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3 years ago

1. I am in group two period three. What element am I? A Ti B, C C. Ca D. WE

Chemistry

2 answers:

disa [49]3 years ago

6 0

It is magnesium (mg) but that is not a answer choice, so I would go with Ca since that’s in group 2

AysviL [449]3 years ago

5 0

Answer:

Explanation:

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Molten gallium reacts with arsenic to form the semiconductor, gallium arsenide, GaAs, used in light-emitting diodes and solar ce

Vsevolod [243]

Answer:

a) 1.2g of arsenic b) 0.64g of arsenic c) 3.481g of gallium d) 3.806g of gallium e) 2.61g arsenic

Explanation:

The balanced equation is:

Ga + As = GaAs, 1:1 mole ratio

a) mass (gallium)/ molar mass of Ga = 4/ 69.723 = 0.0574mol

Mass (arsenic)/ molar mass of As = 5.5/74.9216 = 0.0734, subtracting the moles from each other (knowing already that the ratio is 1:1), arsenic is in excess by 1.2g

b) repeating the procedure (changing the values)

It will be 0.0574 to 0.06593

Arsenic is in excess by 0.00854

0.00854* mass of arsenic (such must be done for the first remaining mole) = 0.64g

c) the mole ratio is 0.0574: 0.00747

Gallium is in excess by 0.05

Mass of excess gallium = 0.05* 69.723 = 3.481g

d) using the mass given, the new ratio is

0.128: 0.0734

Gallium is in excess by 0.054mol

Mass of excess gallium = 0.054*69.723 = 3.806g

e) using the mass again, the new ratio is 0.0574: 0.02,

Gallium is in excess by 0.0374*69.723 = 2.61g

7 0

3 years ago

Match the vocabulary with the definitions.

Fiesta28 [93]

1: Decomposition reaction
2: Combination reaction
3: product
4: Reactant

8 0

2 years ago

Mercury’s atomic emission spectrum is shown below. Estimate the wavelength of the orange line. What is its frequency? What is th

lions [1.4K]

The wavelength of the orange line is 610 nm, the frequency of this emission is 4.92 x 10¹⁴ Hz and the energy of the emitted photon corresponding to this orange line is 3.26 x 10⁻¹⁹ J.

"Your question is not complete, it seems to be missing the diagram of the emission spectrum"

the diagram of the emission spectrum has been added.

From the given chart;

The wavelength of the atomic emission corresponding to the orange line is 610 nm = 610 x 10⁻⁹ m

The frequency of this emission is calculated as follows;

c = fλ

where;

c is the speed of light = 3 x 10⁸ m/s
f is the frequency of the wave
λ is the wavelength

$f = \frac{c}{\lambda } \\\\f = \frac{3\times 10^8}{610 \times 10^{-9}} \\\\f = 4.92 \times 10^{14} \ Hz$

The energy of the emitted photon corresponding to the orange line is calculated as follows;

E = hf

where;

h is Planck's constant = 6.626 x 10⁻³⁴ Js

E = (6.626 x 10⁻³⁴) x (4.92 x 10¹⁴)

E = 3.26 x 10⁻¹⁹ J.

Thus, the wavelength of the orange line is 610 nm, the frequency of this emission is 4.92 x 10¹⁴ Hz and the energy of the emitted photon corresponding to this orange line is 3.26 x 10⁻¹⁹ J.

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