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
Water. Arrhenius acids provide a H+ ion and a base provides a OH- ion, which react to form water.
Answer:
0.129 L = 129.0 mL.
Explanation:
- NaOH neutralizes acetic acid (CH₃COOH) according to the balanced reaction:
<em>NaOH + CH₃COOH → CH₃COONa + H₂O. </em>
- According to the balanced equation: 1.0 mole of NaOH will neutralize 1.0 mole of CH₃COOH.
<em>no. of moles of CH₃COOH = mass/molar mass </em>= (2.0 g)/(60 g/mol) = <em>0.033 mole. </em>
<em>
</em>
no. of moles = (0.258 mol/L)(V)
- At neutralization: no. of moles of NaOH = no. of moles of CH₃COOH
∴ (0.258 mol/L)(V) = 0.033 mole
<em>∴ The volume of NaOH</em> = (0.033 mole)/(0.258 mol/L) = <em>0.129 L = 129.0 mL.</em>
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