Which of the following are effective treatments for some musculoskeletal conditions? Select all that apply. PLEASE ANSWER

Write the balanced equation for the reaction of aqueous Pb ( ClO 3 ) 2 Pb(ClO3)2 with aqueous NaI . NaI. Include phases. chemica

FinnZ [79.3K]

Answer: The mass of precipitate (lead (II) iodide) that will form is 119.89 grams

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Molarity of NaI solution = 0.130 M

Volume of solution = 0.400 L

Putting values in above equation, we get:

$0.130M=\frac{\text{Moles of NaI}}{0.400L}\\\\\text{Moles of NaI}=(0.130mol/L\times 0.400L)=0.52mol$

The balanced chemical equation for the reaction of lead chlorate and sodium iodide follows:

$Pb(ClO_3)_2(aq.)+2NaI(aq.)\rightarrow PbI_2(s)+2NaClO_3(aq.)$

The precipitate (insoluble salt) formed is lead (II) iodide

By Stoichiometry of the reaction:

2 moles of NaI produces 1 mole of lead (II) iodide

So, 0.52 moles of NaI will produce = $\frac{1}{2}\times 0.52=0.26mol$ of lead (II) iodide

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Moles of lead (II) iodide = 0.26 moles

Molar mass of lead (II) iodide = 461.1 g/mol

Putting values in above equation, we get:

$0.26mol=\frac{\text{Mass of lead (II) iodide}}{461.1g/mol}\\\\\text{Mass of lead (II) iodide}=(0.26mol\times 461.1g/mol)=119.89g$

Hence, the mass of precipitate (lead (II) iodide) that will form is 119.89 grams

4 0

3 years ago

True or false if an object displaces 50 ml of water the object's volume is 50mL

andrew11 [14]

Answer:

True

Explanation:

The volume of water displaced by an object completely submerged is its actual volume. It implies that in the container the object create a space of size for itself which is the volume of the object. This approach is used in calculating the density of many irregular solids from their measured masses.

6 0

3 years ago

The equilibrium constant for the gas-phase isomerization of borneol (c10h17oh) to isoborneol at 503 k is 0.106. a mixture consis

Dimas [21]

The solution is as follows:

K = [Partial pressure of isoborneol]/[Partial pressure of borneol] = 0.106

The molar mass of isoborneol/borneol is 154.25 g/mol

Mol isoborneol = 15 g/154.25 = 0.0972 mol
Mol borneol = 7.5 g/154.25 = 0.0486 mol

Use the ICE approach

borneol → isoborneol
I 0.0972 0.0486
C -x +x
E 0.0972 - x 0.0486 + x

Total moles = 0.1458

Using Raoult's Law,
Partial Pressure = Mole fraction*Total Pressure
[Partial pressure of isoborneol] = [(0.0972-x)/0.1458]*P
[Partial pressure of borneol] = [(0.0486+x/0.1458)]*P

0.106 = [(0.0972-x)/0.1458]*P/ [(0.0486+x/0.1458)]*P
Solving for x,
x = 0.0832

Thus,
Mol fraction of borneol = (0.0486+0.0832)/0.1458 = 0.904
Mol fraction of isoborneol = (0.0972-0.0832)/0.1458 = 0.096

6 0

3 years ago

A block of metal has a mass of 0.0694 kg. It displaces 3.6 mL of water. What is it’s density in g/cm^3

Ivan

Answer:

19.28 g/cm^3 to the nearest hundredth.

Explanation:

The volume of water displaced = the volume of the metal.

density = mass / volume

0.0694 kg = 0.0694 * 1000

= 69.4 g.

Density = 69.4 / 3.6

= 19.28 g/cm^3.

4 0

3 years ago

The height of a wave is called its amplitude. From the noises of the mosquito and the air boat, which aspect of sound is related

TEA [102]

The answer is The loudness of sound is related to its amplitude, this is off edmentum exactly so I advise changing up the wording. You can say something about the pitch or you can word it like, The sound of the wave is related to how loud the sound can be. Hope this helped

3 0

3 years ago

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