Answer:
18.0 g H₂O
Explanation:
To find the mass of water (H₂O), you need to (1) convert grams O₂ to moles O₂ (via the molar mass), then (2) convert moles O₂ to moles H₂O (via mole-to-mole ratio from equation coefficients), and then (3) convert moles H₂O to grams H₂O (via the molar mass). It is important to arrange the conversions in a way that allows for the cancellation of units. The final answer should have 3 sig figs to match the sig figs of the given value.
Molar Mass (O₂): 2(15.998 g/mol)
Molar Mass (O₂): 31.996 g/mol
Molar Mass (H₂O): 2(1.008 g/mol) + 15.998 g/mol
Molar Mass (H₂O): 18.014 g/mol
2 H₂ + 1 O₂ -----> 2 H₂O
16.0 g O₂ 1 mole 2 moles H₂O 18.014 g
--------------- x ---------------- x --------------------- x ----------------- = 18.0 g H₂O
31.996 g 1 mole O₂ 1 mole
Answer:
B. products are found on the write side of the arrow in a chemical reaction.
Answer:
5.56 × 10⁻⁸
Explanation:
Step 1: Given data
- Concentration of the weak acid (Ca): 0.187 M
Step 2: Calculate the concentration of H⁺
We will use the following expression.
pH = -log [H⁺]
[H⁺] = antilog -pH = antilog -3.99 = 1.02 × 10⁻⁴ M
Step 3: Calculate the acid dissociation constant (Ka)
We will use the following expression.
![Ka = \frac{[H^{+}]^{2} }{Ca} = \frac{(1.02 \times 10^{-4})^{2} }{0.187} = 5.56 \times 10^{-8}](https://tex.z-dn.net/?f=Ka%20%3D%20%5Cfrac%7B%5BH%5E%7B%2B%7D%5D%5E%7B2%7D%20%7D%7BCa%7D%20%3D%20%5Cfrac%7B%281.02%20%5Ctimes%2010%5E%7B-4%7D%29%5E%7B2%7D%20%7D%7B0.187%7D%20%3D%205.56%20%5Ctimes%2010%5E%7B-8%7D)
C3H8 + 5O2 -----> 3CO2 + 4H2O
This is propane burning in air or pure oxygen.
