All categories

2 years ago

A Compound Contains Two or more _______.

Chemistry

2 answers:

timurjin [86]2 years ago

7 0

A compound contains two or more elements chemically combined

hammer [34]2 years ago

6 0

a compound contains two or more elements chemically combined . hope this helps

You might be interested in

Please get me the right answer

prisoha [69]

Answer:

Explanation:

7 0

2 years ago

Why fluorine is more electronegative than both oxygen and chlorine.

ziro4ka [17]

Why chlorine has highest electron affinity than fluorine?

This is because the atomic radius increases down a group. The electron gained ends up in the outermost shell. ... Fluorine, which is higher up the group then chlorine, has a lower electron affinity. This is because the electrons in the outermost shell of a fluorine atom are closer together.

4 0

1 year ago

What happens when phenol is treated with bromine water?

aleksley [76]

Answer:

Polyhalogen derivatives are given when Phenol is treated with bromine water, in which all the H-atoms present at the o- and p- positions are substituted by Bromine with respect to the -OH group.

hope it helps

thanku

Explanation:

4 0

2 years ago

How to do part (b)??

laila [671]

it has less tightly bound electrons, is able to lose electron easily as compare to metal B at it has 4 unpaired electron in 3d sub-shell.

3 0

2 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