Question

can anyone do needful by solving the above sum

Answer 1

Lets write the balanced reaction first!!

N2 + 3H2 ---> 2NH3

Initial x y 0

Change -1 -3 +2

Final 2moles 2moles 2moles

Now lets solve for x and y ie initial moles of N2 and H2 respectively.

Initial - change = Final

x -1 = 2

x = 3

initial moles of N2 = 3 moles = 3 x 28 = 84g N2 is present initially

y - 3 = 2

y = 5 moles

Initial moles of H2 is 5 moles = 5 x 2 = 10g H2 is present initially

Answer:

84g N2 and 10g H2 is present initially

Answer 2

Answer:

Answer :

A body is taken 32 km above the surface of the earth

Radius of earth = 6400 km

Percentage decrease in weight of the body = ?

\quad ━━━━━━━━━━━━━━━━━━

\begin{gathered}\sf :\implies g' = g\bigg\lgroup 1 + \dfrac{h}{R}\bigg\rgroup{}^{-2}\;\; -eq(1)\\\end{gathered}

:⟹g

′

=g

⎩

⎪

⎪

⎪

⎧

1+

R

h

⎭

⎪

⎪

⎪

⎫

−2

−eq(1)

\qquad\quad\dag\:\small\sf h < < R†h<<R

\qquad\quad\dag\:\small\sf \dfrac{h}{R} < < 1†

R

h

<<1

\begin{gathered}\sf :\implies (1+x)^{-2} = 1 - 2x \;\; x < < 1\\\end{gathered}

:⟹(1+x)

−2

=1−2xx<<1

\sf :\implies \bigg\lgroup 1 + \dfrac{h}{R}\bigg\rgroup{}^{-2} = 1 - \dfrac{2h}{R}:⟹

⎩

⎪

⎪

⎪

⎧

1+

R

h

⎭

⎪

⎪

⎪

⎫

−2

=1−

R

2h

\begin{gathered}\sf :\implies g' = g\bigg\lgroup 1 - \dfrac{2h}{R}\bigg\rgroup\;\; -eq(2)\\\end{gathered}

:⟹g

′

=g

⎩

⎪

⎪

⎪

⎧

1−

R

2h

⎭

⎪

⎪

⎪

⎫

−eq(2)

So then the % decrease in the value of acceleration due to gravity is gonna be,

\begin{gathered}\sf :\implies \delta g = \dfrac{g' - g}{g} \times 100\\\end{gathered}

:⟹δg=

g

g

′

−g

×100

\begin{gathered}\sf :\implies \delta g = \dfrac{g\bigg\lgroup 1 - \dfrac{2h}{R}\bigg\rgroup - g}{g} \times 100\\\end{gathered}

:⟹δg=

g

g

⎩

⎪

⎪

⎪

⎧

1−

R

2h

⎭

⎪

⎪

⎪

⎫

−g

×100

\begin{gathered}\sf :\implies \delta g = \dfrac{g\bigg\{ \bigg\lgroup 1 - \dfrac{2h}{R}\bigg\rgroup - 1 \bigg\} }{g} \times 100\\\end{gathered}

:⟹δg=

g

g{

⎩

⎪

⎪

⎪

⎧

1−

R

2h

⎭

⎪

⎪

⎪

⎫

−1}

×100

\begin{gathered}\sf :\implies \delta g = \bigg\{ 1 - \dfrac{2h}{R} - 1 \bigg\} 100\\\end{gathered}

:⟹δg={1−

R

2h

−1}100

\sf :\implies \delta g = - \dfrac{200h}{R} \;\; -eq(3):⟹δg=−

R

200h

−eq(3)

Finally, the % change in weight will be given by,

\begin{gathered}\sf :\implies \delta W = \delta m + \delta g\\\end{gathered}

:⟹δW=δm+δg

\begin{gathered}\sf :\implies \delta W = - \dfrac{200h}{R}\\\end{gathered}

:⟹δW=−

R

200h

\displaystyle \underline{\bigstar\:\textsf{According to the Question :}}

★According to the Question :

\sf\dashrightarrow \delta W = \dfrac{200\times 32}{6400}⇢δW=

6400

200×32

\underline{\boxed{\pink{\mathfrak {\delta w = -1 \%}}}}

δw=−1%