We can use the heat equation,
Q = mcΔT
where Q is the amount of energy transferred (J), m is the mass of the substance (kg), c is the specific heat (J g⁻¹ °C⁻¹) and ΔT is the temperature difference (°C).
Q = 11.2 kJ = 11200 J
m = <span>145 g
</span>c = ?
ΔT = (67 - 22) °C = 45 °C
By applying the formula,
11200 J = 145 g x c x 45 °C
c = 1.72 J g⁻¹ °C⁻¹
Hence, specific heat of benzene is 1.72 J g⁻¹ °C⁻¹.
It would be 0.341 because if you add 0.229 and 0.112 it will be 0.341
The correct answer is B. Low chemical reactivity
Atomic mass Ar => 39.948 a.m.u
39.948 g --------------- 6.02x10²³ atoms
?? g -------------------- 3.8x10²⁴ atoms
(3.8x10²⁴) x 39.948 / 6.02x10²³ => 250 g
hope this helps!
