Answer:
25.0 g of sodium carbonate are present in 220 ml of the solution.
Explanation:
Hi there!
I have found the complete question on the web:
<em>Calculate the volume in milliliters of a 1.07 M sodium carbonate solution that contains 25.0g of sodium carbonate Na2CO3. Be sure your answer has the correct number of significant digits.</em>
First, let's find how many moles of sodium carbonate have a mass of 25.0 g.
The molar mass of Na₂CO₃ is 106 g/mol.
So, if 106 g of sodium carbonate is equal to 1 mol, then 25.0 g will be:
25.0 g · (1 mol / 106 g) = 0.236 mol Na₂CO₃
The solution contains 1.07 mol sodium carbonate in 1 liter.
So, if 1.07 mol sodium carbonate is present in 1 l solution, then, 0.236 mol will be present in:
0.236 mol · (1000 ml / 1.07 mo) = 221 ml (220 ml without any intermediate rounding).
25.0 g of sodium carbonate are present in 220 ml of the solution.
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Answer:
1.5 moles
Explanation:
Given data:
Number of moles of O₂ consumed = 2.5 mol
Moles of CO₂ produced = ?
Solution:
Chemical equation:
C₃H₈ + 5O₂ → 3CO₂ + 4H₂O
Now we will compare the moles of oxygen with carbon dioxide from balance chemical equation:
O₂ : CO₂
5 : 3
2.5 : 3/5×2.5 = 1.5
Thus from 2.5 moles of oxygen 1.5 moles of carbon dioxide are produced.
1 nm = 1 * 10^-9 m
1mm = 1* 10^-3 m = 10^6 nm.
Thus, 10^6 nm = 1 mm
=> 261 nm = 261 * 1/10^6 mm = 261 * 10^-6 mm = 2.61 * 10^-4 mm. :)