In the following reaction, what element is made?

The diagram above shows the atomic emission line spectrum of hydrogen. The wavelength of each line is shown at the top of the li

Anni [7]

Hello there,

Your correct answer would be "the energy of a photon".

Hope this helps.

~Hottwizzlers

6 0

3 years ago

Read 2 more answers

Compounds have the same properties as the elements they are made up of. Is it true or false

hodyreva [135]

( False ) this the answer

6 0

3 years ago

A neutral atom of gold has the same number of electrons and what?

liraira [26]

Atomic mass!!! the electrons and atomic mass will always be the same most the time

6 0

3 years ago

A solution has a oh- = 1 x 10-5 m what are the h30 and the ph of the solution

Deffense [45]

I think it's easiest to find the pOH from the given [OH-] first.

-log(1x10^-5)

pOH=5

Then find the pH.

pOH+pH=14

5+pH=14

pH=9

Then find the [H+] using the pH.

antilog(-9) (if you dont have an antilog button use 10^-9)

[H+]=1x10^-9

3 0

3 years ago

Read 2 more answers

Write the net ionic equation to show the formation of a solid (insoluble salt) when the following solutions are mixed. Write nor

Brrunno [24]

Answer: The net ionic equations are written below.

Explanation:

Net ionic equation of any reaction does not include any spectator ions.

Spectator ions are defined as the ions which does not get involved in a chemical equation. They are found on both the sides of the chemical reaction when it is present in ionic form.

For 1:

The chemical equation for the reaction of calcium nitrate and sodium sulfate is given as:

$Ca(NO_3)_2(aq.)+Na_2SO_4(aq.)\rightarrow CaSO_4(s)+2NaNO_3(aq.)$

Ionic form of the above equation follows:

$Ca^{2+}(aq.)+2NO_3^-(aq.)+2Na^+(aq.)+SO_4^{2-}(aq.)\rightarrow CaSO_4(s)+2Na^+(aq.)+2NO_3^-(aq.)$

As, sodium and nitrate ions are present on both the sides of the reaction. Thus, it will not be present in the net ionic equation and are spectator ions.

The net ionic equation for the above reaction follows:

$Ca^{2+}(aq.)+SO_4^{2-}(aq.)\rightarrow CaSO_4(s)$

For 2:

The chemical equation for the reaction of lead (II) nitrate and potassium chloride is given as:

$Pb(NO_3)_2(aq.)+2KCl(aq.)\rightarrow PbCl_2(s)+2KNO_3(aq.)$

Ionic form of the above equation follows:

$Pb^{2+}(aq.)+2NO_3^-(aq.)+2K^+(aq.)+2Cl^{-}(aq.)\rightarrow PbCl_2(s)+2K^+(aq.)+2NO_3^-(aq.)$

As, potassium and nitrate ions are present on both the sides of the reaction. Thus, it will not be present in the net ionic equation and are spectator ions.

The net ionic equation for the above reaction follows:

$Pb^{2+}(aq.)+2Cl^{-}(aq.)\rightarrow PbCl_2(s)$

For 3:

The chemical equation for the reaction of calcium chloride and sodium phosphate is given as:

$3CaCl_2(aq.)+2(NH_4)_3PO_4(aq.)\rightarrow Ca_3(PO_4)_2(s)+6NH_4Cl(aq.)$

Ionic form of the above equation follows:

$3Ca^{2+}(aq.)+6Cl^-(aq.)+6NH_4^+(aq.)+2PO_4^{3-}(aq.)\rightarrow Ca_3(PO_4)_2(s)+6NH_4^+(aq.)+6Cl^-(aq.)$

As, ammonium and chloride ions are present on both the sides of the reaction. Thus, it will not be present in the net ionic equation and are spectator ions.

The net ionic equation for the above reaction follows:

$3Ca^{2+}(aq.)+2PO_4^{3-}(aq.)\rightarrow Ca_3(PO_4)_2(s)$

For 4:

The chemical equation for the reaction of barium chloride and sodium sulfate is given as:

$BaCl_2(aq.)+Na_2SO_4(aq.)\rightarrow BaSO_4(s)+2NaCl(aq.)$

Ionic form of the above equation follows:

$Ba^{2+}(aq.)+2Cl^-(aq.)+2Na^+(aq.)+SO_4^{2-}(aq.)\rightarrow BaSO_4(s)+2Na^+(aq.)+2Cl^-(aq.)$

As, sodium and chloride ions are present on both the sides of the reaction. Thus, it will not be present in the net ionic equation and are spectator ions.

The net ionic equation for the above reaction follows:

$Ba^{2+}(aq.)+SO_4^{2-}(aq.)\rightarrow BaSO_4(s)$

Hence, the net ionic equations are written above.

3 0

3 years ago