The first one is 2
The second is 1
The third is 6
And the fourth is 3
Answer:
The molar mass of the metal is 54.9 g/mol.
Explanation:
When we work with gases collected over water, the total pressure (atmospheric pressure) is equal to the sum of the vapor pressure of water and the pressure of the gas.
Patm = Pwater + PH₂
PH₂ = Patm - Pwater = 1.0079 bar - 0.03167 bar = 0.9762 bar
The pressure of H₂ is:
The absolute temperature is:
K = °C + 273 = 25°C + 273 = 298 K
We can calculate the moles of H₂ using the ideal gas equation.
Let's consider the following balanced equation.
M(s) + H₂SO₄(aq) ⟶ MSO₄(aq) + H₂(g)
The molar ratio of M:H₂ is 1:1. So, 9.81 × 10⁻³ moles of M reacted. The molar mass of the metal is:
Hey there!:
Molar mass of Mg(OH)2 = 58.33 g/mol
number of moles Mg(OH)2 :
moles of Mg(OH)2 = 30.6 / 58.33 => 0.5246 moles
Molar mass of H3PO4 = 97.99 g/mol
number of moles H3PO4:
moles of Mg(OH)2 = 63.6 / 97.99 => 0.649 moles
Balanced chemical equation is:
3 Mg(OH)2 + 2 H3PO4 ---> Mg3(PO4)2 + 6 H2O
3 mol of Mg(OH)2 reacts with 2 mol of H3PO4 ,for 0.5246 moles of Mg(OH)2, 0.3498 moles of H3PO4 is required , but we have 0.649 moles of H3PO4, so, Mg(OH)2 is limiting reagent !
Now , we will use Mg(OH)2 in further calculation .
Molar mass of Mg3(PO4)2 = 262.87 g/mol
According to balanced equation :
mol of Mg3(PO4)2 formed = (1/3)* moles of Mg(OH)2
= (1/3)*0.5246
= 0.1749 moles of Mg3(PO4)2
use :
mass of Mg3(PO4)2 = number of mol * molar mass
= 0.1749 * 262.87
= 46 g of Mg3(PO4)2
Therefore:
% yield = actual mass * 100 / theoretical mass
% = 34.7 * 100 / 46
% = 3470 / 46
= 75.5%
Hope that helps!
The element necissary for an eye is plain vision, oxygen,
I would say the answer is emissions. These are the particles that are not supposed to be present in air but due to the production of different substances from humans daily activities these substances go with the air we breath. Hope this helped.
