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3 years ago

Identify two physical properties and two chemical properties of a bag of microwave popcorn before popping and after

2 answers:

4 0

Physical properties of a bag of microwaveable popcorn are the mass of it, the color of it, the size of it, and the weight of it. Two chemical properties of a bag of microwavable popcorn are it changed from seeds to popcorn and it popped.

noname [10]3 years ago

3 0

Answer: The correct answer is:

- Before :

Physical: almost flat & cool to the touch

Chemical: flammability & change from kernels to fluffy popcorn

- After :

Physical: hot & increased volume

Chemical: inability of popcorn to change back into uncooked kernels & flammability

Explanation:

Physical properties are those that can be measured without affecting the composition or identity of the substance. Examples of these properties are density, melting point, boiling point, among others.

There are also chemical properties, which are observed when a substance undergoes a chemical change, that is, a transformation of its internal structure, becoming other new substances. These chemical changes can be reversible or irreversible, when the latter occur in only one direction (as in the combustion of wood).

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2 years ago

A 256 mL sample of HCl gas is in a flask where it exerts a force (pressure) of 67.5 mmHg. What is the pressure of the gas if it

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Answer:

The pressure in the new flask would be $128\; \rm mmHg$ if the $\rm HCl$ here acts like an ideal gas.

Explanation:

Assume that the $\rm HCl$ sample here acts like an ideal gas. By Boyle's Law, the pressure $P$ of the gas should be inversely proportional to its volume $V$ .

For example, let the initial volume and pressure of the sample be $V_1$ and $P_1$ . The new volume $V_2$ and pressure $P_2$ of this sample shall satisfy the equation: $P_1 \cdot V_1 =P_2 \cdot V_2$ .

In this question,

The initial volume of the gas is $V_1= 256\; \rm mL$ .
The initial pressure of the gas is $P_1 = 67.5\; \rm mmHg$ .
The new volume of the gas is $V_2 = 135\; \rm mL$ .

The goal is to find the new pressure of this gas, $P_2$ .

Assume that this sample is indeed an ideal gas. Then the equation $P_1 \cdot V_1 =P_2 \cdot V_2$ should still hold. Rearrange the equation to separate the unknown, $P_2$ . Note: make sure that the units for $V_1$ and $V_2$ are the same before evaluating. That way, the unit of

$\begin{aligned} & P_2\\ &= \frac{P_1 \cdot V_1}{V_2} \\ &= \frac{256\; \rm mL \times 67.5\; \rm mmHg}{135\; \rm mL} \\ & \approx 128\; \rm mmHg\end{aligned}$ .

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3 years ago