Answer:
specific heat of sample = 0.898 j/g.°C
Explanation:
Given data:
Mass of sample = 75 g
Initial temperature of water = 100°C
Mass of water = 100 g
Initial temperature of water = 24.4°C
Final temperature of sample ans water = 34.9°C
Specific heat of sample = ?
Solution:
ΔT(water) = 34.9°C - 24.4°C
ΔT(water) = 10.5°C
ΔT(sample) = 34.9°C - 100°C
ΔT(sample) = - 65.1°C
-Q(sample) = Q(water)
-m×c×ΔT = m×c×ΔT
-[75 g × c × - 65.1°C] = 100 g × 4.18 j/g.°C × 10.5°C
-[-4882.5 g.°C × c = 4389 j
4882.5 g.°C × c = 4389 j
c = 4389 j/ 4882.5 g.°C
c = 0.898 j/g.°C
Cathode rays are a stream of electrons that are negatively charged particles. The particles move towards the anode or positively charged electrode as the posses negative charge. Electric current is the rate of flow of the charge.
The molar enthalpy is 447 kJ and the reaction is endothermic
Answer:
Explanation:
The formula for the force exerted between two ions is
where the Coulomb constant
k = 8.988 × 10⁹ N·m·C⁻²
Data:
z₁ = +2; r₁ = 0.62 nm
z₂ = -1; r₂ = 0.121 nm
Calculations:
r = r₁ + r₂ = 0.62 + 0.121 = 0.741 nm
