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Dmitrij [34]
3 years ago
6

if 10 moles of A and 7.5 moles of B are taken in a flask, identify the value of moles of A4B3 produced in a reaction.​

Chemistry
1 answer:
natta225 [31]3 years ago
3 0

The number of moles of A4B3 produced in a reaction is obtained  by stoichiometry.

The term stoichiometry refers to mass-mole relationship. The first step in solving this problem is by writing the correct reaction equation as follows;

4A + 3B -------> A4B3

If 4 moles of A yields 1 mole of A4B3

10 moles of A yields 10 * 1/4 = 2.5 moles of A4B3

Therefore, 2.5 moles of A4B3  is produced.

Learn more: brainly.com/question/9743981

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Answer:

Option A = 2.2 L

Explanation:

Given data:

volume of one mole of gas = 22.4 L

Volume of 0.1 mole of gas at same condition = ?

Solution:

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For 0.1 mole of methane.

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Does adding 1 mol of NaCl to 1 kg of water lower the vapor pressure of water to the same extent, a lesser extent, or a greater e
igor_vitrenko [27]

Answer:

Adding 1 mol of NaCl to 1 kg of water lower the vapor pressure of water <em><u>to the same extent</u></em>  by adding 1 mol of C_06H_{12}O_6 to 1 kg of water.

Explanation:

1) Moles of NaCl ,n_1=1 mol

Mass of water = m= 1 kg = 1000 g

Moles of water = n_2=\frac{1000 g}{18 g/mol}=55.55 mol

Vapor pressure of the solution = p

Vapor pressure of the pure solvent that is water = p_o=17.5 Torr

Mole fraction of solute(NaCl)= \chi_1=\frac{n_1}{n_1+n_2}

\frac{p_o-p}{p_o}=\frac{n_1}{n_1+n_2}

\frac{17.5 Torr-p}{17.5 Torr}=\frac{1 mol}{1 mol+55.55 mol}

p=17.19 Torr

The vapor pressure for the NaCl solution at 17.19 Torr.

2) Moles of sucrose ,n_1=1mol

Mass of water = m  = 1 kg = 1000 g

Moles of water = n_2=\frac{1000 g}{18 g/mol}=55.55 mol

Vapor pressure of the solution = p'

Vapor pressure of the pure solvent that is water = p_o=17.5 Torr

Mole fraction of solute ( glucose)= \chi_1=\frac{n_1}{n_1+n_2}

\frac{p_o-p}{p_o}=\frac{n_1}{n_1+n_2}

\frac{17.5 Torr-p}{17.5 Torr}=\frac{1 mol}{1 mol+55.55 mol}

p'=17.19 Torr

The vapor pressure for the glucose solution at 17.19 Torr.

p = p' = 17.19 Torr

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Iron(II) sulfide has a primitive cubic unit cell with sulfide ions at the lattice points.
masya89 [10]

We have that the  the density of FeS  is mathematically given as

  •  \phi=2.56h/cm^3

From the question we are told

Iron(II) sulfide has a primitive <em>cubic</em> unit cell with <em>sulfide</em> ions at the <em>lattice points.</em>

The ionic radii of iron(II) ions and sulfide ions are 88 pm and 184 pm, respectively.

What is the density of FeS (in g/cm3)?

<h3>Density</h3>

Generally the equation for the Velocity  is mathematically given as

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For more information on ionic radii visit

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