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NNADVOKAT [17]
3 years ago
7

What is the number on the top left of the carbon element

Chemistry
1 answer:
Fynjy0 [20]3 years ago
3 0
It is 6 which is the atomic number of carbon.
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A certain shade of blue has a frequency of 7.26 × 1014 Hz. What is the energy of exactly one photon of this light?
Gekata [30.6K]
7.20 it the energy because


10\7.26 is 7.20
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3 years ago
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What is the speed of light of an orange light with a wavelength of 600nm and a frequency of 5.00x1014?
vovangra [49]

Answer:

3 × 10^8 m/s

Explanation:

The wavelength, can be calculated by using the following formula;

λ = v/f

Where;

λ = wavelength (m)

v = velocity/speed of light (m/s)

f = frequency (Hz)

According to the provided information in this question, λ = 600nm i.e. 600 × 10^-9m, f = 5.00 x 10^14 Hz

Hence, using λ = v/f

v = λ × f

v = 600 × 10^-9 × 5.00 x 10^14

v = 6 × 10^-7 × 5.00 x 10^14

v = 30 × 10^(-7 + 14)

v = 30 × 10^ (7)

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2 years ago
Calculate the standard free-energy change at 25 ∘C for the following reaction:
lianna [129]

Answer:

Standard free-energy change at 25^{0}\textrm{C} is -3.80\times 10^{2}kJ/mol

Explanation:

Oxidation: Mg(s)-2e^{-}\rightarrow Mg^{2+}(aq.)

Reduction: Fe^{2+}(aq.)+2e^{-}\rightarrow Fe(s)

--------------------------------------------------------------------------------------

Overall: Mg(s)+Fe^{2+}(aq.)\rightarrow Mg^{2+}(aq.)+Fe(s)

Standard cell potential, E_{cell}^{0}=E_{Fe^{2+}\mid Fe}^{0}-E_{Mg^{2+}\mid Mg}^{0}

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We know, standard free energy change at 25^{0}\textrm{C}(\Delta G^{0}): \Delta G^{0}=-nFE_{cell}^{0}

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Here n = 2

So, \Delta G^{0}=-(2)\times (96500C/mol)\times (1.97V)=-380210J/mol=-380.21kJ/mol=-3.80\times 10^{2}kJ/mol

8 0
3 years ago
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What is the current model of atom
leonid [27]

Answer:

The bohr model is the model in use today

7 0
2 years ago
1. The multiplicative inverse of 5/9 is.....<br>a-9/5<br>b-9/5<br>c-3/9<br>d-None of these​
Luda [366]

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None of these cause the correct answer is 9/-5

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