MCQ
$10\ gms$ . each of $CO_2$ , $NH_3$ and $O_2$ were taken in three separate flasks. What is the correct decreasing order of atoms
- A$CO_2$ , $NH_3$ , $O_2$
- B$NH_3$ , $O_2$ , $CO_2$
- C$O_2$ , $NH_3$ , $CO_2$
- ✓$NH_3$ , $CO_2$ , $O_2$
✓
Answer
Correct option: D.
$NH_3$ , $CO_2$ , $O_2$
d
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
The Lassiagne's extract is boiled with dil $\mathrm{HNO}_3$ before testing for halogens because,
→A similarity between optical and geometrical isomerism is that
Assertion : The radius of the first orbit of hydrogen atom is $0.529\,\mathop A\limits^o$
Reason : Radius of each circular orbit $(r_n) - 0.529\,\mathop A\limits^o \,(n^2/Z),$ where $n = 1, 2, 3$ and $Z =$ atomic number.
→Reason : Radius of each circular orbit $(r_n) - 0.529\,\mathop A\limits^o \,(n^2/Z),$ where $n = 1, 2, 3$ and $Z =$ atomic number.
An organic compound containing $C, H$ and $N$ gave following analysis : $C = 40\%$, $H = 13.33\%$ and $N = 46.67\%$. Its empirical formula would be
→$IUPAC$ name of the given compound is
→Find the value of wave number of $(\bar v)$ in terms of Rydberg's constant, when transition of electron takes place between two levels of $He^+$ ion whose sum is $4$ and difference is $2$
→Among the following the correct statement($s$) for electrons in an atom is(are)
→$(A)$ Uncertainty principle rules out the existence of definite paths for electrons.
$(B)$ The energy of an electron in $2 s$ orbital of an atom is lower than the energy of an electron that is infinitely far away from the nucleus.
$(C)$ According to Bohr's model, the most negative energy value for an electron is given by $n=1$, which corresponds to the most stable orbit.
$(D)$ According to Bohr's model, the magnitude of velocity of electrons increases with increase in values of $n$.
Molten sodium is used in nuclear reactors to:
The organic compound used as antiknock agent in petroleum is
The equilibrium constant $K$ for the reaction : $\ce{2HI(g) \rightleftharpoons H_2(g) + I_1(g)}$ at room temp is $2.85$ and that at $698 K$ is $1.4 \times 10 − 2$. This implies that the forward reaction is:
→