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A 99.8 mL sample of a solution that is 12.0% KI by mass (d: 1.093 g/mL) is added to 96.7 mL of another solution that is 14.0% Pb

andre [41]

Answer:

$m_{PbI_2}=18.2gPbI_2$

Explanation:

Hello,

In this case, we write the reaction again:

$Pb(NO_3)_2(aq) + 2 KI(aq)\rightarrow PbI_2(s) + 2 KNO_3(aq)$

In such a way, the first thing we do is to compute the reacting moles of lead (II) nitrate and potassium iodide, by using the concentration, volumes, densities and molar masses, 331.2 g/mol and 166.0 g/mol respectively:

$n_{Pb(NO_3)_2}=\frac{0.14gPb(NO_3)_2}{1g\ sln}*\frac{1molPb(NO_3)_2}{331.2gPb(NO_3)_2} *\frac{1.134g\ sln}{1mL\ sln} *96.7mL\ sln\\\\n_{Pb(NO_3)_2}=0.04635molPb(NO_3)_2\\\\n_{KI}=\frac{0.12gKI}{1g\ sln}*\frac{1molKI}{166.0gKI} *\frac{1.093g\ sln}{1mL\ sln} *99.8mL\ sln\\\\n_{KI}=0.07885molKI$

Next, as lead (II) nitrate and potassium iodide are in a 1:2 molar ratio, 0.04635 mol of lead (II) nitrate will completely react with the following moles of potassium nitrate:

$0.04635molPb(NO_3)_2*\frac{2molKI}{1molPb(NO_3)_2} =0.0927molKI$

But we only have 0.07885 moles, for that reason KI is the limiting reactant, so we compute the yielded grams of lead (II) iodide, whose molar mass is 461.01 g/mol, by using their 2:1 molar ratio:

$m_{PbI_2}=0.07885molKI*\frac{1molPbI_2}{2molKI} *\frac{461.01gPbI_2}{1molPbI_2} \\\\m_{PbI_2}=18.2gPbI_2$

Best regards.

5 0

3 years ago

Read 2 more answers

If 175mL of oxygen is produced at STP, how many grams of hydrogen peroxide, H2O2

Vlad [161]

Answer:

0.53g

Explanation:

We'll begin by converting 175mL to L. This is illustrated below:

1000mL = 1L

Therefore 175mL = 175/1000 = 0.175L

Next, we shall calculate the number of mole of O2 that occupy 0.175L. This is illustrated below:

1 mole of O2 occupy 22.4L at stp.

Therefore, Xmol of O2 will occupy 0.175L i.e

Xmol of O2 = 0.175/22.4

Xmol of O2 = 7.81×10¯³ mole

Therefore, 7.81×10¯³ mole of O2 occupy 175mL.

Next, we shall determine the number of mole of H2O2 that decomposed to produce 7.81×10¯³ mole of O2. This is illustrated below:

2H2O2 —> 2H2O + O2

From the balanced equation above,

2 moles of H2O2 decomposed to produce 1 mole of O2.

Therefore, Xmol of H2O2 will decompose to produce 7.81×10¯³ mole of O2 i.e

Xmol of H2O2 = 2 x 7.81×10¯³

Xmol of H2O2 = 1.562×10¯² mole

Therefore, 1.562×10¯² mole of H2O2 decomposed in the reaction.

Finally, we shall convert 1.562×10¯² mole of H2O2 to grams. This is illustrated below:

Molar mass of H2O2 = (2x1) + (16x2) = 34g/mol

Mole of H2O2 = 1.562×10¯² mole

Mass of H2O2 =..?

Mole = mass /Molar mass

1.562×10¯² = mass /34

Cross multiply

Mass of H2O2 = 1.562×10¯² x 34

Mass of H2O2 = 0.53g

Therefore, 0.53g of Hydrogen peroxide, H2O2 were decomposition in the reaction.

3 0

3 years ago

Which of the following actions does not require power

Kitty [74]

Answer:

Eating an apple

Explanation:

No electricity

4 0

3 years ago

How does carbon 14 work

Verdich [7]

Carbon-14 dating is a way of determining the age of certain archeological artifacts of a biological origin up to about 50,000 years old. It is used in dating things such as bone, cloth, wood and plant fibers that were created in the relatively recent past by human activities.

6 0

3 years ago

Why might people add a base to an acidic lake?

spayn [35]

Answer:

To decrease the algae

Explanation:

Acid lakes have more algae than other lakes

6 0

3 years ago