Gallium changes from a solid to a liquid when the temperature increases from 2 5 ∘ C to 3 3 ∘ C .
Nitrogen becomes a liquid when methane and nitrogen are cooled from − 17 0 ∘ C to − 20 0 ∘ C .
Gold changes from a liquid to a gas when heated to 2 , 85 6 ∘ C .
Explanation
Problem Analysis We are given a table of melting and boiling points for several substances and asked to determine the state changes that occur under specific temperature changes.
Finding the First Substance First, we need to identify which substance changes from solid to liquid when the temperature increases from 2 5 ∘ C to 3 3 ∘ C . This means we are looking for a substance with a melting point between 2 5 ∘ C and 3 3 ∘ C . Looking at the table, Gallium (Ga) has a melting point of 3 0 ∘ C , which falls within this range.
Analyzing Methane and Nitrogen Next, we analyze what happens when methane and nitrogen are cooled from − 17 0 ∘ C to − 20 0 ∘ C . We are told methane freezes, which is consistent with its melting point of − 18 3 ∘ C . Now, let's consider nitrogen. Its melting point is − 21 0 ∘ C and its boiling point is − 19 6 ∘ C . Since − 20 0 ∘ C is between − 21 0 ∘ C and − 19 6 ∘ C , nitrogen will be in the liquid state.
Analyzing Gold Finally, we need to determine what happens when gold is heated to 2 , 85 6 ∘ C . Gold has a melting point of 1 , 06 4 ∘ C and a boiling point of 2 , 85 6 ∘ C . Since the gold is heated to its boiling point, it will change from a liquid to a gas.
Examples
Understanding phase changes is crucial in many real-world applications. For example, in metallurgy, controlling the temperature of metals is essential for shaping and hardening them. Similarly, in cryogenics, the study of extremely low temperatures, understanding the behavior of gases like nitrogen is vital for applications such as MRI machines and the preservation of biological samples. Even in cooking, knowing the melting points of substances like butter or chocolate helps us achieve the desired texture and consistency in our dishes.
To find the number of electrons flowing through a device with a current of 15.0 A for 30 seconds, we first calculate the total charge using the formula Q = I × t , resulting in 450 C. Then, by dividing this charge by the charge of an electron (approximately 1.6 × 1 0 − 19 C ), we find approximately 2.8125 × 1 0 21 electrons flow through the device.
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