The final temperature, t₂ = 30.9 °C
<h3>Further explanation</h3>
Given
24.0 kJ of heat = 24,000 J
Mass of calorimeter = 1.3 kg = 1300 g
Cs = 3.41 J/g°C
t₁= 25.5 °C
Required
The final temperature, t₂
Solution
Q = m.Cs.Δt
Q out (combustion of compound) = Q in (calorimeter)
24,000 = 1300 x 3.41 x (t₂-25.5)
t₂ = 30.9 °C
I think it would be homogeneous and heterogeneous
Vanadium(V) oxide is the inorganic compound with the formula V₂O₅. Commonly known as vanadium pentoxide, it is a brown/yellow solid, although when freshly precipitated from aqueous solution, its colour is deep orange. Because of its high oxidation state, it is both an amphoteric oxide and an oxidizing agent.
formula:V2O5
pls mark as brainliest
Mass of CO₂ evolved : 0.108 g
<h3>Further explanation</h3>
Given
1.205g sample, 36% MgCO3 and 44% K2CO3
Required
mass of CO2
Solution
0.36 x 1.205 g=0.4338 g
mass C in MgCO₃(MW MgCO₃=84 g/mol, Ar C = 12/gmol)
= (12/84) x 0.4338
= 0.062 g
0.44 x 1.205 g = 0.5302 g
Mass C in K₂CO₃(MW=138 g/mol) :
= (12/138) x 0.5302
= 0.046 g
Total mass Of CO₂ :
= 0.062 + 0.046
= 0.108 g
Answer: 51.9961 g/mol, don't know if it helps :)
Explanation:
