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Andrej [43]
3 years ago
14

A soluble iodide was dissolved in water. Then an excess of silver nitrate, AgNO3, was added to precipitate all of the io- dide i

on as silver iodide, AgI. If 1.545 g of the soluble iodide gave 2.185 g of silver iodide, how many grams of iodine are in the sample of soluble iodide
Chemistry
1 answer:
Sergeeva-Olga [200]3 years ago
7 0

Answer:

m_{I^-}=1.18gI^-

Explanation:

Hello,

In this case, the reaction is given as:

I^-+AgNO_3\rightarrow AgI+NO_3^-

Thus, starting by the yielded grams of silver iodide, we obtain:

n_{I^-}=2.185gAgI*\frac{1molAgI}{234.77gAgI}*\frac{1molI}{1molAgI}=9.31x10^{-3}molI^-

Which correspond to the iodide grams in the silver iodide. In such a way, by means of the law of the conservation of mass, it is known that the grams of each atom MUST remain constant before and after the chemical reaction whereas the moles do not, therefore, the mass of iodine from the silver iodide will equal the mass of iodine present in the soluble iodide, thereby:

m_{I^-}=9.31x10^{-3}molI^-\frac{127gI^-}{1molI^-} =1.18gI^-

And the rest, correspond to the iodide's metallic cation which is unknown. Such value has sense since it is lower than the initial mass of the soluble iodide which is 1.454g, so 0.272 grams correspond to the unknown cation.

Best regards.

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A chemist must prepare of hydrochloric acid solution with a pH of at . He will do this in three steps: Fill a volumetric flask a
Eva8 [605]

Answer:

1.7 mL

Explanation:

<em>A chemist must prepare 550.0 mL of hydrochloric acid solution with a pH of 1.60 at 25 °C. He will do this in three steps: Fill a 550.0 mL volumetric flask about halfway with distilled water. Measure out a small volume of concentrated (8.0 M) stock hydrochloric acid solution and add it to the flask. Fill the flask to the mark with distilled water. Calculate the volume of concentrated hydrochloric acid that the chemist must measure out in the second step. Round your answer to 2 significant digits.</em>

Step 1: Calculate [H⁺] in the dilute solution

We will use the following expresion.

pH = -log [H⁺]

[H⁺] = antilog - pH = antilog -1.60 = 0.0251 M

Since HCl is a strong monoprotic acid, the concentration of HCl in the dilute solution is 0.0251 M.

Step 2: Calculate the volume of the concentrated HCl solution

We want to prepare 550.0 mL of a 0.0251 M HCl solution. We can calculate the volume of the 8.0 M solution using the dilution rule.

C₁ × V₁ = C₂ × V₂

V₁ = C₂ × V₂/C₁

V₁ = 0.0251 M × 550.0 mL/8.0 M = 1.7 mL

3 0
3 years ago
How many grams of CaCl2 are in 250 mL of 2.0 M CaCl2?
Tom [10]
The answer is:  " 56 g CaCl₂ " .
__________________________________________________________

Explanation:
__________________________________________________________
2.0 M CaCl₂  = 2.0 mol CaCl₂ / L  ; 

Since: "M" = "Molarity" (measurement of concentration); 

                  = moles of solute per L {"Liter"} of solution.
__________________________________________________________
Note the exact conversion:  1000 mL = 1 L . 

Given: 250 mL ;   

250 mL = ?  L  ?  ;  


250 mL * (1 L / 1000 L) =  (250/1000) L = 0.25 L . 
___________________________________________________________
 
(2.0 mol CaCl₂ / L ) * (0.25L) = (2.0) * (0.25) mol  = 0.50 mol CaCl₂ ;

We have: 0.50 mol CaCl₂ ;  Convert to "g" (grams):

→ 0.50 mol CaCl₂  .
___________________________________________________________
1 mol CaCl₂ = ? g ?

From the Periodic Table of Elements:

1 mol Ca = 40.08 g

1 mol Cl  =  <span>35.45 g .
</span>
There are 2 atoms of Cl in " CaCl₂ " ;  

→ Note the subscript, "2", in the " Cl₂ " ; 
__________________________________________________________
So, to calculate the molar mass of "CaCl₂" :

40.08 g  +  2(35.45 g) = 

40.08 g  +  70.90 g = 110.98 g ;  round to 4 significant figures; 

                                 → round to 111 g/mol .
__________________________________________________________
So:

→  0.50 mol CaCl₂  = ? g CaCl₂  ? ; 

→  0.50 mol CaCl₂ * (111 g CaCl₂ / mol CaCl₂) ;

                                             = (0.50) * (111 g) CaCl₂ ;

                                             =  55.5 g CaCl₂  ;

                                                → round to 2 significant figures; 

                                                →  56 g CaCl₂ .
___________________________________________________________
The answer is:  " 56 g CaCl₂ " .
___________________________________________________________
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3 years ago
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Fluoride is moderately basic, by far the most basic of the halides. It is a terrible leaving group. Fluoride is many orders of m
TEA [102]

Answer:   Bromide is many orders of magnitude better than fluoride in leaving group ability

Explanation:

As Size of an atom  Increases,  the Basicity Decreases this is because  if we move downwards  from the top of the periodic table to the bottom of the periodic table, the size of an atom increases. As size increases, basicity will decrease, meaning the element  will be less likely to act as a base implying that the element will be less likely to share its electrons.

in the same vein. With an increase in size, basicity decreases, making the ability of the leaving group to leave increase to increase . This can be seen in the halogens going down the group  from

F--- worst

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3 0
3 years ago
What is the pH of a 0.0042 M hydrochloric acid solution?
babymother [125]

Answer:

E) 2.38

Explanation:

The pH of any solution , helps to determine the acidic strength of the solution ,

i.e. ,

  • Lower the value of pH , higher is its acidic strength

and ,

  • Higher the value of pH , lower is its acidic strength .

pH is given as the negative log of the concentration of H⁺ ions ,

hence ,

pH = - log H⁺

From the question ,

the concentration of the solution is 0.0042 M , and being it a strong acid , dissociates completely to its respective ions ,

Therefore , the concentration of H⁺ = 0.0042 M .

Hence , using the above equation , the value of pH can be calculated as follows -

pH = - log H⁺

pH = - log ( 0.0042 M )

pH =  2.38 .

4 0
3 years ago
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Explanation:

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