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zloy xaker [14]
3 years ago
5

The following pairs of soluble solutions can be mixed. In some cases, this leads to the formation of an insoluble precipitate. D

ecide, in each case, whether or not an insoluble precipitate is formed.
a. K2S and NH4Cl
b. CaCl2 and NH4CO3
c. Li2S and MnBr2
d. Ba(NO3)2 and Ag2SO4
e. RbCO3 and NaCl
Chemistry
1 answer:
Westkost [7]3 years ago
6 0

Answer:

a) K_{2} S and NH_{4} Cl :

There are no insoluble precipitate forms.

b) Ca Cl_{2} and (NH_{4} )_{2} Co_{3} :

There are the insoluble precipitates of CaCo_{3}  forms.

c) Li_{2}S and MnBr_{2} :

There are the insoluble precipitates of MnS  forms.

d) Ba(No_{3} )_{2} and Ag_{2} So_{4} :                        

As Ag_{2} So_{4} is insoluble, therefore no precipitate forms.

e) Rb_{2}Co_{3} and NaCl:

There are no insoluble precipitates forms.

Explanation:

a)

Solubility rule suggests:- K_{2} S ⇒ soluble, NH_{4} Cl ⇒ soluble.

                                          KCl ⇒ soluble, (NH_{4})_{2} S  ⇒ soluble.

There are no insoluble precipitate forms.

b)

Solubility rule suggests:- Ca Cl_{2} ⇒ soluble, (NH_{4} )_{2} Co_{3} ⇒ soluble.

                                        CaCo_{3} ⇒ insoluble, NH_{4} Cl  ⇒ soluble.

There are the insoluble precipitates of CaCo_{3}  forms.

c)

Solubility rule suggests:- Li_{2}S ⇒ soluble, MnBr_{2} ⇒ soluble.

                                        LiBr ⇒ soluble, MnS  ⇒ insoluble.

There are the insoluble precipitates of MnS  forms.

d)

Solubility rule suggests:- Ba(No_{3} )_{2} ⇒ soluble, Ag_{2} So_{4} ⇒insoluble.

                                     

As Ag_{2} So_{4} is insoluble, therefore no precipitate forms.

e)

Solubility rule suggests:- Rb_{2}Co_{3} ⇒ soluble, NaCl ⇒ soluble.

                                        RbCl ⇒ soluble, Na_{2} Co_{3}  ⇒ soluble.

There are no insoluble precipitates forms.

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(i) Based on the graph, determine the order of the decomposition reaction of cyclobutane at 1270 K. Justify your answer.
Leni [432]

Answer:

(c)(i) The order of the reaction based on the graph provided is first order.

(ii) 99% of the cyclobutane would have decomposed in 53.15 milliseconds.

d) Rate = K [Cl₂]

K = rate constant

The justification is presented in the Explanation provided below.

e) A catalyst is a substance that alters the rate of a reaction without participating or being used up in the reaction.

Cl₂ is one of the reactants in the reaction, hence, it participates actively and is used up in the process of the reaction, hence, it cannot be termed as a catalyst for the reaction.

So, this shows why the student's claim is false.

Explanation:

To investigate the order of a reaction, a method of trial and error is usually employed as the general equations for the amount of reactant left for various orders are known.

So, the behaviour of the plot of maybe the concentration of reactant with time, or the plot of the natural logarithm of the concentration of reactant with time.

The graph given is evidently an exponential function. It is a graph of the concentration of cyclobutane declining exponentially with time. This aligns with the gemeral expression of the concentration of reactants for a first order reaction.

C(t) = C₀ e⁻ᵏᵗ

where C(t) = concentration of the reactant at any time

C₀ = Initial concentration of cyclobutane = 1.60 mol/L

k = rate constant

The rate constant for a first order reaction is given

k = (In 2)/T

where T = half life of the reaction. It is the time taken for the concentration of the reactant to fall to half of its initial concentration.

From the graph, when the concentration of reactant reaches half of its initial concentration, that is, when C(t) = 0.80 mol/L, time = 8.0 milliseconds = 0.008 s

k = (In 2)/0.008 = (0.693/0.008) = 86.64 /s

(ii) Calculate the time, in milliseconds, that it would take for 99 percent of the original cyclobutane at 1270 K to decompose

C(t) = C₀ e⁻ᵏᵗ

when 99% of the cyclobutane has decomposed, there's only 1% left

C(t) = 0.01C₀

k = 86.64 /s

t = ?

0.01C₀ = C₀ e⁻ᵏᵗ

e⁻ᵏᵗ = 0.01

In e⁻ᵏᵗ = In 0.01 = -4.605

-kt = -4.605

t = (4.605/k) = (4.605/86.64) = 0.05315 s = 53.15 milliseconds.

d) The reaction mechanism for the reaction of cyclopentane and chlorine gas is given as

Cl₂ → 2Cl (slow)

Cl + C₅H₁₀ → HCl + C₅H₉ (fast)

C₅H₉ + Cl → C₅H₉Cl (fast)

The rate law for a reaction is obtained from the slow step amongst the the elementary reactions or reaction mechanism for the reaction. After writing the rate law from the slow step, any intermediates that appear in the rate law is then substituted for, using the other reaction steps.

For This reaction, the slow step is the first elementary reaction where Chlorine gas dissociates into 2 Chlorine atoms. Hence, the rate law is

Rate = K [Cl₂]

K = rate constant

Since, no intermediates appear in this rate law, no further simplification is necessary.

The obtained rate law indicates that the reaction is first order with respect to the concentration of the Chlorine gas and zero order with respect to cyclopentane.

e) A catalyst is a substance that alters the rate of a reaction without participating or being used up in the reaction.

Cl₂ is one of the reactants in the reaction, hence, it participates actively and is used up in the process of the reaction, hence, it cannot be termed as a catalyst for the reaction.

So, this shows why the student's claim is false.

Hope this Helps!!!

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Red #40 has an acute oral LD50 of roughly 5000 mg dye/1 kg body weight. This means if you had a mass of 1 kg, ingesting 5000 mg
FrozenT [24]

Answer:

350 g dye

0.705 mol

2.9 × 10⁴ L

Explanation:

The lethal dose 50 (LD50) for the dye is 5000 mg dye/ 1 kg body weight. The amount of dye that would be needed to reach the LD50 of a 70 kg person is:

70 kg body weight × (5000 mg dye/ 1 kg body weight) = 3.5 × 10⁵ mg dye = 350 g dye

The molar mass of the dye is 496.42 g/mol. The moles represented by 350 g are:

350 g × (1 mol / 496.42 g) = 0.705 mol

The concentration of Red #40 dye in a sports drink is around 12 mg/L. The volume of drink required to achieve this mass of the dye is:

3.5 × 10⁵ mg × (1 L / 12 mg) = 2.9 × 10⁴ L

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PilotLPTM [1.2K]

Answer:

17.4

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Thus metric is used to determine a large weight of compounds which is a SI unit and handable.    

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