1) we can calculate the molecular weight of H₃PO₄
atomic mass (H)=1 amu
atomic mass (P)=31 amu
atomic mass (O)=16 amu
molecular weight (H₃PO₄)=3(1)+31+4(16)=98 amu.
1 mol (H₃PO₄)=98 g
1 mol= 6.022 * 10²³ molecules.
2) we calculate the mass of 4.00*10²³ molecules.
98 g-------------------6.022*10²³ molecules
x------------------------4.00*10²³ molecules
x=(98 g * 4.00*10²³ molecules) / 6.022*10²³ molecules≈65 g
Answer: 65 g
When the student crushes sugar there there occurs an increase in the number of molecules or particles of sugar. This means that now more number of sugar particles (solute) are able to interact with water (solvent).
Therefore, there will be more number of collisions taking place between the solute and solvent molecules.
As a result, sugar will dissolve readily into water as there is an increase in interaction between solute and solvent particles.
Thus, we can conclude that when he crushes the sugar and then stirs it into the water then it will work as this leads to more interaction or collisions between the solute and solvent particles.
Answer:
1=C
2=B
3=E
4=A
5=D
Explanation:
Closer to 0 is more acidic, closer to 14 is more basic. 7 is neutral and the cut off point between if acidic or basic.
The enthalpy of the solution is <u>positive </u>and the entropy is <u>positive</u>.
Potassium trioxonitrate (V) KNO₃(s) is a strong oxidizing solid substance that when dissolved in water changes to aqueous solution.
In its aqueous solution state, the randomness of molecules increases as a result of that the entropy will also increase leading to the positive state of the entropy.
Similarly, provided that the solution becomes quite cold to the touch, the enthalpy is also in it positive state.
Therefore, we can conclude that the enthalpy of the solution is <u>positive </u>and the entropy is <u>positive</u>.
Learn more about Potassium trioxonitrate (V) KNO₃(s) here:
brainly.com/question/25303112
Answer: This is called moles
Explanation: This is described to be the number of molecules, atoms or ions in a substance :)