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2 years ago

How much heat is required to change the temperature of a 15 g aluminum can with 100 g of water from 24.5°C to 55°C?

1 answer:

3 0

Heat capacity of aluminium = 0.900 J/g°C
While heat capacity of water = 4.186 J/g°C
Heat = heat gained by water + heat gained by aluminium
Heat gained by water = 100 × 4.186 × 30.5
= 12767.3 Joules
Heat gained by aluminium = 15 × 0.9 × 30.5
= 411.75 Joules
Heat required = 13179.05 Joules or 13.179 kJoules

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Which is a base unit used in the metric system?

DaniilM [7]

Answer:

<h2> Quarts </h2><h2>Litres </h2><h2>Pints </h2><h2>Degrees Fahrenheit. </h2>

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4 0

3 years ago

Calculate the pKa of lactic acid (CH3CH(OH)COOH) given the following information. 3.005 grams of potassium lactate are added to

snow_lady [41]

Answer:

$\displaystyle \text{p} K_a \approx 3.856$

Explanation:

Because 3.005 grams of potassium lactate is added to 100. mL of solution, its concentration is:

$\displaystyle \begin{aligned} \left[ \text{KC$_3$H_$_5$O$_3$}\right] & = \frac{3.005\text{ g KC$_3$H_$_5$O$_3$}}{100.\text{ mL}} \cdot \frac{1\text{ mol KC$_3$H_$_5$O$_3$}}{128.17 \text{ g KC$_3$H_$_5$O$_3$}} \cdot \frac{1000\text{ mL}}{1\text{ L}} \\ \\ &= 0.234\text{ M}\end{aligned}$

By solubility rules, potassium is completely soluble, so the compound will dissociate completely into potassium and lactate ions. Therefore, [KC₃H₅O₃] = [C₃H₅O₃⁺]. Note that lactate is the conjugate base of lactic acid.

Recall the Henderson-Hasselbalch equation:

$\displaystyle \begin{aligned}\text{pH} = \text{p}K_a + \log \frac{\left[\text{Base}\right]}{\left[\text{Acid}\right]} \end{aligned}$

[Base] = 0.234 M and [Acid] = 0.500 M. We are given that the resulting pH is 3.526. Substitute and solve for p<em>Kₐ</em>:

$\displaystyle \begin{aligned} (3.526) & = \text{p}K_a + \log \frac{(0.234)}{(0.500)} \\ \\ 3.526 & = \text{p}K_a + (-0.330) \\ \\ \text{p}K_a & = 3.856\end{aligned}$

In conclusion, the p<em>Kₐ </em>value of lactic acid is about 3.856.

5 0

1 year ago

Find [cu2+] in a solution saturated with cu4(oh)6(so4) if [oh − ] is fixed at 2.3 ✕ 10−6 m. note that cu4(oh)6(so4) gives 1 mol

ki77a [65]

<span>

</span> $Cu_4(OH)_6(SO_4)$
<span>
You have OH- conc = </span>2.3 ✕ 10−6 m
From the formula, you can observe the ratio of Cu2+ to OH- is 4 : 6 = 2:3

So, for 2.3 ✕ 10−6 m OH-
[Cu2+] = $\frac{2}{3} \times 2.3 \times 10^{-6}$

$= 1.53 \times 10^{-6} 
$

6 0

3 years ago

The alkali metals play virtually the same general chemical role in all their reactions. (b) How is it based on atomic properties

vfiekz [6]

b) It is based on atomic properties as alkali metals requires 7 more electrons to complete their outer orbit. And they try to give those electrons to other elements to obtain noble gas configuration.

Noble gases are the gases which do not react easily with anything. They are also called as Inert gases, and belongs to group 18 of the periodic table.

Alkali metals are the substances which are found in Group I of a periodic table. Mostly the elements which are present are:

Lithium (Li)

Sodium (Na)

Potassium (K)

Rubidium (Rb)

Cesium (Cs)

Francium (Fr)

Properties of alkali metals are: Soft, shiny reactive metals. They are soft enough to cut with knife. Metals react with water and air quickly and gets tarnish, so pure metals are stored in container by dipping them in oil to prevent oxidation.

To know more about Alkali metals, refer to this link:

brainly.com/question/18153051

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1 year ago

Students set up three trials of an experiment to test the

Dvinal [7]

Need the answer to this question

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2 years ago