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1 month ago

Limes have a [H3O+] of 1.3 x 10-2 mol/L. Their pOH is

1 answer:

4 0

To calculate the pOH with water as the universal solvent, take the value of 10 ^ -14, divide it by the hydronium ion concentration, and then apply the negative logarithm to find the hydroxide ion concentration in the solution.

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After heating the solution, you decant the liquid and wash the remaining solid with distilled water. What you are removing by th

lorasvet [2795]

What is being removed during the wash is the solvent.

7 0

2 months ago

Which of the following refers to a chemical property?. . a. At room temperature, mercury is a liquid, but gold is a solid.. . b.

alisha [2963]

Among the listed options, the one indicating a chemical property is B. Water does not ignite, whereas gasoline is capable of burning. This distinction arises because both combustion and flammability are categorized as chemical properties.

6 0

2 months ago

What salt is produced from mixing csoh and h2co3

Anarel [2989]

Answer:

The resulting salt is Cesium Carbonate

Explanation:

2CsOH + H2CO3 → Cs2CO3 + 2H2O

3 0

2 months ago

A metallic object holds a charge of −4.8 × 10−6 C. What total number of electrons does this represent? (e = 1.6 × 10−19 C is the

Anarel [2989]

Answer:

$n=3.0\times 10^{13}$

Explanation:

Charge of one electron = $-1.6\times 10^{-19}\ C$

The formula for calculating charge is:

$Charge=n\times q_e$

Given that: Charge = $-4.8\times 10^{-6}\ C$

$-4.8\times 10^{-6}=n\times (-1.6\times 10^{-19})$

$n=\frac{4.8\times 10^{-6}}{1.6\times 10^{-19}}=3.0\times 10^{13}$

Total electrons, n = $3.0\times 10^{13}$

5 0

3 months ago

The vapor pressure of benzene at 298 K is 94.4 mm of Hg. The standard molar Gibbs free energy of formation of liquid benzene at

Alekssandra [3086]

Answer:

ΔfG°(C₆H₆(g)) = 129.7kJ/mol

Explanation:

Identifying the given parameters from the question;

Vapor pressure = 94.4 mm of Hg

The reaction for vaporization is expressed as;

C₆H₆(l) ⇄ C₆H₆(g)

The equilibrium in terms of activities can be defined as:

K = a(C₆H₆(g)) / a(C₆H₆(l))

The activity for pure substances equals one:

a(C₆H₆(l)) = 1

For an ideal gas phase, activity is approximated as the ratio of partial pressure to total pressure. Under standard conditions:

K = p(C₆H₆(g)) / p°

Where p° = 1atm = 760mmHg is the standard pressure

Thus, we find;

K = 94mmHg / 760mmHg = 0.12421

The formula for Gibbs free energy is:

ΔG = - R·T·ln(K)

Here, R represents the gas constant = 8.314472J/molK

Consequently, the ΔG° for the vaporization of benzene is calculated as:

ΔvG° = - 8.314472 · 298.15 · ln(0.12421)

ΔvG° = 5171J/mol = 5.2kJ/mol

The change in Gibbs free energy for the reaction is determined by the difference between the Gibbs free energy of formation of the products and reactants:

ΔvG° = ΔfG°(C₆H₆(g)) - ΔfG°(C₆H₆(l))

<pThus:

ΔfG°(C₆H₆(g)) = ΔvG° + ΔfG°(C₆H₆(l))

ΔfG°(C₆H₆(g)) = 5.2kJ/mol + 124.5kJ/mol

ΔfG°(C₆H₆(g)) = 129.7kJ/mol

6 0

3 months ago

Limes have a [H3O+] of 1.3 x 10-2 mol/L. Their pOH is​

Limes have a [H3O+] of 1.3 x 10-2 mol/L. Their pOH is