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3 years ago

Which phase change in an example of an exothermic process?

Chemistry

2 answers:

photoshop1234 [79]3 years ago

5 0

C. liquid to solid i hope this helps

Rina8888 [55]3 years ago

5 0

I believe it would be A.

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To save time, you can approximate the initial mass of the solid to the nearest ±1 g. For example, if you are asked to add 14.3 g

tensa zangetsu [6.8K]

Answer: Set with equal densities are:

15.2 g copper and 50.0 g copper ; 11.2 g gold and 14.9 g gold

Explanation:

Density is defined as mass of the substance present in unit volume of the substance.

$Density=\frac{Mass}{Volume}$

It is an intensive property which means that two different masses of same substance will have same value of density under same conditions.

So, from the given samples with equal density are:

15.2 g copper and 50.0 g copper will have the same value density.

11.2 g gold and 14.9 g gold will have the same value density.

6 0

3 years ago

Read 2 more answers

Un profesor desea preparar una disolución de agua oxigenada al 20%v/v y dispone solamente de 50 cm3 h2o2 ¿qué volumen d disoluci

Sergeeva-Olga [200]

Answer:

250cm³ de disolución se deben preparar

Explanation:

El porcentaje volumen volumen (%v/v) se define como el volumen de soluto -En este caso, H2O2- en 100cm³ de disolución.

Se desea preparar una solución que contenga 20cm³ de H2O2 por cada 100cm³

Como solamente se cuenta con 50cm³ de H2O2, el volumen que se debe preparar es:

50cm³ H2O2 * (100cm³ Disolución / 20cm³ H2O2) =

<h3>250cm³ de disolución se deben preparar</h3>

5 0

3 years ago

Which element has the smallest atomic radius

kupik [55]

Helium give me a thanks or brainiest answer if this helps!

6 0

3 years ago

Read 2 more answers

50 mL of 0.1 M acetic acid is mixed with 50 mL of 0.1 M sodium acetate (the conjugate base). The Ka of acetic acid is approximat

trapecia [35]

Answer:

4.76

Explanation:

In this case, we have to start with the buffer system:

$CH_3COOH~->~CH_3COO^-~+~H^+$

We have an acid ( $CH_3COOH$ ) and a base ( $CH_3COO^-$ ). Therefore we can write the henderson-hasselbach reaction:

$pH~=~pKa+Log\frac{[CH_3COO^-]}{[CH_3COOH]}$

If we want to calculate the pH, we have to calculate the pKa:

$pH=-Log~Ka=4.76$

According to the problem, we have the same concentration for the acid and the base 0.1M. Therefore:

$[CH_3COO^-]=[CH_3COOH]$

If we divide:

$\frac{[CH_3COO^-]}{[CH_3COOH]}~=~1$

If we do the Log of 1:

$Log~1=~zero$

So:

$pH~=~pKa$

With this in mind, the pH is 4.76.

I hope it helps!

8 0

3 years ago

Which is true about each element in a period or horizontal row of the periodic table?

Ber [7]

The true about element in period or horizontal row of periodic table is

Each element has one more outer electron than the element before it. (answer B)

Explanation

In periodic table there are four horizontal rows(periods) that is period 1,2,3 and 4.

Across the period each element has one more electron than element before it.

For example Na and Mg which are in periodic 3.

Na has 2.8.1 or [Ne] 3s¹ electron configuration while Mg has 2.8.2 or [Ne] 3S² electron configuration.

Since Mg has 2 outer electrons while Na has 1 outer electron, Mg has more electron than Na.

3 0

3 years ago