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3 years ago

The temperature at which a solid melts is the same as the temperature at which its liquid form solidifies (true or false)

Chemistry

2 answers:

Dafna11 [192]3 years ago

6 0

Answer:

True

Explanation:

This is true because, upon cooling the particles in a liquid loses energy, stops moving and remains constant and the forms a solid.

Therefore we can say the freezing point and melting point occurs at the same temperature

Vanyuwa [196]3 years ago

4 0

its false not true................

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What is the slope of the line?

pychu [463]

Answer:

The slope of the line is $-\frac{10}{7}$ .

Explanation:

The slope of the line ( $m$ ) is the change in dependent variable ( $y$ ) divided by the change in independent variable ( $x$ ):

$m = \frac{y_{f}-y_{o}}{x_{f}-x_{o}}$ (1)

If we know that $(x_{o}, y_{o}) = (0, 2)$ and $(x_{f}, y_{f}) = (1.4, 0)$ , then the slope of the line is:

$m = \frac{0-2}{1.4-0}$

$m = -\frac{10}{7}$

The slope of the line is $-\frac{10}{7}$ .

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3 years ago

Identify the examples of diffusion in the list below

Sergeu [11.5K]

Answer:

movement of particles of object from one place to another

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spreading of ink in water

Explanation:

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4 years ago

Read 2 more answers

Suppose you are working with a NaOH stock solution but you need a solution with a lower concentration for your experiment. Calcu

Monica [59]

Answer: The volume of the 1.224 M NaOH solution needed is 26.16 mL

Explanation:

In order to prepare the dilute NaOH solution, solvent is added to a given amount of the NaOH stock solution up to a final volume of 250.0 mL.

Since only solvent is added, the amount of the solute, NaOH, in the dilute solution is the same as in the volume taken from the stock solution.

Molarity (M) is calculated from the following equation:

M = n ÷ V

where n is the number of moles of the solute in the solution, and V is the volume of the solution.

Accordingly, the number of moles of the solute is given by

n = M x V

Now, let's designate the stock NaOH solution and the dilute solution as (1) and (2), respectively . The number of moles of NaOH in each of these solutions is:

n (1) = M (1) x V (1)

n (2) = M (2) x V (2)

As the amount of NaOH in the dilute solution is the same as in the volume taken from the stock solution,

n (1) = n (2)

and

M (1) x V (1) = M (2) x V (2)

For the stock solution, M (1) = 1.244 M, and V (1) is the volume needed. For the dilute solution, M (2) = 0,1281 M, and V (2) = 250.0 mL.

The volume of the stock solution needed, V (1), is calculated as follows:

V (1) = M (2) x V (2) ÷ M (1)

V (1) = 0.1281 M x 250.0 mL ÷ 1.224 M

V (1) = 26.16 mL

The volume of the 1.224 M NaOH solution needed is 26.16 mL.

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3 years ago

Stacy made the following table to compare the functions of plant and animal structures, but she is missing a row title. Which of

fredd [130]

Answer:

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2 years ago

What is the % dissociation of a solution of acetic acid if at equilibrium the solution has a pH = 4.74 and a pKa = 4.74?

Ymorist [56]

Answer:

$\% diss = 50\%$

Explanation:

Hello there!

In this case, when considering weak acids which have an associated percent dissociation, we first need to set up the ionization reaction and the equilibrium expression:

$HA\rightleftharpoons H^++A^-\\\\Ka=\frac{[H^+][A^-]}{[HA]}$

Now, by introducing x as the reaction extent which also represents the concentration of both H+ and A-, we have:

$Ka=\frac{x^2}{[HA]_0-x} =10^{-4.74}=1.82x10^{-5}$

Thus, it is possible to find x given the pH as shown below:

$x=10^{-pH}=10^{-4.74}=1.82x10^{-5}M$

So that we can calculate the initial concentration of the acid:

$\frac{(1.82x10^{-5})^2}{[HA]_0-1.82x10^{-5}} =1.82x10^{-5}\\\\\frac{1.82x10^{-5}}{[HA]_0-1.82x10^{-5}} =1\\\\$

$[HA]_0=3.64x10^{-5}M$

Therefore, the percent dissociation turns out to be:

$\% diss=\frac{x}{[HA]_0}*100\% \\\\\% diss=\frac{1.82x10^{-5}M}{3.64x10^{-5}M}*100\% \\\\\% diss = 50\%$

Best regards!

6 0

3 years ago

The temperature at which a solid melts is the same as the temperature at which its liquid form solidifies (true or false)​

The temperature at which a solid melts is the same as the temperature at which its liquid form solidifies (true or false)