confused with dual ac/dc adapter
It is most likely an unregulated supply which uses a mains transformer, rectifier, and filter capacitor to convert AC to DC. 6.5VDC will be the nominal voltage at full load (150mA). The voltage waveforms at input and output of the rectifier typically look like this:-



There are several reasons for the unloaded voltage being so much higher. 


The filter capacitor charges up to the peak voltage of the rectified sine wave, which is 1.4 times the rms voltage. 
The rectifier diodes may each drop 1V or more under full load, but closer to 0.5V at very low current. If a bridge rectifier is used then the current goes through 2 diodes in series, so the the total voltage drop could vary by 1V or more.   
The transformer has leakage inductance and winding resistance, which causes its output voltage to drop under load. Since the rated voltage is for full load, the unloaded voltage will be higher. 
The mains voltage could vary by 10% or more, and the power supply has to deliver its rated output at minimum mains voltage. The transformer reduces voltage by a fixed ratio, so if the mains voltage is higher then the transformer's output voltage will also be higher. 


Under load the transformer's output voltage will reduce, the diodes will drop more voltage, and the filter capacitor will partially discharge between peaks of the rectified sine wave. As a result of all these losses, the average DC voltage at rated output current should be close to the rated 6.5VDC. To prove this you need to connect a load which draws ~150mA when measuring the DC output voltage.

If your application requires exactly 6.5V then this power supply may not be suitable. However devices which are designed to work with unregulated supplies can usually handle the higher unloaded voltage.