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AlladinOne [14]
3 years ago
7

PLZ HELP QUICKLY!!

Chemistry
1 answer:
frutty [35]3 years ago
6 0

Answer:

Robert Hooke.

Anton van Leeuwenhoek.

Matthias Schleiden.

Theodor Schwann.

Rudolf Virchow.

Explanation:

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b

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2 years ago
g Enter your answer in the provided box. If 30.8 mL of lead(II) nitrate solution reacts completely with excess sodium iodide sol
Ronch [10]

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Explanation:

Hello,

In this case, the undergoing chemical reaction is:

Pb(NO_3)_2(aq)+2NaI(aq)\rightarrow PbI_2(s)+2NaNO_3(aq)

Thus, for 0.904 g of precipitate, that is lead (II) iodide, we can compute the initial moles of lead (II) ions in lead (II) nitrate:

n_{Pb^{2+}}=0.904gPbI_2*\frac{1molPbI_2}{461gPbI_2}*\frac{1molPb(NO_3)_2}{1molPbI_2}  *\frac{1molPb^{2+}}{1molPb(NO_3)_2} =1.96x10^{-3}molPb^{2+}

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Regards.

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3 years ago
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