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dimaraw [331]
2 years ago
7

In which condition does barium chloride conduct electricity? And why

Chemistry
1 answer:
Anuta_ua [19.1K]2 years ago
3 0

Answer:

When barium chloride (BaCl 2) is dissolved in water, the water conducts electricity. In what form will the dissolved BaCl 2 be found? a. as Ba 2+ and Cl - ions b. as Ba atoms and Cl 2 molecules

Explanation:

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1. In_____particles are in contact with each other.
Musya8 [376]

Answer:

condensed matter should be your answer

Explanation:

7 0
3 years ago
Hannah added a silvery white metal to a beaker of cold water and noticed that
Schach [20]

Answer:

A. Hot water increases the collision rate of the molecules, causing the

reaction to occur faster.

Explanation:

Hot water molecules move faster than cold water molecules.

3 0
3 years ago
How does the periodic table make it easier for the scientist to work with and study elements
ololo11 [35]

Answer:

The periodic table has gone through many changes since Dmitri Mendeleev drew up its original design in 1869, yet both the first table and the modern periodic table are important for the same reason: The periodic table organizes elements according to similar properties so you can tell the characteristics of an element just by looking at its location on the table

Explanation:

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3 years ago
Read 2 more answers
A 0.21 M solution of a weak acid HA dissociates such that 99.1% of the weak acid remains intact (i.e., remains as HA). To the ne
bixtya [17]

Answer:

pKa = - 1.36012

Explanation:

  • HA ↔ H+ + A-

∴ Ka = [H+] [A-] / [HA]

  • pKa = - Log Ka
  • % dissociates (%α) = ([A-] / [A-] + [HA])×100

∴ %α = 99.1%

∴ <em>C</em> HA = 0.21 M

Mass Balance:

⇒ <em>C </em>HA = [HA] + [A-] = 0.21 M............(1)

Charge balance:

⇒ [H+] = [A-] + [OH-].....[OH-] is neglected, it come from the water

⇒ [H+] = [A-]...............(2)

(2) in (1):

⇒ [HA] = 0.21 - [H+]

replacing in %α:

∴ %α = 99.1% = ([A-]/([A-]+[HA]))×100

⇒ 0.991 = [A-] / [A-] + [HA] = [A-] / 0.21 M

⇒ [A-] = (0.21 M)*(0.991) = 0.20811 M

Replacing in Ka:

⇒ Ka = [H+]² / (0.21 - [H+])

∴ [H+] = [A-] = 0.20811 M

⇒ Ka = (0.20811)² / (0.21 - 0.20811)

⇒ Ka = 22.9152

⇒ pKa = - Log (22.9152)

⇒ pKa = - 1.3601

4 0
3 years ago
Find the enthalpy of neutralization of HCl and NaOH. 137 cm3 of 2.6 mol dm-3 hydrochloric acid was neutralized by 137 cm3 of 2.6
Lerok [7]

Answer:

Approximately \rm -89 \; kJ\cdot mol^{-1}.

Assumption: the density of the solution is equal to the density of pure water.

Explanation:

The enthalpy of neutralization is defined as the enthalpy change for each moles of water produced. (Clark, Physical & Theoretical Chemistry, Chemistry Libretexts.)

Each mole of \rm NaOH formula units will neutralize one mole of \rm HCl to produce one mole of water. \rm HCl and \rm NaOH are available at equal volume and concentration. In other words, there's an equal number of both reactants. All \rm HCl and \rm NaOH will react to form water.

V(\mathrm{HCl}) = \rm 137\; cm^{3} = 0.137\;dm^{3}.

V(\mathrm{NaOH}) = \rm 137\; cm^{3} = 0.137\;dm^{3}.

n = c\cdot V = \rm 0.137\;dm^{3} \times 2.6\;mol\cdot dm^{-3} = 0.3652\; mol.

In other words, there are \rm 0.3652\; mol of \rm HCl and \rm NaOH each. The two will react to produce \rm 0.3652\; mol of water.

How much heat is released?

Assume that the volume of the liquid is equal to the volume of the \rm HCl solution plus the volume of the \rm NaOH solution. That's \rm 0.274\;dm^{3}. Assume that the density of the solution is equal to that of water under room temperature. \rho(\text{water}) = \rm 1.000\; kg\cdot dm^{-3}. The mass of the liquid will be m = \rho \cdot V = \rm 0.274\; dm^{3} \times 1.000\; kg\cdot dm^{-3} = 0.274\; kg = 274\;g.

Change in temperature:

\Delta T = \rm 325.8 - 298 = 27.8\; K.

Heat that the solution absorbed:

Q = c\cdot m \cdot \Delta T = \rm 4.18\;J\cdot K^{-1}\cdot g^{-1} \times 274\; g\times 27.8\;K = 36410.216\; J = 36.410216\; kJ.

That will also be the amount of heat released from the reaction if there's no energy loss.

\displaystyle \Delta H(\text{Neutralization}) = \frac{-Q}{n(\text{water produced})} = \rm \frac{36.410216\; kJ}{0.3652\; mol} \approx 89\; kJ\cdot mol^{-1}.

6 0
3 years ago
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