A, O2 has to be a reactant for combustion to burn
Answer:
It will be reported too low.
Explanation:
To measure the specific heat of the metal (s), the calorimeter may be used. In it, the metal will exchange heat with the water, and they will reach thermal equilibrium. Because it can be considered an isolated system (there're aren't dissipations) the total amount of heat (lost by metal + gained by water) must be 0.
Qmetal + Qwater = 0
Qmetal = -Qwater
The heat is the mass multiplied by the specific heat multiplied by the temperature change. If c is the specific heat of the water:
m_metal*s*ΔT_metal = - m_water *c*ΔT_water
s = -m_water *c*ΔT_water / m_metal*ΔT_metal
So, if m_water is now less than it was supposed to be, s will be reported too low, because they are directly proportional.
The correct answer is A. Alkali metals.
I hope this helped
Answer:
Part A. The half-cell B is the cathode and the half-cell A is the anode
Part B. 0.017V
Explanation:
Part A
The electrons must go from the anode to the cathode. At the anode oxidation takes place, and at the cathode a reduction, so the flow of electrons must go from the less concentrated solution to the most one (at oxidation the concentration intends to increase, and at the reduction, the concentration intends to decrease).
So, the half-cell B is the cathode and the half-cell A is the anode.
Part B
By the Nersnt equation:
E°cell = E° - (0.0592/n)*log[anode]/[cathode]
Where n is the number of electrons being changed in the reaction, in this case, n = 2 (Sn goes from S⁺²). Because the half-reactions are the same, the reduction potential of the anode is equal to the cathode, and E° = 0 V.
E°cell = 0 - (0.0592/2)*log(0.23/0.87)
E°cell = 0.017V
Answer:
Diamond.
Explanation:
Having thermal conductivity of more than 2,000 WMK
WMK = Watts per Meter per Kelvin
