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2 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]2 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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The decomposition of copper(II) nitrate on heating is endothermic reaction. 2Cu(NO3)2(s) → 2C10(s) + 4NO2(g) + O2(g) Calculate t

Basile [38]

Answer:

The enthalpy change for the given reaction is 424 kJ.

Explanation:

$2Cu(NO_3)_2(s)\rightarrow 2CuO(s) + 4NO_2(g) + O_2(g),\Delta H_{rxn}=?$

We have :

Enthalpy changes of formation of following s:

$\Delta H_{f,Cu(NO_3)_2}=-302.9 kJ/mol$

$\Delta H_{f,CuO}=-157.3 kJ/mol$

$\Delta H_{f,NO_2}= 33.2 kJ/mol$

$\Delta H_{f,O_2}= 0 kJ/mol$ (standard state)

$\Delta H_{rxn}=\sum [\Delta H_f(product)]-\sum [\Delta H_f(reactant)]$

The equation for the enthalpy change of the given reaction is:

$\Delta H_{rxn}$ =

$=(2 mol\times \Delta H_{f,CuO}+4\times \Delta H_{f,NO_2}+1 mol\times \Delta H_{f,O_2})-(2mol\times \Delta H_{f,Cu(NO_3)_2})$

$\Delta H_{rxn}$ =

$(2mol\times (-157.3 kJ/mol)+4\times 33.2 kJ/mol=1 mol\times 0 kJ/mol)-(2 mol\times (-302.9 kJ/mol)$

$\Delta H_{rxn}=424 kJ$

The enthalpy change for the given reaction is 424 kJ.

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2 years ago

What happened to the volume of gas when the syringe was exposed to various temperature conditions? Using the concepts explored i

rewona [7]

Answer:

Thus, when the volume of the gas is exposed to a temperature above -273.15 K, the volume increases linearly with the temperature.

Explanation:

The expression for Charles's Law is shown below:

$\frac {V_1}{T_1}=\frac {V_2}{T_2}$

This states that the volume of the gas is directly proportional to the absolute temperature keeping the pressure conditions and the moles of the gas constant.

Thus, when the volume of the gas is exposed to a temperature above -273.15 K, the volume increases linearly with the temperature.

For example , if the temperature of the gas is reduced to half, the volume also reduced to half.

At -273.15 K, according to Charles's law, it is possible to make the volume of an ideal gas = 0.

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2 years ago

How many grams of calcium cyanide (Ca(CN)2) are contained in 0.79 mol of calcium cyanide?

Allushta [10]

I dont know but do you know da wae brudda?

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2 years ago

A 2135 cm3 sample of dry air has a pressure of 98.4 kpa at 127 degrees Celsius. What is the volume of the sample if the Temperat

kykrilka [37]

the equation is p1 x v1 divided by T1 = p1 x v2 = T2 but since the pressure is kept constant you do not even need it so the equation would now be v1 divided by t1 = v2 divided by t2

2135 cm3 divided by 127 degrees celcius = x divided by 206

answer: 3460 cm3

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2 years ago

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Brian's aunt has cats. When Brian recently visited her, he started sneezing badly and believes that it was because

kifflom [539]

Answer:

By visiting other households with cats.

Explanation:

This will give Brian a variety of other houses and determine if it is truly cats or just alleries from other items. This is the most direct way to get Brian the answer he is looking for.

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2 years ago