Answer:
Water (H2O) is polar because of the bent shape of the molecule. The shape means most of the negative charge from the oxygen on side of the molecule and the positive charge of the hydrogen atoms is on the other side of the molecule. When solutes are added to water, they may be affected by the charge distribution
Explanation:
Answer #1 is "there is 2.5 grams of solute in every 100 g of solution."
We calculate for 2.5% by mass solution by dividing the mass of the solute by the mass of the solution and then multiply by 100.
Answer #2 is "that mass ratio would be 2.5/100 or 2.5 grams of solute/100 grams of solution."
We weigh out 2.5 grams of solute and then add 97.5 grams of solvent to make a total of 100 gram solution, that is,
mass of solute / mass of solution = 2.5g solute / (2.5g solute + 97.5g solvent)
= 2.5g solute / 100g solution
Answer#3 is "a solution mass of 1 kg is 10 times greater than 100 g, thus one kilogram (1 kg) of a 2.5% ki solution would contain 25 grams of ki."
We multiply 10 to each mass so that 100 grams becomes 1000grams since 1000 grams is equal to 1 kg:
mass of solute / mass of solution = 2.5g*10/[(2.5g*10) + (97.5g*10)]
= 25g solute/(25g solute + 975g solvent)
= 25g solute/1000g solution
= 25g solute/1kg solution
Answer:
Magnetic property
Explanation:
Iron filling is a magnetic compound, unlike the salt crystals. This means they are attracted by magnets.
To separate a mixture of iron filling s and salt crystals, a magnet should be used to remove the iron fillings from the mixture.
The answer for the following problem is described below.
<em><u> Therefore the standard enthalpy of combustion is -2800 kJ</u></em>
Explanation:
Given:
enthalpy of combustion of glucose(Δ
of 
) =-1275.0
enthalpy of combustion of oxygen(Δ of 
) = zero
enthalpy of combustion of carbon dioxide(Δ of 
) = -393.5
enthalpy of combustion of water(Δ of 
) = -285.8
To solve :
standard enthalpy of combustion
We know;
Δ = ∈Δ (products) - ∈Δ (reactants)
(s) +6 (g) → 6 (g)+ 6 (l)
Δ = [6 (-393.5) + 6(-285.8)] - [6 (0) + (-1275)]
Δ = [6 (-393.5) + 6(-285.8)] - [0 - 1275]
Δ = 6 (-393.5) + 6(-285.8) - 0 + 1275
Δ = -2361 - 1714 - 0 + 1275
Δ =-2800 kJ
<em><u> Therefore the standard enthalpy of combustion is -2800 kJ</u></em>
