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3 years ago

An initial mixture of nitrogen gas and hydrogen gas is reacted in a rigid container at a certain temperature as follows:

Chemistry

1 answer:

krok68 [10]3 years ago

3 0

Answer:

The concentration of nitrogen gas and hydrogen gas that reacted initially are 10.0 M and 11.0 M respectively.

Note: The question is incomplete. The complete question is as follows:

An initial mixture of nitrogen gas and hydrogen gas is reacted in a rigid container at a certain temperature as follows:

3H2(g)+N2(g)⇌2NH3(g)

At equilibrium, the concentrations are

[H2] = 5.0 M, [N2] = 8.0 M, and [NH3] = 4.0 M.

What were the concentrations of nitrogen gas and hydrogen gas that were reacted initially?

Explanation:

Given the following data:

Concentration of H2 at equilibrium = 5.0 M

Concentration of N2 at equilibrium = 8.0 M

Concentration of NH3 at equilibrium = 4.0 M

The given equilibrium reaction is,

Initial a b c

Change -3x -x +2x

Equilibrium (a-3x) (b-x) +2x

The value of x is calculated from the equilibrium of NH3

Concentration of NH3 at equilibrium = 4.0 M = 2x

x = 2.0 M

Substituting the value of x above

Initial Concentration of H2:

a - 3 × 2.0 M = 5.0 M

a = 6.0 M + 5 = 11.0 M

Initial Concentration of N2:

b - 2.0 M = 8 M

b = 8.0 M + 2.0 M

b = 10.0 M

Therefore, the concentration of nitrogen gas and hydrogen gas that reacted initially are 10.0 M and 11.0 M respectively.

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Sugar dissolving in warm water is chemical or phyiscal change

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3 years ago

If a 100. -g sample of a hydrated compound contains 37.07-g sodium, 48.39-g carbonate and 14.54-g water, find the empirical form

Mumz [18]

he required empirical formula based on the data provided is Na2CO3.H2O.

<h3>What is empirical formula?</h3>

The term empirical formula refers to the formula of a compound which shows the ratio of each specie present.

We have the following;

Mass of sodium = 37.07-g

Mass of carbonate = 48.39 g

Mass of water = 14.54-g

Number of moles of sodium = 37.07-g/23 g/mol = 2 moles

Number of moles of carbonate = 48.39 g/61 g/mol = 1 mole

Number of moles of water = 14.54/18 g/mol = 1 mole

The mole ratio is 2 : 1: 1

Hence, the required empirical formula is Na2CO3.H2O

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2 years ago

To change an object from a liquid to a gas _________ must happen

nignag [31]

Evaporation!
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3 years ago

If a solution contains 3 moles/liter of sodium chloride (nacl, made of sodium ions and chloride ions), what is the osmolarity of

valkas [14]

Answer: Osmolarity of the sodium chloride solution is 18 Osmol/L.

Explanation:

Osmolarity is defined as the the concentration of the solution which is expressed as the total number of solution particles present in one liter of solvent.

We are given the molarity of the solution which is 3mol/L and to convert it into osmolarity, we will multiply the number of osmoles that are produced by the solute.

Osmole is defined as the particles that contribute to the osmotic pressure of a solution.

The solute given here is sodium chloride (NaCl). Number of osmoles can be determined by the dissociation of this solvent into ions.

The equation given by the dissociation of NaCl:

$NaCl\rightarrow Na^++Cl^-$

1 mole of sodium chloride produces 2 moles of ions.

So, 3 moles of sodium chloride will produce = (3 × 2) = 6 moles of ions.

Hence, osmolarity of the sodium chloride solution will be $6\times 3mol/L=18Osmol/L$

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3 years ago

Molarity to percent by mass. Convert 1.672 mol/L MgCl2(aq) solution to percent by mass of MgCl2 in the solution. The solution de

nignag [31]

Answer:

$\%m/m=14\%$

Explanation:

Hello!

In this case, since the molarity of magnesium chloride (molar mass = 95.211 g/mol) is 1.672 mol/L and we know the density of the solution, we can first compute the concentration in g/L as shown below:

$[MgCl_2]=1.672\frac{molMgCl_2}{L}*\frac{95.211gMgCl_2}{1molMgCl_2}=159.2\frac{gMgCl_2}{L}$

Next, since the density of the solution is 1.137 g/mL, we can compute the concentration in g/g as shown below:

$[MgCl_2]=159.2\frac{gMgCl_2}{L}*\frac{1L}{1000mL}*\frac{1mL}{1.137g}=0.14$

Which is also the by-mass fraction and in percent it turns out:

$\%m/m=0.14*100\%\\\\\%m/m=14\%$

Best regards!

6 0

3 years ago