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3 years ago

How does the addition of salt to solid ice affect the melting transition from solid to liquid?

Chemistry

2 answers:

serious [3.7K]3 years ago

7 0

Answer:

Im sure the answer is c

Hope this helped have a good day :) please mark me the brainliest answer?

Vadim26 [7]3 years ago

4 0

<h2>Answer:</h2>

The water would need to <u>B. absorb less energy during the plateau in order to melt</u>.

<h2>Explanation:</h2>

Salt when added to ice lowers the freezing point of the ice. So, if ice is added to the solid ice it doesn't let it to freeze rather the temperature of water may fall but it won't freeze.

For melting, it needs less energy after adding salt because salt itself absorbs the energy from the surroundings to help the phase transition of water from solid to liquid.

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Why knowing the density of water important?? explain why please so I can understand it don’t copy for the internet I won’t under

Mkey [24]

Answer:

it allows us to determine whether objects will float or sink when placed in a liquid or even a gas.

Explanation:

For example, In an oil spill in the ocean, the oil rises to the top because it is less dense than water, creating an oil slick on the surface of the ocean. A Styrofoam cup is less dense than a ceramic cup, so the Styrofoam cup will float in water and the ceramic cup will sink.

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2 years ago

What is the composition, in atom percent, of an alloy that consists of a) 5.5 wt% Pb and b) 94.5 wt% of Sn? Assume that the atom

Anastaziya [24]

Answer : The percent composition of Pb and Sn in atom is, 3.21 % and 96.8 % respectively.

Explanation :

First we have to calculate the number of atoms in 5.5 wt% Pb and 94.5 wt% of Sn.

As, 207.2 g of lead contains $6.022\times 10^{23}$ atoms

So, 5.5 g of lead contains $\frac{5.5}{207.2}\times 6.022\times 10^{23}=1.59\times 10^{22}$ atoms

and,

As, 118.71 g of lead contains $6.022\times 10^{23}$ atoms

So, 94.5 g of lead contains $\frac{94.5}{118.71}\times 6.022\times 10^{23}=4.79\times 10^{23}$ atoms

Now we have to calculate the percent composition of Pb and Sn in atom.

$\% \text{Composition of Pb}=\frac{\text{Atoms of Pb}}{\text{Atoms of Pb}+\text{Atoms of Sn}}\times 100$

$\% \text{Composition of Pb}=\frac{1.59\times 10^{22}}{(1.59\times 10^{22})+(4.79\times 10^{23})}\times 100=3.21\%$

and,

$\% \text{Composition of Sn}=\frac{\text{Atoms of Sn}}{\text{Atoms of Pb}+\text{Atoms of Sn}}\times 100$

$\% \text{Composition of Sn}=\frac{4.79\times 10^{23}}{(1.59\times 10^{22})+(4.79\times 10^{23})}\times 100=96.8\%$

Thus, the percent composition of Pb and Sn in atom is, 3.21 % and 96.8 % respectively.

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