(B) The total pressure is given by Dalton's Law: $P_{Total} = P_{H_2O} + P_{O_2}$.Given $P_{Total} = 760\, mm$ and $P_{H_2O} = 355.5\, mm$ at $80\,^oC

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(B) The total pressure is given by Dalton's Law: $P_{Total} = P_{H_2O} + P_{O_2}$.Given $P_{Total} = 760\, mm$ and $P_{H_2O} = 355.5\, mm$ at $80\,^oC$.Therefore,the initial partial pressure of $O_2$ is $P_{O_2(1)} = 760 - 355.5 = 404.5\, mm$.When the gas is transferred to a $50\, mL$ vessel at constant temperature,we apply Boyle's Law: $P_1V_1 = P_2V_2$.Here,$P_1 = 404.5\, mm$,$V_1 = 100\, mL$,$V_2 = 50\, mL$.$P_2 = \frac{404.5 \times 100}{50} = 809\, mm$.Thus,the partial pressure of $O_2$ in the $50\, mL$ vessel is $809\, mm$.

Class 11 Chemistry Data Entry - Chemistry Level 0 of all question bank

The vapor pressure of water at $80\,^oC$ is $355.5\, mm$ of $Hg$. $A$ $100\, mL$ vessel contains $O_2$ gas at $80\,^oC$ saturated with water vapor,and the total pressure is $760\, mm$ of $Hg$. If the contents of the vessel are transferred to a $50\, mL$ vessel at the same temperature,what will be the partial pressure of $O_2$ in $mm$?

A
$1115$
B
$809$
C
$405$
D
$3555$

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If $\overrightarrow{a}=\hat{i}+\hat{j}+\hat{k}$,$\overrightarrow{b}=\hat{i}+\hat{j}$,$\overrightarrow{c}=\hat{i}$ and $(\overrightarrow{a} \times \overrightarrow{b}) \times \overrightarrow{c}=\lambda \overrightarrow{a}+\mu \overrightarrow{b}$,then $\lambda+\mu$ is equal to:

TS EAMCET 2003

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AIPMT 1998

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$A$ point source emits sound equally in all directions in a non-absorbing medium. Two points $P$ and $Q$ are at a distance of $9 \ m$ and $25 \ m$ respectively from the source. The ratio of the amplitudes of the waves at $P$ and $Q$ is

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$A$ light beam is traveling from Region $I$ to Region $IV$ (Refer Figure). The refractive indices in Region $I, II, III$ and $IV$ are $n_0, \frac{n_0}{2}, \frac{n_0}{6}$ and $\frac{n_0}{8}$,respectively. The angle of incidence $\theta$ for which the beam just misses entering Region $IV$ is

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$A$ tension of $20 ~N$ is applied to a copper wire of cross-sectional area $0.01 ~cm^2$. The Young's modulus of copper is $1.1 \times 10^{11} ~N/m^2$ and the Poisson's ratio is $0.32$. The decrease in the cross-sectional area of the wire is:

AP EAMCET 2013

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If $\overrightarrow{a}=\hat{i}+\hat{j}+\hat{k}$,$\overrightarrow{b}=\hat{i}+\hat{j}$,$\overrightarrow{c}=\hat{i}$ and $(\overrightarrow{a} \times \overrightarrow{b}) \times \overrightarrow{c}=\lambda \overrightarrow{a}+\mu \overrightarrow{b}$,then $\lambda+\mu$ is equal to:

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