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3 years ago

What is the symbol for copper

Chemistry

2 answers:

bekas [8.4K]3 years ago

7 0

The symbol for copper is CU, which means copper cuprum, the Latin word for copper.

Juli2301 [7.4K]3 years ago

4 0

Cu is the symbol and you just add 35 plus 29 to find the atomic mass and 29 is the atomic number because the atomic number is the same as the protons.

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Determine the a) energy (in eV) and b) wavelength (in cm) corresponding to blue light of frequency 670 THz

Lesechka [4]

Answer :

(a) The energy of blue light (in eV) is 2.77 eV

(b) The wavelength of blue light is $4\times 10^{-5}cm$

Explanation:

The relation between the energy and frequency is:

$Energy=h\times Frequency$

where,

h = Plank's constant = $6.626\times 10^{-34}J.s$

Given :

Frequency = $670THz=670\times 10^{12}s^{-1}$

Conversion used :

$1THz=10^{12}Hz\\1Hz=1s^{-1}\\1THz=10^{12}s^{-1}$

So,

$Energy=(6.626\times 10^{-34}J.s)\times (670\times 10^{12}s^{-1})$

$Energy=4.44\times 10^{-19}J$

Also,

$1J=6.24\times 10^{18}eV$

So,

$Energy=(4.44\times 10^{-19})\times (6.24\times 10^{18}eV)$

$Energy=2.77eV$

The energy of blue light (in eV) is 2.77 eV

The relation between frequency and wavelength is shown below as:

$Frequency=\frac{c}{Wavelength}$

Where,

c = the speed of light = $3\times 10^8m/s$

Frequency = $670\times 10^{12}s^{-1}$

So, Wavelength is:

$670\times 10^{12}s^{-1}=\frac{3\times 10^8m/s}{Wavelength}$

$Wavelength=\frac{3\times 10^8m/s}{670\times 10^{12}s^{-1}}=4\times 10^{-7}m=4\times 10^{-5}cm$

Conversion used : $1m=100cm$

The wavelength of blue light is $4\times 10^{-5}cm$

7 0

3 years ago

Find the standard enthalpy of formation of ethylene, C2H4(g), given the following data: C2H4(g) + 3 O2(g) --> 2CO2(g) + 2 H2O

Brums [2.3K]

Answer : The standard enthalpy of formation of ethylene is, 52.4 kJ

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The formation reaction of $C_2H_4$ will be,