Consider the water gas equilibrium reaction
Cs) + H20(g) = CO(g) + H2(g) . Which of
the following statement is true at equilibrium?
If the pressure on the system is
increased by halving the
volume, more CO and H2
would be formed.
If the amount of C(s) is
increased, less water would be
formed
If the amount of C(s) is
increased, more CO and H,
would be formed
If the pressure on the system is
increased by halving the volume, more
water would be formed.
Answers
Explanation:
\huge \fbox \red{\fbox{Answer}}
Answer
\large \bf{Given:}Given:
\sf \mapsto{Mass \: of \: baby(m)=500g= \frac{500}{1000} kg = 0.5kg}↦Massofbaby(m)=500g=
1000
500
kg=0.5kg
\sf \mapsto{Initial \: velocity(u)=54km/h=54 \times \frac{5}{18} m/s = 15m/s}↦Initialvelocity(u)=54km/h=54×
18
5
m/s=15m/s
\sf \mapsto{Time \: taken(t)=6sec}↦Timetaken(t)=6sec
\large \bf{To \: find:}Tofind:
\sf \leadsto{Force \: acting \: on \: baby. }⇝Forceactingonbaby.
\large \bf{Concept \: used:}Conceptused:
\sf \rightarrow{Final \: velocity(v) \: of \: baby \: is \: 0m/s.}→Finalvelocity(v)ofbabyis0m/s.
\large \bf{Formula \: used:}Formulaused:
\boxed {\bf{Acceleration \: of \: baby(a) = \frac{Final \: velocity - Initial \: velocity}{Time} = \frac{v - u}{t} }}
Accelerationofbaby(a)=
Time
Finalvelocity−Initialvelocity
=
t
v−u
\boxed {\bf{Force=Mass×Acceleration = ma}}
Force=Mass×Acceleration=ma
\large \bf{According \: to \: Question:}AccordingtoQuestion:
\sf \implies{Acceleration \: of \: baby(a) = \frac{Final \: velocity - Initial \: velocity}{Time} = \frac{v - u}{t} }⟹Accelerationofbaby(a)=
Time
Finalvelocity−Initialvelocity
=
t
v−u
\sf \implies{Acceleration \: of \: baby(a) = \frac{0- 15}{6} = \frac{- 15}{6} = - 2.5m/ {s}^{2} }⟹Accelerationofbaby(a)=
6
0−15
=
6
−15
=−2.5m/s
2
\bf{Then,}Then,
\sf \implies{Force=Mass×Acceleration}⟹Force=Mass×Acceleration
\sf \implies{Force=0.5×2.5=1.25N}⟹Force=0.5×2.5=1.25N
\bf{Hence, Force \: acting \: on \: baby \: is \: 1.25 \: newton.}Hence,Forceactingonbabyis1.25newton.
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