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3 years ago

10

During the cold winter months the surface Of a lake becomes hard Ice this is due to which change of state

1 answer:

nikitadnepr [17]3 years ago

8 0

Freezing

Explanation:

This change in state when surface of a lake changes to hard ice is known as freezing.

When a solid is heated, it melts and changes to a liquid. On further heating, it changes to vapor and the randomness increases.
When a liquid is cooled below its freezing point, it changes to a solid such as ice for water.
The phase change is known as freezing.

Learn more:

Phase change brainly.com/question/1875234

#learnwithBrainly

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The nucleus is incredibly dense--it contains nearly all of the mass of the atom within a tiny volume. TRUE FALSE

const2013 [10]

Answer: True

The nucleus is a small, dense region at the center of the atom. It consists of positive protons and neutral neutrons, so it has an overall positive charge. The nucleus is just a tiny part of the atom, but it contains virtually all of the atom's mass.

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3 years ago

You transfer a sample of a gas at 17°C from a volume of 5.67 L and 1.10 atm to a container at 37°C that has a pressure of 1.10 a

laila [671]

Explanation:

V1T1 = V2T2

V2=V1T1/T2

= 5.67*17/37= 2.6

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4 years ago

The two isotopes of chlorine are LaTeX: \begin{matrix}35\\17\end{matrix}Cl35 17 C l and LaTeX: \begin{matrix}37\\17\end{matrix}C

Kay [80]

<u>Answer:</u> The percentage abundance of $_{17}^{35}\textrm{Cl}$ and $_{17}^{37}\textrm{Cl}$ isotopes are 77.5% and 22.5% respectively.

<u>Explanation:</u>

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

$\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i$ .....(1)

Let the fractional abundance of $_{17}^{35}\textrm{Cl}$ isotope be 'x'. So, fractional abundance of $_{17}^{37}\textrm{Cl}$ isotope will be '1 - x'

<u>For $_{17}^{35}\textrm{Cl}$ isotope:</u>

Mass of $_{17}^{35}\textrm{Cl}$ isotope = 35 amu

Fractional abundance of $_{17}^{35}\textrm{Cl}$ isotope = x

<u>For $_{17}^{37}\textrm{Cl}$ isotope:</u>

Mass of $_{17}^{37}\textrm{Cl}$ isotope = 37 amu

Fractional abundance of $_{17}^{37}\textrm{Cl}$ isotope = 1 - x

Average atomic mass of chlorine = 35.45 amu

Putting values in equation 1, we get:

$35.45=[(35\times x)+(37\times (1-x))]\\\\x=0.775$

Percentage abundance of $_{17}^{35}\textrm{Cl}$ isotope = $0.775\times 100=77.5\%$

Percentage abundance of $_{17}^{37}\textrm{Cl}$ isotope = $(1-0.775)=0.225\times 100=22.5\%$

Hence, the percentage abundance of $_{17}^{35}\textrm{Cl}$ and $_{17}^{37}\textrm{Cl}$ isotopes are 77.5% and 22.5% respectively.

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3 years ago

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natulia [17]

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geniusboy [140]

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4 years ago