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2 years ago

Sample of gas occupies 10.0 L at 50 ˚C. Assuming that the pressure is constant what volume will

Chemistry

1 answer:

scZoUnD [109]2 years ago

8 0

Answer:

20 Liters

Explanation:

If the pressure is supposed to be constant, one of Charles - Gay Lussac's laws can be used to solve the exercise. His statement says that "the volume of the gas is directly proportional to its temperature at constant pressure", mathematically it would be:

$\frac{V1}{T1} = \frac{V2}{T2}$

Considering T₁ = 50 ° C; V₁ = 10.0 L; and T₂ = 100 ° C, we can calculate the volume V₂ Clearing it from the equation and replacing the values to perform the calculation:

V2= (V1 x T2) / T1 = (10.0 L x 100°C) / 50 °C = 20 L

Therefore, the gas at 100 ° C will occupy a volume of 20.0 L.

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How many are molecules ( or formula) in each sample?

andre [41]

Answer:

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

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

What is the difference between atom and it’s ion ?

Sati [7]

Answer:
D. Number of electrons

8 0

3 years ago

How do different types of waves make particles of matter move

Alborosie

That depends on the wave, if you're talking about sound, it makes matter move in a similar wavelength as them, a mountainous shape. Light however would make whatever matter it hits start to move in the same direction as the light's angle of approach.

6 0

2 years ago

The molarity of concentrated acetic acid is 17.45 M. When acetic acid is diluted in water it is commonly called vinegar with a m

Sladkaya [172]

M1 = 17.45 M

M2 = 0.83 M

V2 = 250 ml

M1. V1= M2. V2

V1 = (M2. V2)/M1 = (0.83× 250)/ 17.45= 11.89 ml

7 0

2 years ago

How many molecules are there in 2.30g of NH3

Vinil7 [7]

Answer:

8.13x10^22 molecules

Explanation:

We can use the Avogadro's number(6.022 x 10^23 units / mole)

2.30 g NH3 (1 mol / 17.03 g ) (6.022 x 10^23 molecules / 1 mol ) = 8.13x10^22 molecules

Hope this helps! Feel free to ask any questions!

6 0

2 years ago