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3 years ago

An unknown solution has a ph of 8. how would you classify this solution?

1 answer:

6 0

I believe the correct answer from the choices listed above is option C. You would classify this solution as basic since it has the pH of more than 7 which signifies a basic solution. Hope this answers the question. Have a nice day. Feel free to ask more questions.

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The student placed 10 mL of PbCl2 (saturated solution) in the test tube and added a pinch of NaCl. A white precipitate of PbCl2

NNADVOKAT [17]

PbCl₂(aq) → Pb²⁺(aq) + 2Cl⁻(aq)

NaCl(aq) → Na⁺(aq) + Cl⁻(aq)

Pb²⁺(aq) + 2Cl⁻(aq) ⇄ PbCl₂(s)

At increase the concentration of chloride ions - concentration of lead ions decreases, the lead chloride is formed.

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3 years ago

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What is a wind turbine often used for?

kirill115 [55]

Answer:

Explanation:

Converting thermal energy into electrical energy

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2 years ago

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What is the volume of 60 g of ether if the density of ether is 70 g/mL

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3 years ago

Question - Complete and balance the following chemical equations:

Mekhanik [1.2K]

Answer is because
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2 years ago

The energy E of the electron in a hydrogen atom can be calculated from the Bohr formula: E = R_y/n^2 In this equation R_y stands

Katarina [22]

Answer:

The wavelength of the line in the emission line spectrum of hydrogen caused by the transition of the electron for the given energy levels is $5.23\times 10^{-5} m$

Explanation:

Given :

The energy E of the electron in a hydrogen atom can be calculated from the Bohr formula:

$E=\frac{R_y}{n^2}$

$R_y=2.18\times 10^{-18} J$ = Rydberg energy

n = principal quantum number of the orbital

Energy of 11th orbit = $E_{11}$

$E_{11}=\frac{2.18\times 10^{-18} J}{11^2}=1.80\times 10^{-20} J$

Energy of 10th orbit = $E_{10}$

$E_{10}=\frac{2.18\times 10^{-18} J}{10^2}=2.18\times 10^{-20} J$

Energy difference between both the levels will corresponds to the energy of the wavelength of the line which can be calculated by using Planck's equation.

$E'=E_{10}-E_{11}=2.18\times 10^{-20} J-1.80\times 10^{-20} J$

$=E'=0.38\times 10^{-20} J$

$\lambda =\frac{hc}{E'}$ (Planck's' equation)

$\lambda = \frac{6.626\times 10^{-34} Js\times 3\times 10^8 m/s}{0.38\times 10^{-20} J}$

$\lambda = 5.2310\times 10^{-5} m\approx 5.23\times 10^{-5} m$

The wavelength of the line in the emission line spectrum of hydrogen caused by the transition of the electron for the given energy levels is $5.23\times 10^{-5} m$

3 0

2 years ago