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4 years ago

How do water molecules dissolve a salt?

Chemistry

1 answer:

Artist 52 [7]4 years ago

5 0

Answer:

Water can dissolve salt because the positive part of water molecules attracts the negative chloride ions and the negative part of water molecules attracts the positive sodium ions. The amount of a substance that can dissolve in a liquid (at a particular temperature) is called the solubility of the substance.

Explanation:

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Which best describes a difference between prokaryotic and eukaryotic cells

tekilochka [14]

Answer:Please find the attachment

Explanation:

7 0

4 years ago

Phosphoric acid has 3 pka values, which are 2.1, 6.9, and 12.4. draw the protonated form of phosphoric acid associated with the

Margaret [11]

As mentioned above, phosphoric acid has 3 pKa values, and after 3 ionization it gives 3 types of ions at different pKa values:

H₃PO₄(aq) + H₂O(l) ⇌ H₃O⁺(aq) + H₂PO₄⁻ (aq) pKₐ₁


H₂PO₄⁻(aq) + H₂O(l) ⇌ H₃O⁺(aq) + HPO₄²⁻ (aq) pKₐ₂

HPO₄²⁻(aq) + H₂O(l) ⇌ H₃O⁺(aq) + PO₄³⁻ (aq) pKₐ₃

At the highest pKa value (12.4) of phosphoric acid, the last OH group will lose its hydrogen. On the picture I attached, it is shown required protonated form of phosphoric acid before reaction whose pKa value is 12.4.

6 0

4 years ago

Natural gas is a mixture of many substances, primarily CH₄, C₂H₆, C3H8, and C4₄H₁₀. assuming that the total pressure of the gase

olchik [2.2K]

Answer:

A. The partial pressure for CH4 = 0.0925atm

B. The partial pressure for C2H6 = 0.925atm

C. The partial pressure for C3H8 = 0.346atm

D. The partial pressure for C4H10 = 0.115atm

Explanation:

Total pressure = 1.48atm

Total mole = 0.4+4+1.5+0.5=6.4

A. Mole fraction of CH4 = 0.4/6.4 = 0.0625

The partial pressure for CH4 = 0.0625 x 1.48 = 0.0925atm

B. Mole fraction of C2H6 = 4/6.4 = 0.625

The partial pressure for C2H6 = 0.625 x 1.48 = 0.925atm

C. Mole fraction of C3H8 = 1.5/6.4 = 0.234

The partial pressure for C3H8 = 0.234 x 1.48 = 0.346atm

D. Mole fraction of C4H10 = 0.5/6.4 = 0.078

The partial pressure for C4H10 = 0.078 x 1.48 = 0.115atm

7 0

3 years ago

For a gaseous reaction, standard conditions are 298 K and a partial pressure of 1 atm for all species. For the reaction 2 NO ( g

enyata [817]

Answer:

ΔG = -61.5 kJ/mol (Spontaneous process)

Explanation:

2 NO (g) + O₂ (g) ⇄ 2NO₂ (g)

Let's apply the thermodynamic formula to calculate the ΔG

ΔG = ΔG° + R .T . lnQ

We don't know if the gases are at equilibrium, that's why we apply Q (reaction quotient)

ΔG = - 69 kJ/mol + 8.31x10⁻³ kJ/K.mol . 298K . ln Q

How can we know Q? By the partial pressures (Qp)

P NO = 0.450atm

PO₂ = 0.1 atm

PNO₂ = 0.650 atm

Qp = [NO₂]² / [NO]² . [O₂]

Qp = 0.650² / 0.450² . 0.1 = 20.86

ΔG = - 69 kJ/mol + 8.31x10⁻³ kJ/K.mol . 298K . ln 20.86

ΔG = -61.5 kJ/mol (Spontaneous process)

5 0

3 years ago

A gas stream containing n-hexane in nitrogen with a relative saturation of 0.58 (as a fraction, multiply by 100% if you prefer %

kondor19780726 [428]

This problem is describing a gas mixture whose mole fraction of hexane in nitrogen is 0.58 and which is being fed to a condenser at 75 °C and 3.0 atm, obtaining a product at 3.0 atm and 20 °C, so that the removed heat from the system is required.

In this case, it is recommended to write the enthalpy for each substance as follows:

$H_{C-6}=y_{C-6}C_v(T_b-Ti)+\Delta _vH+C_v(T_f-Tb)\\\\H_{N_2}=y_{N_2}C_v(T_f-Ti)$

Whereas the specific heat of liquid and gaseous n-hexane are about 200 J/(mol*K) and 160 J/(mol*K) respectively, its condensation enthalpy is 31.5 kJ/mol, boiling point is 69 °C and the specific heat of gaseous nitrogen is about 29.1 J/(mol*K) according to the NIST data tables and $y_{C-6}$ and $y_{N_2}$ are the mole fractions in the gaseous mixture. Next, we proceed to the calculation of both heat terms as shown below:

$H_{C-6}=0.58*200(69-75)+(-31500)+160(20-69)=-40036J/mol\\\\H_{N_2}=0.42*29.1(20-75)=-672.21J/mol$

It is seen that the heat released by the nitrogen is neglectable in comparison to n-hexanes, however, a rigorous calculation is being presented. Then, we add the previously calculated enthalpies to compute the amount of heat that is removed by the condenser:

$Q=-40036+(-672.21)=-40708.21J$

Finally we convert this result to kJ:

$Q=-40708.21J*\frac{1kJ}{1000J}\\\\Q=-40.7kJ$

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6 0

2 years ago