Anode- oxidization
Cathode-reduction
Answer:
It can help identify elements quicker.
Explanation:
Answer:
64799.4 J
Explanation:
The following data were obtained from the question:
Mass (M) = 1.05 kg = 1.05 x 1000 = 1050g
Specific heat capacity (C) = 0.9211 J/g°C
Initial temperature (T1) = 23°C
Final temperature (T2) = 90°C
Change in temperature (ΔT) = T2 – T1 =
90°C – 23°C = 67°C
Heat required (Q) =....?
The heat required to increase the temperature of the kettle can b obtain as follow:
Q = MCΔT
Q = 1050 x 0.9211 x 67
Q = 64799.4 J
Therefore, 64799.4 J of heat is required to increase th temperature of the kettle from 23°C to 90°C.
From the coefficients of the equation, we know that for every 3 moles of water consumed, 1 mole of diphosphorus trioxide is consumed.
This means we need to find the mass of 0.75 moles of diphosphorus trioxide.
- The atomic mass of phosphorous is 30.973761998 g/mol.
- The atomic mass of oxygen is 15.9994 g/mol.
So, the formula mass of diphosphorus trioxide is:
- 2(30.973761998)+3(15.9994)=109.945723996 g/mol.
Thus, 0.75 moles have a mass of:
- 0.75(109.945723996), which is about 82.5 g (to 3 sf)
