To determine the specific heat capacity of the metal and assist in its identification, the heat absorbed by the calorimeter can be computed using: Energy = mass * specific heat capacity * temperature change Q = 250 * 1.035 * (11.08 - 10) Q = 279.45 cal/g. Next, we employ the same formula for the metal because the heat taken in by the calorimeter should equal the heat expelled by the metal. -279.45 = 50 * c * (11.08 - 45) [the minus sign indicates energy release] solving for c gives us 0.165. Therefore, the specific heat capacity of the metal amounts to 0.165 cal/g°C.
Students dealing with ionic bonds comprehend better how to convey what the model should showcase.
Explanation:
- Upon dissolving ionic compounds in water, the compounds separate into their constituent ions via a process called dissociation.
- The ions become attracted to water molecules, which carry a polar charge.
- If the pull between the ions and the water molecules is strong enough to disband the ionic bonds, the compound dissolves.
- The ions disperse in the solution, each surrounded by water molecules to inhibit reattachment.
- The ionic solution forms an electrolyte, allowing it to conduct electricity.
- In contrast, while covalent compounds do dissolve in water, they separate into molecules, not individual atoms.
- Water acts as a polar solvent, yet covalent compounds are generally nonpolar.
- This implies that covalent compounds often do not dissolve in water and instead form a distinct layer on top of the water.
Answer:
The correct options include choice 2, 3, and 6.
Explanation:
Density is identified as the mass of a substance per unit volume occupied by that substance.
The density remains constant for a given substance, regardless of variations in mass and volume hence it is considered an intensive property.
2. 20.2 g of silver in 21.6 mL of water and 12.0 g of silver also in 21.6 mL of water.
3. 15.2 g of copper in 21.6 mL of water and 50.0 g of copper in 23.4 mL of water.
6. 11.2 g of gold in 21.6 mL of water and 14.9 g of gold in 23.4 mL of water.
The same metals in both instances will yield consistent densities due to the fixed density of the metal.
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The result is:
The new volume is: 
Rationale:
Because the temperature remains constant, we can apply Boyle's Law to solve this issue.
Boyle's Law stipulates that:
Where,
P is the gas's pressure.
V is the gas's volume.
According to the information provided:
Let's put the values into the equation:
Consequently, the new volume is:
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