Answer:
Option C. 251 kJ
Explanation:
The activation energy (Ea) of a given reaction is the minimum energy that must be overcomed for reactant to proceed to product.
The activation energy (Ea) can be obtained from an energy profile diagram by simply calculating the difference between the energy of the activation complex (i.e the peak) and the energy of the reactant.
Thus, we can obtain the activation energy for the reaction above as follow:
Activation complex = 332.6 kJ
Energy of reactant = 81.6 kJ
Activation energy =?
Activation energy = Activation complex – Energy of reactant
Activation energy = 332.6 – 81.6
Activation energy = 251 kJ
Therefore, the activation energy of the reaction is 251 kJ
Chemical Bond
because they are chemically bonding
Answer:
17188 years
Explanation:
Recall the formula;
0.693/t1/2 = 2.303/t log(No/N)
t1/2 = half life of 14C
No = initial activity of 14C
N = activity of 14C at time t
t = age of the sample
Substituting values
0.693/5730= 2.303/t log(160/ 20)
1.21 * 10^-4= 2.0798/t
t = 2.0798/1.21 * 10^-4
t = 17188 years
Answer:
V₂ = 4.0L
Explanation:
Decreasing temperature => Decreasing Volume (Charles Law)
For a given volume, use a temperature ratio that will give a smaller volume.
Volume at lower temp = 4.6L(70K/82K) = 4.0L ... Using (82K/70K) would give a larger volume => contrary to temperature effects on gas volumes when pressure and mass are kept constant.
Pressure effects on Gas Volumes:
Note: The same idea is applied to pressure effects on gas volumes also except that changes in pressure affect gas volumes indirectly. That is, an increase in pressure => decrease in volume, or a decrease in pressure => increase in volume. Boyles Law => V ∝ 1/P.
Given a gas volume of 4.60L at 760mmHg, what is volume at 848mmHg?
Increasing pressure => Decreases Volume (Boyles Law)
For the given volume use a pressure ratio that will give a smaller volume.
Volume at higher pressure = 4.6L(760mm/848mm) =4.1L. Using (848mm/760mm) would give a larger volume => contrary to pressure effects on gas volume when temperature and mass of gas are kept constant.
Answer:
Explanation:
3BaCl₂(aq) + 2Na₃PO₄(aq) = Ba₃(PO₄)₂ + 6NaCl(aq)
3 x 208 g = 624 g 2 x 164 g = 328 g
328 g of Na₃PO₄ reacts with 624 g of BaCl₂
.611 g of Na₃PO₄ reacts with 624 x .611/328 g of BaCl₂
624 x .611/328 g = 1.16 g of BaCl₂
BaCl₂ available is .504 g which is less than required .
Hence BaCl₂ is limiting reagent .
