Select the correct answer. Which statement is true for a heating curve?

The ksp of calcium carbonate, caco3, is 3.36 × 10-9 m2. calculate the solubility of this compound in g/l.

maw [93]

CaCO₃ partially dissociates in water as Ca²⁺ and CO₃²⁻. The balanced equation is,
CaCO₃(s) ⇄ Ca²⁺(aq) + CO₃²⁻(aq)
Initial Y - -
Change -X +X +X
Equilibrium Y-X X X

Ksp for the CaCO₃(s) is 3.36 x 10⁻⁹ M²

Ksp = [Ca²⁺(aq)][CO₃²⁻(aq)]
3.36 x 10⁻⁹ M² = X * X
3.36 x 10⁻⁹ M² = X²
X = 5.79 x 10⁻⁵ M

Hence the solubility of CaCO₃(s) = 5.79 x 10⁻⁵ M
= 5.79 x 10⁻⁵ mol/L

Molar mass of CaCO₃ = 100 g mol⁻¹

Hence the solubility of CaCO₃ = 5.79 x 10⁻⁵ mol/L x 100 g mol⁻¹
= 5.79 x 10⁻³ g/L

7 0

3 years ago

According to boyle’s law, when the pressure on a gas in an enclosed container increases, the volume of that gas

Doss [256]

Decreases. Hope I helped

3 0

3 years ago

The carbon atoms in graphite and the carbon atoms in diamond have different

kicyunya [14]

Answer:

<u>structural arrangements</u>

_______________________________________

<h2>properties of daimond: </h2><h3>appearance: transparent</h3><h3>hardness: very hard</h3><h3>thermal conductivity :very poor</h3><h3>electric conductivity: poor</h3><h3>density:</h3>

$3510 {kgm}^{3}$

<h3>uses: jewellery and drilling</h3>

_______________________________________

<h2>properties of graphite:</h2>

<h3>appearance: black shiny</h3><h3>hardness: soft ,slippery to touch</h3><h3>thermal conductivity : moderate</h3><h3>electric conductivity: good</h3><h3>density:</h3>

$2250 {kgm}^{3}$

<h3>uses:dry cell, electric arc, pencil lead, lubricant</h3>

_______________________________________

<h2>How Diamond and Graphite are chemically identical?</h2>

On heating diamond or graphite in the air, they burn completely to form carbon dioxide.
- Equal quantities of diamond and graphite when burned, produce exactly the same amount of carbon dioxide.

_______________________________________

<h2>Why the physical properties of diamond and graphite are so different?</h2>

Due to the difference in the arrangement of carbon atoms in diamond and graphite

_______________________________________

<h2><em><u>hope</u></em><em><u> it</u></em><em><u> helps</u></em><em><u> you</u></em><em><u><</u></em><em><u>3</u></em></h2>

7 0

3 years ago

Consider the elements: Na, Mg, Al, Si, P.

Olegator [25]

Answer:

The elements mentioned in the series are Na, Mg, Al, Si, and P. It can be seen that all these elements are located in the same period. The atomic number of the mentioned elements are Na-11, Mg-12, Al-13, Si-14, and P-15.

a) There will be an increase in the ionization energy with the increase in the element's atomic number across a period. More energy is needed to withdraw an electron from a completely occupied shell in comparison to an incompletely occupied shell.

The atomic number of Na is 11. When one electron is withdrawn from Na, it gets converted into the inert gas configuration of Ne. Thus, it will require more energy to withdraw the second electron from Na. Hence, Na exhibits the lowest second ionization energy.

b) Across the period, with an increase in the element's atomic number, the atomic radii reduces from left to right. Thus, P exhibits the smallest atomic radius.

c) The metallic nature of the elements reduces from left to right across a period in the periodic table. Thus, P is the least metallic element.

d) Diamagnetic signifies towards the element that exhibits pair electrons in its sub-shells. The electronic configuration of Mg is,

1s2 2s2 2p6 3s2

In Mg, no unpaired electrons are present, while all the remaining elements mentioned exhibit unpaired electrons in their valence shell. Thus, Mg is the diamagnetic element.

8 0

3 years ago

Completely reacting 150.0 g of a substance with oxygen releases 395.1 J of energy. How much energy would be released if 450.0 g

MrMuchimi

To solve this problem we just need to use the rule of three:
150g..................395.1J
450g................xJ

x = 450*395.1/150 = 1185,3J

450.0 g of the substance completely reacted with oxygen will produce 1.1853 kJ(<span>kiloJoule</span>)

4 0

4 years ago

Read 2 more answers