Adding solvent or removing solute from a solution is called diluting. And a solution is said to be concentrated if it has more solute. The opposite of diluting is called concentrating. The measure of the amount of solute in a solution is expressed in concentration.
The high surface tension helps the paper clip - with much higher density - float on the water. The property of the surface of a liquid that allows it to resist an external force, due to the cohesive nature of its molecules.
Basically it means that there is a sort of skin on the surface of water where the water molecules hold on tight together. If the conditions are right, they can hold tight enough to support your paper clip. The paperclip is not truly floating, it is being held up by the surface tension.
> 2,000
mL of a 5.0 × 10–5% (w/v) sucrose solution
5.0 × 10–3
g/mL * 2000 mL * (1 mol / 342.30 g) = 0.0292 mol
<span>
> 2,000 mL of a 5.0 ppm sucrose solution</span>
5 grams /
1000000 mL * 2000 mL* (1 mol / 342.30 g) = 0.0000292 mol
<span>
> 20 mL of a 5.0 M sucrose solution </span>
5.0 M *
0.020 L = 0.1 mol
Answer:
<span>2,000 mL
of a 5.0 ppm sucrose solution</span>
Answer:
Ammonia is limiting reactant
Amount of oxygen left = 0.035 mol
Explanation:
Masa of ammonia = 2.00 g
Mass of oxygen = 4.00 g
Which is limiting reactant = ?
Balance chemical equation:
4NH₃ + 3O₂ → 2N₂ + 6H₂O
Number of moles of ammonia:
Number of moles = mass/molar mass
Number of moles = 2.00 g/ 17 g/mol
Number of moles = 0.12 mol
Number of moles of oxygen:
Number of moles = mass/molar mass
Number of moles = 4.00 g/ 32 g/mol
Number of moles = 0.125 mol
Now we will compare the moles of ammonia and oxygen with water and nitrogen.
NH₃ : N₂
4 : 2
0.12 : 2/4×0.12 = 0.06
NH₃ : H₂O
4 : 6
0.12 : 6/4×0.12 = 0.18
O₂ : N₂
3 : 2
0.125 : 2/3×0.125 = 0.08
O₂ : H₂O
3 : 6
0.125 : 6/3×0.125 = 0.25
The number of moles of water and nitrogen formed by ammonia are less thus ammonia will be limiting reactant.
Amount of oxygen left:
NH₃ : O₂
4 : 3
0.12 : 3/4×0.12= 0.09
Amount of oxygen react = 0.09 mol
Amount of oxygen left = 0.125 - 0.09 = 0.035 mol
<span>Name of type of mechanism </span>initiation step<span> first </span>propagation step<span> second </span>propagation step<span>(ii) </span>write<span> an overall </span>equation<span> for the </span>formation of dichloromethane<span> from ... Best Answer: i) This is a </span>free-radical<span> substitution mechanism.</span>
