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3 years ago

Calculate the number of molecules in 6.00 moles h2s?.

2 answers:

6 0

Use the equation: Nm = M x Av
Nm = Moles x Avogadro constant
Nm = 6 x 6.02x10²³
Nm = 3.612x10²⁴

so in six moles of H2S there are 3.612x10²⁴ molecules

hope that helps

LenKa [72]3 years ago

5 0

To calculate the number of molecules in 6.00 moles of hydrogen sulfide, H2S, the equivalence factor used is Avogadro's number equal to (6.022 x 10^23 molecules/mole). The answer is 6.00 moles of hydrogen sulfide * (6.022 x 10^23 molecules/mole) equal to 3.61 x10^24 molecules.

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mixture of N 2 And H2 Gases weighs 13.22 g and occupies a volume of 24.62 L at 300 K and 1.00 atm.Calculate the mass percent of

anygoal [31]

Answer: The mass percent of nitrogen gas and hydrogen gas is 91.41 % and 8.59 % respectively.

Explanation:

To calculate the number of moles, we use the equation given by ideal gas equation:

PV = nRT

where,

P = Pressure of the gaseous mixture = 1.00 atm

V = Volume of the gaseous mixture = 24.62 L

n = number of moles of the gaseous mixture = ?

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = Temperature of the gaseous mixture = 300 K

Putting values in above equation, we get:

$1.00atm\times 24.62L=n_{mix}\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 300K\\\\n_{mix}=\frac{1.00\times 24.62}{0.0821\times 300}=0.9996mol$

We are given:

Total mass of the mixture = 13.22 grams

Let the mass of nitrogen gas be 'x' grams and that of hydrogen gas be '(13.22 - x)' grams

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

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

For nitrogen gas:

Molar mass of nitrogen gas = 28 g/mol

$\text{Moles of nitrogen gas}=\frac{x}{28}mol$

For hydrogen gas:

Molar mass of hydrogen gas = 2 g/mol

$\text{Moles of hydrogen gas}=\frac{(13.22-x)}{2}mol$

Equating the moles of the individual gases to the moles of mixture:

$0.9996=\frac{x}{28}+\frac{(13.22-x)}{2}\\\\x=12.084g$

To calculate the mass percentage of substance in mixture we use the equation:

$\text{Mass percent of substance}=\frac{\text{Mass of substance}}{\text{Mass of mixture}}\times 100$

Mass of the mixture = 13.22 g

For nitrogen gas:

Mass of nitrogen gas = x = 12.084 g

Putting values in above equation, we get:

$\text{Mass percent of nitrogen gas}=\frac{12.084g}{13.22g}\times 100=91.41\%$

For hydrogen gas:

Mass of hydrogen gas = (13.22 - x) = (13.22 - 12.084) g = 1.136 g

Putting values in above equation, we get:

$\text{Mass percent of hydrogen gas}=\frac{1.136g}{13.22g}\times 100=8.59\%$

Hence, the mass percent of nitrogen gas and hydrogen gas is 91.41 % and 8.59 % respectively.

5 0

3 years ago

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Assoli18 [71]

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3 years ago

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kodGreya [7K]

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3 years ago

Calculate the percentage of each element in acetic acid, hc2h3o2, and glucose, c6h12o6.

My name is Ann [436]

Acetic acid, HC2H3O2

First, calculate for the molar mass of acetic acid as shown below.
M = 1 + 2(12) + 3(1) + 2(16) = 60 g

Then, calculating for the percentages of each element.
 Hydrogen:
P1 = ((4)(1)/60)(100%) = 6.67%

 Carbon:
P2 = ((2)(12)/60)(100%) = 40%

Oxygen
P3 =((2)(16) / 60)(100%) = 53.33%

Glucose, C6H12O6

The molar mass of glucose is as calculated below,
6(12) + 12(1) + 6(16) = 180

The percentages of the elements are as follow,
 Hydrogen:
P1 = (12/180)(100%) = 6.67%

Carbon:
P2 = ((6)(12) / 180)(100%) = 40%

Oxygen:
P3 = ((6)(16) / 180)(100%) = 53.33%

b. Since the empirical formula of the given substances are just the same and can be written as CH2O then, the percentages of each element composing them will just be equal.

6 0

3 years ago

Graph of the relationship between mass and thermal energy

elixir [45]

Thermal energy is the sum of the kinetic and potential energies of all the particles in an object.

Assume that you have 250 gL of water and 1 kg of water at the same temperature.

Then, each water molecule has the same kinetic energy.

The larger sample contains four times as many molecules, so it contains four times as much thermal energy.

Thus, thermal energy is directly proportional to mass.

In symbols, KE ∝ m or KE = km.

The graph of a direct proportion is a straight line passing trough the origin.

It should look something like the graph below.

6 0

4 years ago