6050 - Platform Manager 2: Why does I2C slave address shows 0x78?
0x78 will be acknowledged because the device would think that it is going to get a 10-bit slave address because of the 10bit address scheme In the 10bit address scheme, you require two bytes to transfer the address. After the start condition, a leading '11110' would tell the slave that it's a 10-bit addressing scheme.
10-Bit Addressing Scheme: The 10-bit slave address is formed from the first two bytes following a START condition (S) or a repeated START condition (Sr). The first seven bits of the first byte are the combination 1111 0XX of which the last two bits (XX) are the two Most Significant Bits (MSBs) of the 10-bit address. The entire second byte will form the remaining 8 Bits of your 10-Bit address. The eighth bit of the first byte is the R/W bit that determines the direction of the message. Although there are eight possible combinations of the reserved address bits 1111 XXX, only the four combinations 1111 0XX are used for 10-bit addressing. The remaining four combinations 1111 1XX are reserved for future I2C-bus enhancements.
Here's a link that explains about 10-bit addressing scheme:
http://www.i2c-bus.org/addressing/10-bit-addressing/