Where are the right-handed leptons? So the weak force only works on left-handed leptons and quarks. Does this mean there are no right-handed leptons and quarks?
If there are, why don't we see an abundance of right-handed particles that do not decay? If there are no right handed leptons and quarks, why not?
 A: There are right handed quarks and leptons, they just don't interact weakly. The matter content of the Standard Model is often written in the "QUDLE" form as:
$$\begin{array}
{} & SU(3)_C & SU(2)_L & U(1)_Y & U(1)_Q \\ \hline
q_L^i = \begin{pmatrix} u_L^i \\ d_L^i \end{pmatrix}  & 3 & 2 & 1/6  & \begin{pmatrix} \frac{2}{3} \\ -\frac{1}{3} \end{pmatrix} \\ \hline
u_R^i & 3 & 1 & 2/3 & 2/3\\ \hline
d_R^i & 3 & 1 & -1/3 & -1/3     \\ \hline
l_L^i = \begin{pmatrix} \nu^i_L \\ e^i_L \end{pmatrix}  & 1 & 2 & -1/2  & \begin{pmatrix} 0 \\ -1 \end{pmatrix}    \\ \hline
e_R^i & 1 & 1 & -1 & -1\\ \hline
\phi   & 1 & 2 & 1/2 & \begin{pmatrix} - \\ 0 \end{pmatrix} \\ \hline
\end{array}$$
where it is understood that the Standard Model is a gauge theory with gauge group $SU(3) \times SU(2) \times U(1)_Y$ that gets spontaneously broken to $SU(3) \times U(1)_Q$. In particular, when people say that only left-handed quarks/leptons participate in weak force, they are talking about the fact that the up/down type left-handed quarks, as well as the left-handed leptons transform as the '2' representation under $SU(2)$, whereas the right-handed counterparts transform under the '1' (singlet) representation. This means that in the broken theory, the $W^\pm$ bosons which mediate decays don't interact with the right-handed quarks and leptons. (Note that the right-handed quarks and leptons still interact with Z bosons though, due to their $U(1)_Y$ charge)
The right handed quarks and leptons exist though. Note that after spontaneous symmetry breaking, the quarks and leptons gain mass terms that look like $m_u (\bar{u}_L u_R + \bar{u}_R u_L)$ (I am ignoring all the issues about flavor basis/mass basis not being same), these mass terms 'mix' the left and right handed modes, in the sense that you can't really treat them as separate.
In the table above, I also omit the neutrinos. Years ago, I believe people thought they could potentially be massless, and thus there could truly be no right-handed neutrinos. However, we have come to find out via experiment that neutrinos are massive, hence the simplest explanation would be that right-handed neutrinos do exist. I haven't included this in the table because

*

*I think theres still many open questions about neutrinos

*I don't know much about neutrinos personally, if anyone wants to come fix this answer feel free

