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juin [17]
3 years ago
9

Two changes are described below. Ice melts to form water. Sugar cubes dissolve in hot coffee. Which statement is true about the

two changes? Both are chemical changes because there is a change in the composition of ice and sugar. Both are physical changes because there is a change in the physical states of ice and sugar. A is a physical change due to a change of state, but B is a chemical change because new molecules are formed. A is a chemical change due to a change of state, but B is a physical change because new molecules are formed.
Chemistry
2 answers:
almond37 [142]3 years ago
7 0

Answer:

both are physical changes

Explanation:

mrs_skeptik [129]3 years ago
6 0

Answer:

Both are physical changes because there is a change in the physical states of ice and sugar

Explanation:

1. melting of ice to water is a change of state from solid to liquid, although there is a change in the physical state but the chemical nature of the water remain unchanged. no new substance is produced since ice and water is the same chemically. since the reaction can be reversed by freezing. This shows a physical change

2. Dissolving sugar cubes in hot coffee is also a physical change since the sugar is dissolved into the coffee but the chemical natures of the sugar and the coffee remain the same. no new substance is formed.  that is a chemical change

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A food substance kept at 0°C becomes rotten (as determined by a good quantitative test) in 8.3 days. The same food rots in 10.6
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Answer:

1.   67.2 kJ/mol

Explanation:

Using the derived expression from Arrhenius Equation

In \ (\frac{k_2}{k_1}) = \frac{E_a}{R}(\frac{T_2-T_1}{T_2*T_1})

Given that:

time t_1 = 8.3 days = (8.3 × 24 ) hours = 199.2 hours

time t_2 = 10.6 hours

Temperature T_1 = 0° C = (0+273 )K = 273 K

Temperature T_2 = 30° C = (30+ 273) = 303 K

Rate = 8.314 J / mol

Since (\frac{k_2}{k_1}=\frac{t_2}{t_1})

Then we can rewrite the above expression as:

In \ (\frac{t_2}{t_1}) = \frac{E_a}{R}(\frac{T_2-T_1}{T_2*T_1})

In \ (\frac{199.2}{10.6}) = \frac{E_a}{8.314}(\frac{303-273}{273*303})

2.934 = \frac{E_a}{8.314}(\frac{30}{82719})

2.934 = \frac{30E_a}{687725.766}

30E_a = 2.934 *687725.766

E_a = \frac{2.934 *687725.766}{30}

E_a =67255.58 \ J/mol

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