<span>M(HCl) * </span><span>V(HCl) </span>= <span>M(NaOH) * </span><span>V(<span>NaO<span>H)
</span></span></span>
M(HCl) = 0.35
<span>V(HCl) = 45mL
</span>M(NaOH)= 0.35
now, solne for V(NaOH) by putting these values in the above equation.
M(HCl) * <span>V(HCl) </span>= <span>M(NaOH) * </span><span>V(NaOH)</span>
<span>0.35 * 45 = 0.35 * V(NaOH)</span>
<span>V(NaOH) = 45 mL</span>
Answer:
The assumption is quite reasonable.........
A lightbulb contains Ar gas at a temperature of 295K and at a pressure of 75kPa. The light bulb is switched on, and after 30 minutes its temperature is 418 K. What is a numerical setup for calculating the pressure of the gas inside the light bulb at 418K?
Explanation:
P
1
T
1
=
P
2
T
2
given constant
n
, and constant
V
, conditions that certainly obtain with a fixed volume light bulb.
And so
P
2
=
P
1
T
1
×
T
2
=
75
⋅
k
P
a
295
⋅
K
×
418
⋅
K
≅
100
⋅
k
P
a
.
Had the light bulb been sealed at normal pressure during its manufacture, what do you think might occur when it is operated?
Answer:
A) The reaction is exothermic reaction
B) 46 gm CH3CH2OH = 1236KJ
15.3 gm ch3ch2OH = 1236/46 x 15.3
= 411.10 KJ..........released
therefore 1 KJ = 0.239 K cal
so, 411.10 x 0.239 kcal
= 98.2529 .................. released
c) 54 gm of H2O produced = 1236KJ
so, 42.7 gm H2O produced = 1236/54 x 42.7KJ
= 977.35 KJ released.
Explanation:
