What do lightning and stars have in common?

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

Carvone (C10H14O) is the main component of spearmint oil. It has a pleasant aroma and mint flavor. Carvone is often added to che

marta [7]

Answer: The mass of carbon in given amount of carvone is 47.52 grams

Explanation:

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

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

Given mass of carvone = 55.4 g

Molar mass of carvone = 150.22 g/mol

Putting values in equation 1, we get:

$\text{Moles of carvone}=\frac{55.4g}{150.22g/mol}=0.369mol$

The chemical formula of carvone is $C_{10}H_{14}O$

This contains 10 moles of carbon atom, 14 moles of hydrogen atom and 1 mole of oxygen atom

Now, calculating the mass of carbon in carvone from equation 1, we get:

Molar mass of carbon = 12 g/mol

Moles of carbon = (10 × 0.396) = 3.96 moles

Putting values in equation 1, we get:

$3.96mol=\frac{\text{Mass of carbon}}{12g/mol}\\\\\text{Mass of carbon}=(3.96mol\times 12g/mol)=47.52g$

Hence, the mass of carbon in given amount of carvone is 47.52 grams

6 0

3 years ago

What do you think is meant by the statement "the digestive system is very different

Kryger [21]

It could possibly refer to “the small intestines absorb most of the nutrients in food, and your circulatory system passes them on to other parts of your body to store or use. Special cells help absorbed nutrients cross intestinal lining into your bloodstream” hope this helps Good luck

7 0

3 years ago

Water Erosion: Cause and Effect

Cloud [144]

Answer:

Definitely add more points

Explanation:

I'm saying this because this is worth way more points

3 0

3 years ago

Write the isotopic symbol for the following (show your work) a) An isotope of iodine whose atoms have 78 neutrons b) An isotope

morpeh [17]

Answer:

For a: The isotopic representation of iodine is $_{53}^{131}\textrm{I}$

For b: The isotopic representation of cesium is $_{55}^{137}\textrm{Cs}$

For c: The isotopic representation of strontium is $_{38}^{52}\textrm{Sr}$

Explanation:

The isotopic representation of an atom is: $_Z^A\textrm{X}$

where,

Z = Atomic number of the atom

A = Mass number of the atom

X = Symbol of the atom

For a:

We are given:

Number of neutrons = 78

Atomic number of iodine = 53 = Number of protons

Mass number = 53 + 78 = 131

Thus, the isotopic representation of iodine is $_{53}^{131}\textrm{I}$

For b:

We are given:

Number of neutrons = 82

Atomic number of cesium = 55 = Number of protons

Mass number = 55 + 82 = 137

Thus, the isotopic representation of cesium is $_{55}^{137}\textrm{Cs}$

For c:

We are given:

Number of neutrons = 52

Atomic number of strontium = 38 = Number of protons

Mass number = 38 + 52 = 90

Thus, the isotopic representation of strontium is $_{38}^{52}\textrm{Sr}$

3 0

3 years ago