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dimaraw [331]
3 years ago
9

How many atoms are in 2.85x10^3 g of gallium

Chemistry
1 answer:
OLga [1]3 years ago
8 0

Answer:

2850 grams I hope this help

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The figure below demonstrates what happens to white light when it passes through a prism. Which colors of lights has the highest
Yuri [45]

Answer:

Blues

Explanation:

Colors of Rainbow => ROYGBIV

Red    Orange   Yellow   Green   Blue   Indigo   Violet

low energy  => => => => => => => => => => high energy

low frequency => => => => => => => => =>  high frequency

long wavelengths => => => => => => => => short wavelengths

Note: All L's on the Left => low, low, long; other end are opposites.

7 0
3 years ago
I have doubts about the air pollution puzzle. Can anyone help? Sarah replied, but I can't see her response
Eva8 [605]

Answer:

wdym exactly?

Explanation:

7 0
3 years ago
Whats the meaning<br> Speed ?<br> and the<br> meaning of<br> of Motion?
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5 0
3 years ago
Perform the following operation and express the answer in scientific notation. 1.26x104 + 2.50x104 [?]x10​
ss7ja [257]
Since your multiplying by the same number, 104, both terms, you can group them like this:
104(1.26+2.50)
104(3.76)
391.04

Knowing that scientific notation needs to be a number x, that’s greater or equal to 1, and less then 10.
3.9104 * 10^2
4 0
3 years ago
Given the wavelength of the corresponding emission line, calculate the equivalent radiated energy from n = 4 to n = 2 in both jo
lions [1.4K]

Explanation:

It is given that,

Initial orbit of electrons, n_i=4

Final orbit of electrons, n_f=2

We need to find energy, wavelength and frequency of the wave.

When atom make transition from one orbit to another, the energy of wave is given by :

E=-13.6(\dfrac{1}{n_f^2}-\dfrac{1}{n_i^2})

Putting all the values we get :

E=-13.6(\dfrac{1}{(4)^2}-\dfrac{1}{(2)^2})\\\\E=2.55\ eV

We know that : 1\ eV=1.6\times 10^{-19}\ J

So,

E=2.55\times 1.6\times 10^{-19}\\\\E=4.08\times 10^{-19}\ J

Energy of wave in terms of frequency is given by :

E=hf

f=\dfrac{E}{h}\\\\f=\dfrac{4.08\times 10^{-19}}{6.63\times 10^{-34}}\\\\f=6.14\times 10^{14}\ Hz

Also, c=f\lambda

\lambda is wavelength

So,

\lambda=\dfrac{c}{f}\\\\\lambda=\dfrac{3\times 10^8}{6.14\times 10^{14}}\\\\\lambda=4.88\times 10^{-7}\ m\\\\\lambda=488\ nm

Hence, this is the required solution.

4 0
3 years ago
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