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
Answer:
Cells are uncountable becasue they move around your body, make up your skin and other organs as well. And because when you grow, the cells multiply, and that makes it very hard for scientists to count cells in a human's body.
Explanation:
Hope this helped!
<span><span>When you write down the electronic configuration of bromine and sodium, you get this
Na:
Br: </span></span>
<span><span />So here we the know the valence electrons for each;</span>
<span><span>Na: (2e)
Br: (7e, you don't count for the d orbitals)
Then, once you know this, you can deduce how many bonds each can do and you discover that bromine can do one bond since he has one electron missing in his p orbital, but that weirdly, since the s orbital of sodium is full and thus, should not make any bond.
However, it is possible for sodium to come in an excited state in wich he will have sent one of its electrons on an higher shell to have this valence configuration:</span></span>
<span><span /></span><span><span>
</span>where here now it has two lonely valence electrons, one on the s and the other on the p, so that it can do a total of two bonds.</span><span>That's why bromine and sodium can form </span>
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</span>
Answer:
a. Methanol remains the same
b. Methanol decreases
c. Methanol increases
d. Methanol remains the same
e. Methanol increases
Explanation:
Methanol is produced by the reaction of carbon monoxide and hydrogen in the presence of a catalyst as follows; 2H2+CO→CH3OH.
a) The presence or absence of a catalyst makes no difference on the equilibrium position of the system hence the methanol remains constant.
b) The amount of methanol decreases because the equilibrium position shifts towards the left and more reactants are formed since the reaction is exothermic.
c) If the volume is decreased, there will be more methanol in the system because the equilibrium position will shift towards the right hand side.
d) Addition of helium gas has no effect on the equilibrium position since it does not participate in the reaction system.
e) if more CO is added the amount of methanol increases since the equilibrium position will shift towards the right hand side.