Yes. First, fix a programming language; for example, C. Now, the halting problem asks for a given C program and a given stdin, will the program terminate. For simplicity, let's just say that we close stdin and give the program an empty input.
It turns out no algorithm can solve the halting problem for all C programs. The explanation is that, if you had such an algorithm, you could write a C program that applies the algorithm to itself (this is a special quine), and then does the opposite of whatever the halting algorithm says the program would do (if it says the program halts, the program will just enter an infinite loop; if it says it does not halt, the program calls exit).
Now, imagine a list of all valid C programs in alphabetical order, beginning with the empty string (which, oddly enough, is a valid C program). Consider a number where the Nth digit after the decimal point is 0 if the Nth C program halts, and 1 if it does not; for example, the first digit, corresponding to the empty string, is 0, while a few digits later, the digit corresponding to "int main() {main(); return 0;}" will be 1.
Putting it all together, we know there are C programs for which we cannot compute the solution to the halting problem, and we have a number whose digits are determined by the solution to the halting problem for every C program. So that number is not computable (because we cannot compute every digit, so we cannot compute it to arbitrary precision).