Answer:
-2.86x10³ kJ
Explanation:
The enthalpy of a reaction (ΔH) is defined as the heat produced or consumed by a reaction. In the reaction:
2 C₂H₆(g) + 7 O₂(g) → 4 CO₂(g) + 6 H₂O(g)
The ΔH is the heat envolved in the reaction per 2 moles of C₂H₆. 1.43x10³ kJ are involved when 1 mole reacts. Thus, when 2 moles react, involved heat is:
1.43x10³ kJ ₓ 2 = <em>2.86x10³ kJ</em>. As the reaction is a combustion reaction (Produce CO₂ and H₂O), the heat involved in the reaction is <em>PRODUCED, </em>that means ΔH is negative, <em>-2.86x10³ kJ</em>
Answer:
0.0187 M
Explanation:
Step 1: Write the balanced neutralization reaction
NaOH + HCl ⇒ NaCl + H₂O
Step 2: Calculate the reacting moles of HCl
18.7 mL of 0.01500 M HCl react.
0.0187 L × 0.01500 mol/L = 2.81 × 10⁻⁴ mol
Step 3: Calculate the reacting moles of NaOH
The molar ratio of HCl to NaOH is 1:1. The reacting moles of NaOH are 1/1 × 2.81 × 10⁻⁴ mol = 2.81 × 10⁻⁴ mol.
Step 4: Calculate the molarity of NaOH
2.81 × 10⁻⁴ moles are in 15.00 mL of NaOH.
[NaOH] = 2.81 × 10⁻⁴ mol/0.01500 L = 0.0187 M
Answer:
81 L gas
General Formulas and Concepts:
<u>Ideal Gas Law</u>
- STP (Standard Conditions for Temperature and Pressure) = 22.4 L per mole at 1 atm, 273 K
<u>Stoichiometry</u>
- Using Dimensional Analysis
Explanation:
<u>Step 1: Define</u>
[Given] 3.6 mols gas at STP
[Solve] volume (L) of gas
<u>Step 2: Convert</u>
- [DA] Set up:
- [DA] Multiply [Cancel out units]:
<u>Step 3: Check</u>
<em>Follow sig fig rules and round. We are given 2 sig figs.</em>
80.64 L gas ≈ 81 L gas
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Let the mass of the solute be x
So, the equation would be
x/2.5+x ×100 = 23.22
x/2.5+x = 23.22/100
100x = 58.05 + 23.22x
100x - 23.22x = 58.05
76.78x = 58.05
x =0.756 ≈ 0.76 litres
