PH= log[H3O+]
10.25=log [H3O+]
[H3O+] = 10^10.25
[H3O+]= 1.778 ×10^10
Answer:
1. C₄H₁₀ + ¹³/₂O₂ → 4CO₂ + 5H₂O
2. V = 596L
Explanation:
Butane (C₄H₁₀) reacts with oxygen (O₂) to produce carbon dioxide (CO₂) and water (H₂O) thus:
C₄H₁₀ + O₂ → CO₂ + H₂O
1. The balanced chemical equation is:
C₄H₁₀ + ¹³/₂O₂ → 4CO₂ + 5H₂O
2. 0,360kg of butane are:
360g×
=<em>6,19moles of butane</em>
These moles of butane are:
6,19moles of butane×
= <em>24,8 moles CO₂</em>
Using V=nRT/P
Where:
n are moles (24,8 moles CO₂); R is gas constant (0,082atmL/molK); T is temperature, 20°C (293,15K); and P is pressure (1atm).
Volume (V) is:
<em>V = 596L</em>
I hope it helps!
Answer:
see explanation
Explanation:
The process of ionization to produce cations is endothermic. For formation of Ca⁺² two ionization steps need be illustrated as follows...
1st ionization step: Ca° + 590Kj => Ca⁺ + e⁻
2nd ionization step: Ca⁺ + 1151Kj => Ca⁺² + e⁻
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Net Ionization Rxn: Ca° + 1741Kj => Ca⁺² + 2e⁻
Answer:
The mass of PbSO4 formed 15.163 gram
Explanation:
mole of Pb(NO₃)₂ = 1.25 x 0.05 = 0.0625
mole of Na₂SO₄ = 2 x 0.025 = 0.05
Pb(NO₃)₂ + Na₂SO₄ → PbSO₄ + 2 NaNO₃
( Mole/Stoichiometry )
= 0.0625 = 0.05
From (Mole/ Stoichiometry ) we can conclude that Na₂SO₄ is limiting reagent.
Mass of PbSO₄ precipitate = 0.05 x Molecular mass of PbSO₄
= 0.05 x 303.26 g
= 15.163 g
Answer:
Yes, yield.
Explanation:
N2(g) + 3 H2(g) → 2 NH3 (g) balanced equation
First, find limiting reactant:
Moles H2 = 1.83 g x 1 mole/2 g = 0.915 moles H2
Moles N2 = 9.84 g N2 x 1 mole/28 g = 0.351 moles N2
The mole ratio of H2: N2 is 3:1, so H2 is limiting (0.915 is less than 3 x 0.351)
Theoretical yield of NH3 = 0.915 mol H2 x 2 mol NH3/3 mol H2 = 0.61 moles NH3