MNaHCO₃: 23+1+12+(48×3) = 84g
mCH₃COOH: 12+(1×3)+12+(16×2)+1 = 60g
.................
84g NaHCO₃ react with 60g CH₃COOH
83g NaHCO₃ react with...........
84g ----- 60g
83g ----- X
X = 59,29g CH₃COOH
We used 70g CH₃COOH, it' too much.
So, acetic acid is excess reagent, and sodium bicarbonate is limiting reagent.
_______________________________
B) Amount of CH3COOH is in excess.
:•)
Answer:
The answer to the question above is
The energy required to heat 87.1 g acetone from a solid at -154.0°C to a liquid at -42.0°C = 29.36 kJ
Explanation:
The given variables are
ΔHfus = 7.27 kJ/mol
Cliq = 2.16 J/g°C
Cgas = 1.29 J/g°C
Csol = 1.65 J/g°C
Tmelting = -95.0°C.
Initial temperature = -154.0°C
Final temperature = -42.0°C?
Mass of acetone = 87.1 g
Molar mass of acetone = 58.08 g/mol
Solution
Heat required to raise the temperature of solid acetone from -154 °C to -95 °C or 59 °C is given by
H = mCsolT = 87.1 g* 1.65 J/g°C* 59 °C = 8479.185 J
Heat required to melt the acetone at -95 °C = ΔHfus*number of moles =
But number of moles = mass÷(molar mass) = 87.1÷58.08 = 1.5
Heat required to melt the acetone at -95 °C =1.5 moles*7.27 kJ/mol = 10.905 kJ
The heat required to raise the temperature to -42 degrees is
H = m*Cliq*T = 87.1 g* 2.16 J/g°C * 53 °C = 9971.21 J
Total heat = 9971.21 J + 10.905 kJ + 8479.185 J = 29355.393 J = 29.36 kJ
The energy required to heat 87.1 g acetone from a solid at -154.0°C to a liquid at -42.0°C is 29.36 kJ
Answer:
Deep ocean trenches, volcanoes, island arcs, submarine mountain ranges, and fault lines are examples of features that can form along plate tectonic boundaries. Volcanoes are one kind of feature that forms along convergent plate boundaries, where two tectonic plates collide and one moves beneath the other.
Explanation:
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Please brainiest
Answer:
sorry i dont know the answer
Explanation:
sorry i dont know the answer
