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3 years ago

GIVING BRAINLIEST!

Chemistry

1 answer:

KATRIN_1 [288]3 years ago

5 0

Answer:

No new matter can be created, and no matter can be destroyed. Matter can only change in state or chemical composition. The amount of mass before the change will equal the amount after the change.

Explanation:

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What is the osmolarity of a solution made by dissolving 0.50 mole of the strong electrolyte cacl2 in 1.0 l of solution?

valina [46]

Osmolarity=osmole of the solute/litres of the solution
ionic equation for dissociation of CaCl2 is
CaCl2--->Ca2+ +2Cl-
total osmoles for reaction are 1(Ca2+) + 2(Cl-)= 3 osmoles
therefore

0.50 moles of CaCl2 x 3 osmoles/ 1mole of CaCl2 = 1.5osmoles
osmolarity=1.5 /1.0 L=1.5 osmol/l

6 0

2 years ago

Express the van der Waals equation of state as a virial expansion in powers of 1/Vm and obtain expressions for B and C in terms

nirvana33 [79]

Answer:

$PV_{m} = RT[1 + (b-\frac{a}{RT})\frac{1}{V_{m} } + \frac{b^{2} }{V^{2} _{m} } + ...]$

B = b -a/RT

C = b^2

a = 1.263 atm*L^2/mol^2

b = 0.03464 L/mol

Explanation:

In the given question, we need to express the van der Waals equation of state as a virial expansion in powers of 1/Vm and obtain expressions for B and C in terms of the parameters a and b. Therefore:

Using the van deer Waals equation of state:

$P = \frac{RT}{V_{m}-b } - \frac{a}{V_{m} ^{2} }$

With further simplification, we have:

$P = RT[\frac{1}{V_{m}-b } - \frac{a}{RTV_{m} ^{2} }]$

Then, we have:

$P = \frac{RT}{V_{m} } [\frac{1}{1-\frac{b}{V_{m} } } - \frac{a}{RTV_{m} }]$

Therefore,

$PV_{m} = RT[(1-\frac{b}{V_{m} }) ^{-1} - \frac{a}{RTV_{m} }]$

Using the expansion:

$(1-x)^{-1} = 1 + x + x^{2} + ....$

Therefore,

$PV_{m} = RT[1+\frac{b}{V_{m} }+\frac{b^{2} }{V_{m} ^{2} } + ... -\frac{a}{RTV_{m} }]$

Thus:

$PV_{m} = RT[1 + (b-\frac{a}{RT})\frac{1}{V_{m} } + \frac{b^{2} }{V^{2} _{m} } + ...]$ equation (1)

Using the virial equation of state:

$P = RT[\frac{1}{V_{m} }+ \frac{B}{V_{m} ^{2}}+\frac{C}{V_{m} ^{3} }+ ...]$

Thus:

$PV_{m} = RT[1+ \frac{B}{V_{m} }+ \frac{C}{V_{m} ^{2} } + ...]$ equation (2)

Comparing equations (1) and (2), we have:

B = b -a/RT

C = b^2

Using the measurements on argon gave B = −21.7 cm3 mol−1 and C = 1200 cm6 mol−2 for the virial coefficients at 273 K.

$b = \sqrt{C} = \sqrt{1200} = 34.64[tex]cm^{3}/mol$ [/tex] = 0.03464 L/mol

a = (b-B)*RT = (34.64+21.7)*(1L/1000cm^3)*(0.0821)*(273) = 1.263 atm*L^2/mol^2

3 0

2 years ago

For each of the following compounds, state whether it is ionic or covalent, and if it is ionic, write the symbols for the ions i

kupik [55]

Answer :

Covalent compound : It is defined as the compound which is formed by the sharing of electrons between the atoms forming a compound.

The covalent compound are usually formed when two non-metals react.

Ionic compound : It is defined as the compound which is formed when electron gets transferred from one atom to another atom.

All the polyatomic ions always form an ionic compound.

Polyatomic ions : It is a charged species that composed of two or more atoms and these charged species are bonded by the covalent bond.

For the given options:

Option A: $KClO_4$

This compound is formed by the combination of potassium, $K^+$ which is a metal and $ClO_4^-$ ion which is a polyatomic ion. Thus, it will form an ionic compound.

Option B: $Mg(C_2H_3O_2)_2$

This compound is formed by the combination of magnesium, $Mg^{2+}$ which is a metal and $C_2H_3O_2^{-}$ ion which is a polyatomic ion. Thus, it will form an ionic compound.

Option C: $H_2S$

Hydrogen and sulfur, both are non-metals and they will form a covalent compound.

Option D: $Ag_2S$

This compound is formed by the combination of silver, $Ag^{+}$ which is a metal and sulfur, $S^{2-}$ which is a non-metal. Thus, it will form an ionic compound.

Option E: $N_2Cl_4$

Nitrogen and chlorine, both are non-metals and they will form a covalent compound.

Option F: $Co(NO_3)_2$

This compound is formed by the combination of cobalt, $Co^{2+}$ which is a metal and $NO_3^{-}$ ion which is a polyatomic ion. Thus, it will form an ionic compound.

6 0

2 years ago

When the following reaction comes to equilibrium, will the concentrations of the reactants or products be greater? Does the answ

Nutka1998 [239]

Answer:

At equilibrium, the concentration of the reactants will be greater than the concentration of the products. This does not depend on the initial concentrations of the reactants and products.

Explanation:

The value of Kc gives us an idea of the extent of the reaction. A big Kc (Kc > 1) means that in the equilibrium there are more products than reactants, and the opposite happens for a small Kc (Kc < 1). The equilibrium is reached no matter what the initial concentrations are.

The value of the equilibrium constant is relatively SMALL; therefore, the concentration of reactants will be GREATER THAN the concentration of products. This result is INDEPENDENT OF the initial concentration of the reactants and products.

4 0

3 years ago

Models need to be changed when _______.

dedylja [7]

B. When scientific understanding changes.

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2 years ago