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3 years ago

How many equivalents are in a solution that contains 0.25 moles Mg2+ and 0.50 moles Cl–?

Chemistry

1 answer:

Vinil7 [7]3 years ago

7 0

Answer:

Explanation:

Mg²⁺ is divalent , hence

Molecular weight / 2 = equivalent weight .

.25 moles = 2 x .25 equivalents = .5 equivalents .

Cl⁻ is monovalent so

molecular weight = equivalent weight

.50 mole = .50 equivalent

Total equivalent = .50 of Mg²⁺ + .50 of Cl⁻

= 1 equivalent .

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How much energy must be removed from a 125 g sample of benzene (molar mass= 78.11 g/mol) at 425.0 K to liquify the sample and lo

riadik2000 [5.3K]

Answer : The energy removed must be, -67.7 kJ

Solution :

The process involved in this problem are :

$(1):C_6H_6(g)(425.0K)\rightarrow C_6H_6(g)(353.0K)\\\\(2):C_6H_6(g)(353.0K)\rightarrow C_6H_6(l)(353.0K)\\\\(3):C_6H_6(l)(353.0K)\rightarrow C_6H_6(l)(335.0K)$

The expression used will be:

$\Delta H=[m\times c_{p,g}\times (T_{final}-T_{initial})]+m\times \Delta H_{vap}+[m\times c_{p,l}\times (T_{final}-T_{initial})]$

where,

$\Delta H$ = heat released by the reaction = ?

m = mass of benzene = 125 g

$c_{p,g}$ = specific heat of gaseous benzene = $1.06J/g^oC$

$c_{p,l}$ = specific heat of liquid benzene = $1.73J/g^oC$

$\Delta H_{vap}$ = enthalpy change for vaporization = $33.9kJ/mole=33900J/mole=\frac{33900J/mole}{78.11g/mole}J/g=434.0J/g$

Molar mass of benzene = 78.11 g/mole

Now put all the given values in the above expression, we get:

$\Delta H=[125g\times 1.06J/g.K\times (353.0-(425.0))K]+125g\times -434.0J/g+[125g\times 1.73J/g.K\times (335.0-353.0)K]$

$\Delta H=-67682.5J=-67.7kJ$

Therefore, the energy removed must be, -67.7 kJ

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3 years ago

Cells which form pathways to and from the brain are <br> .epiltheilal neurons connnective

nikdorinn [45]

Answer:

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2 years ago

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3 years ago

Is the process of a liquid changing into a solid when cooling a chemical reaction?

Alona [7]

Answer: The bulk of liquids separate from the standard substance through crystallization, creating a crystalline block.

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3 years ago

For the reaction: 3 H2(g) + N2(g) <--> 2 NH3(g), the concentrations at equilbrium were [H2] = 0.10 M, [N2] = 0.10 M, and [

masya89 [10]

The equilibrium constant, k of the reaction in which case, the concentrations of the given reactants and products are as indicated is; Choice A; K = 3.1 x 10⁵

<h3>What is the equilibrium constant , k of the reaction as described in the task content?</h3>

It follows from above that the concentrations of the reactants and products are as follows; [H2] = 0.10 M, [N2] = 0.10 M, and [NH3] = 5.6 M at equilibrium.

Hence, the equilibrium constant of the reaction in discuss is;

K = [5.6]²/[0.10]³[0.10]

k = 5.6² × 10⁴

k = 3.136 × 10⁵

K = 3.1 × 10⁵.