Answer:
Q1: 728.6 J.
Q2:
a) 668.8 J.
b) 0.3495 J/g°C.
Explanation:
<em>Q1: Calculate the energy change (q) of the surroundings (water) using the enthalpy equation:</em>
- The amount of heat absorbed by water = Q = m.c.ΔT.
where, m is the mass of water (m = d x V = (1.0 g/mL)(24.9 mL) = 24.9 g).
c is the specific heat capacity of liquid water = 4.18 J/g°C.
ΔT is the temperature difference = (final T - initial T = 32.2°C - 25.2°C = 7.0°C).
<em>∴ The amount of heat absorbed by water = Q = m.c.ΔT</em> = (24.9 g)(4.18 J/g°C)(7.0°C) = 728.6 J.
<em>Q2: Calculate the energy change (q) of the surroundings (water) using the enthalpy equation
</em>
<em>qwater = m × c × ΔT. </em>
<em>We can assume that the specific heat capacity of water is 4.18 J / (g × °C) and the density of water is 1.00 g/mL. calculate the specific heat of the metal. Use the data from your experiment for the unknown metal in your calculation.</em>
<em></em>
a) First part: the energy change (q) of the surroundings (water):
- The amount of heat absorbed by water = Q = m.c.ΔT.
where, m is the mass of water (m = d x V = (1.0 g/mL)(25 mL) = 25 g).
c is the specific heat capacity of liquid water = 4.18 J/g°C.
ΔT is the temperature difference = (final T - initial T = 31.6°C - 25.2°C = 6.4°C).
<em>∴ The amount of heat absorbed by water = Q = m.c.ΔT</em> = (25 g)(4.18 J/g°C)(6.4°C) = <em>668.8 J.</em>
<em>b) second part:</em>
<em>Q water = Q unknown metal. </em>
<em>Q unknown metal = - </em>668.8 J. (negative sign due to the heat is released from the metal to the surrounding water).
<em>Q unknown metal = - </em>668.8 J = m.c.ΔT.
m = 27.776 g, c = ??? J/g°C, ΔT = (final T - initial T = 31.6°C - 100.5°C = - 68.9°C).
<em>- </em>668.8 J = m.c.ΔT = (27.776 g)(c)( - 68.9°C) = - 1914 c.
∴ c = (<em>- </em>668.8)/(- 1914) = 0.3495 J/g°C.
<em></em>
is formed is 0.066 M/s</h2><h2>b) The rate at which molecular oxygen 
is reacting is 0.033 M/s</h2>