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2 years ago

What would be the predominant state of matter if there were no intermolecular forces?

1 answer:

4 0

Stronger Intermolecular forces (greater attractions) hold molecules together better. Meaning the molecules don't want to leave whatever liquid they are part of. for example, chloroform(the knock out gas), which has only weak intermolecular forces evaporates very quickly at room temp. However, water which participates in H-bonding, strong intermolecular forces, evaporates slowly at room temp. The colder a liquid is, the harder it is for the molecules to overcome their intermolecular bonds and break free. a.k.a. slow evaporation. but when a liquid is heated the molecules have enough energy to overcome their brotherly bonds and go into the gas phase.

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According to Table I, which equation represents a change resulting in the greatest quantity of energy released?

love history [14]

The answer is 3. The releasing of energy means exothermic reaction. So the ΔH should be negative. And the greatest quantity of energy released means that the greatest number. So according to the table I, the answer is 3.

8 0

3 years ago

Read 2 more answers

(02.02 LC) The two main types of weathering are Select one:

Firlakuza [10]

Answer:

Explanation:

4 0

2 years ago

In the laboratory you dissolve 13.9 g of potassium phosphatein a volumetric flask and add water to a total volume of 250mL.

tensa zangetsu [6.8K]

Answer:

Molarity of the solution? 0,262 M.

Concentration of the potassium cation? 0,786 M.

Concentration of the phosphate anion? 0,262 M.

Explanation:

Potassium phosphate (K₃PO₄; 212,27 g/mol) dissolves in water thus:

K₃PO₄ → 3 K⁺ + PO₄³⁻ (1)

Molarity is an unit of chemical concentration given in moles of solute (K₃PO₄) per liters of solution.

There are 250 mL of solution≡0,25 L

The moles of K₃PO₄ are:

13,9 g of K₃PO₄ × $\frac{1mol}{212,27 g}$ = 0,0655 moles of K₃PO₄

The molarity of the solution is:

$\frac{0,0655 moles}{0,25L}$ = 0,262 M

In (1) you can see that 1 mole of K₃PO₄ produces 3 moles of potassium cation. The moles of potassium cation are:

0,0655 moles×3 = 0,1965 moles

The concentration is:

$\frac{0,1965 moles}{0,25L}$ = 0,786 M

The moles of K₃PO₄ are the same than moles of PO₄³⁻, thus, concentration of phosphate anion is the same than concentration of K₃PO₄. 0,262 M

I hope it helps!

4 0

3 years ago

The concentration of a solution can be expressed in

puteri [66]

The concentration of a solution can be expressed in (4) moles per liter~

3 0

2 years ago

Read 2 more answers

Calculate the standard reaction Gibbs free energy for the following cell reactions: (a) 2 Ce41(aq) 1 3 I2(aq) S 2 Ce31(aq) 1 I32

Law Incorporation [45]

Answer:

For a: The standard Gibbs free energy of the reaction is -347.4 kJ

For b: The standard Gibbs free energy of the reaction is 746.91 kJ

Explanation:

Relationship between standard Gibbs free energy and standard electrode potential follows:

$\Delta G^o=-nFE^o_{cell}$ ............(1)

For a:

The given chemical equation follows:

$2Ce^{4+}(aq.)+3I^{-}(aq.)\rightarrow 2Ce^{3+}(aq.)+I_3^-(aq.)$

Oxidation half reaction: $Ce^{4+}(aq.)\rightarrow Ce^{3+}(aq.)+e^-$ ( × 2)

Reduction half reaction: $3I^_(aq.)+2e^-\rightarrow I_3^-(aq.)$

We are given:

$n=2\\E^o_{cell}=+1.08V\\F=96500$

Putting values in equation 1, we get:

$\Delta G^o=-2\times 96500\times (+1.80)=-347,400J=-347.4kJ$

Hence, the standard Gibbs free energy of the reaction is -347.4 kJ

For b:

The given chemical equation follows:

$6Fe^{3+}(aq.)+2Cr^{3+}+7H_2O(l)(aq.)\rightarrow 6Fe^{2+}(aq.)+Cr_2O_7^{2-}(aq.)+14H^+(aq.)$

Oxidation half reaction: $Fe^{3+}(aq.)\rightarrow Fe^{2+}(aq.)+e^-$ ( × 6)

Reduction half reaction: $2Cr^{2+}(aq.)+7H_2O(l)+6e^-\rightarrow Cr_2O_7^{2-}(aq.)+14H^+(aq.)$

We are given:

$n=6\\E^o_{cell}=-1.29V\\F=96500$

Putting values in equation 1, we get:

$\Delta G^o=-6\times 96500\times (-1.29)=746,910J=746.91kJ$

Hence, the standard Gibbs free energy of the reaction is 746.91 kJ

7 0

3 years ago