A turn-of-the-century chemist isolated an aromatic compound of molecular formula C6H4Br2. He carefully nitrated this compound and purified three isomers of formula C6H3Br2NO2. Propose structures for the original compound and the three nitrated derivatives.

Answer:

[tex]C-C> C=C> C\equiv C[/tex]

So, 2 [tex]>[/tex] 1 [tex]>[/tex] 3

Explanation:

Carbon is a  non-metal and has 6 electrons. In the excited state, it contains 4 electrons in its outermost shell and so 4 electrons take place in covalent bond formation. In a molecule , 2 carbon atoms can be bound by a single bond, the double bond and by a triple bond. Since the bond length of [tex]C\equiv C[/tex] is 120 pm, the bond length of [tex]C=C[/tex] is 134 pm and the bond length of [tex]C-C[/tex] is 154 pm. Thus in the given compound, [tex]C= C-C\equiv C[/tex] , in which [tex]C= C[/tex] is labeled as 1,  [tex]C-C[/tex] is labeled as 2 and [tex]C\equiv C[/tex] is labeled as 3. So the bond length of carbon-carbon bonds in the molecule, in decreasing order is [tex]2 > 1 >3[/tex].

Answer: Option (A) is the correct answer.

Explanation:

Chemical equation for the given reaction is as follows.

      [tex]HCOO^{-}(aq) + H^{+}(aq) \rightarrow HCOOH(aq)[/tex]

And, the expression to calculate pH of this reaction is as follows.

      pH = [tex]pk_{a} + log \frac{[HCOO^{-}]}{[HCOOH]}[/tex]

As the concentration of [tex]HCOO^{-}[/tex] is directly proportional to pH. Hence, when there occurs a decrease in the pH of the solution the [tex][HCOO^{-}][/tex] will also decrease.

Thus, we can conclude that the statement, HCOO will accept a proton from HCl to produce more HCOOH and [tex]H_{2}O[/tex], best supports the student's claim.

The correct option is (A) HCOO will accept a proton from HCl to produce more HCOOH and H.

To understand why option (A) best supports the student's claim, let's consider the buffer system and the effect of adding a small amount of HCl.

 A buffer solution consists of a weak acid and its conjugate base, or a weak base and its conjugate acid. In this case, the buffer is composed of formic acid (HCOOH), a weak acid, and its conjugate base, the formate ion (HCOO), along with the potassium salt (HCOOK). The Henderson-Hasselbalch equation relates the pH of the buffer to the concentrations of the weak acid and its conjugate base:

[tex]\[ \text{pH} = \text{pK}_a + \log \left( \frac{[\text{HCOO}^-]}{[\text{HCOOH}]} \right) \][/tex]

Given that the pH of the buffer is 3.75 and the concentrations of HCOOH and HCOO are both 0.100 M, we can calculate the pKa of formic acid:

[tex]\[ 3.75 = \text{pK}_a + \log \left( \frac{0.100}{0.100} \right) \][/tex]

[tex]\[ 3.75 = \text{pK}_a + \log(1) \][/tex]

[tex]\[ 3.75 = \text{pK}_a \][/tex]

 When a small amount of 0.100 M HCl is added to the buffer, the HCl (a strong acid) will dissociate completely into H and Cl ions. The H ions will then react with the formate ions (HCOO) in the buffer, as they are the base component of the buffer system:

[tex]\[ \text{HCOO}^- (aq) + \text{H}^+ (aq) \rightarrow \text{HCOOH} (aq) \[/tex]

This reaction will produce more HCOOH and consume some of the added H ions, thus minimizing the change in pH. The formic acid (HCOOH) will not accept a proton from HCl because it is already protonated; instead, it can only donate a proton, which is not favorable in this case since HCl is a stronger acid.

Option (B) is incorrect because HCOOH cannot accept a proton; it is the acid component of the buffer.

Option (C) is incorrect because HCOO does not donate a proton to HCl; rather, it accepts a proton from HCl.

Option (D) is incorrect because HCOOH does not donate a proton to HCl; HCOOH is a weaker acid than HCl, so it does not donate a proton to HCl's protons.

 Therefore, the student's claim that the pH will decrease by a very small amount upon the addition of a small amount of HCl is supported by the fact that the formate ions (HCOO) will accept protons from HCl to produce more formic acid (HCOOH), thus resisting a large change in pH, which is the defining characteristic of a buffer solution.

Answer:

For a: The standard Gibbs free energy of the reaction is -347.4 kJ

For b: The standard Gibbs free energy of the reaction is 746.91 kJ

Explanation:

Relationship between standard Gibbs free energy and standard electrode potential follows:

[tex]\Delta G^o=-nFE^o_{cell}[/tex]           ............(1)

The given chemical equation follows:

[tex]2Ce^{4+}(aq.)+3I^{-}(aq.)\rightarrow 2Ce^{3+}(aq.)+I_3^-(aq.)[/tex]

Oxidation half reaction:   [tex]Ce^{4+}(aq.)\rightarrow Ce^{3+}(aq.)+e^-[/tex]       ( × 2)

Reduction half reaction:   [tex]3I^_(aq.)+2e^-\rightarrow I_3^-(aq.)[/tex]

We are given:

[tex]n=2\\E^o_{cell}=+1.08V\\F=96500[/tex]

Putting values in equation 1, we get:

[tex]\Delta G^o=-2\times 96500\times (+1.80)=-347,400J=-347.4kJ[/tex]

Hence, the standard Gibbs free energy of the reaction is -347.4 kJ

The given chemical equation follows:

[tex]6Fe^{3+}(aq.)+2Cr^{3+}+7H_2O(l)(aq.)\rightarrow 6Fe^{2+}(aq.)+Cr_2O_7^{2-}(aq.)+14H^+(aq.)[/tex]

Oxidation half reaction:   [tex]Fe^{3+}(aq.)\rightarrow Fe^{2+}(aq.)+e^-[/tex]       ( × 6)

Reduction half reaction:   [tex]2Cr^{2+}(aq.)+7H_2O(l)+6e^-\rightarrow Cr_2O_7^{2-}(aq.)+14H^+(aq.)[/tex]

We are given:

[tex]n=6\\E^o_{cell}=-1.29V\\F=96500[/tex]

Putting values in equation 1, we get:

[tex]\Delta G^o=-6\times 96500\times (-1.29)=746,910J=746.91kJ[/tex]

Hence, the standard Gibbs free energy of the reaction is 746.91 kJ

Final answer:

The standard reaction Gibbs free energy for an electrochemical cell reaction is calculated using the formula ΔG° = -nFE°cell, where n is the number of moles of electrons transferred, F is the Faraday constant, and E°cell is the standard cell potential.

Explanation:

To calculate the standard reaction Gibbs free energy (ΔG°) for an electrochemical cell reaction, you can use the following equation that relates ΔG° to the standard cell potential (E°cell):

ΔG° = -nFE°cell

Where:

n is the number of moles of electrons transferred in the reaction,

F is the Faraday constant (96,485 C/mol e⁻), and

E°cell is the standard cell potential.

To perform the calculations for each reaction:

Identify n, the number of electrons transferred in the balanced reaction.

Use the given E°cell values to calculate ΔG° using the formula above.

It's important to remember that these equations should be applied using standard conditions, which means all solutes are at 1 M concentration, all gases are at 1 atm pressure, and the temperature is 298 K (25 °C).

The question is incomplete, here is the complete question:

A(aq) + B(aq) → 2C(aq)

If the value of Kc for the reaction is 434, what is the concentration of C at equilibrium if initial concentrations of A and B are both 0.500 M. (Hint: Everything is squared after you set-up the equilibrium expression with the values given.

Answer: The concentration of C at equilibrium is 0.912 M

Explanation:

We are given:

Initial concentration of A = 0.500 M

Initial concentration of B = 0.500 M

The given equation follows:

                     [tex]A(aq.)+B(aq.)\rightarrow 2C(aq.)[/tex]

Initial:               0.5   0.5

At eqllm:        0.5-x   0.5-x         2x

The expression of [tex]K_c[/tex] for above equation follows:

[tex]K_c=\frac{[C]^2}{[A][B]}[/tex]

We are given:

[tex]K_c=434[/tex]

Putting values in above equation, we get:

[tex]434=\frac{(2x)^2}{(0.5-x)\times (0.5-x)}\\\\x=0.456,0.553[/tex]

Neglecting the value of x = 0.553 M because the equilibrium concentration of A and B will become negative, which is not possible

So, equilibrium concentration of C = 2x = 2(0.456) = 0.912 M

Hence, the concentration of C at equilibrium is 0.912 M

The question is incomplete, here is the complete question:

A 30.0-mL sample of an unknown strong base is neutralized after the addition of 12.0 mL of a 0.150 M HNO₃ solution. If the unknown base concentration is 0.0300 M, give some possible identities for the unknown base.  (Select all that apply)

A.)  Ca(OH)₂

B.)  LiOH

C.)  Sr(OH)₂

D.)  Al(OH)₃

E.)  NaOH

F.)  Ba(OH)₂

Answer: The unknown base could be [tex]Ca(OH)_2,Sr(OH)_2\text{ or }Ba(OH)_2[/tex]

Explanation:

To calculate the number of moles for given molarity of solution, we use the equation:

       .......(1)

Molarity of solution = 0.150 M

Volume of solution = 12.0 mL

Putting values in equation 1, we get:

[tex]0.150M=\frac{\text{Moles of nitric acid}\times 1000}{12.00}\\\\\text{Moles of nitric acid}=\frac{0.150\times 12.00}{1000}=1.8\times 10^{-3}moles[/tex]

Molarity of solution = 0.0300 M

Volume of solution = 30.0 mL

Putting values in equation 1, we get:

[tex]0.0300M=\frac{\text{Moles of unknown base}\times 1000}{30.00}\\\\\text{Moles of unknown base}=\frac{0.0300\times 30.00}{1000}=0.9\times 10^{-3}moles[/tex]

Mole ratio of acid and base is calculated as: [tex]\frac{\text{Moles of unknown base}}{\text{Moles of nitric acid}}=\frac{0.9\times 10^{-3}}{1.8\times 10^{-3}}=\frac{2}{1}[/tex]

Number of [tex]OH^-[/tex] = 2 × number of [tex]H^+[/tex] ions

So, the unknown base is diprotic in nature.

Hence, the unknown base could be [tex]Ca(OH)_2,Sr(OH)_2\text{ or }Ba(OH)_2[/tex]

The unknown base could be a strong base such as sodium hydroxide (NaOH), potassium hydroxide (KOH), barium hydroxide (Ba(OH)2), or calcium hydroxide (Ca(OH)2), as these compounds release two hydroxide ions per molecule.

To find the possible identities of the unknown strong base, we first need to understand the neutralization process which involves the reaction of an acid and a base to produce water and salts.

The neutralization reaction is given by: HNO3 + BOH -> H2O + BNO3 where B is the cation from the unknown base. From the stoichiometry of the reaction, we can see that 1 mole of HNO3 reacts with 1 mole of the unknown base.

The number of moles of HNO3 = volume (in L) x molarity = 0.012 L x 0.150 mol/L = 0.0018 mol. So, the number of moles of the unknown base would also be 0.0018 mol.

The molarity of the unknown base = number of moles / volume (in L) = 0.0018 mol / 0.030 L = 0.060 M. But the problem states that the concentration of the unknown base is 0.030 M, which means for every molecule of base, there are two hydroxide ions.

Therefore, the unknown base could be a strong base such as sodium hydroxide (NaOH), potassium hydroxide (KOH), barium hydroxide (Ba(OH)2), or calcium hydroxide (Ca(OH)2), etc., which are all examples of sets of strong bases that release two hydroxide ions per molecule.

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Answer: This will result in the increase of concentration of NO and decrease in the concentrations of nitrogen gas and oxygen gas.

Explanation:

The given chemical equation follows:

[tex]N_2(g)+O_2(g)\rightleftharpoons 2NO(g);\Delta H=182.6kJ[/tex]

As, enthalpy of the reaction is positive. So, it is an endothermic reaction.

For an endothermic reaction, heat is getting absorbed during a chemical reaction and is written on the reactant side.

[tex]A+\text{heat}\rightleftharpoons B[/tex]

Any change in the equilibrium is studied on the basis of Le-Chatelier's principle. This principle states that if there is any change in the variables of the reaction, the equilibrium will shift in the direction to minimize the effect.

If the concentration of oxygen gas is increased, the equilibrium will shift in the direction where the concentration of oxygen gas will be decreased. Thus, the reaction will shift in the right direction or forward direction.

Hence, this will result in the increase of concentration of NO and decrease in the concentrations of nitrogen gas and oxygen gas.

Final answer:

Adding more O2 to the equilibrium mixture of N2 and O2 reacting to form NO causes the equilibrium to shift right, increasing NO concentration while decreasing N2 and O2 concentrations.

Explanation:

When more O2 is added to the equilibrium mixture of the reaction N2(g) + O2(g) <=> 2NO(g), the system responds by shifting the equilibrium to the right, according to Le Châtelier's Principle, to counteract the increase in O2 concentration. This results in an increase in the concentration of NO and a decrease in the concentrations of N2 and O2. This shift is a direct consequence of the system's attempt to re-establish equilibrium by consuming the added O2 and producing more NO.

Answer:

We will have 7.30 grams lead(II) oxide

Explanation:

Step 1: Data given

Mass of lead (II)carbonate = 8.75 grams

Molar mass PbCO3 = 267.21 g/mol

Step 2: The balanced equation

PbCO3 (s) ⇆ PbO(s) + CO2(g)

Step 3: Calculate moles PbCO3

Moles PbCO3 = mass / molar mass

Moles PbCO3 = 8.75 grams / 267.21 g/mol

Moles PbCO3 = 0.0327 moles

Step 4: Calculate moles PbO

For 1 mol PbCO3 we'll have 1 mol PbO and 1 mol CO2

For 0.0327 moles PbCO3 we'll have 0.0327 moles PbO

Step 5: Calculate mass PbO

Mass PbO = moles PbO * molar mass PbO

Mass PbO = 0.0327 moles * 223.2 g/mol

Mass PbO = 7.30 grams

We will have 7.30 grams lead(II) oxide

Answer: The time taken is 9764.4 seconds

Explanation:

Rate law expression for first order kinetics is given by the equation:

[tex]k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}[/tex]

where,

k = rate constant  = [tex]1.42\times 10^{-4}s^{-1}[/tex]

t = time taken for decay process = ? sec

[tex][A_o][/tex] = initial amount of the reactant = 100 grams

[A] = amount left after decay process =  25 grams

Putting values in above equation, we get:

[tex]1.42\times 10^{-4}=\frac{2.303}{t}\log\frac{100}{25}\\\\t=9764.4s[/tex]

Hence, the time taken is 9764.4 seconds

Answer:

The answer is: False.

Explanation:

Six Sigma is a methodology for continuous improvement that increases performance by controlling variability and focusing the process on customer specifications. It will be used within other methods such as Rummler: Brache, Scrum and TQM.

JumpStart is an educational media franchise for children, consisting mainly of educational games, produced by JumpStart Games.

The answer is: False.

Jumpstart is a fast-paced method that can be used within other methods such as Rummler-Brache, Scrum, and TQM to identify problems and solutions in a single session.

The statement is true. Jumpstart is a fast-paced method that can be used within other methods such as Rummler-Brache, Scrum, and TQM to identify problems and solutions in a single session. Jumpstart involves bringing together a diverse group of stakeholders who have knowledge and expertise relevant to the problem at hand. The session is structured, time-boxed, and facilitated to engage participants in generating ideas, analyzing data, and developing potential solutions.

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Answer: the more effective is using three 0.5-mL extractions with ether

Explanation:

Using 1.5 ml of ether

Q = [Vaq/(K×Vorg +Vaq)]^n

Q= remaining fraction

Vaq= volume of aqueous phase

Vorg= volume of organic phase

K= distribution coefficient

Using 1.5 ml ether once

Vaq= 1ml, K= 10, n=1

Q = [Vaq/(K×Vorg +Vaq)]^n

Q= [1/10×1.5+1)]^1 = 0.0625

Using 0.5ml ether 3 times

Q = [1/(10×0.5 +1)]^3= 0.0046

Hence using 0.5ml of ether 3 times is more effective

Answer : The fraction of the molecules in the neutral (imidazole) form are, 0.799

Explanation : Given,

pH = 6.6

[tex]p_{K_a}=6.0[/tex]

Using Henderson Hesselbach equation :

[tex]pH=pK_a+\log \frac{[Salt]}{[Acid]}[/tex]

[tex]pH=pK_a+\log \frac{[\text{Imidazole}]}{[\text{Imidazolium ion}]}[/tex]

Now put all the given values in this expression, we get:

[tex]6.6=6.0+\log \frac{[\text{Imidazole}]}{[\text{Imidazolium ion}]}[/tex]

[tex]\frac{[\text{Imidazole}]}{[\text{Imidazolium ion}]}=10^{6.6-6.0}[/tex]

[tex]\frac{[\text{Imidazole}]}{[\text{Imidazolium ion}]}=10^{0.6}[/tex]

[tex]\frac{[\text{Imidazole}]}{[\text{Imidazolium ion}]}=3.98[/tex]

[tex][\text{Imidazole}]=3.98[\text{Imidazolium ion}][/tex]     ...........(1)

Now we have to determine the fraction of the molecules are in the neutral (imidazole) form.

Fraction of neutral imidazole = [tex]\frac{[\text{Imidazole}]}{[\text{Imidazole}]+[\text{Imidazolium ion}]}[/tex]

Now put the expression 1 in this expression, we get:

Fraction of neutral imidazole = [tex]\frac{3.98[\text{Imidazolium ion}]}{3.98[\text{Imidazolium ion}]+[\text{Imidazolium ion}]}[/tex]

Fraction of neutral imidazole = [tex]\frac{3.98[\text{Imidazolium ion}]}{4.98[\text{Imidazolium ion}]}[/tex]

Fraction of neutral imidazole = [tex]\frac{3.98}{4.98}[/tex]

Fraction of neutral imidazole = 0.799

Thus, the fraction of the molecules in the neutral (imidazole) form are, 0.799

Final answer:

To find the fraction of imidazole molecules in their neutral form in a 0.10 M solution at pH 6.6, given a pKa of 6.0, we use the Henderson-Hasselbalch equation. The calculation shows that for every 4 molecules of neutral imidazole, there's about 1 deprotonated molecule, demonstrating a significant proportion of neutral molecules in the solution.

Explanation:

The question asks for the fraction of imidazole molecules in the neutral form when a 0.10 M solution of imidazole has a pH of 6.6, given that the pKa of the imidazolium ion is 6.0. This involves understanding the dissociation of acids in solution and applying the Henderson-Hasselbalch equation. The equation is: pH = pKa + log([A-]/[HA]), where [A-] is the concentration of the deprotonated form and [HA] is the concentration of the protonated (neutral) form. Solving for the ratio [HA]/[A-] gives us the fraction of molecules in the neutral form.

In this scenario, rearranging the Henderson-Hasselbalch equation and substituting the given values (pH = 6.6 and pKa = 6.0) allows us to solve for the fraction of neutral imidazole molecules: 6.6 = 6.0 + log([HA]/[A-]). This results in a log([HA]/[A-]) of 0.6, which corresponds to a ratio of [HA]/[A-] equal to about 4. Thus, for every 4 molecules of neutral imidazole, there's approximately 1 deprotonated molecule, indicating a large fraction of the molecules remain in the neutral form at pH 6.6.

Answer:

0.04M

Explanation:

The following were obtained from the question:

C1 = 0.8M

V1 = 5mL

V2 = 100mL

C2 =?

The final concentration of the diluted solution can be obtained by doing the following:

C1V1 = C2V2

0.8 x 5 = C2 x 100

Divide both side by 100

C2 = (0.8 x 5) /100

C2 = 0.04M

The concentration of the diluted solution is 0.04M

Answer:

The new concentration of [tex]PCl_5[/tex] will be 0.9 M.

Explanation:

[tex]PCl_5(g)\rightleftharpoons PCl_3(g) + Cl_2(g)[/tex]

The equilibrium concentration of all species:

[tex][PCl_5]=0.40 M[/tex]

[tex][PCl_3]=[Cl_2]=0.20 M[/tex]

The equilibrium constant's expression can be written as:

[tex]K_c=\frac{[PCl_3][Cl_2]}{[PCl_5]}[/tex]

[tex]K_c=\frac{0.20 M\times 0.20 M}{0.40 M}=0.1[/tex]

On halving the volume of the container, the concentration will get doubled;

[tex]PCl_5(g)\rightleftharpoons PCl_3(g) + Cl_2(g)[/tex]

0.80 M                                         0.40M   0.40 M

[tex]Q_c=\frac{0.40 M\times 0.40 M}{0.80 M}=0.2[/tex]

[tex]Q_c>K_c[/tex]

Reaction will go backward.

[tex]PCl_5(g)\rightleftharpoons PCl_3(g) + Cl_2(g)[/tex]

Initially

0.80 M                                         0.40M   0.40 M

After reestablishment of an  equilibrium

(0.80+x) M                                         (0.40-x)M   (0.40-x) M

So, the equilibrium expression for above reaction can be written as :

[tex]K_c=\frac{(0.40-x) M\times (0.40-x) M}{(0.80+x) M}[/tex]

[tex]0.1=\frac{(0.40-x) M\times (0.40-x) M}{(0.80+x) M}[/tex]

x = 0.1 M

The new concentration of [tex]PCl_5[/tex] will be:

[tex][PCl_5]=(0.80+x) M=(0.80+0.1) M = 0.90[/tex]

Answer: Option (b) is the correct answer.

Explanation:

Chemical properties are defined as the properties which tend to chemical composition of a substance. Neutral elements with similar number of valence electrons tend to show similar chemical properties as these have same reactivity which actually affects their chemical properties.

This means that elements of the same group tend to show similar chemical properties.

For example, oxygen and sulfur atom are both group 16 elements and they have 6 valence electrons. Therefore, they show similar chemical properties.

Thus, we can conclude that a pair of O and S elements is expected to be most similar in their chemical properties.

Final answer:

The Haber process is a reaction between N2 and H2 to form NH3. The disappearance rate of H2 is three times that of N2 (Disappearance rate of H2 = 3 (Disappearance rate of N2)), and the reaction rate can be described as -Δ[N2]/Δt. The activation energy for this reaction must be positive.

Explanation:

The question relates to the Haber process, a reaction between nitrogen (N2) and hydrogen (H2) to form ammonia (NH3). Analyzing the given chemical equation N2 + 3H2 → 2NH3, we can infer several things without any experiments:

That duration is 11.775 seconds.

The area of an alveolus, represented as a sphere with a diameter of 0.1 mm, is A = 4.π.r². For an alveolus, r would be: R = 5.10^(-5)m or r = 0.00005m

Locating the region:

4.3.14.(5.10^(-5)) = AA equals 3.14.10^(-8)m³.

Since 90% of the final concentration is to be changed, c = 0.9.3.14.10^(-8)

c = 28.26.10^(-9)

With an oxygen diffusivity of 2.4.10^(-9)m²/s, one alveolus may spread 2.4.10^(-9) oxygen molecules in a second. Thus:

= 2.4.10^(-9)/secondm²t seconds equals (-9)±28.26.10^(28.26.10^(-9) )/(2.4.10^(-9) ) = m² t

t is 11.775 seconds.

It will take 11.775 seconds for the concentration change at the centre to reach 90%.

The radioisotope will take 219 days to decay from 28 g to 0.875 g.

Explanation:

Any radioactive isotope is tend to decay with time. So the rate of decay of the radioactive isotopes is termed as disintegration constant. Since, the initial mass of the radioactive isotope is given along with the reducing mass. In order to determine the time required to reduce the mass of the radioisotope from 28 g to 0.875 g, first the disintegration constant is need to be determined. The disintegration constant can be obtained from half life time of the isotope. As half life time is the measure of time required to reduce half of the concentration of the isotope.

Half life time = 0.6932/disintegration constant

44 = 0.6932/λ

λ = 0.6932/44=0.0158

So, with this values of disintegration constant, initial mass and final mass, the time required to reduce from initial to final mass can be obtained using law of disintegration constant as follows.

N = Noe^(-λt)

[tex]t= -\frac{1}{disintegration constant}* ln(\frac{N}{N_{0} } )\\ \\t = -\frac{1}{0.0158}*ln(\frac{0.875}{28})\\ \\t = -\frac{1}{0.0158}*ln(0.03125)\\\\t = -63.29*(-3.466)\\\\t=219 days.[/tex]

Thus, the radioisotope will take 219 days to decay from 28 g to 0.875 g.

It would take 220 days for 28.0 g sample of iron‑59 to decay to 0.875 g

The half life of a substance is the amount of time it takes to decay to half of the initial value. It is given by:

[tex]N(t)=N_o(\frac{1}{2}) ^\frac{t}{t_\frac{1}{2} } \\\\N(t)=amount\ after\ t\ years, N_o=original\ value,t_\frac{1}{2} =half\ life\\\\Given\ that:\\\\N(t)=0.875g,N_o=28g,t_\frac{1}{2} =44\ days. hence:\\\\\\0.875=28(\frac{1}{2} )^\frac{t}{44} \\\\t=220\ days[/tex]

Hence it would take 220 days for 28.0 g sample of iron‑59 to decay to 0.875 g

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Answer:

a.  pNO₂ = 1 atm    pN₂O₄ = 4 atm

b.  pNO₂ = 1 atm    pN₂O₄ = 4 atm

c. It does not matter.

Explanation:

From the information given in this question we know the equilibrium involved is

N₂O₄ (g) ⇄ 2 NO₂ (g)

with Kp  given by

Kp = p NO₂²/ p N₂O₄ = 0.25

We know that if we place 4.5 atm of N₂O₄ is placed in a flask, a quantity x is going to be consumed producing 2x atm of NO₂  and we can setup the following equation:

0.25 =  p NO₂²/ p N₂O₄ =  (2x)² / (4.5 - x)

0.25 x  (4.5 - x) = 4x²

4x² + 0.25 x - 1.125 = 0

after solving this quadratic equation, we get two roots

x₁ = 0.5

x₂ = -0.56

the second root is physically impossible, and the partial pressures for x₁ = 0.5  will be

pNO₂ = 2 x 0.5 atm = 1.0 atm

pN₂O₄ = (4.5 - 0.5) atm = 4.0 atm

Similarly for part b, we get the equilibrium equation

0.25 = (9- 2x)² / x

0.25x = 81 - 36x + 4x²

the roots of this equation are:

x₁ = 5.0625

x₂ = 4

the first root is physically impossible since it will give us a negative partial pressure of N₂O₄ :

p N₂O₄ = 9 - 2(5.0625) = -1.13

the second root give us the following partial pressures:

p N₂O₄ = (9 - 2x4) atm = 1 atm

p NO₂ = 4 atm

The partial pressures are the same, it does not matter from which direction an equilibrium position is reached since what is essential is that the partial pressures of the gasses N₂O₄ and  NO₂ obey the equilibrium equation.

Answer: Partial pressure of He that would give a solubility of 0.730 M is 15.5 atm

Explanation:

Henry's law states that the amount of gas dissolved or molar solubility of gas is directly proportional to the partial pressure of the liquid.

To calculate the molar solubility, we use the equation given by Henry's law, which is:

[tex]C_{He}=K_H\times p_{liquid}[/tex]

where,

[tex]K_H[/tex] = Henry's constant =?

[tex]p_{He}[/tex] = partial pressure = 1.7 atm

Putting values in above equation, we get:

[tex]0.080=K_H\times 1.7atm\\\\K_H=0.047Matm^{-1}[/tex]

To find partial pressure of He would give a solubility of 0.730 M

[tex]0.730=0.047Matm^{-1}\times p_{liquid}[/tex]

[tex]p_{liquid}=15.5atm[/tex]

Thus partial pressure of He that would give a solubility of 0.730 M is 15.5 atm

We have that from the Question, it can be said that   The partial pressure of He would give a solubility of 0.730 M is

P_2=4.7atm

From the Question we are told

At a particular temperature, the solubility of He in water is 0.080 M when the partial pressure is 1.7 atm. What partial pressure of He would give a solubility of 0.730 M

Generally the equation for constant temperature  is mathematically given as

[tex]\frac{C_2}{C_1}=\frac{P_2}{P_1}\\\\Therefore\\\\P_2=\frac{P_1C_1}{C_1}\\\\P_2=\frac{0.22*1.7}{0.080}\\\\P_2=4.7atm\\\\[/tex]

Therefore

The partial pressure of He would give a solubility of 0.730 M is

P_2=4.7atm

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To calculate the number of grams of Mg needed for this reaction, use the equation q = mcΔT. By rearranging the equation and substituting the values, you can determine the energy released by the reaction, and then convert it to grams of Mg using the molar mass.

To calculate the number of grams of Mg needed for this reaction, we need to use the equation q = mcΔT. We know the initial and final temperatures of the water, as well as the mass and specific heat capacity of water. By rearranging the equation and substituting the values, we can calculate the energy released by the reaction, and then use the molar mass of Mg to determine the number of grams needed.

Let's go through the steps:

After performing these calculations, the number of grams of Mg needed for this reaction to release enough energy to increase the temperature of the water from 29 °C to 78 °C is approximately 0.24 g.

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To increase the temperature of 78 mL of water from 29°C to 78°C, 1.36 grams of Magnesium is required given the specific heat capacity of water and the heat of combustion of Magnesium.

Firstly, we need to calculate the amount of energy required to heat water from 29°C to 78°C. The specific heat capacity of water is 4.184 J/g°C, and the volume of water is 78 mL, which is equivalent to 78 grams (1 ml of water = 1 gram). To find the energy required, we use the formula Q = m × c × ∆T, where Q is the heat energy required, m is the mass, c is the specific heat capacity, and ∆T is the change in temperature. So,

Q = (78g) × (4.184 J/g°C) × (78°C - 29°C) = 15950.688 Joules

This is the amount of energy we need. Now we need to know how much Mg is needed to produce this energy. Given that Mg reacts to release 11.7 kJ (or 11700 Joules) per gram, we can set up the following equation to solve for mass:

Energy = mass × heat of combustion

Therefore, mass = Energy/Heat of combustion = 15950.688 J / 11700 J/g = 1.36 g

Thus, 1.36 grams of Mg are needed to increase the temperature of 78 mL of water from 29°C to 78°C.

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Answer:

Good buffer systems are:  

Option B)  Potassium acetate (KCH3COO) + acetic acid (CH3COOH).

Option C: Hydrofluoric acid (HF) + sodium fluoride (NaF)

Option D: Hypochlorous acid (HClO) + potassium hypochlorite (KClO):

Option E: Ammonium bromide (NH4Br) + ammonia (NH3)

Explanation:

Buffers consist in a mixture of a weak acid with its salt or a weak alkaly with its salt. All buffer systems are conformed by:

1) Weak acid + salt formed with conjugated base of the acid

or

2) Weak alkaly + salt formed with conjugated acid of the alkaly

It is very important these salts come from the weak acid or weak alkaly. It means, the anion of the acid must be the anion in the salt which is going to be part of the buffer system. On the other hand, the cation of the weak alkaly must be the cation of the salt which is going to form the salt in the buffer system.

Then, when we evaluate all options in this exercise, answers are the following:

Option A) Perchloric acid (HClO4)  + potassium perchlorate (KClO4).

It is not a buffer system because HClO4 is a strong acid. A buffer requires a weak acid or weak alkaly.  KClO4 is a salt formed by the conjugated base of HClO4, but a buffer requires two condition:  Weak acid or alkaly + its salt.

Option B)  Potassium acetate (KCH3COO) + acetic acid (CH3COOH).

This mixture is a buffer because it is formed by a weak acid (acetic acid)  and its salt (KCH3COO is a salt coming from weak acid ---CH3COOH---).

Buffer component reactions:

Reaction weak acid:     CH3COOH + H2O <-----> H3O+ + CH3COO-

Reaction salt in water:  KCH3COO ---> K+ + CH3COO-

CH3COO- is the anion of the weak acid so it must be part of the salt in the buffer system. Then KCH3COO is a salt from CH3COOH.

Option C: Hydrofluoric acid (HF) + sodium fluoride (NaF)

This mixture is a buffer because it is formed by a weak acid (Hydrofluoric acid)  and its salt (NaF is a salt coming from weak acid ---HF---).

Buffer component reactions:

Reaction weak acid:     HF + H2O <-----> H3O+ + F-

Reaction salt in water:  NaF ---> Na+ + F-

F- is the anion of the weak acid so it must be part of the salt in the buffer system. Then NaF is a salt from HF.

Option D: Hypochlorous acid (HClO) + potassium hypochlorite (KClO):

It is a buffer because it is formed by a weak acid (hypochlorous acid)  and its salt (KClO is a salt coming from weak acid ---HClO---).

Buffer component reactions:

Reaction weak acid:     HClO + H2O <-----> H3O+ + ClO-

Reaction salt in water:  KClO ---> K+ + ClO-

ClO- is the anion of the weak acid so it must be part of the salt in the buffer system. Then KClO is a salt from HClO.

Option E: Ammonium bromide (NH4Br) + ammonia (NH3)

The combination of this weak alkaly (NH3) and the salt (ammonium bromide, NH4Br) is a buffer becuase it is formed by a weak compound and its salt.

Buffer component reactions:

Reaction weak alkaly:    NH3 + H2O <-----> NH4+ + OH-  

Reaction salt in water:    NH4Br ---> NH4+ + OH-

NH4+ is the cation of the weak alkaly so it must be part of the salt in the buffer system. Then NH4Br is a salt from NH3.

Remember a buffer is formed by the combination of two different chemical sustances:  Weak acid or Weak alkaly plus its salt.

Final answer:

Good buffer systems consist of a weak acid and its conjugate base or a weak base and its conjugate acid, capable of resisting changes in pH when acids or bases are added. Examples include mixtures of acetic acid with potassium acetate, hydrofluoric acid with sodium fluoride, hypochlorous acid with potassium hypochlorite, and ammonium bromide with ammonia.

Explanation:

Good buffer systems are made of a combination of a weak acid and its conjugate base, or a weak base and its conjugate acid. From the given options, a buffer system can be represented as follows:

0.16 M potassium acetate + 0.26 M acetic acid

0.18 M hydrofluoric acid + 0.12 M sodium fluoride

0.31 M hypochlorous acid + 0.28 M potassium hypochlorite

0.35 M ammonium bromide + 0.32 M ammonia

Each of these systems includes a weak acid or base along with its salt, which is made up of the conjugate base or acid. This combination allows the buffer to resist changes in pH when small amounts of acids or bases are added. For example, when considering acetic acid and sodium acetate as components of the buffer, the addition of a strong acid like HCl will react mainly with the acetate ions rather than causing a large change in pH.

The question is incomplete. So, the complete question is:

For each of the following situations, state the probability rule or rules that you would use and apply it or them. Write a sentence explaining how the situation illustartes the use of the probability rules. (a) The Probability of event A is 0.417. What is the probability that even A does not occur? (b) A coin is tossed 4 times. The probability of zero heads is 1/16 and the probability of zero tails is 1/16. What's the probability that all four tosses result in the same outcomes? (c) Refer to part b, what's the probability that there is at least one head and at least one tail? (d) The probability of event A is 0.4 and the probability of event B is 0.8. Events A and B are disjoint. Can this happen? (e) Event A is rare. Its probability is -0.04. Can this happen?

Answer and Explanation:

(a) P(A)=0.417

Since it's wanted to known the probability of not occuring,

P(A') = 1 - P(A)

P(A') = 1 - 0.417

P(A') = 0.583

The probability of event A not occuring is 0.583.

(b) If the coin is fair, the probability of either head or tail is 1/2. Since, there were 4 tosses, the probability for total outcomes would be 1/16.

For all heads: P(H) = 1/16;

For all tail: P(T) = 1/16:

As it wants "either all heads OR all tails", it will be

P = P(H)+P(T)

P = 1/16+1/16

P = 1/8 = 0.125

The probability of all heads or all tails is 0.125.

(c) The probability of at least one head or one tail is the probability of NOT being all heads and all tails, so:

1 - 0.125 = 0.875

The probability of at least one of each is 0.875.

(d) Yes, events A and B can be disjoint if their intersection is zero:

P(A∩B) = 0 and it also means the events are mutually exclusive.

(e) No, probability can not be negative, because an event has to occur to calculate probability and a set of numbers fo the event is positive.

Final answer:

To calculate the probability of event A not occurring when P(A) = 0.417, the complement rule is applied: P(~A) = 1 - P(A), yielding P(~A) = 0.583. This exemplifies the complement rule where the probabilities of an event and its complement must sum to 1.

Explanation:

The probability that event A occurs is given as P(A) = 0.417. To find the probability that event A does not occur, we use the complement rule, which is based on the principle that the sum of the probabilities of all possible outcomes in a sample space is 1. Therefore, P(~A) = 1 - P(A).

By applying this rule, we can calculate the probability that event A does not occur as follows:

This situation illustrates the use of the complement rule because event A and event 'not A' are mutually exclusive and collectively exhaustive; thus, they must sum up to a probability of 1. The complement rule is particularly useful when it is easier to calculate the probability of an event not happening than the event happening.

Explanation:

According to the law of conservation of mass, mass can neither be created nor it can be destroyed. But it can be simply transformed from one form to another.

Therefore, when [tex]AgNO_{3}[/tex] is added to NaCl then the compound formed will have same mass as that of reactants.

       [tex]AgNO_{3} + NaCl \rightarrow AgCl + NaNO_{3}[/tex]

Total mass of reactants is (169.87 + 58.44) g/mol = 228.31 g/mol

Total mass of products is (143.32 + 84.99) g/mol = 228.31 g/mol

Thus, we can conclude that mass of the new mixture will stay the same.

Answer:

The overall equilibrium constant K in terms of the equilibrium constants [tex]K_1\& K_2[/tex]:

[tex]K=K_1\times K_2[/tex]

Explanation:

[tex]CH_4(g) H_2O(g)\rightleftharpoons CO(g)+3H_2(g)[/tex]

Equilibrium constant of reaction :

[tex]K_1=\frac{[CO][H_2]^3}{[CH_4][H_2O]}[/tex]

[tex]CO(g) +H_2O(g)\rightleftharpoons CO_2(g) + H_2(g)[/tex]

Equilibrium constant of reaction :

[tex]K_2=\frac{[CO_2][H_2]}{[CO][H_2O]}[/tex]

The net reaction is:

[tex] CH_4(g)+2H_2O(g)\rightleftharpoons CO_2(g)+4H_2(g) [/tex]

Equilibrium constant of reaction :

[tex]K=\frac{[CO_2][H_2]^4}{[CH_4][H_2O]^2}[/tex]

[tex]K=\frac{[CO_2][H_2]^4}{[CH_4][H_2O]^2}\times \frac{[CO]}{[CO]}[/tex]

Rearranging the equation :

[tex]K=\frac{[CO][H_2]^3}{[CH_4][H_2O]}\times \frac{[CO_2][H_2]}{[CO][H_2O]}[/tex]

[tex]K=K_1\times K_2[/tex]

The overall equilibrium constant K in terms of the equilibrium constants [tex]K_1\& K_2[/tex]:

[tex]K=K_1\times K_2[/tex]

Answer:

The other geometry is square planar and this is because the four coordinate complexes forms d⁸ configuration.

Explanation:

(1) The other geometry is square planar

(2) This absorption spectrum (square planar) strongly suggests a tetrahedral geometry because  in the molecular geometry, a central atom is surrounded by constituent atoms, which form the corners of a square on the same plane.

This square planar is as a results of d⁸ electronic configuration and this is common for transition metal complexes with d⁸ configuration.

Four-coordinate nickel complexes can have either a tetrahedral or square planar geometry. The absorption spectrum suggests a tetrahedral geometry because it allows for more transitions and therefore a more intense color. This is due to the less energy splitting of d orbitals in the tetrahedral geometry.

Four-coordinate nickel complexes can adopt either a tetrahedral or a square planar geometry. These different geometries result from the different spatial arrangements of the ligands surrounding the nickel atom.

In a tetrahedral geometry, each of the ligand pairs forms an angle of 109.5°, such as in the complex [Zn(CN)4]²¯. In contrast, square planar complexes, like [Pt(NH3)2Cl2], have the ligands oriented at right angles to each other.

From your absorption spectrum, the presence of a tetrahedral geometry is suggested as opposed to the square planar one due to the allowedness of the transitions. Transition metals with a tetrahedral complex usually allow for more transitions and therefore more bands in the absorption spectrum.

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Answer:

(a) Barium is produced at the negative electrode

(b) Iodine is produced at the positive electrode

Explanation:

When  an electric current is passed through a solution containing electrolyte, a non spontaneous reaction is stimulated. This results in the flow of positively charged ions to negatively charged electrodes(cathode) and negatively charged ions to positively charged electrodes(anode)

When an electric current is passed through molten [tex]BaI_{2}[/tex] in the electrolytic cell, the following reactions takes place:

                                            [tex]BaI_{2}[/tex] → [tex]Ba^{2+}[/tex] + 2[tex]I^{-}[/tex]

At the anode;

Iodine ions will lose an electron and will be oxidized to iodine

[tex]2I^{-}[/tex] → [tex]I_{2}[/tex] + [tex]e^{-}[/tex]

At  the cathode;

Barium ions gains electrons and its reduced to barium metal

[tex]Ba^{2+}[/tex] + [tex]2e^{-}[/tex] → Ba

In the electrolysis of molten BaI2, Barium forms at the negative electrode, and Iodine forms at the positive electrode.

In the electrolysis of molten BaI2, different products form at the negative and positive electrodes. At the negative electrode, also known as the cathode, reduction occurs and Barium (Ba) is formed as a result of Ba2+ ions gaining electrons. On the other hand, at the positive electrode, known as the anode, oxidation happens and Iodine (I2) is formed as a result of I- ions losing electrons.

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Answer: The mass of water required is 1.79 grams

Explanation:

To calculate the amount of oxygen gas, we use the equation given by ideal gas which follows:

[tex]PV=nRT[/tex]

where,

P = pressure of the gas = 0.926 atm

V = Volume of the gas = 1.3 L

T = Temperature of the gas = 295 K

R = Gas constant = [tex]0.0821\text{ L. atm }mol^{-1}K^{-1}[/tex]

n = number of moles of oxygen gas = ?

Putting values in above equation, we get:

[tex]0.926atm\times 1.3L=n\times 0.0821\text{ L. atm }mol^{-1}K^{-1}\times 295K\\\\n=\frac{0.926\times 1.3}{0.0821\times 295}=0.0497mol[/tex]

For the given chemical equation:

[tex]2H_2O(l)\rightarrow 2H_2(g)+O_2(g)[/tex]

By Stoichiometry of the reaction:

1 mole of oxygen gas is produced from 2 moles of water

So, 0.0497 moles of oxygen gas will be produced by = [tex]\frac{2}[1}\times 0.0497=0.0994mol[/tex] of water

To calculate the number of moles, we use the equation:

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

Moles of water = 0.0994 moles

Molar mass of water = 18 g/mol

Putting values in above equation, we get:

[tex]0.0994mol=\frac{\text{Mass of water}}{18g/mol}\\\\\text{Mass of water}=(0.0994mol\times 18g/mol)=1.79g[/tex]

Hence, the mass of water required is 1.79 grams

The mass of H2O required to produce 1.3 L of O2 at 295 K and 0.926 atm can be calculated using the ideal gas law. The volume of O2 is converted to moles using the gas law, which is then doubled to account for the reaction stoichiometry. The result is multiplied by the molar mass of water to find the required mass.

To find the mass of H2O required to form 1.3 L of O2 at a temperature of 295 K and a pressure of 0.926 atm, we can use the ideal gas law, PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.

First, calculate the moles of O2 gas using the ideal gas law:

To solve the problem, the calculation steps should be performed with the respective numerical values, which were not provided here because this is only a guideline for how to approach the problem.

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2.38×10^-3

Explanation:

from the question,the we calculate the latent heat of vaporization with the difference in temperature being put into consideration

The possible structures for C4H8O with an alcohol group and no carbonyl groups, as indicated by the given IR spectrum criteria, include butan-1-ol, butan-2-ol, 2-methylpropan-1-ol, and 2-methylpropan-2-ol.

The IR spectrum criteria given indicate the presence of an alcohol group due to the broad signal between 3200 and 3600 cm⁻¹, which signifies O-H stretching. The absence of signals between 1600 and 1850 cm⁻¹ suggests that there are no carbonyl (C=O) groups in the molecule. Considering these constraints, the molecules of C4H8O that fit these characteristics are butan-1-ol, butan-2-ol, 2-methylpropan-1-ol, and 2-methylpropan-2-ol. These structures illustrate the versatility of alcohol functionalization within a four-carbon skeleton without the presence of carbonyl groups, thus fitting the IR spectrum restrictions provided.

The question is incomplete, here is the complete question:

A chemist fills a reaction vessel with 9.47 atm nitrogen monoxide gas, 2.61 atm chlorine gas, and 8.64 atm nitrosyl chloride gas at a temperature of 25°C. Under these conditions, calculate the reaction free energy for the following chemical reaction:

[tex]2NO(g)+Cl_2(g)\rightleftharpoons 2NOCl(g)[/tex]

Use the thermodynamic information in the ALEKS Data tab. Round your answer to the nearest kilojoule.

Answer: The Gibbs free energy change of the reaction is -44.0 kJ

Explanation:

The equation used to calculate standard Gibbs free change of a reaction is:

[tex]\Delta G^o_{rxn}=\sum [n\times \Delta G^o_f_{(product)}]-\sum [n\times \Delta G^o_f_{(reactant)}][/tex]

For the given chemical reaction:

[tex]2NO(g)+Cl_2(g)\rightarrow 2NOCl(g)[/tex]

[tex]\Delta G^o_{rxn}=[(2\times \Delta G^o_f_{(NOCl(g))})]-[(2\times \Delta G^o_f_{(NO(g))})+(2\times \Delta G^o_f_{(Cl_2)})][/tex]

We are given:

[tex]\Delta G^o_f_{(NOCl(g))}=66.08kJ/mol\\\Delta G^o_f_{(Cl_2(g))}=0kJ/mol\\\Delta G^o_f_{(NO(g))}=86.55kJ/mol[/tex]

Putting values in above equation, we get:

[tex]\Delta G^o_{rxn}=[(2\times (66.08))]-[(2\times (86.55))+(1\times (0))]\\\\\Delta G^o_{rxn}=-40.94kJ[/tex]

[tex]Q_p=\frac{(p_{NOCl})^2}{(p_{NO})^2\times p_{Cl_2}}[/tex]

We are given:

[tex]p_{NOCl}=8.64atm\\p_{NO}=9.47atm\\p_{Cl_2}=2.61atm[/tex]

Putting values in above expression, we get:

[tex]Q_p=\frac{(8.64)^2}{(9.47)^2\times 2.61}=0.319[/tex]

To calculate the Gibbs free energy change of the reaction, we use the equation:

[tex]\Delta G=\Delta G^o+RT\ln Q_p[/tex]

where,

[tex]\Delta G^o[/tex] = standard Gibbs free energy change = -40.94 kJ

R = Gas constant = 8.314 J/mol.K = 0.008314 kJ/mol.K

T = temperature = [tex]25^oC=[25+273]K=298K[/tex]

[tex]Q_p[/tex] = reaction coefficient = 0.319

Putting values in above equation, we get:

[tex]\Delta G=-40.94+(0.008314J/mol.K\times 298K\times \ln (0.319)\\\\\Delta G=-43.77kJ[/tex]

Hence, the Gibbs free energy change of the reaction is -44.0 kJ