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dlinn [17]
3 years ago
11

Help please. I only need the concepts.

Chemistry
1 answer:
melomori [17]3 years ago
5 0

Boyle’s = increase as volume decreases

Charles = increases and pressure increases

Gay-lussacs = increases as pressure increases

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How many are molecules ( or formula) in each sample?
andre [41]

Answer:

  • 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}

<u>Explanation</u>:

<u>Number of molecules for 55.93 \mathrm{kg} \text { of } \mathrm{NaHCO}_{3}</u>

\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}

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

\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
Acetylene gas (c2h2) reacts with oxygen gas (o2) to produce carbon dioxide (co2) and water vapor (h2o). how many liters of c2h2
MissTica
According to the balanced equation of the reaction:

2C2H2 + 5O2 → 4CO2 + 2H2O

So we can mention all as liters,

A) as we see that 2 liters of C2H2 react with 5 liters of oxygen to produce 4 liters of CO4 and 2 liters of H2O 

So, when we have 75L of CO2

and when we have 2 L of C2H2 reacts and gives 4 L of CO2

 2C2H2 → 4CO2 

∴ The volume of C2H2 required is:

                    = 75L / 2

                    = 37.5 L

B) and, when we have 75 L of CO2


and 4CO2 → 2H2O 

∴ the volume of H2O required is: 
                               
                            = 75 L /2
                            = 37.5 L 


C) and from the balanced equation and by  the same way:

when 5 liters O2  reacts to give 4 liters of CO2

and we have 75 L of CO2:

5 O2 → 4 CO2 

   ??  ←  75 L

∴ the volume of O2 required is:

                              = 75 *(5/4)

                              = 93.75 L



D) about the using of the number of moles the answer is: 

no, there is no need to find the number of moles as we called everything in the balanced equation by liters and use it as a liter unit to get the volume, without the need to get the number of moles.
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