Answer: 0.2885J/g°c
Explanation:
Loss of heat of metal = Gain of heat by the water
Therefore
-Qm = +Qw
Where
Q = mΔTCp
Q = heat
M= mass
ΔT = T.f - Ti
Ti= initial temperature
T.f= final temperature
Cp= Specific heat (m is metal, w is water)
-Qm = +Qw
-[m(T.f-Ti)Cpm] = m(T.f-Ti)Cpw
Therefore
-[83.2g(28.42-90.31)Cpm] = 41.82g(28.42-
19.93)4.194J/g°c
Cpm = 1485.54÷ 5149.25
Cpm = 0.2885J/g°c
Therefore , the specific heat of the metal is 0.2885J/g°c
Mol of Kr gas = 1.244
<h3>Further explanation</h3>
In general, the gas equation can be written
<h3> PV=nRT
</h3>
where
P = pressure, atm
V = volume, liter
n = number of moles
R = gas constant = 0.08205 L.atm / mol K
T = temperature, Kelvin
P=1.31 atm
V=23.3 L
T=26+273=299 K
mol of sample :
Answer:
The answer to your question is No, it is not.
Explanation:
Data
C₄H₁₀ + 13O₂ ⇒ 8CO₂ + 10H₂O
In a double replacement reaction, two reactants interchange cations an example of these reactions are neutralization reactions. In neutralization reactions, an acid and a base react to form a salt and water.
The reaction of this problem is not a double replacement reaction because the products are carbon dioxide and water, not a salt and water.
The reaction of this problem is a combustion reaction.
