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3 years ago

Briefly explain why adding salt to icy roads helps melt the ice and prevent it from freezing again.

Chemistry

1 answer:

g100num [7]3 years ago

3 0

Answer:

See explanation below

Explanation:

The melting point (the temperature that it changes from solid to liquid) of a pure substance depends on its mass, the forces between its molecules, and the pressure of the system.

The melting point, or the freezing point, of a mixture, is a temperature between the melting point of its components. Salt has a huge negative melting point, so when it is added to the water, the melting point decreases from 0ºC to -21ºC approximately.

So, the water will only freeze again if the temperature becomes below -21ºC which is very difficult to happen.

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How do you determine if something is extensive or intensive

Shalnov [3]

Answer:

Intensive properties do not depend on the quantity of matter. Examples include density, state of matter, and temperature. Extensive properties do depend on sample size. Examples include volume, mass, and size.

Explanation:

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3 years ago

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A cough syrup contains 5.0% ethyl alcohol, c2h5oh, by mass. if the density of the solution is 0.9928 g/ml, determine the molarit

WARRIOR [948]

To answer the question above, let us a basis of the 1000 mL or 1 L.
volume = (0.9928 g/mL)(1000mL) = 992.8 g
Then, determine the mass of the alcohol by multiplying the total mass by the decimal equivalent of 5%.
mass of alcohol = 0.05(992.8 g) = 49.64 g
Then, determine the number of moles of ethyl alcohol by dividing the mass of alcohol by the molar mass (46 g/mol).
n = 49.64 g/ (46 g/mol) = 1.08 mol
Then, divide the number of moles by the volume (our basis is 1 L)
molarity = 1.08 mol/ 1 L = 1.08 M

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3 years ago

Creating a prototype and troubleshooting o

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4 years ago

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What is the maximum concentration of Ag⁺ that can be added to a 0.00750 M solution of Na₂CO₃ before a precipitate will form? (Ks

Firlakuza [10]

Answer:

$\large \boxed{1.64\times 10^{-5}\text{ mol/L }}$

Explanation:

Ag₂CO₃(s) ⇌2Ag⁺(aq) + CO₃²⁻(aq); Ksp = 8.10 × 10⁻¹²

2x 0.007 50 + x

$K_{sp} =\text{[Ag$^{+}$]$^{2}$[CO$_{3}^{2-}$]} = (2x)^{2}\times 0.00750 = 8.10 \times 10^{-12}\\0.0300x^{2} = 8.10 \times 10^{-12}\\x^{2} = 2.70 \times 10^{-10}\\x = \sqrt{2.70 \times 10^{-10}} = \mathbf{1.64\times 10^{5}} \textbf{ mol/L}\\\text{The maximum concentration of Ag$^{+}$ is $\large \boxed{\mathbf{1.64\times 10^{-5}}\textbf{ mol/L }}$}$

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3 years ago

Using the balanced equation below, how many grams of cesium fluoride would be required to make 73.1 g of cesium xenon heptafluor

jenyasd209 [6]

Answer:

27.9 g

Explanation:

CsF + XeF₆ → CsXeF₇

First we convert 73.1 g of cesium xenon heptafluoride (CsXeF₇) into moles, using its molar mass:

Molar mass of CsXeF₇ = 397.193 g/mol

73.1 g CsXeF₇ ÷ 397.193 g/mol = 0.184 mol CsXeF₇

As 1 mol of cesium fluoride (CsF) produces 1 mol of CsXeF₇, in order to produce 0.184 moles of CsXeF₇ we would need 0.184 moles of CsF.

Now we convert 0.184 moles of CsF to moles, using the molar mass of CsF:

Molar mass of CsF = 151.9 g/mol

0.184 mol * 151.9 g/mol = 27.9 g

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3 years ago