Calculate the mass of dry ice that should be added to the water so that the dry ice completely sublimes away when the water reac
hes 16 ∘c. assume no heat loss to the surroundings.
1 answer:
There are a lot of missing information in this problem. I've found a similar problem which is shown in the attached picture. Let's just use the information there that we don't have here.
From the principle of conservation of energy:
Heat of dry ice + Heat of water = 0
Heat of dry ice = - Heat of water
(m of dry ice)ΔH = -(m of water)(Cp)(ΔT)
where Cp for water is 4.187 kJ/kg·°C
Hence,
(m of dry ice)(ΔH°g - ΔH°s) = -(Density*Volume)(Cp)(ΔT)
where the density of water is 1 kg/L and the molar mass of dry ice is 44 g/mol.
Then,
(m of dry ice)(-393.5 - -427.4 kJ/mol)(44 g/mol) = -(1 kg/L*12L)(4.187 kJ/kg·°C)(16 - 88 °C)
Solving for m of dry ice,
<em>Mass = 2.43 g of dry ice</em>
From the principle of conservation of energy:
Heat of dry ice + Heat of water = 0
Heat of dry ice = - Heat of water
(m of dry ice)ΔH = -(m of water)(Cp)(ΔT)
where Cp for water is 4.187 kJ/kg·°C
Hence,
(m of dry ice)(ΔH°g - ΔH°s) = -(Density*Volume)(Cp)(ΔT)
where the density of water is 1 kg/L and the molar mass of dry ice is 44 g/mol.
Then,
(m of dry ice)(-393.5 - -427.4 kJ/mol)(44 g/mol) = -(1 kg/L*12L)(4.187 kJ/kg·°C)(16 - 88 °C)
Solving for m of dry ice,
<em>Mass = 2.43 g of dry ice</em>
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Explanation:
Those substances are said to be flexible which can be bent without breaking. There are many substances which are hard in nature but still can be bent. The hardness of such materials is due to strong interactions between the molecules and the flexibility comes due to their amorphous backbone. Therefore, greater the crystalline level of macromolecules lesser is the flexibility and greater the amorphous character greater is the flexibility and vice versa. Also, the flexibility of polymers is increased by adding plastisizers in it. Plastisizers make the hard polymers flexible by breaking the crosslinkers and enabling the macromolecules to move past one another.