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A biochemist studying breakdown of the insecticide DDT finds that it decomposes by a first-order reaction with a half-life of 12.0 yr. How long does it take DDT in a soil sample to decompose from 809 ppbm to 10.0 ppbm (parts per billion by mass)

Answers

Answer 1

Answer:

Answer:

76.0 years

Explanation:

Step 1: Given data

Half-life (t1/2): 12.0 y

Initial concentration of DDT ([A]₀): 809 ppbm

Final concentration of DDT ([A]): 10.0 ppbm

Time elapsed (t): ?

Step 2: Calculate the rate constant (k)

We will use the following expression.

k = ln 2/ t1/2 = ln 2 / 12.0 y = 0.0578 y⁻¹

Step 3: Calculate the time elapsed

For a first-reaction order, we will use the following equation.

ln [A]/[A]₀ = -k × t

t = ln [A]/[A]₀ / (-k)

t = (ln 10.0 ppbm/809 ppbm) / (-0.0578 y⁻¹)

t = 76.0 y

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A compound has a molecular formular of C12H24O6.What is the compound's empirical formula

Answers

Answer:

We are given the formula of the compound:

C12H24O6

The empirical formula of a molecular formula is the lowest whole number ratio between the number of atoms of each element

The ratio of C to H to O in the given formula is :

12 : 24 : 6

we notice that all 3 of the numbers have 6 in common. Dividing all three of the numbers by 6, we get:

2 : 4 : 1

Hence, the ratio of Carbon to Hydrogen to Oxygen in the empirical formula of the given compound is 2 : 4 : 1 ,

Empirical Formula = C2H4O

Answer:

The empirical formula for C12 H24 O6 is C2 H4 O.

What is the net ionic equation for the reaction that occurs when aqueous solutions of naoh and hno3 are mixed?

Answers

Answer:-

H+ + OH- --> H2O

Explanation:-

The chemical equation is NaOH + HNO3 --> NaNO3 + H2O

Now for the ionic compounds

HNO3 --> H+ + NO3 -

NaOH--> Na+ + OH-

NaNO3 --> Na+ + NO3-

Water being covalent will remain as H2O,

Hence

HNO3 + NaOH--> NaNO3 + H2O

H+ + NO3 - + Na+ + OH- --> Na+ + No3 - + H2O.

Crossing out common terms

H+ + OH- --> H2O

Arrange the following substances in order of increasing solubility of water. C6H14, C6H13Br, C6H13OH, C6H12(OH)2.

Answers

Answer:

C6H14 < C6H13Br < C6H13OH < C6H12(OH)2

Explanation:

Hello,

In this case, since the solubility in water is related with the presence of polar bonds in the given molecules we can see that C6H12(OH)2 has the presence two O-H bonds which promote the highest solubility via hydrogen bonds as well as the C6H13OH but in a lower degree as only on O-H bond is present. Next since the bond C-Br in is slightly close to the polar bond C6H13Br rather than the C-C bonds only had by C6H14 we can infer that C6H13Br is more soluble in water than C6H14, therefore the required order is:

C6H14 < C6H13Br < C6H13OH < C6H12(OH)2

Whereas C6H12(OH)2 is the most soluble and C6H14 the least soluble in water.

Best regards.

Write the balanced equation for the equilibrium reaction for the dissociation ofsilver chloride in water, and write the K expression for this reaction. Then create an ICE chart. Since we know the equilibrium concentration of the silver ion, we can solve for Ksp.Does it agree with the literature value

Answers

Answer:

See explanation

Explanation:

Hello there!

In this case, since the the concentrations are not given, and not even the Ksp, we can solve this problem by setting up the chemical equation, the equilibrium constant expression and the ICE table only:

$AgCl(s)\rightleftharpoons Ag^+(aq)+Cl^-(aq)$

Next, the equilibrium expression according to the produced aqueous species as the solid silver chloride is not involved in there:

$Ksp=[Ag^+][Cl^-]$

And therefore, the ICE table, in which x stands for the molar solubility of the silver chloride:

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ AgCl(s)\rightleftharpoons Ag^+(aq)+Cl^-(aq)$

I - 0 0

C - +x +x

E - x x

Which leads to the following modified equilibrium expression:

$Ksp=x^2$

Unfortunately, values were not given, and they cannot be arbitrarily assigned or assumed.

Regards!

Consider the titration of 30 mL of 0.030 M NH3 with 0.025 M HCl. Calculate the pH after the following volumes of titrant have been added: a) 0 mL; b) 10 mL; c) 20 mL; d)35 mL; e) 36 mL; f) 37 mL.

Answers

The pH of the solution in the titration of 30 mL of 0.030 M NH₃ with 0.025 M HCl, is:

a) pH = 10.86

b) pH = 9.66

c) pH = 9.15

d) pH = 7.70

e) pH = 5.56

f) pH = 3.43

Calculating the pH

a) 0 mL

Initially, the pH of the solution is given by the dissociation of NH₃ in water.

NH₃ + H₂O ⇄ NH₄⁺ + OH⁻ (1)

The constant of the above reaction is:

$Kb = ([NH_(4)^(+)][OH^(-)])/([NH_(3)]) = 1.76\cdot 10^(-5)$ (2)

At the equilibrium, we have:

NH₃ + H₂O ⇄ NH₄⁺ + OH⁻ (3)

0.030 M - x x x

$1.76\cdot 10^(-5)*(0.030 - x) - x^(2) = 0$

After solving for x and taking the positive value:

x = 7.18x10⁻⁴ = [OH⁻]

Now, we can calculate the pH of the solution as follows:

$pH = 14 - pOH = 14 + log(7.18\cdot 10^(-4)) = 10.86$

Hence, the initial pH is 10.86.

b) 10 mL

After the addition of HCl, the following reaction takes place:

NH₃ + HCl ⇄ NH₄⁺ + Cl⁻ (4)

We can calculate the pH of the solution from the equilibrium reaction (3).

$1.76\cdot 10^(-5)(Cb - x) - (Ca + x)*x = 0$ (5)

Finding the number of moles of NH₃ and NH₄⁺

The number of moles of NH₃ (nb) and NH₄⁺ (na) are given by:

$n_(b) = n_(i) - n_(HCl)$ (6)

$n_(b) = 0.030 mol/L*0.030 L - 0.025 mol/L*0.010 L = 6.5\cdot 10^(-4) moles$

$n_(a) = n_(HCl)$ (7)

$n_(a) = 0.025 mol/L*0.010 L = 2.5 \cdot 10^(-4) moles$

Calculating the concentrations of NH₃ and NH₄⁺

The concentrations are given by:

$Cb = (6.5\cdot 10^(-4) moles)/((0.030 L + 0.010 L)) = 0.0163 M$ (8)

$Ca = (2.5 \cdot 10^(-4) mole)/((0.030 L + 0.010 L)) = 6.25 \cdot 10^(-3) M$ (9)

Calculating the pH

After entering the values of Ca and Cb into equation (5) and solving for x, we have:

$1.76\cdot 10^(-5)(0.0163 - x) - (6.25 \cdot 10^(-3) + x)*x = 0$

x = 4.54x10⁻⁵ = [OH⁻]

Then, the pH is:

$pH = 14 + log(4.54\cdot 10^(-5)) = 9.66$

Hence, the pH is 9.66.

c) 20 mL

We can find the pH of the solution from the reaction of equilibrium (3).

Calculating the concentrations of NH₃ and NH₄⁺

The concentrations are (eq 8 and 9):

$Cb = (0.030 mol/L*0.030 L - 0.025 mol/L*0.020 L)/((0.030 L + 0.020 L)) = 8.0\cdot 10^(-3) M$

$Ca = (0.025 mol/L*0.020 L)/((0.030 L + 0.020 L)) = 0.01 M$

Calculating the pH

After solving the equation (5) for x, we have:

$1.76\cdot 10^(-5)(8.0\cdot 10^(-3) - x) - (0.01 + x)*x = 0$

x = 1.40x10⁻⁵ = [OH⁻]

Then, the pH is:

$pH = 14 + log(1.40\cdot 10^(-5)) = 9.15$

So, the pH is 9.15.

d) 35 mL

We can find the pH of the solution from reaction (3).

Calculating the concentrations of NH₃ and NH₄⁺

$Cb = (0.030 mol/L*0.030 L - 0.025 mol/L*0.035 L)/((0.030 L + 0.035 L)) = 3.85\cdot 10^(-4) M$

$Ca = (0.025 mol/L*0.035 L)/((0.030 L + 0.035 L)) = 0.0135 M$

Calculating the pH

After solving the equation (5) for x, we have:

$1.76\cdot 10^(-5)(3.85\cdot 10^(-4) - x) - (0.0135 + x)*x = 0$

x = 5.013x10⁻⁷ = [OH⁻]

Then, the pH is:

$pH = 14 + log(5.013\cdot 10^(-7)) = 7.70$

So, the pH is 7.70.

e) 36 mL

Finding the number of moles of NH₃ and NH₄⁺

$n_(b) = 0.030 mol/L*0.030 L - 0.025 mol/L*0.036 L = 0$

$n_(a) = 0.025 mol/L*0.036 L = 9.0 \cdot 10^(-4) moles$

Since all the NH₃ reacts with the HCl added, the pH of the solution is given by the dissociation reaction of the NH₄⁺ produced in water.

At the equilibrium, we have:

NH₄⁺ + H₂O ⇄ NH₃ + H₃O⁺

Ca - x x x

$Ka = (x^(2))/(Ca - x)$

$Ka(Ca - x) - x^(2) = 0$ (10)

Calculating the acid constant of NH₄⁺

We can find the acid constant as follows:

$Kw = Ka*Kb$

Where Kw is the constant of water = 10⁻¹⁴

$Ka = (1\cdot 10^(-14))/(1.76 \cdot 10^(-5)) = 5.68 \cdot 10^(-10)$

Calculating the pH

The concentration of NH₄⁺ is:

$Ca = (9.0 \cdot 10^(-4) moles)/((0.030 L + 0.036 L)) = 0.0136 M$

After solving the equation (10) for x, we have:

x = 2.78x10⁻⁶ = [H₃O⁺]

Then, the pH is:

$pH = -log(H_(3)O^(+)) = -log(2.78\cdot 10^(-6)) = 5.56$

Hence, the pH is 5.56.

f) 37 mL

Now, the pH is given by the concentration of HCl that remain in solution after reacting with NH₃ (HCl is in excess).

Calculating the concentration of HCl

$C_(HCl) = (0.025 mol/L*0.037 L - 0.030 mol/L*0.030 L)/((0.030 L + 0.037 L)) = 3.73 \cdot 10^(-4) M = [H_(3)O^(+)]$

Calculating the pH

$pH = -log(H_(3)O^(+)) = -log(3.73 \cdot 10^(-4)) = 3.43$

Therefore, the pH is 3.43.

Find more about pH here:

brainly.com/question/491373

I hope it helps you!

Answer:

a)10.87

b)9.66

c)9.15

d)7.71

e) 5.56

f) 3.43

Explanation:

tep 1: Data given

Volume of 0.030 M NH3 solution = 30 mL = 0.030 L

Molarity of the HCl solution = 0.025 M

Step 2: Adding 0 mL of HCl

The reaction: NH3 + H2O ⇔ NH4+ + OH-

The initial concentration:

[NH3] = 0.030M [NH4+] = 0M [OH-] = OM

The concentration at the equilibrium:

[NH3] = 0.030 - XM

[NH4+] = [OH-] = XM

Kb = ([NH4+][OH-])/[NH3]

1.8*10^-5 = x² / 0.030-x

1.8*10^-5 = x² / 0.030

x = 7.35 * 10^-4 = [OH-]

pOH = -log [7.35 * 10^-4]

pOH = 3.13

pH = 14-3.13 = 10.87

Step 3: After adding 10 mL of HCl

The reaction:

NH3 + HCl ⇔ NH4+ + Cl-

NH3 + H3O+ ⇔ NH4+ + H2O

Calculate numbers of moles:

Moles of NH3 = 0.030 M * 0.030 L = 0.0009 moles

Moles HCl = 0.025 M * 0.010 L = 0.00025 moles

Moles NH4+ = 0 moles

Number of moles at the equilibrium:

Moles NH3 = 0.0009 -0.00025 =0.00065 moles

Moles HCl = 0

Moles NH4+ = 0.00025 moles

Concentration at the equilibrium:

[NH3]= 0.00065 moles / 0.040 L = 0.01625M

[NH4+] = 0.00625 M

pOH = pKb + log [NH4+]/[NH3]

pOH = 4.75 + log (0.00625/0.01625)

pOH = 4.34

pH = 9.66

Step 3: Adding 20 mL of HCl

Calculate numbers of moles:

Moles of NH3 = 0.030 M * 0.030 L = 0.0009 moles

Moles HCl = 0.025 M * 0.020 L = 0.00050 moles

Moles NH4+ = 0 moles

Number of moles at the equilibrium:

Moles NH3 = 0.0009 -0.00050 =0.00040 moles

Moles HCl = 0

Moles NH4+ = 0.00050 moles

Concentration at the equilibrium:

[NH3]= 0.00040 moles / 0.050 L = 0.008M

[NH4+] = 0.01 M

pOH = pKb + log [NH4+]/[NH3]

pOH = 4.75 + log (0.01/0.008)

pOH = 4.85

pH = 14 - 4.85 = 9.15

Step 4: Adding 35 mL of HCl

Calculate numbers of moles:

Moles of NH3 = 0.030 M * 0.030 L = 0.0009 moles

Moles HCl = 0.025 M * 0.035 L = 0.000875 moles

Moles NH4+ = 0 moles

Number of moles at the equilibrium:

Moles NH3 = 0.0009 -0.000875 =0.000025 moles

Moles HCl = 0

Moles NH4+ = 0.000875 moles

Concentration at the equilibrium:

[NH3]= 0.000025 moles / 0.065 L = 3.85*10^-4M

[NH4+] = 0.000875 M / 0.065 L = 0.0135 M

pOH = pKb + log [NH4+]/[NH3]

pOH = 4.75 + log (0.0135/3.85*10^-4)

pOH = 6.29

pH = 14 - 6.29 = 7.71

Step 5: adding 36 mL HCl

Calculate numbers of moles:

Moles of NH3 = 0.030 M * 0.030 L = 0.0009 moles

Moles HCl = 0.025 M * 0.036 L = 0.0009 moles

Moles NH4+ = 0 moles

Number of moles at the equilibrium:

Moles NH3 = 0.0009 -0.0009 =0 moles

Moles HCl = 0

Moles NH4+ = 0.0009 moles

[NH4+] = 0.0009 moles / 0.066 L = 0.0136 M

Kw = Ka * Kb

Ka = 10^-14 / 1.8*10^-5

Ka = 5.6 * 10^-10

Ka = [NH3][H3O+] / [NH4+]

Ka =5.6 * 10^-10 = x² / 0.0136

x = 2.76 * 10^-6 = [H3O+]

pH = -log(2.76 * 10^-6)

pH = 5.56

Step 6: Adding 37 mL of HCl

Calculate numbers of moles:

Moles of NH3 = 0.030 M * 0.030 L = 0.0009 moles

Moles HCl = 0.025 M * 0.037 L = 0.000925 moles

Moles NH4+ = 0 moles

Number of moles at the equilibrium:

Moles NH3 = 0.0009 -0.000925 =0 moles

Moles HCl = 0.000025 moles

Concentration of HCl = 0.000025 moles / 0.067 L = 3.73 * 10^-4 M

pH = -log 3.73*10^-4= 3.43

When an electron moves up to higher energy levels, the atom Choose... a photon of light whereas the atom Choose... a photon of light when an electron drops to a lower energy level. The photons emitted from an atom appear as

Answers

Answer:

Explanation:

When an electron moves from a lower energy level to a higher energy level, energy is absorbed by the atom. When an electron moves from a higher to a lower energy level, energy is released and photon is emitted.

this emitted photon is depicted as a small wave-packet being expelled by the atom in a well-defined direction.

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