What is the transcendence degree of Q_p and C over Q? Is the tr.deg of Q_p over Q 1? and what about C over Q?
 A: In both cases the transcendence degree is the cardinality of the continuum.  CH is not needed.
This is a corollary of the following result: let $K$ be any infinite field, and let $L/K$ be any extension.  Then 
$\# L = \operatorname{max} (\# K, \operatorname{trdeg}_K L)$.
To prove this, in turn it suffices to establish the following two results (each of which is straightforward):
1) If $K$ is infinite and $L/K$ is algebraic, then $\# L = \# K$.
2) If $K$ is any infinite field, $T = \{t_i\}_{i \in I}$ is an arbitrary set of indeterminates and $K(T)$ is a purely transcendental function field in the indeterminates $T$, then  $ \# K(T) \leq \# T + \# K$.  
A: The transcendence degree of either $\mathbb{C}$ or $\mathbb{Q}_p$ over $\mathbb{Q}$ is exactly the cardinality of the continuum.  Certainly it can't be countable, since any field with countable transcendence degree over a countable field is countable.  On the other hand, either transcendence degree is at most that of the continuum.  So we're already done assuming CH.  But I think the result holds even without CH; can anyone confirm / deny?  (Edit:  Yes, as the other answer shows.)
