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nydimaria [60]
2 years ago
5

Write the balanced molecular equation for the neutralization reaction between HI and Ba(OH)2 in aqueous solution. Include physic

al states.
molecular equation
Chemistry
1 answer:
Marta_Voda [28]2 years ago
5 0

neutralization reaction = acid+base = salt + water

molecular equation = balanced equation

2HI(aq) + Ba(OH)₂(aq) ⇒ BaI₂(aq) + 2H₂O(l)

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8 0
3 years ago
15A.4(a) What is the temperature of a two-level system of energy separation equivalent to 400 cm−1 when the population of the up
maksim [4K]

Answer:

T = 525K    

Explanation:

The temperature of the two-level system can be calculated using the equation of Boltzmann distribution:

\frac{N_{i}}{N} = e^{-\Delta E/kT}  (1)

<em>where Ni: is the number of particles in the state i, N: is the total number of particles, ΔE: is the energy separation between the two levels, k: is the Boltzmann constant, and T: is the temperature of the system </em>         

The energy between the two levels (ΔE) is:

\Delta E = hck    

<em>where h: is the Planck constant, c: is the speed of light and k: is the wavenumber</em>      

\Delta E = 6.63\cdot 10^{-34} J.s \cdot 3\cdot 10^{8}m/s \cdot 4 \cdot 10^{4}m^{-1} = 7.96 \cdot 10^{-21}J  

Solving the equation (1) for T:

T = \frac{-\Delta E}{k \cdot Ln(N_{i}/N)}  

<em>With Ni = N/3 and k = 1.38x10⁻²³ J/K, </em><em>the temperature of the two-level system is:</em><em> </em>

T = \frac{-7.96 \cdot 10^{-21}J}{1.38 \cdot 10^{-23} J/K \cdot Ln(N/3N)} = 525K                                  

I hope it helps you!

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What is the [OH-] of a solution prepared by dissolving 0.0912 g of hydrogen chloride in sufficient pure water to prepare 250.0 m
velikii [3]

Answer: The [OH^-] of a solution is 10^{-12} M

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.

Molarity=\frac{n\times 1000}{V_s}

where,

n = moles of solute

V_s = volume of solution in ml

moles of HCl = \frac{\text {given mass}}{\text {Molar mass}}=\frac{0.0912g}{36.5g/mol}=0.0025mol

Now put all the given values in the formula of molality, we get

Molarity=\frac{0.0025\times 1000}{250}=0.01

pH or pOH is the measure of acidity or alkalinity of a solution.

HCl\rightarrow H^++Cl^{-}

According to stoichiometry,

1 mole of HCl gives 1 mole of H^+

Thus 0.01 moles of HCl gives =\frac{1}{1}\times 0.01=0.01 moles of H^+

Putting in the values:

[H^+][OH^-]=10^{-14}

[0.01][OH^-]=10^{-14}

[OH^-]=10^{-12}

Thus the [OH^-] of a solution prepared by dissolving 0.0912 g of hydrogen chloride in sufficient pure water to prepare 250.0 ml of solution is 10^{-12} M

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