It’s C. 0.31 atm
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Au^2S^3+ 3H^2 = 2Au + 3H^2S
Answer:
190.4g
Explanation:
1.6mol of KBr (119.002g KBr/1 mol) = 190.4g
since you want to find grams, take the molar mass of KBr (119.002) per 1 mol and use it as your conversion factor (119.002g KBr/1 mol) which will then cancel out mols and leave you with grams.
Answer:
Option D. 230 J
Explanation:
We'll begin by calculating the temperature change of the iron. This can be obtained as follow:
Initial temperature (T₁) = 50 °C
Final temperature (T₂) = 75 °C
Change in temperature (ΔT) =?
ΔT = T₂ – T₁
ΔT = 75 – 50
ΔT = 25 °C
Thus, the temperature change of the iron is 25 °C.
Finally, we shall determine the amount of heat energy used. This can be obtained as follow:
Mass (M) = 20 g
Change in temperature (ΔT) = 25 °C
Specific heat capacity (C) = 0.46 J/gºC
Heat (Q) =?
Q = MCΔT
Q = 20 × 0.46 × 25
Q = 230 J
Thus, the amount of heat used was 230 J
