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2 years ago

HELP ME PLEASE I NEED HELP SO BADLY

Chemistry

1 answer:

crimeas [40]2 years ago

5 0

State the given:

Moles of Sulfur = 5 moles

Molar mass of Sulfur = 32.06g/mol

Look through the formulas:

Moles = Mass/Molar Mass

Rearrange the equation:

Mass = Moles x Molar mass

Plug in your given:

5 moles Sulfur x 32.06g

1 mol

= 160.3g of Sulfur

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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]

Answer:

$M=0.0637M$

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:

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

Finally, the resulting molarity in 30.8 mL (0.0308 L):

$M=\frac{1.96x10^{-3}molPb^{2+}}{0.0308L}\\ \\M=0.0637M$

Regards.

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