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2 years ago

A science teacher demonstrated physical and chemical changes by using a fork to hold a marshmallow over a lit candle

1 answer:

3 0

Physical and chemical changes are the property of the reactions. The best observation evident of the chemical change is the deposition of the brown crust over the marshmallow.

<h3>What is a chemical change?</h3>

A chemical change is a change that results in the formation of a new substance that cannot be changed back into its original form. They are irreversible and include decomposition, combination, and addition reactions.

The color change of the marshmallow from white to brown shows the chemical change that cannot be reversed back to the original form while the melting of the candle and heating of the metallic fork are physical changes.

Therefore, the burning of marshmallows is a chemical change.

Learn more about chemical change here:

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How many times do u have to run back and forth across a football field to be meter

Taya2010 [7]

You would have to run 100 kilometers across it once. Then you will have to run 10 meters across it because a football field is 110 meters according to the NFL’s website

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3 years ago

Which is a possible result of rising ocean temperatures caused by global warming?

goldfiish [28.3K]

Pretty sure it’s 2, increase in strength of hurricanes

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3 years ago

The iodide ion reacts with hypochlorite ion (the active ingredient in chlorine bleaches) in the following way:

LiRa [457]

Explanation:

(a) As the given chemical reaction equation is as follows.

$OCl^{-} + I^{-} \rightarrow OI^{-1} + Cl^{-1}$

So, when we double the amount of hypochlorite or iodine then the rate of the reaction will also get double. And, this reaction is "first order" with respect to hypochlorite and iodine.

Hence, equation for rate law of reaction will be as follows.

Rate = $K \times [OCl^{-}] \times [l^{-}]$

(b) Since, the rate equation is as follows.

Rate = $K [OCl^{-}][l^{-}]$

Let us assume that ( $[OCl^{-}] = [l^{-}]$ )

Putting the given values into the above equation as follows.

$1.36 \times 10^{-4} = K \times (1.5 \times 10^{-3})^2$

$1.36 \times 10^{-4} = K \times (2.25 \times 10^{-6})$

K = $\frac{1.36 \times 10^{-4}}{2.25 \times 10^{-6}}$

= $60.4 M^{-1}sec^{-1}$

Hence, the value of rate constant for the given reaction is $60.4 M^{-1}sec^{-1}$ .

(c) Now, we will calculate the rate as follows.

Rate = $K [OCl^{-}][l^{-}]$

= $60.4 \times (1.8 \times 10^{3}) \times (6.0 \times 10^{4})$

= $6.52 \times 10^{5}$

Therefore, rate when $[OCl^{-}] = 1.8 \times 10^{3}$ M and $[I^{-}]= 6.0 \times 10^{4}$ M is $6.52 \times 10^{5}$ .

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3 years ago

Function of leaf in points

klasskru [66]

Explanation:

The main function of a leaf is to produce food for the plant by photosynthesis. Chlorophyll, the substance that gives plants their characteristic green colour, absorbs light energy.

Photosynthesis.

Transpiration.

Guttation.
Storage.

Defense.

I will be describing what each of these do to the plant.

Photosynthesis: Photosynthesis is a process used by plants and other organisms to convert light energy into chemical energy that can later be released to fuel the organisms' activities.
Transpiration: Transpiration is the process of water movement through a plant and its evaporation from aerial parts, such as leaves, stems and flowers. Water is necessary for plants but only a small amount of water taken up by the roots is used for growth and metabolism. The remaining 97–99.5% is lost by transpiration and guttation.
Guttation: Guttation is the exudation of drops of xylem sap on the tips or edges of leaves of some vascular plants, such as grasses, and a number of fungi. Guttation is not to be confused with dew, which condenses from the atmosphere onto the plant surface. Guttation generally happens during the night time.
Storage: A storage organ is a part of a plant specifically modified for storage of energy (generally in the form of carbohydrates) or water. Storage organs often grow underground, where they are better protected from attack by herbivores.
Defense: The first line of defense in plants is an intact and impenetrable barrier composed of bark and a waxy cuticle. Both protect plants against herbivores. Other adaptations against herbivores include hard shells, thorns (modified branches), and spines (modified leaves).

Hope this helps!

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3 years ago

2. What major alkene product is produced by the dehydration of the following alcohols?

weeeeeb [17]

2-Dimethylcyclohexanol major alkene product is produced by the dehydration of the alcohols.

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3 years ago