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3 years ago

PLEASE explain your answer!!!!

Chemistry

1 answer:

Aleks04 [339]3 years ago

6 0

Answer:

Explanation: confusing

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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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4 years ago

How many significant figures are in 0.0000003 g?

stiv31 [10]

Answer:

Therefore, none of the zeros is a significant figure and hence there is only one significant figure in this measurement in support G. The measurement is 0.7 minutes. Here again, zero is not a significant figure and therefore the only significant figure is one which is seven in support Edge.

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3 years ago

POSS<br> Explain when gold is considered an element and when it is considered a solution.

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Answer:

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3 years ago

PLEASE explain your answer!!!!​

PLEASE explain your answer!!!!