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antiseptic1488 [7]

3 years ago

6

Calculate the mass in grams of 202 atoms of iron

1 answer:

Sergeeva-Olga [200]3 years ago

4 0

Answer:

To calculate the number of atoms in a sample, divide its weight in grams by the amu atomic mass from the periodic table, then multiply the result by Avogadro's number: 6.02 x 10^23. Set up Equation Express the relationship of the three pieces of information you need to calculate the number of atoms in the sample in the form of an equation.

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Radioactive is the most penetrating nuclear radiation

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3 years ago

Which property is shared by both acids and bases?

Evgen [1.6K]

$\huge{\textbf{\textsf{{\color{navy}{An}}{\purple{sw}}{\pink{er}} {\color{pink}{:}}}}}$

3. Conducts electricity in aqueous solution.

Thanks
Hope it helps.

3 0

3 years ago

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]

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

<u>Explanation:</u>

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

When iron reacts with oxygen what substance is produced?

zhannawk [14.2K]

It will produce iron oxide

8 0

3 years ago

Use the periodic table to identify the molar mass of each element below. Answer without doing any calculations.

Delicious77 [7]

Answer:

Beryllium (Be) : 9.01 g/mol

Silicon (Si) : 28.09 g/mol

Calcium (Ca) : 40.08 g/mol

Rhodium (Rh) : 102.91 g/mol

Explanation:

8 0

4 years ago

Read 2 more answers