1
TG EAPCET 2025 (Online) 2nd May Morning Shift
MCQ (Single Correct Answer)
+1
-0

The RMS velocity of dihydrogen is $\sqrt{7}$ times more than that of dinitrogen. If $T_{\mathrm{H}_2}$ and $T_{\mathrm{N}_2}$ are the temperatures of dihydrogen and dinitrogen, then the correct relationship between them is

A

$T_{\mathrm{H}_2}=T_{\mathrm{N}_2}$

B

$T_{\mathrm{H}_2}>T_{\mathrm{N}_2}$

C

$T_{\mathrm{H}_2}=\sqrt{7} T_{\mathrm{N}_2}$

D

$T_{\mathrm{H}_2}=\frac{T_{\mathrm{N}_2}}{2}$

2
TG EAPCET 2025 (Online) 2nd May Morning Shift
MCQ (Single Correct Answer)
+1
-0

Which of the following solution has highest amount of solute?

A

1.0 L of $0.25 \mathrm{M} \mathrm{Na}_2 \mathrm{CO}_3(106 \mathrm{u})$

B

0.25 L of $0.2 \mathrm{M} \mathrm{Na}_2 \mathrm{SO}_4(142 \mathrm{u})$

C

0.5 L of $1.0 \mathrm{M} \mathrm{KMnO}_4(158 \mathrm{u})$

D

0.75 L of $0.5 \mathrm{M}\left(\mathrm{NH}_2\right)_2 \mathrm{CO}(60 \mathrm{u})$

3
TG EAPCET 2025 (Online) 2nd May Morning Shift
MCQ (Single Correct Answer)
+1
-0

At 298 K , the enthalpy change ( in kJ ) for the reaction given below is

$$ \mathrm{CH}_4(g)+\mathrm{O}_2(g) \longrightarrow \mathrm{C}(s)+2 \mathrm{H}_2 \mathrm{O}(l) $$

$$ \begin{aligned} Given:\mathrm{H}_2(g)+\frac{1}{2} \mathrm{O}_2(g) & \longrightarrow \mathrm{H}_2 \mathrm{O}(l) ; \Delta H^{\ominus}=-286 \mathrm{~kJ} \\ \mathrm{C}(s)+\mathrm{O}_2(g) & \longrightarrow \mathrm{CO}_2(g) ; \Delta H^{\ominus}=-394 \mathrm{~kJ} \\ \mathrm{CH}_4(g)+2 \mathrm{O}_2(g) & \longrightarrow \mathrm{CO}_2(g)+2 \mathrm{H}_2 \mathrm{O}(l) \Delta H^{\ominus}=-890 \mathrm{~kJ}\end{aligned} $$

A

+496

B

-496

C

-1284

D

+680

4
TG EAPCET 2025 (Online) 2nd May Morning Shift
MCQ (Single Correct Answer)
+1
-0

For the reaction $\mathrm{N}_2 \mathrm{O}_4(g) \rightleftharpoons 2 \mathrm{NO}_2(g)$, the correct relation between degree of dissociation $(\alpha)$ of $\mathrm{N}_2 \mathrm{O}_4(g)$ and equilibrium constant, $K_p$ is ( $p=$ total pressure of mixture)

A

$\alpha=\frac{K_p / p}{4+\frac{K_p}{p}}$

B

$\alpha=\frac{K_p}{4+K_p}$

C

$\alpha=\left(\frac{K_p / p}{4+\frac{K_p}{p}}\right)^{\frac{1}{2}}$

D

$\alpha=\left(\frac{K_p}{4+K_p}\right)^{\frac{1}{2}}$

TS EAMCET Papers

All year-wise previous year question papers