Answer:
20 g Ag
General Formulas and Concepts:
<u>Chemistry - Stoichiometry</u>
- Using Dimensional Analysis
<u>Chemistry - Atomic Structure</u>
Explanation:
<u>Step 1: Define</u>
[RxN] Cu (s) + AgNO₃ (aq) → CuNO₃ (aq) + Ag (s)
[Given] 10 g Cu
<u>Step 2: Identify Conversions</u>
[RxN] 1 mol Cu = 1 mol Ag
Molar Mass of Cu - 63.55 g/mol
Molar Mass of Ag - 197.87 g/mol
<u>Step 3: Stoichiometry</u>
<u /> = 16.974 g Ag
<u>Step 4: Check</u>
<em>We are given 1 sig fig. Follow sig fig rules and round.</em>
16.974 g Ag ≈ 20 g Ag
Answer:
D.
Explanation:
-log(1.0x10^-5) = pH
pH + pOH = 14 (rearrange it)
OH- = 10^-pOH = 1.0 x 10^-9
- Hope that helped! Let me know if you need further explantion.
Answer:
The answer to your question is it is not at equilibrium, it will move to the products.
Explanation:
Data
Keq = 2400
Volume = 1 L
moles of NO = 0.024
moles of N₂ = 2
moles of O₂ = 2.6
Process
1.- Determine the concentration of reactants and products
[NO] = 0.024 / 1 = 0.024
[N₂] = 2/1 = 2
[O₂] = 2.6/ 1= 2.6
2.- Balanced chemical reaction
N₂ + O₂ ⇒ 2NO
3.- Write the equation for the equilibrium of this reaction
Keq = [NO]²/[N₂][O₂]
- Substitution
Keq = [0.024]² / [2][2.6]
-Simplification
Keq = 0.000576 / 5.2
-Result
Keq = 1.11 x 10⁻⁴
Conclusion
It is not at equilibrium, it will move to the products because the experimental Keq was lower than the Keq theoretical-
1.11 x 10⁻⁴ < 2400
C. The thermometer reads 100
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<h3>
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