Answer:
Q = -811440 J
Explanation:
Given data:
Mass of oil = 2.76 Kg (2.76× 1000 = 2760 g)
Initial temperature = 191 °C
Final temperature = 23°C
Specific heat capacity of oil = 1.75 J/g.°C
Solution:
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
ΔT = 23°C - 191 °C
ΔT = -168°C
Q = 2760 g ×1.75 J/g.°C ×-168°C
Q = -811440 J
Negative sign show heat is released.
The chemical formula for copper sulfate is CuSO4
Answer:
The reaction will be slower because the leaving group will be poorer.
Explanation:
One Important thing to put at the back of our mind is that weak bases are good leaving groups. Another thing to take note is that in the halogen series(which is our main subject in this question) as you go down the group the greater or the more heavy the halide is and the heavier halides are more stable that is Bromine will be more stable than chlorine.
Now, we are now told that the bromo halide is been replaced by the chloro halide which means that the rate of Reaction will surely decrease because the leaving group. The cl^- is a stronger base.
D
This feature is formed at destructive boundaries where the denser plate (usually the oceanic plate) is subducted underneath the less dense plate (usually the continental plate).
Explanation:
the stress in the boundary between the two plates causes them to warp at the boundary forming a trench. This forced bending and the friction between the two plates (remember tectonic plates are very rugged) causes fissures to develop at the boundary. As the denser plate dives into the mantle, it begins to melt and the molten rock rises through the fissures. The magma erupts at the surface in several fissures forming volcanic mountains ranges along the convergent boundary.
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Answer:
b) pH = 9.25
Explanation:
- NH4+(aq) + H2O(l) ↔ NH3(aq) + H3O+(aq)
- NH3 + H2O ↔ NH4+ + OH-
- 2 H2O ↔ H3O+ + OH-
⇒ Kb = [ NH4+ ] * [ OH- ] / [ NH3 ] = 1.86 E-5......from literature
mass balance NH4+:
⇒ M NH4+ = [ NH4+ ] - [ OH- ]
∴ [ NH3 ] ≅ M NH4+ = 0.26 M
⇒ Kb = (( 0.26 + [ OH- ] )) * [ OH- ] / 0.26 = 1.86 E-5
⇒ 0.26 [ OH-] + [ OH- ]² = 4.836 E-6
⇒ [ OH- ]² + 0.26 [ OH- ] - 4.836 E-6 = 0
⇒ [ OH- ] = 1.859 E-5 M
⇒ pOH = - Log ( 1.859 E-5 )
⇒ pOH = 4.7305
⇒ pH = 14 - pOH = 9.269
