Consider two aqueous nonvolatile and nonelectrolyte solutions, each with a solute concentration of 1 M. One contains glucose, while the other contains an unidentified covalent solid. Which of the following are sure to be identical in each solution?
A) Their freezing points
B) The identity of the solvent
C) The identity of the solute
D) Their densities

B) The identity of the solvent

Explanation:

Basically, the solvent is the liquid in which a solute is dissolved in. But the solute is the material to be dissolved.

Now in this case, the solute in the first solution is glucose and the solute in the second solution is an unidentified covalent solid material.

This means that:

• the identity of the solute cannot be identical in each solutions, which also means that the freezing points and densities of the solutions cannot be identical too.

• the only thing that is sure to be identical in the solution is the identity of the solvent.

2fe + 3h2so4 -> fe2(so4)3 + 3h2

it is necessary to balance equations in order to get the proper amount of reactants as well as products in order not to make a miscalculation.

you cannot change the subscripts as that would change the substance itself. for instance, diatomic elements such as h2 only have the subscript two when a gas. another example would be so4, if you were to change the subscript, the element would no longer be sulfate.

we change the coefficients instead as that only changes the number of molecules rather than the substance itself.

hope this !

given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

                                  = 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g