$4.010 \times 10^{25} \text { molecules of } \mathrm{NaHCO}_{3} \text { present in } 55.93 \mathrm{kg} \text { of } \mathrm{NaHCO}_{3}$
$16.86 \times 10^{26} \text { molecules of } \mathrm{Na}_{3} \mathrm{PO}_{4} \text { present in } 459 \mathrm{kg}\left(4.59 \times 10^{5} \mathrm{gm}\right) \text { of } \mathrm{Na}_{3} \mathrm{PO}_{4}$

Explanation:

Number of molecules for $55.93 \mathrm{kg} \text { of } \mathrm{NaHCO}_{3}$

$\text { Firstly molar mass is calculated of } \mathrm{NaHCO}_{3}:$

Atomic mass of Na + H + C + 3(O) = 22.99 + 1.008 + 12.01 + 3 × 16.00 = 84.00 g/mol

$\text { Number of molecules of } \mathrm{NaHCO}_{3} \text { in } 55.93 \text { kg are as follows: }$

$55.93 \times\left(10^{3} \mathrm{gm}\right) \times \frac{1 \mathrm{mol} \mathrm{NaHCO}_{3}}{84.00 \mathrm{gm} \mathrm{NaHCO}_{3}} \times\left(6.022 \times 10^{23} \mathrm{molecules} \text { i.e Avogadro number }\right)$

$=4.010 \times 10^{26} \text { molecules of } \mathrm{NaHCO}_{3} \text { present in } 55.93 \mathrm{kg} \text { of } \mathrm{NaHCO}_{3}$

Number of molecules for for $\left(4.59 \times 10^{5} \mathrm{gm}\right) \text { of } \mathrm{Na}_{3} \mathrm{PO}_{4}$

$\text { Firstly molar mass is calculated of } \mathrm{Na}_{3} \mathrm{PO}_{4}$

= Atomic mass of 3(Na) + P + 4(O)

= 3(22.99) + 30.97 + 4(16.00) = 163.94 g/mol

$459 \times\left(10^{3} \mathrm{gm}\right) \times \frac{1 \mathrm{mol} N a_{3} P O_{4}}{163.94 \mathrm{gm} N a_{3} P O_{4}} \times\left(6.022 \times 10^{23} \mathrm{molecules} \text { i.e Avogadro number) } / 1 \mathrm{mol}\right.$

$=16.86 \times 10^{26} \text { molecules of } \mathrm{Na}_{3} \mathrm{PO}_{4} \text { present in } 459 \mathrm{kg}\left(4.59 \times 10^{5} \mathrm{gm}\right) \text { of } \mathrm{Na}_{3} \mathrm{PO}_{4}$

8 0

3 years ago

A dehydration reaction starting with 3.8 g cyclohexanol produces 2.6 g cyclohexene. Calculate the theoretical yield for this rea

diamong [38]

Answer:

Theoretical yield of C6H10 = 3.2 g.

Explanation:

Defining Theoretical yield as the quantity of product obtained from the complete conversion of the limiting reactant in a chemical reaction. It can be expressed as grams or moles.

Equation of the reaction

C6H11OH --> C6H10 + H2O

Moles of C6H11OH:

Molar mass of C6H110H = (12*6) + (1*12) + 16

= 100 g/mol

Mass of C6H10 = 3.8 g

number of moles = mass/molar mass

=3.8/100

= 0.038 mol.

Using stoichoimetry, 1 moles of C6H110H was dehydrated to form 1 mole of C6H10 and 1 mole of water.

Therefore, 0.038 moles of C6H10 was produced.

Mass of C6H10 = molar mass * number of moles

Molar mass of C6H10 = (12*6) + (1*10)

= 82 g/mol.

Mass = 82 * 0.038

= 3.116 g of C6H10.

Theoretical yield of C6H10 = 3.2 g

4 0

4 years ago

Use the periodic table to select which type of bond is present and which of the listed properties is most likely for each substa

jenyasd209 [6]

Answer:

A = Metallic Bond

B = Strong bonding, strong conductor, high melting and boiling points

Explanation:

Since the bond is between two metals (located in groups 11 and 12), they would experience metallic bonding. Metallically bonded molecules have high melting and boiling points due to the strength of the metallic bond. They also experience strong electrical current due to the there delocalized electrons.

3 0

3 years ago