Answer: The empirical formula for the given compound is 
Explanation : Given,
Mass of O = 0.370 g
Mass of N = 0.130 g
To formulate the empirical formula, we need to follow some steps:
Step 1: Converting the given masses into moles.
Moles of Oxygen = 
Moles of Nitrogen = 
Step 2: Calculating the mole ratio of the given elements.
For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.00928 moles.
For Oxygen = 
For Nitrogen = 
Step 3: Taking the mole ratio as their subscripts.
The ratio of O : N = 2 : 1
Hence, the empirical formula for the given compound is 
Answer:
25mL
Explanation:
Please see the step-by-step solution in the picture attached below.
Hope this can help you. Have a nice day!
Answer:
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