Answer:
643g of methane will there be in the room
Explanation:
To solve this question we must, as first, find the volume of methane after 1h = 3600s. With the volume we can find the moles of methane using PV = nRT -<em>Assuming STP-</em>. With the moles and the molar mass of methane (16g/mol) we can find the mass of methane gas after 1 hour as follows:
<em>Volume Methane:</em>
3600s * (0.25L / s) = 900L Methane
<em>Moles methane:</em>
PV = nRT; PV / RT = n
<em>Where P = 1atm at STP, V is volume = 900L; R is gas constant = 0.082atmL/molK; T is absolute temperature = 273.15K at sTP</em>
Replacing:
PV / RT = n
1atm*900L / 0.082atmL/molK*273.15 = n
n = 40.18mol methane
<em>Mass methane:</em>
40.18 moles * (16g/mol) =
<h3>643g of methane will there be in the room</h3>
Answer:
0.231 mol/L
Explanation:
The first step is to write the balanced equation for this reaction:
The second step is to find the number of moles in the acid:
number of moles = volume * concentration
= 0.035 L * 0.275 mol/L
= 0.009625 mol
The third step is to use the molar ratio from the balanced chemical equation to find the number of moles of NaOH that can neutralize 0.009625 mol of sulphuric acid.
n(sulphuric acid) : n(sodium hydroxide)
1 : 2
0.009625 mol : x
x = 0.01925 mol
Fourth step is to calculate the concentration of sodium hydroxide:
Answer:
Sample A contains the smaller volume
Answer:
SO₄²⁻(aq) + Ba²⁺(aq) ⇒ BaSO₄(s)
Explanation:
Let's consider the molecular equation that occurs when aqueous solutions of ammonium sulfate and barium nitrate are mixed.
(NH₄)₂SO₄(aq) + Ba(NO₃)₂(aq) ⇒ BaSO₄(s) + 2 NH₄NO₃(aq)
The complete ionic equation includes all the ions and insoluble species.
2 NH₄⁺(aq) + SO₄²⁻(aq) + Ba²⁺(aq) + 2 NO₃⁻(aq) ⇒ BaSO₄(s) + 2 NH₄⁺(aq) + 2 NO₃⁻(aq)
The net ionic equation includes only the ions that participate in the reaction and the insoluble species.
SO₄²⁻(aq) + Ba²⁺(aq) ⇒ BaSO₄(s)
