Answer:
Solutions in water are especially common, and are called aqueous solutions. Non-aqueous solutions are when the liquid solvent involved is not water.
Answer: CrO₄⁻ and Ba²⁺
Explanation:
1) Chemical equation given:
2H⁺ + CrO₄⁻ + Ba²⁺ + 2OH⁻ → Ba²⁺ + CrO₄⁻ + 2H₂O
2) Analysis
That is an oxidation-reduction equation (some species are been oxidized and others are being reduced).
The given equation is known as total ionic equation, because it shows all the species as ions that are part of the reaction.
2) Specator ions
Spectator ions are the ions that do not change their oxidation state and are easily identified as they are the same in the reactant and product sides.
Here the ions that are the same in the reactant and product sides are:
CrO₄⁻ and Ba²⁺
3) Addtitional explanation.
Once you identify the spectator ions you can delete them from the equation to obtain the net ionic equation , which in this case turns to be:
2H⁺ + 2OH⁻ → 2H₂O
But this is not part of the question; it is some context to help you understand the use of the spectator ions concept.
Answer:
The correct answer is 32.2 grams.
Explanation:
Based on the given information, the enthalpy of formation for aluminum oxide is 1676 kJ/mol. It signifies towards the energy that is required to generate aluminum and oxygen, and both of these exhibit zero enthalpy of formation. Therefore, the ΔHreaction is the required energy to generate 2 moles of aluminum. Thus, the energy needed for the formation of single mole of aluminum is,
ΔHrxn = 1676/2 = 838 kJ/mol
Q or the energy input mentioned in the given case is 1000 kJ. Therefore, the number of moles of Al generated is,
(1000 kJ) / (838 kJ/Al mole) = 1.19 moles of Aluminum
The grams of aluminum produced can be obtained by using the formula,
mass = moles * molecular mass
= 1.19 * 26.98
= 32.2 grams.
Answer: 4.22 grams of solute is there in 278 ml of 0.038 M 
Explanation:
Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.
where,
n = moles of solute
= volume of solution in L
Now put all the given values in the formula of molality, we get
mass of = 
Thus 4.22 grams of solute is there in 278 ml of 0.038 M
