It takes 348 kJ/mol to break a carbon-carbon single bond. Calculate the maximum wavelength of light for which a carbon-carbon single bond could be broken by absorbing a single photon. Round your answer to correct number of significant digits

Answer: 344 nm

Explanation:

$E=\frac{Nhc}{\lambda}$

E= energy = 348kJ= 348000 J (1kJ=1000J)

N = avogadro's number = $6.023\times 10^{23}$

h = Planck's constant = $6.626\times 10^{-34}Js
$

c = speed of light = $3\times 10^8ms^{-1}$

$348000=\frac{6.023\times 10^{23}\times 6.626\times 10^{-34}\times 3\times 10^8}{\lambda}$

$\lambda=\frac{6.023\times 10^{23}\times 6.626\times 10^{-34}\times 3\times 10^8}{348000}$

$\lambda=3.44\times 10^{-7}m=344nm$ $1nm=10^{-9}m$

Thus the maximum wavelength of light for which a carbon-carbon single bond could be broken by absorbing a single photon is 344 nm

5 0

2 years ago

PLEASE HELP ME URGENT!!! How do i find mass when not given in chemistry

Arisa [49]

Answer:

It depends what other values you have. Can you give more info? If they give density then you can solve for m.

3 0

2 years ago

Why did earths early atmosphere lose its hydrogen and helium gases

Masteriza [31]

these very light gases from escaping into space.

4 0

2 years ago

Read 2 more answers

What masses of monobasic and dibasic sodium phosphate will you use to make 250 mL of 0.1 M sodium phosphate buffer, pH = 7?

Oduvanchick [21]

Answer:

Mass of monobasic sodium phosphate = 1.857 g
Mass of dibasic sodium phosphate = 1.352 g

Explanation:

The equilibrium that takes place is:

H₂PO₄⁻ ↔ HPO₄⁻² + H⁺ pka= 7.21 (we know this from literature)

To solve this problem we use the Henderson–Hasselbalch (H-H) equation:

pH = pka + $log\frac{[A^{-} ]}{[HA]}$

In this case [A⁻] is [HPO₄⁻²], [HA] is [H₂PO₄⁻], pH=7.0, and pka = 7.21

If we use put data in the H-H equation, and solve for [HPO₄⁻²], we're left with:

$7.0=7.21+log\frac{[HPO4^{-2} ]}{[H2PO4^{-} ]}\\ -0.21=log\frac{[HPO4^{-2} ]}{[H2PO4^{-} ]}\\\\10^{-0.21} =\frac{[HPO4^{-2} ]}{[H2PO4^{-} ]}\\0.616 * [H2PO4^{-}] = [HPO4^{-2}]$

From the problem, we know that [HPO₄⁻²] + [H₂PO₄⁻] = 0.1 M

We replace the value of [HPO₄⁻²] in this equation:

0.616 * [H₂PO₄⁻] + [H₂PO₄⁻] = 0.1 M

1.616 * [H₂PO₄⁻] = 0.1 M

[H₂PO₄⁻] = 0.0619 M

With the value of [H₂PO₄⁻] we can calculate [HPO₄⁻²]:

[HPO₄⁻²] + 0.0619 M = 0.1 M

[HPO₄⁻²] = 0.0381 M

With the concentrations, the volume and the molecular weights, we can calculate the masses:

Molecular weight of monobasic sodium phosphate (NaH₂PO₄)= 120 g/mol.
Molecular weight of dibasic sodium phosphate (Na₂HPO₄)= 142 g/mol.

mass of NaH₂PO₄ = 0.0619 M * 0.250 L * 120 g/mol = 1.857 g

mass of Na₂HPO₄ = 0.0381 M * 0.250 L * 142 g/mol = 1.352 g

5 0

3 years ago