Answer:
1.53x10^22 atoms of Au
Explanation:
To find the atoms of gold we need first, to convert the mass of gold to moles using molar mass of gold (196.97g/mol). Then, these moles must be converted to number of atoms based on definition of moles (1 mole = 6.022x10²³ atoms).
<em>Moles Au:</em>
5.00g Au * (1mol / 196.97g) = 0.0254 moles of Au
<em>Atoms of Au:</em>
0.0254 moles * (6.022x10²³ atoms / 1 mole) =
<h3>1.53x10^22 atoms of Au</h3>
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The question is incomplete, here is the complete question:
Write a balanced chemical equation for each single replacement reaction that takes place in aqueous solution. write no reaction if a reaction does not occur
1.) Zn + PbCl₂
2.) Cu + Fe(NO₃)₂
<u>Answer:</u>
<u>For 1:</u> The reaction does occur.
<u>For 2:</u> The reaction does not occur.
<u>Explanation:</u>
Single displacement reaction is defined as the reaction in which more reactive element displaces a less reactive element.
The reactivity of metal is determined by a series known as reactivity series. The metals lying above in the series are more reactive than the metals which lie below in the series.

For the given options:
Zinc is more reactive than lead as it lies above in the series. So, it will displace lead from its chemical equation.
The chemical equation for the reaction of zinc and lead chloride follows:

Copper is less reactive than iron as it lies below in the series. So, it will not displace iron from its chemical equation.
The chemical equation for the reaction of copper and iron (II) nitrate follows:

The question is incomplete, here is the complete question:
What volume (mL) of the partially neutralized stomach acid having concentration 2 M was neutralized by 0.1 M NaOH during the titration? (portion of 25.00 mL NaOH sample was used; this was the HCl remaining after the antacid tablet did it's job)
<u>Answer:</u> The volume of HCl neutralized is 1.25 mL
<u>Explanation:</u>
To calculate the volume of acid, we use the equation given by neutralization reaction:

where,
are the n-factor, molarity and volume of stomach acid which is HCl
are the n-factor, molarity and volume of base which is NaOH.
We are given:

Putting values in above equation, we get:

Hence, the volume of HCl neutralized is 1.25 mL