The activation energy Ea can be related to rate constant (k) at temperature (T) through the equation:
ln(k2/k1) = Ea/R[1/T1 - 1/T2]
where :
k1 is the rate constant at temperature T1
k2 is the rate constant at temperature T2
R = gas constant = 8.314 J/K-mol
Given data:
k1 = 0.543 s-1; T1 = 25 C = 25+273 = 298 K
k2 = 6.47 s-1; T = 47 C = 47+273 = 320 K
ln(6.47/0.543) = Ea/8.314 [1/298 - 1/320]
2.478 = 2.774 *10^-5 Ea
Ea = 0.8934*10^5 J = 89.3 kJ
Answer:
54 g
Explanation:
1 mole of water = H2O
mass of 1 mole of H2O= mass of h2 + mass of o
= 2× mass of h +mass of o
= 2×1+16 =18 g
1 mole of water = 18g
3moles of water = 18×3g= 54g
Answer:
covalent bonds............
<u>Answer:</u> The solubility of
in water is 
<u>Explanation:</u>
The balanced equilibrium reaction for the ionization of cadmium phosphate follows:

3s 2s
The expression for solubility constant for this reaction will be:
![K_{sp}=[Cd^{2+}]^3[PO_4^{3-}]^2](https://tex.z-dn.net/?f=K_%7Bsp%7D%3D%5BCd%5E%7B2%2B%7D%5D%5E3%5BPO_4%5E%7B3-%7D%5D%5E2)
We are given:

Putting values in above equation, we get:

Hence, the solubility of
in water is 