K_{a}[/tex] + logC)

pH =  [tex]\frac{1}{2}[/tex] (1.8× + log0.05)

pH =  [tex]\frac{1}{2}[/tex] [1.8× + (-1.30)]

pH =  [tex]\frac{1}{2}[/tex] (1.8× -1.30)

pH = -0.64

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Considering the definition of pH, the hydronium ion concentration [H₃O⁺] of the solution is  1×10⁻³ M.

pH is the Hydrogen Potential. It is a measure of acidity or alkalinity that indicates the amount of hydrogen ions present in a solution or substance.

Mathematically, pH is calculated as the negative base 10 logarithm of the activity of hydrogen ions:

pH= - log [H⁺]= -log [H₃O⁺]

The numerical scale that measures the pH of substances includes the numbers from 0 to 14. The pH value 7 corresponds to neutral substances. Acidic substances are those with a pH lower than 7, while basic substances have a pH higher than 7.

In this case, being pH= 3, you can replace this value in the definition of pH:

3= -log [H₃O⁺]

Solving:

[H₃O⁺]= 10⁻³ M

[H₃O⁺]= 1×10⁻³ M

Finally, the hydronium ion concentration [H₃O⁺] of the solution is  1×10⁻³ M.

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The mass of AgCl precipitated is 0.215g.

Precipitate is to form an insoluble compound either by reacting two salts or by changing the temperature to affect the solubility of the compound.

To calculate mass of AgCl precipitate -

The first step is to make a balanced chemical equation.

2AgNO3 + CaCl2 ---> 2AgCl + Ca(NO3)2

Molecular Weights:

CaCl2 = 110.98 g/mol

AgNO3 =170.01g/mol

AgCl= 143.45 g/mol

Volume:

CaCl2: 30.0mL=0.03L

AgNO3: 15.0mL=0.015 L

mole  CaCl2 = 30/1000 x 0.2 =0.006

mole  AgNO3 =15/1000 x 0.1 = 0.0015

CaCl2 is in excess so 0.0015 mole AgCl  will form

0.0015 x 143.5 = 0.215g

The 0.215 mass in grams of AgCl will precipitate.

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Ksp(the solubility product constant) = [Cu⁺] [I⁻]

So, the Ksp for Cui would be:

Ksp = (2.26 × 10⁻⁶) (2.26 × 10⁻⁶) = 5.11 x 10⁻¹²

Formula used:

K = [tex]K_{sp} = [A^+]^a [B^-]^b[/tex], where

Ksp = solubility product constant

A⁺    = cation in an aquious solution

B⁻ = anion in an aqueous solution

a, b = relative concentrations of a and b

The equilibrium constant for a solid material dissolving in an aqueous solution is the SOLUBILITY PRODUCT CONSTANT, Ksp. It stands for the degree of solute dissolution in solution. A substance's Ksp value increases with how soluble it is.

Take into account the general dissolving response (in aqueous solutions) below:

                           aA(s)⇔cC(aq)+dD(aq)

The molarities or concentrations of the products (cC and dD) must be multiplied in order to find the Ksp. Any product that has a coefficient in front of it must be raised to the power of that coefficient (and also multiply the concentration by that coefficient).

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The equation described by the kb value is 5.21 x [tex]10^-^3[/tex].

The potential of Hydrogen is what pH is formally known as. The negative logarithm of the concentration of H+ ions is known as pH. Thus, the definition of pH as the amount of hydrogen is provided. The hydrogen ion concentration in a solution is described by the pH scale, which also serves as a gauge for the solution's acidity or basicity.

Assuming  PO₄³⁻ (the phosphate anion).

PO₄³- + H₂O ==> HPO₄²⁻ + OH⁻

Kb = [HPO₄²⁻][OH⁻] / [PO₄³⁻]

We can find the [OH⁻] from the pH of 12.70.

pH + pOH = 14

14.0 - 12.7 = 1.3 = pOH

[OH] = 1x[tex]10^-^1^.^3[/tex]

[OH-] = 5.0x[tex]10^-^2[/tex] M

[HPO₄²⁻] = 5.0x[tex]10^-^2[/tex] M

Kb = (5.0x[tex]10^-^2[/tex])2 / 0.48

     = 2.5x[tex]10^-^3[/tex] / 0.48

Kb = 5.21 x [tex]10^-^3[/tex]

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Divide the number of moles of NaOH by the number of liters of NaOH solution needed to reach the titration's endpoint to determine the molarity of the NaOH solution.

"The process of calculating the quantity of a material A by adding measured increments of substance B, the titrant, with which it reacts until exact chemical equivalency is obtained (the equivalence point)" is the definition of titration.

Titration is a method of chemical analysis where the amount of a sample's ingredient is determined by adding an exact known amount of a different substance to the measured sample in which the desired constituent interacts in a specific, known proportion.

The equivalence point, or the point at which chemically equivalent amounts of the reactants have been combined, is to be detected by the titration. The stoichiometry of the reaction determines how many reactants have been combined at the equivalence point.

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Density of ZnS given that the zn-s distance and bond angle are 0. 234 nm and 109. 5o, respectively if the atomic weights of zinc and sulfur are 65. 41 g/mol and 32. 06 g/mol. is 4.1109 g/cm³

For density In the crystal structure, we determine the angle's value for one set of ZnS bonds. θ θ + ∅ + 90° = 180° θ = 90° - ∅ θ = 90° - (109.5° / 2) After determining that = 35.25°, we derive the value of x from the geometry: distance angle d = 0.234 x = dsin = 0.234 sin35.25°) = 0.135 nm = 0.135 107 cm; following this, we get the length of the unit cell: a = 4x a = 4(0.135) a = 0.54 nm = 0.54 107 cm; and The formula for n' is (number of corner atoms in unit ell contribution of each corner atom in unit cell) + (number of face center atoms in unit cell contribution of each face center atom in unit cell). n' = 8 × 1/8) + (6 × 1/2) = 1 + 3 = 4

Density = P = [4( 65.41 + 32.06)] / [ ( 0.54 × 10⁻⁷ )³ × (6.023 × 10²³)] = 389.88 / 94.84 = 4.1109 g/cm³

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