Answer:
D
Explanation:
1.) First convert to molar mass
18.1*1/231.533*18.01528=
Answer:
0.03127 fraction of a sample of this nuclide would remain after 12.5 hrs.
Explanation:
Initial mass of the isotope =
Time taken by the sample, t = 12.5 hour
Formula used :
where,
= initial mass of isotope
N = mass of the parent isotope left after the time, (t)
= half life of the isotope = 2.5 hour
= rate constant
Now put all the given values in this formula, we get
0.03127 fraction of a sample of this nuclide would remain after 12.5 hrs.
Answer : The enthalpy change is, 7.205 KJ
Solution :
The conversions involved in this process are :
Now we have to calculate the enthalpy change.
where,
= enthalpy change = ?
m = mass of water = 10 g
= specific heat of solid water = 2.09 J/gk
= specific heat of liquid water = 4.18 J/gk
n = number of moles of water =
= enthalpy change for fusion = 6.01 KJ/mole = 6010 J/mole
Now put all the given values in the above expression, we get
(1 KJ = 1000 J)
Therefore, the enthalpy change is, 7.205 KJ
Molar mass NH3 = 17.031 g/mol
1 mole NH3 -------------- 17.031 g
?? moles NH3 ---------- 346 g
346 x 1 / 17.031
=> 20.31 moles of NH3
B= CH2O
empirical formula using the simplest ratio of atoms in the compound