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2 years ago

Name the phase change in each of these events: (c) Wet clothes dry on a summer day

1 answer:

5 0

The phase change occurs during evaporation. On a hot day, wet clothing is drying. A liquid becomes a gas when it evaporations.

During the process of evaporation, a liquid transforms into a gaseous phase that is not saturated with the evaporating substance. The term "vaporization of a liquid" refers to this process. Clothes start to dry as water vapor escapes from the fabric's surface.

<h3>What is evaporation?</h3>

A liquid turns into a gas through the process of evaporation. Raindrops that "vanish" on a hot day or damp clothing that dries in the sun are good examples of the phenomena. In these instances, the liquid water is evaporating into a gas known as water vapor rather than really dissipating. Global evaporation takes place.

Learn more about evaporation: brainly.com/question/5019199

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4. Which of the following is not an example of a disturbance that might jumpstart

BigorU [14]

Answer: D: a volcano forming a new island in a chain

Explanation:

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3 years ago

A proton is also known as this symbol, _____ ion.

mr Goodwill [35]

Hydrogen ion, strictly, the nucleus of a hydrogen atom separated from its accompanying electron. The hydrogen nucleus is made up of a particle carrying a unit positive electric charge, called a proton. The isolated hydrogen ion, represented by the symbol H+, is therefore customarily used to represent a proton.

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3 years ago

Find how many milliliters of NaOH should be used to reach the half-equivalence point during the titration of 20.00 mL 0.888 M bu

Varvara68 [4.7K]

<u>Answer:</u> The volume of NaOH solution required to reach the half-equivalence point is 0.09 mL

<u>Explanation:</u>

The chemical equation for the dissociation of butanoic acid follows:

$CH_3CH_2CH_2COOH\rightleftharpoons CH_3CH_2CH_2COO^-+H^+$

The expression of $K_a$ for above equation follows:

$K_a=\frac{[CH_3CH_2CH_2COO^-][H^+]}{[CH_3CH_2CH_2COOH]}$

We are given:

$[CH_3CH_2CH_2COOH]=0.888M\\K_a=1.54\times 10^{-5}$

$[CH_3CH_2CH_2COO^-]=[H^+]$

Putting values in above expression, we get:

$1.54\times 10^{-5}=\frac{[H^+]^2}{0.888}$

$[H^+]=-0.0037,0.0037$

Neglecting the negative value because concentration cannot be negative

To calculate the volume of base, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is butanoic acid

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is NaOH.

We are given:

$n_1=1\\M_1=\frac{0.0037}{2}M\text{ (half equivalence)}\\V_1=20.00mL\\n_2=1\\M_2=0.425M\\V_2=?mL$

Putting values in above equation, we get:

$1\times \frac{0.0037}{2}\times 20.00=1\times 0.425\times V_2\\\\V_2=\frac{1\times 0.0037\times 20}{1\times 0.425\times 2}=0.087mL$

Hence, the volume of NaOH solution required to reach the half-equivalence point is 0.09 mL

5 0

3 years ago

The transfer of a phosphate group to a molecule or compound is called The transfer of a phosphate group to a molecule or compoun

ki77a [65]

Answer:

Phosphorylation

Explanation:

Phosphorylation is the term used to describe the transfer of a phosphate group (PO34-) from a donor to a receptor molecule or compound. The process of phosphorylation is usually catalysed by a biological enzyme called KINASE.

Phosphorylation can either be addition of an inorganic phosphate to a molecule. For example, addition of phosphate group to ADP to form ATP, or the donation of a phosphate group to another molecule, e.g transfer of phosphate from ATP to glucose to form glucose-6-phosphate during glycolysis.

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3 years ago

WILL CHOSE THE BRAINLIEST!!! NEED HELP FAST!!!

skelet666 [1.2K]

Its not always accurate because you dont know what they could have done to it before they try to sell it

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3 years ago