Copper II sulfate solution is blue.
Answer: option D. the ability of a base to react with a soluble metal salt.
Justification:
NaOH is a strong base, which means that in water it will dissociate according to this reaction:
- NaOH(aq) → Na⁺ (aq) + OH⁻ (aq)
On the other hand, CuSO₄ is a soluble ionic salt which in water will dissociate into its ions according to this other reaction:
Hence, in solution, the sodium ion (Na⁺) will react with the metal salt in a double replacement reaction, where the highly reactive sodium ion (Na⁺) will substitute the Cu²⁺ in the CuSO₄ to form the sodium sulfate salt, Na₂SO₄ (water soluble), and the copper(II) hydroxide, Cu(OH)₂ (insoluble).
That is what the given reaction represents:
CuSO₄ (aq) + 2NaOH(aq) → Cu(OH)₂(s) + Na₂SO₄(aq)
↑ ↑ ↑ ↑
soluble metal salt strong base insoluble base solube salt
Answer:
8.3 kJ
Explanation:
In this problem we have to consider that both water and the calorimeter absorb the heat of combustion, so we will calculate them:
q for water:
q H₂O = m x c x ΔT where m: mass of water = 944 mL x 1 g/mL = 944 g
c: specific heat of water = 4.186 J/gºC
ΔT : change in temperature = 2.06 ºC
so solving for q :
q H₂O = 944 g x 4.186 J/gºC x 2.06 ºC = 8,140 J
For calorimeter
q calorimeter = C x ΔT where C: heat capacity of calorimeter = 69.6 ºC
ΔT : change in temperature = 2.06 ºC
q calorimeter = 69.60J x 2.06 ºC = 143.4 J
Total heat released = 8,140 J + 143.4 J = 8,2836 J
Converting into kilojoules by dividing by 1000 we will have answered the question:
8,2836 J x 1 kJ/J = 8.3 kJ
C. the number of protons and neutrons an element has in its nucleus
Answer : When we consider the atmospheric pressure as 1 atm then according to the ideal gas equation we can find out the molar mass of any unknown by this formula ;
PV=nRT
so if the pressure increases than 1 atm then we can see from the above equation that it will result in greater value for the number of moles (n) in the above equation.
While n = m/M where m is mass of the unknown in g and M is molecular mass.
So, if pressure is higher then it will result in molar mass of unknown which is much smaller.
