All categories

2 years ago

I could really use some help with this question, thank you in advance!

Chemistry

1 answer:

Taya2010 [7]2 years ago

6 0

C because
Explanation Plato

You might be interested in

Which scenario ensures that Earth’s global temperature is maintained?

Angelina_Jolie [31]

Answer:

The energy that flows into the Earth system is equal to the energy that flows out of the Earth system.

Explanation:

Because I geussed it right

3 0

2 years ago

Read 2 more answers

How many moles is 2.2 x 1023 atoms of Fe?

kow [346]

Answer:

it's 3.77 moles

Explanation:

Pls give brainliest :)

4 0

2 years ago

13.0 g NaHSO4 is dissolved in water to make a 2.00 L solution what is the molarité

Sati [7]

Answer:

0.054 M

Explanation:

1 mol NaHSO4 -> 120 g

x ->13 g

x= 0.108 mol NaHSO4

M= mol solute/ L solution

M= 0.108 mol NaHSO4/ 2.00L

M= 0.054 M

8 0

3 years ago

Octasulfur reacts with pure oxygen to produce sulfur dioxide. Determine the percent

Margaret [11]

Answer:

Octasulfur is just S8. Eight S atoms in

a sort of crown shape. Sulfur Dioxide is a gas, SO2. Does that help?

Explanation:

5 0

3 years ago

The energy E of the electron in a hydrogen atom can be calculated from the Bohr formula: E = R_y/n^2 In this equation R_y stands

Katarina [22]

Answer:

The wavelength of the line in the emission line spectrum of hydrogen caused by the transition of the electron for the given energy levels is $5.23\times 10^{-5} m$

Explanation:

Given :

The energy E of the electron in a hydrogen atom can be calculated from the Bohr formula:

$E=\frac{R_y}{n^2}$

$R_y=2.18\times 10^{-18} J$ = Rydberg energy

n = principal quantum number of the orbital

Energy of 11th orbit = $E_{11}$

$E_{11}=\frac{2.18\times 10^{-18} J}{11^2}=1.80\times 10^{-20} J$

Energy of 10th orbit = $E_{10}$

$E_{10}=\frac{2.18\times 10^{-18} J}{10^2}=2.18\times 10^{-20} J$

Energy difference between both the levels will corresponds to the energy of the wavelength of the line which can be calculated by using Planck's equation.

$E'=E_{10}-E_{11}=2.18\times 10^{-20} J-1.80\times 10^{-20} J$

$=E'=0.38\times 10^{-20} J$

$\lambda =\frac{hc}{E'}$ (Planck's' equation)

$\lambda = \frac{6.626\times 10^{-34} Js\times 3\times 10^8 m/s}{0.38\times 10^{-20} J}$

$\lambda = 5.2310\times 10^{-5} m\approx 5.23\times 10^{-5} m$

The wavelength of the line in the emission line spectrum of hydrogen caused by the transition of the electron for the given energy levels is $5.23\times 10^{-5} m$

3 0

2 years ago

I could really use some help with this question, thank you in advance!​

I could really use some help with this question, thank you in advance!