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Helga [31]
3 years ago
8

In an experiment, potassium chlorate decomposed according to the following chemical equation.

Chemistry
1 answer:
Arada [10]3 years ago
7 0

Answer:  

(240 × 3 × 31.998)/(122.5 × 2)   g

Step-by-step explanation:

We know we will need a balanced equation with masses and molar masses, so let’s gather all the information in one place.  

M_r:          122.5                31.998

               2KClO₃ ⟶ 2KCl + 3O₂

Mass/g:     240  

Mass of O₂ = 240 g KClO₃ × (1 mol KClO₃/122.5 g KClO₃) × (3 mol O₂/2 mol KClO₃)  × (31.998 g O₂/1 mol O₂) = 94.0 g O₂

Mass of O₂= (240 × 3 × 31.998)/(2 × 122.5) = 94.0 g O₂


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1 year ago
A) A metal ruler has a<br> mass of 42g and a<br> volume of 10cm". What<br> is the density?
nataly862011 [7]

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4.2g/cm^3

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4 years ago
A lab technician mixed a 550 ml solution of water and alcohol. if​ 3% of the solution is​ alcohol, how many milliliters of alcoh
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A soluble iodide was dissolved in water. Then an excess of silver nitrate, AgNO3, was added to precipitate all of the io- dide i
Sergeeva-Olga [200]

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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