B. base
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Answer:
Silver, 0.239 J/(g °C)
Explanation:
- The heat change is related to specific heat as given by the formula;
Heat change = mass of substance × specific heat × change in temperature
- Therefore; considering same amount of substance or equal masses and have the same initial temperature.
- The change in temperature will be inversely proportional to the specific heat.
- Therefore; the higher the specific heat lower the temperature change.
- Hence, the change in temperature will be highest for the substance with the lowest specific heat.
Therefore; the one that will increase in temperature the most is Silver
Your answer is B: The outer electron of Adam B has moved to a higher energy state
The reaction;
O(g) +O2(g)→O3(g), ΔH = sum of bond enthalpy of reactants-sum of food enthalpy of products.
ΔH = ( bond enthalpy of O(g)+bond enthalpy of O2 (g) - bond enthalpy of O3(g)
-107.2 kJ/mol = O+487.7kJ/mol =O+487.7 kJ/mol +487.7kJ/mol =594.9 kJ/mol
Bond enthalpy (BE) of O3(g) is equals to 2× bond enthalpy of O3(g) because, O3(g) has two types of bonds from its lewis structure (0-0=0).
∴2BE of O3(g) = 594.9kJ/mol
Average bond enthalpy = 594.9kJ/mol/2
=297.45kJ/mol
∴ Averange bond enthalpy of O3(g) is 297.45kJ/mol.
Remember that:
number of moles = mass/molar mass
First, we get the molar mass of the nitrogen gas molecule:
It is known the the nitrogen gas is composed of two nitrogen atoms, each with molar mass 14 gm (from the periodic table)
Therefore, molar mass of nitrogen gas = 14 x 2 = 28 gm
Second we calculate the mass of the precipitate:
we have number of moles = 0.03 moles (given)
and molar mass = 28 gm (calculated)
Using the equation mentioned before,
mass = number of moles x molar mass = 0.03 x 28 = 0.84 gm