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Chemistry 3 Marks Question

Model Paper 5 · Gujarat Board (GSEB) STD 11 Science (GSEB - English Medium). 7 questions.

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3 Marks Question

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7 questions · self-marked practice — reveal the answer and mark yourself.

Question 13 Marks
Balance the following equation,
$
Fe+H_2 O \longrightarrow Fe_3 O_4+H_2
$
Answer
Step (i): Write the skeleton of equation : $Fe + H _2 O \longrightarrow Fe _3 O _4+ H _2$
Step (ii): Changing the $H _2$ to its nascent form $2[ H ]$.
$
Fe+H_2 O \longrightarrow Fe_3 O_4+2 H
$
Step (iii): $Fe _3 O _4$ has the largest number of atoms. The balance this, multiply $H _2 O$ by 4 to balance oxygen atoms. In 4 molecules of $H _2 O$, there are 8 atoms of H which are balanced by multiplying H on the RHS by 8 .
$
3 Fe+4 H_2 O \longrightarrow Fe_3 O_4+8 H
$
Step (iv): Converting $[ H ]$ to its molecule form $H _2$.
$
3 Fe+4 H_2 O \longrightarrow Fe_3 O_4+4 H_2
$
The equation obtained is a balanced chemical equation.
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Question 23 Marks
Define electron gain enthalpy. What are its units?
Answer
The energy which is released by an isoated gaseous atom in gaining an electron from outside atom or ion to form negative ion (or anion) is called electron gain enthalpy $(\Delta e g H)$. Unit of electron gain enthalpy is $kJ / mol$.
In some cases like in noble gas atoms do not have any attraction to gain an electron in that case energy has to be supplied.
For example
$\begin{aligned} & N e(g)+e^{-} \rightarrow N e^{-}(g) \\ & \Delta e g H=+116 k J mol^{-1}\end{aligned}$
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Question 33 Marks
In astronomical observations, signals observed from the distant starts are generally weak. If the photon detector receives a total of $=3.15 \times 10^{-18} J$ from the radiations of 600 nm . Calculate the number of photons received by the detector.
Answer
Total energy received $=3.15 \times 10^{-18}$
$
\lambda=600 nm=600 \times 10^{-9} m=6 \times 10^{-7} m
$
The energy of one photon, $E =\frac{h c}{\lambda}=\frac{\left(6.626 \times 10^{-34} Js \right) \times\left(3 \times 10^8 ms^{-1}\right)}{\left(6 \times 10^{-7} m\right)}$
$
=3.3125 \times 10^{-19} J
$
$\therefore$ No. of photons $=\frac{3.15 \times 10^{-18} J}{3.3125 \times 10^{-19} J}=10$
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Question 43 Marks
Which method can be used to find out the strength of reductant/oxidant in a solution? Explain with an example.
Answer
The redox system is used. In redox systems, the titration method can be adapted to determine the strength of a reductant/oxidant using a redox-sensitive indicator. The usage of indicators in redox titration is illustrated below: In one situation, the reagent itself is intensely coloured, e.g. permanganate ion $MnO _4^{-}$. Here, $MnO _4^{-}$acts as the self indicator. The visible endpoint, in this case, is achieved after the last of the reductant $\left( Fe _2{ }^{+}\right.$or $\left.C _2 O _4^{2-}\right)$ is oxidised and the first lasting tinge of pink colour appears at $MnO _4^{-}$concentration as low as $10^{-6} mol dm ^{-3}$ and $\left(10^{-6} mol L ^{-1}\right)$. This ensures a minimal 'overshoot' in colour beyond the equivalence point, the point where the reductant and the oxidant are equal in terms of their mole stoichiometry.
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Question 53 Marks
Express the change in internal energy of a system when
1. No heat is absorbed by the system from the surroundings, but work (w) is done on the system. What type of wall does the system have?
2. No work is done on the system, but q amount of heat is taken out from the system and given to the surroundings. What type of wall does the system have?
3. w amount of work is done by the system and q amount of heat is supplied to the system. What type of system would it be?
Answer
1. $\Delta U = w _{ ad }$, wall is adiabatic
2. $\Delta U =- q$, thermally conducting walls
3. $\Delta U = q - w$, closed system.
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Question 63 Marks
1. Two ideal gases under the same pressure and temperature are allowed to mix in an isolated system.
What will be the sign of entropy change?
2. Water can be lifted into the water tank at the top of the house with the help of a pump. Then why is it not considered to be spontaneous?
3. What is the enthalpy change for an adiabatic process?
Answer
1. When two ideal gases at the same pressure and temperature are allowed to mix in an isolated system, the entropy change is positive because degrees of freedom increases on mixing the gases.
2. A spontaneous process should occur continuously by itself after initiation. But this is not so in the given case because water will go up so long as the pump is working.
So, this process is not considered as a spontaneous process.
3. For an adiabatic process,
$
\Delta H=0
$
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