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
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<h3>Answer:</h3>
89.6 L of O₂
<h3>Solution:</h3>
The balanced chemical equation is as,
CH₄ + 2 O₂ → CO₂ + 2 H₂O
As at STP, one mole of any gas (Ideal gas) occupies exactly 22.4 L of Volume. Therefore, According to equation,
44 g ( 1 mol) CO₂ is produced by = 44.8 L (2 mol) of O₂
So,
88 g CO₂ will be produced by = X L of O₂
Solving for X,
X = (88 g × 44.8 L) ÷ 44 g
X = 89.6 L of O₂
The partial pressure of oxygen given the total barometric pressure is : 108.15 mmHg
<u>Given data : </u>
Total barometric pressure = 515 mmHg
Assuming oxygen percentage = 21%
Barometric pressure dry at 37°C
<h3 /><h3>Determine the partial pressure of oxygen </h3>
Applying the relation below
Partial pressure = oxygen percentage * Barometric pressure
= 21% * 515 mmHg
= 108.15 mmHg
Hence we can conclude that the partial pressure of oxygen is 108.15 mmHg.
Learn more about Partial pressure : brainly.com/question/1835226
