½H2(g) + ½I2(g) → HI(g) ΔH = +6.2 kcal/mol
or...
½H2(g) + ½I2(g) + 6,2kcal/mole → HI(g)
________
21.0 kcal/mole + C(s) + 2S(s) → CS2(l)
or...
C(s) + 2S(s) → CS2(l) ΔH = +2,1 kcal/mole
_________
ΔH > 0 ----------->>> ENDOTHERMIC REACTIONS
Answer:
15.0 µm
Step-by-step explanation:
Density = mass/volume
D = m/V Multiply each side by V
DV = m Divide each side by D
V = m/D
Data:
m = 1.091 g
D = 7.28 g/cm³
l = 10.0 cm
w = 10.0 cm
Calculation:
<em>(a) Volume of foil
</em>
V = 1.091 g × (1 cm³/7.28 g)
= 0.1499 cm³
(b) <em>Thickness of foil
</em>
The foil is a rectangular solid.
V = lwh Divide each side by lw
h = V/(lw)
= 0.1499/(10 × 10)
= 1.50 × 10⁻³ cm Convert to millimetres
= 0.015 mm Convert to micrometres
= 15.0 µm
The foil is 15.0 µm thick.
Answer:
The significance of "Er" in the diagram is :
B.) Threshold energy for reaction
Explanation:
Threshold energy : It is total amount of energy required by the reactant molecule to reach the transition state .
Activation energy : It is the excess energy absorbed by the molecules to reach the transition state.
<u>Activation Energy = Threshold Energy - Average Kinetic Energy</u>
<u>This means Activation energy decreases on increasing kinetic energy</u>
On increasing Temperature average kinetic energy of the molecule increases which reduces the activation energy and the reaction occur faster in that case.
Catalyst also reduces the Activation energy.
<u>Er = Threshshold energy for reaction at 30 degree</u>
<u>Ea = Activation Energy</u>
<u>The given figure shows that the threshold energy decreases on increasing the temperature</u>
<u>Only the molecule having energy greater than Er can react to form product</u>