64KB
8 bits
64kb
8
Yes and no. All memory location from 0H to 0FFFFH are addressable, but some of them are needed for the program, interrupt vectors, and the stack, so you would need to pay attention to where things are located in memory to design an appropriate program. In addition, if your system is using memory mapped I/O, some locations will be reserved.
The data size in the 8085 is 8 bits.
its the memory size required for coding
RAM size is important because it is required for keeping all application and program data in your computer system. It provides enough memory between processes to help you avoid waiting for things to progress or load.
128kb
The maximum memory that can be dynamically allocated depends on the size of the heap memory. Dynamic blocks of memory can be allocated in system heap until it is not full.
128Kb
Virtual memory is used to increase the size of working memory in the system main memory by using the locations in secondary storage such as harddisk.
The 8085 is an 8 bit processor, so its word length is 8 bits.
That depends on the operating system and computer architecture.
The 8085 is an 8-bit microprocessor. Even though there are some 16-bit registers (BC, DE, HL, SP, PC), with some 16-bit operations that can be performed on them, and a 16-bit address bus, the accumulator (A), the arithmetic logic unit (ALU), and the data bus are 8-bits in size, making the 8085 an 8-bit computer.
Any modern mobile (cellular) phone has built-in memory. This may be sufficient for some downloads, subject to amount of spare built-in memory and size of the downloaded item. Memory cards are not required for downloads per se, but supply additional memory (which may be required to allow for large downloaded items).