Answer:
See explanation below
Explanation:
The question is incomplete. However, here's the missing part of the question:
<em>"For the following reaction, Kp = 0.455 at 945 °C: </em>
<em>C(s) + 2H2(g) <--> CH4(g). </em>
<em>At equilibrium the partial pressure of H2 is 1.78 atm. What is the equilibrium partial pressure of CH4(g)?"</em>
With these question, and knowing the value of equilibrium of this reaction we can calculate the partial pressure of CH4.
The expression of Kp for this reaction is:
Kp = PpCH4 / (PpH2)²
We know the value of Kp and pressure of hydrogen, so, let's solve for CH4:
PpCH4 = Kp * PpH2²
*: You should note that we don't use Carbon here, because it's solid, and solids and liquids do not contribute in the expression of equilibrium, mainly because their concentration is constant and near to 1.
Now solving for PpCH4:
PpCH4 = 0.455 * (1.78)²
<u><em>PpCH4 = 1.44 atm</em></u>
Answer:
Mg(s) +<em> 2</em> HCl (aq) → H₂(g) + MgCl₂
0.415g of H₂(g) <em>-Assuming mass of Mg(s) = 10.0g-</em>
Explanation:
Balancing the reaction:
Mg(s) + HCl (aq) → H₂(g) + MgCl₂
There are in products two atoms of H and Cl, the balancing equation is:
Mg(s) +<em> 2</em> HCl (aq) → H₂(g) + MgCl₂
<em>Assuming you add 10g of Mg(s) -Limiting reactant-</em>
<em />
10g of Mg are (Atomic mass: 24.305g/mol):
10g × (1 mol / 24.305g) = <em>0.411 moles of Mg</em>
<em>-Theoretical yield is the amount of product you would have after a chemical reaction occurs completely-</em>
Assuming theoretical yield, as 1 mole of Mg(s) produce 1 mole of H₂(g), theoretical yield of H₂(g) is 0.411moles H₂(g). In grams:
0.411mol H₂(g) × (1.01g / mol) = <em>0.415g of H₂(g)</em>
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i hope that helps</span>
