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My name is Ann [436]
3 years ago
15

How many grams of water at 50◦c must be added to 16 grams of ice at −12◦c to result in only liquid water at 0◦c?

Chemistry
1 answer:
melisa1 [442]3 years ago
4 0

When water at 50 C is added to ice at -12 C, heat is transferred from hot water to ice.

- Heat given out by water = Heat absorbed by ice

Calculating the heat released by hot water:

q = m CwaterΔTq = m (4.184\frac{J}{(g.^{0}C)} )(0^{0}C-50^{0}C)

Calculating heat absorbed by 16 g of ice: Ice at -12^{0}C is converted to ice at O^{0}C and then ice at O^{0} C to water at 0^{0}C

q = m CiceΔT + m (Heat of fusion)

q = 16 g(2.11\frac{J}{g.^{0}C})(0^{0}C - (-12^{0}C)) + 16 g (333.55\frac{J}{g})

q = 405.12 J +5336.8 J =5741.92 J

- Heat given out by water = Heat absorbed by ice

-(m(4.184\frac{J}{g.^{0}C})(0^{0}C-50^{0}C) = 5741.92 J

m = 27.4 g

Therefore, 27.4 g water at 50^{0}C must be added to 16 g of ice at -12^{0}C to convert to liquid water at 0^{0}C

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Answer:

Equilibrium shifts to the right

Explanation:

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8 0
3 years ago
(h) The student made observations related to the contents of the Erlenmeyer flask during the titration. Identify an observation
katen-ka-za [31]

Answer:

Explanation:

During titration indicators are often used to identify chemical changes between reacting species.

For colorless solutions in which no noticeable changes can easily be seen, indicators are the best bet. Most titration processes involves a combination of acids and bases to an end point.

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5 0
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Answer:

The standard reaction enthalpy for the given reaction is 235.15 kJ/mol.

Explanation:

O_2(g) \rightarrow \frac{2}{3}O_3(g),\Delta H^o_{1}=142.7 kJ/mol..[1]

O_2(g) \rightarrow 2 O(g),\Delta H^o_{2}=498.4 kJ/mol..[2]

NO(g) + O_3(g)\rightarrow NO_2(g) + O_2(g) ,\Delta H^o_{3} = -200 kJ/mol..[3]

NO_2(g)\rightarrow NO(g) + O(g),\Delta H^o_{4}=?..[4]

Using Hess's law:

Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.

2 × [4] = [2]- (3 ) × [1] - (2) × [3]

2\times \Delta H^o_{4}=\Delta H^o_{2} -3\times \Delta H^o_{1}-2\times \Delta H^o_{3}

2\times \Delta H^o_{4}=498.4 kJ/mol-3\times 142.7 kJ/mol-2\times -200 kJ/mol

2\times \Delta H^o_{4}=470.3 kJ/mol

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The standard reaction enthalpy for the given reaction is 235.15 kJ/mol.

7 0
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