Answer:
Explanation:
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In this case, since the by-mass percent of a solution is a measure of the mass of the solute over the mass of the solution:
As we know the mass of the solution and the by-mass percent, we can compute the mass of glucose in the 480 g of solution:
Thus, by plugging in the data, we obtain:
Finally, since the solution is made up of glucose and water, we compute the mass of water as follows:
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M C C O R M I C K T E A M ’ S O P E N Q U A N T U M M AT E R I A L S D ATA B A S E
O F F E R S U N L I M I T E D A C C E S S T O
ANALYSES OF NEARLY 300,000 COMPOUNDS
Answer:
4.3 × 10⁻⁵ M s⁻¹
Explanation:
Step 1: Given data
- Rate constant (k): 2.20 × 10⁷ M⁻¹s⁻¹
- Concentration of NO ([NO]): 3.3 × 10⁻⁶ M
- Concentration of O₃ ([O₃]): 5.9 × 10⁻⁷ M
- First order with respect to both NO and O₃
Step 2: Write the balanced reaction
NO + O₃ ⇒ NO₂ + O₂
Step 3: Calculate the reaction rate
The rate law is:
rate = k × [NO] × [O₃]
rate = 2.20 × 10⁷ M⁻¹s⁻¹ × 3.3 × 10⁻⁶ M × 5.9 × 10⁻⁷ M
rate = 4.3 × 10⁻⁵ M s⁻¹
