Answer:
mass of HCl = 243.5426 grams
Explanation:
1- we will get the mass of the reacting gold:
volume of gold = length * width * height
volume of gold = 3.2 * 3.8 * 2.8 = 34.048 cm^3 = 34.048 ml<span>
density = mass / volume
Therefore:
mass = density * volume
mass of gold = </span>19.3 * 34.048 = 657.1264 grams
2- we will get the number of moles of the reacting gold:
number of moles = mass / molar mass
number of moles = 657.1264 / 196.96657
number of moles = 3.3362 moles
3- we will get the number of moles of the HCl:
First, we will balanced the given equation. The balanced equation will be as follows:
Au + 2HCl ......> AuCl2 + H2
This means that one mole of Au reacts with 2 moles of HCl.
Therefore 3.3362 moles will react with 2*3.3362 = 6.6724 moles of HCL
4- we will get the mass of the HCl:
From the periodic table:
molar mass of H = 1 gram
molar mass of Cl = 35.5 grams
Therefore:
molar mass of HCl = 1 + 35.5 = 36.5 grams/mole
number of moles = mass / molar mass
Therefore:
mass = number of moles * molar mass
mass of HCl = 6.6724 * 36.5
mass of HCl = 243.5426 grams
Hope this helps :)
answer
2500
because it's mm
and you just have to add 2 more place values
The reaction is not balanced
<h3>Further explanation</h3>
Given
Reaction
2Fe(s)+3O₂(g)⇒2Fe₂O₃(s)
Required
The number of atoms
Solution
In a balanced chemical equation, the number of atoms in the compound that reacts (the reactants and products) will have the same number
Reactants : Fe(s)+O₂(g)
Fe = 2 atoms
O = 3 x 2 = 6 atoms
Products : Fe₂O₃(s)
Fe = 2 x 2 = 4 atoms
O = 2 x 3 = 6 atoms
The reaction is not balanced because the number of Fe atoms is not the same
The balanced reaction should be:
4Fe(s)+3O₂(g)⇒2Fe₂O₃(s)
The answer would be A) sand, it is not soluble in water
Answer:
Investigating a data breach
Explanation:
A data breach usually involves data exfiltration over a computer network. the other options involve data being stored on a device locally which isn't volatile data like text messages, photos or rearranging data in defragmentation all of which does not require a network.
