Using the ideal gas law equation, we can find the number of H₂ moles produced.
PV = nRT
Where P - pressure - 0.811 atm x 101 325 Pa/atm = 82 175 Pa
V - volume - 58.0 x 10⁻³ m³
R - universal gas constant - 8.314 Jmol⁻¹K⁻¹
T - temperature - 32 °C + 273 = 305 K
substituting these values in the equation,
82 175 Pa x 58.0 x 10⁻³ m³ = n x 8.314 Jmol⁻¹K⁻¹ x 305 K
n = 1.88 mol
The balanced equation for the reaction is as follows;
CaH₂(s) + 2H₂O(l) --> Ca(OH)₂(aq) + 2H₂(g)
stoichiometry of CaH₂ to H₂ is 1:2
When 1.88 mol of H₂ is formed , number of CaH₂ moles reacted = 1.88/2 mol
therefore number of CaH₂ moles reacted = 0.94 mol
Mass of CaH₂ reacted - 0.94 mol x 42 g/mol = 39.48 g of CaH₂ are needed
So let's convert this amount of mL to grams:
Then we need to convert to moles using the molar weight found on the periodic table for mercury (Hg):
Then we need to convert moles to atoms using Avogadro's number:
![\frac{6.022*10^{23}atoms}{1mole} *[8.135*10^{-2}mol]=4.90*10^{22}atoms](https://tex.z-dn.net/?f=%20%5Cfrac%7B6.022%2A10%5E%7B23%7Datoms%7D%7B1mole%7D%20%2A%5B8.135%2A10%5E%7B-2%7Dmol%5D%3D4.90%2A10%5E%7B22%7Datoms%20)
So now we know that in 1.2 mL of liquid mercury, there are 
present.
The most common organic acids are the carboxylic acids
Answer:
22.9g of H₂O are produced
Explanation:
The reaction of hydrogen with oxygen is:
2H₂(g) + O₂(g) → 2H₂O(g)
<em>Where 2 moles of hydrogen react with 1 mole of oxygen to produce 2 moles of water</em>
<em />
Moles of hydrogen are:
2.56g × (1mole / 2.01g) = <em>1.27moles H₂</em>
Moles of oxygen are:
20.3g × (1mole / 32g) = <em>0.634moles O₂</em>
These moles of oxygen react with:
0.634moles O₂ × (2mol H₂ / 1mol O₂) =<em> 1.27mol H₂</em>
<em>-All hydrogen in reaction-</em>
And produce:
0.634moles O₂ × (2mol H₂O / 1mol O₂) =<em> 1.27mol H₂O</em>
In grams:
1.27mol <em>H₂O </em>× (18.01g / 1mol) = <em>22.9g of H₂O are produced</em>
