What is the ph of a solution with a hydrogen ion h+ concentration of 10-8 m?

SpyIntel [72]

Double indemnity protection provides double benefits if death is the result of an accident.

the answer is d)

4 0

3 years ago

In the mass spectrum of the molecule phenol, C6H5OH, the approximate intensity of the peak at m/z 95, relative to the molecular

klasskru [66]

The correct answer to thi

6 0

4 years ago

what electronic transition in a hydrogen atom starting from n=7 energy level will produce infrared radiation with energy of 55.1

Lostsunrise [7]

Answer:

(n = 7) ⟶ (n = 4)

Explanation:

1. Convert the energy to joules per mole of electrons.

E = 55.1 × 1000 = 55 100 J/mol

2. Convert the energy to joules per electron

E = 55 100/(6.022 × 10²³)

E = 9.150 × 10⁻²⁰ J/electron

3. Use the Rydberg equation to calculate the transition

Rydberg's original formula was in terms of wavelengths, but we can rewrite it to have the units of energy. The formula then becomes

$\Delta E = R_{\text{H}} (\frac{ 1}{ n_{f}^{2}} - \frac{ 1}{ n_{i}^{2}})$

where

$R_{\text{H}}$ = the Rydberg constant = 2.178 × 10⁻¹⁸ J

$n_{i}$ and $n_{f}$ are the initial and final energy levels.

$9.150 \times 10^{-20} = 2.178 \times 10^{-18}(\frac{ 1}{ n_{f}^{2}} - \frac{ 1}{ 7^{2}})$

$\frac{9.150 \times 10^{-20} }{2.178 \times 10^{-18}} = \frac{ 1}{ n_{f}^{2}} - \frac{ 1}{ 49}$

$\frac{ 1}{ n_{f}^{2}} = \text{0.042 01} + \frac{1 }{49 }$

$\frac{ 1}{ n_{f}^{2}} = \text{0.042 01 + 0.020 41}$

$\frac{ 1}{ n_{f}^{2}} = \text{0.062 42}$

$n_{f}^{2} = \frac{1 }{ \text{0.062 42}}$

$n_{f}^{2} = 16.02$

$n_{f} = \sqrt{16.02}$

$n_{f} = 4.003 \approx 4$

4 0

3 years ago

What does the rate of a reaction determine?

mafiozo [28]

Answer:

Reaction Rate is the measure of the change in concentration of the disappearance of reactants or the change in concentration of the appearance of products per unit time

6 0

3 years ago

An element has a metallic-gray appearance. It also has the properties of both a metal and a nonmetal. In which section of the ta

katrin2010 [14]

Answer:

you would find it in section 3 as these are semi-metals

Explanation:

3 0

3 years ago

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