Answer:
nitrogen oxides and sulphur oxides
Explanation:
Factors responsible for antropogenic pollution are:
- burning of fossil fuels
- deforestation
- mining
- sewage
- industrial effluent
- pesticides, fertilizers, etc.
The primary air pollutants released from burning of fossil fuels are oxides of nitrogen, sulfur oxides and carbon monoxide.
Out of which the main pollutants that are responsible for acidic precipitation are oxides of nitrogen and sulfur oxides.
Sulfur oxides and Oxides of nitrogen reacts with moisture present in the air to form sulfuric acid and nitric acid respectively.
These acids get mixed with rain and cause acidic precipitation.
Therefore, the correct option is oxides of nitrogen and sulfur oxides.
Answer:
Explanation:
H = 0.44/1.01 = 0.4356
O = 6.92/16 = 0.4319
This gives a 1:1 ratio. So the closest thing you could say is the formula is 0H
Going to your chemical storage room, you could justify that it is H2O2 or hydrogen peroxide. The question needs one more fact to make the answer certainty.
Answer:
The test solution is acidified using a few drops of dilute nitric acid, and then a few drops of silver nitrate solution are added. Different coloured silver halide precipitates form, depending on the halide ions present: ... iodide ions give a yellow precipitate of silver iodide.
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.