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3 years ago

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In a photoelectron spectroscopy experiment, photons with wavelength 130 nm cause the ejection of 6 s electrons from gold, but ph

otons with wavelength 200 nm do not. A student claims this is because the photons with wavelength 200 nm are lower in energy than photons with wavelength 130 nm . Do you agree or disagree with the student's claim? Justify your answer in terms of the energies of the photons.

1 answer:

Veronika [31]3 years ago

3 0

Answer:

See explanation

Explanation:

The energy of a photon refers to the energy carried by each photon. The energy of a photon is inversely related to its wavelength.

This simply means that the longer the wavelength of a photon, the lower the energy of the photon.

200nm is longer than 130nm, therefore, 130nm photon has more energy than 200nm photon.

This explains the fact that photons with wavelength 130 nm cause the ejection of 6 s electrons from gold, but photons with wavelength 200 nm do not.

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Identify the power of ten that defines each of these prefixes. Input your answers as 10 x where x is the power of ten. nano- ___

Gemiola [76]

<u>Answer:</u> The quantity of every prefix is written below as a power of ten.

<u>Explanation:</u>

In the metric system of measurement, the name of multiples and subdivision of any unit is done by combining the name of the unit with the prefixes.

<u>For Example:</u> deka, hecto and kilo means 10, 100 and 1000 respectively. Deci, centi and milli means one-tenth, one-hundredth, and one-thousandth respectively.

The quantity of these prefixes are written as the power of 10.

For the given prefixes:

<u>Nano:</u> The quantity will be $10^{-9}$

<u>Kilo:</u> The quantity will be $10^3$

<u>Centi:</u> The quantity will be $10^{-2}$

<u>Micro:</u> The quantity will be $10^{-6}$

<u>Milli:</u> The quantity will be $10^{-3}$

<u>Mega:</u> The quantity will be $10^6$

Hence, the quantity of every prefix is written above as a power of ten.

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3 years ago

lesantik [10]

Answer:

d. 0.208 M NaOH

Explanation:

M[NaOH] = 23+16+1= 40g/mol

2.40L = 2.4dm3

M=m/Mv

M=20.0g/40g/mol×2.4dm3

M=20.0g ÷ 96

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Given the following data:

bagirrra123 [75]

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Let the three equations with $\Delta H$ given be denoted as (1), (2), (3), and the last equation (4). Let $a$ , $b$ , and $c$ be letters such that $a \times (1) + b \times (2) + c \times (3) = (4)$ . This relationship shall hold for all chemicals involved.

There are three unknowns; it would thus take at least three equations to find their values. Species present on both sides of the equation would cancel out. Thus, let coefficients on the reactant side be positive and those on the product side be negative, such that duplicates would cancel out arithmetically. For instance, $3 + (-1) = 2$ shall resemble the number of $\text{H}_2$ left on the product side when the second equation is directly added to the third. Similarly

$\text{NH}_4 \text{Cl} \; (s)$ : $-2 \; a = 1$
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Thus

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Verify this conclusion against a fourth species involved- $\text{N}_2 \; (g)$ for instance. Nitrogen isn't present in the net equation. The sum of its coefficient shall, therefore, be zero.

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Apply the Hess's Law based on the coefficients to find the enthalpy change of the last equation.

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