height doesn't affect the gravitational potential energy of an object.
Answer: -
15.55 M
35.325 molal
Explanation: -
Let the volume of the solution be 1000 mL.
Density of nitric acid = 1.42 g/ mL
Total Mass of nitric acid Solution = Volume of nitric acid x Density of nitric acid
= 1000 mL x 1.42 g/ mL
= 1420 g.
Percentage of HNO₃ = 69%
Amount of HNO₃ = 
= 979.8 g
Molar mass of HNO₃ = 1 x 1 + 14 x 1 + 16 x 3 = 63 g /mol
Number of moles of HNO₃ = 
= 15.55 mol
Molarity is defined as number of moles per 1000 mL
We had taken 1000 mL as volume and found it to contain 15.55 moles.
Molarity of HNO₃ = 15.55 M
Mass of water = Total mass of nitric acid solution - mass of nitric acid
= 1420 - 979.8
= 440.2 g
So we see that 440.2 g of water contains 15.55 moles of HNO₃
Molality is defined as number of moles of HNO₃ present per 1000 g of water.
Molality of HNO₃ = 
= 35.325 molal
Answer:
The least whole number coefficient for HNO₃ is 6
Explanation:
The chemical equation above is the reaction between calcium orthophosphate and nitric acid.
To balance a chemical equation, we have to consider law of conservation of matter which states that matter can neither be created nor destroyed.
What this law implies is that, whatever we have at the reactant side must be equal to whatever is obtainable at the product side.
The above equation is
Ca₃(PO₄)₂ + HNO₃ → Ca(NO₃)₂ + H₃PO₄
To balance the equation, we'll have to check the number of atoms at each side and possibly balance the equation with the number of moles.
The balanced equation is
Ca₃(PO₄)₂ + 6HNO₃ → 3Ca(NO₃)₂ + 2H₃PO₄
From the balanced equation above, we can see that the number of calcium (Ca), Phosphorus (P), Oxygen(O), Nitrogen(N) and hydrogen (H) are balanced at both sides of the equation.
The least number coefficient for HNO₃ is 6
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Answer:
The oxidation number of the metal decreases
2 Al + Fe₂O₃ → Al₂O₃ + 2 FeO
The metal element iron, is reduced from Fe⁺³ in Fe₂O₃ to Fe⁺² in FeO
Explanation:
When an element gains electron, the element becomes reduced, hence when a metal is reduced, the metal gains electrons, which reduces the oxidation number of the metal
An example of a metal being reduced is;
2 Al + Fe₂O₃ → Al₂O₃ + 2 FeO
In the above reaction, the iron (III) oxide is reduced to iron (II) oxide by aluminium metal.
