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swat32
3 years ago
12

Two of the simplest compounds containing just carbon and hydrogen are methane and ethane. Methane contains 0.3357 g of hydrogen

for every 1.00 g of carbon. The ratio of grams of hydrogen per gram of carbon in methane to grams of hydrogen per gram of carbon in ethane is 4:3. Determine the number of grams of hydrogen per gram of carbon in ethane.
Chemistry
1 answer:
Natasha_Volkova [10]3 years ago
5 0
Let the ratio of grams of hydrogen per gram of carbon in methane be M, we know that:
M = 0.3357 g / 1 g

Next, lets represent the grams of hydrogen per gram of carbon in ethane be E. The final piece of information we have is:

M / E = 4/3

If we cross multiply,

3M = 4E

Now, substituting the value of M from earlier and solving for E,

E = (3 * 0.3357) / 4
E = 0.2518

There are 0.2518 grams of hydrogen per gram of carbon in ethane.
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4 0
2 years ago
Write the net ionic equation for the reaction of aqueous solutions of ammonium chloride and iron(III) hydroxide.
vagabundo [1.1K]

Answer:

Fe(OH)_3(s)\rightarrow Fe^{3+}(aq)+3OH^-(aq)

Explanation:

Hello.

In this case, for the reaction between aqueous solutions of ammonium chloride and iron (III) hydroxide, we have the following complete molecular reaction:

3NH_4Cl(aq)+Fe(OH)_3(s)\rightarrow 3NH_4OH+FeCl_3

And the full ionic equation, taking into account that the iron (III) hydroxide cannot be dissolved as it is insoluble in water:

3NH_4^+(aq)+3Cl^-(aq)+Fe(OH)_3(s)\rightarrow 3NH_4^+(aq)+3OH^-(aq)+Fe^{3+}(aq)+3Cl^-(aq)

Finally, the net ionic equation, considering that spectator ions are NH₄⁺, Cl⁻ as they are both the left and right side, therefore, the net ionic equation is:

Fe(OH)_3(s)\rightarrow Fe^{3+}(aq)+3OH^-(aq)

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8 0
2 years ago
An aqueous solution of glucose has molality of 6.81 and a density of 1.2 g/ml what is the molarity of glucose in the solution
andriy [413]

Answer:

the molarity is 3.68 moles/L

Explanation:

the molality of the solution of sucrose is

m= moles of glucose / Kg of solvent (water)= 6.81 ,

since the molecular weight of glucose is 180.156 gr/mole , then per each kilogram of solvent there is

6.81 moles*180.156 gr/mole + 1000 gr of water = 2226.86 gr of solution

from the density

volume of solution = mass of solution/density = 2286.86 gr / 1.2 gr/ml = 1855.71 ml

therefore there is 1000 gr of water in 1855.71 ml

then the molarity M is

M= moles of glucose / L of solution = (moles of glucose /  Kg of solvent) *  (Kg of solvent/L of solution) = 6.81 moles/Kg * 1Kg/1.85 L = 3.68 moles/L

M= 3.68 moles/L

Note:

- Would be wrong in this case to assume density of water = 1 Kg/L since the solution is heavily concentrated in glucose and therefore the density of water deviates from its pure value.

3 0
3 years ago
explain the relationship between the rate of effusion of a gas and its molar mass. methane gas (ch4) effuses 3.4 times faster th
Musya8 [376]

The molar mass of the unknown gas is 184.96 g/mol

<h3>Graham's law of diffusion </h3>

This states that the rate of diffusion of a gas is inversely proportional to the square root of the molar mass i.e

R ∝ 1/ √M

R₁/R₂ = √(M₂/M₁)

<h3>How to determine the molar mass of the unknown gas </h3>

The following data were obtained from the question:

  • Rate of unknown gas (R₁) = R
  • Rate of CH₄ (R₂) = 3.4R
  • Molar mass of CH₄ (M₂) = 16 g/mol
  • Molar mass of unknown gas (M₁) =?

The molar mass of the unknown gas can be obtained as follow:

R₁/R₂ = √(M₂/M₁)

R / 3.4R = √(16 / M₁)

1 / 3.4 = √(16 / M₁)

Square both side

(1 / 3.4)² = 16 / M₁

Cross multiply

(1 / 3.4)² × M₁ = 16

Divide both side by (1 / 3.4)²

M₁ = 16 / (1 / 3.4)²

M₁ = 184.96 g/mol

Learn more about Graham's law of diffusion:

brainly.com/question/14004529

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