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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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C. hydrogen bonding

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A gas at 928 kpa, 129 C occupies a volume of 569 L. Calculate the volume at 319 kpa and<br> 32 C.
lisabon 2012 [21]

Answer:

1255.4L

Explanation:

Given parameters:

P₁  = 928kpa

T₁  = 129°C

V₁  = 569L

P₂ = 319kpa

T₂  = 32°C

Unknown:

V₂  = ?

Solution:

The combined gas law application to this problem can help us solve it. It is mathematically expressed as;

           \frac{P_{1} V_{1} }{T_{1} }   = \frac{P_{2} V_{2} }{T_{2} }

P, V and T are pressure, volume and temperature

where 1 and 2 are initial and final states.

Now,

 take the units to the appropriate ones;

             kpa to atm,  °C to K

P₂ = 319kpa in atm gives 3.15atm

P₁  = 928kpa gives 9.16atm

T₂  = 32°C gives 273 + 32  = 305K

T₁  = 129°C gives 129 + 273  = 402K

Input the values in the equation and solve for V₂;

        \frac{9.16  x 569}{402}   = \frac{3.15 x V_{2} }{305}

       V₂   = 1255.4L

4 0
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Answer: Option (B) is the correct answer.

Explanation:

Expression for the given decomposition reaction is as follows.

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Let us assume that x concentration of N_{2}O_{4} is present at the initial stage. Therefore, according to the ICE table,

                    N_{2}O_{4} \rightarrow 2NO_{2}

Initial :               x                   0

Change :       - 0.1        2 \times 0.1

Equilibrium : (x - 0.1)             0.2

Now, expression for K_{p} of this reaction is as follows.

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Putting the given values into the above formula as follows.

          K_{p} = \frac{P^{2}_{NO_{2}}}{P_{N_{2}O_{4}}}

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Answer: The given statement is true.

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Reactivity of elements is the tendency of the elements to gain or loose electrons. The reactivity decreases down the group in a periodic table.

In the given reaction, Iodine and chlorine are the elements of the same group in the periodic table and iodine lies below chlorine in the group. So, the reactivity of iodine is less than the reactivity of chlorine.

Hence, in the given reaction, iodine will not replace chlorine because it lies below in the periodic table.

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