Answer:
The freezing point of the solution is - 4.39 °C.
Explanation:
We can solve this problem using the relation:
<em>ΔTf = (Kf)(m),</em>
where, ΔTf is the depression in the freezing point.
Kf is the molal freezing point depression constant of water = -1.86 °C/m,
density of water = 1 g/mL.
<em>So, the mass of 575 mL is 575 g = 0.575 kg.</em>
m is the molality of the solution (m = moles of solute / kg of solvent = (465 g / 342.3 g/mol)/(0.575 kg) = 2.36 m.
<em>∴ ΔTf = (Kf)(m</em>) = (-1.86 °C/m)(2.36 m) = <em>- 4.39 °C.</em>
<em>∵ The freezing point if water is 0.0 °C and it is depressed by - 4.39 °C.</em>
<em>∴ The freezing point of the solution is - 4.39 °C.</em>
Answer:
hexane C6H14
IS YOUR ANSWER
Explanation:
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Answer:
The attractive force is negative and MgO has a higher melting point
Explanation:
From Couloumb's law:
Energy of interaction, E = k 
where q1 and q2 are the charges of the ions, k is Coulomb's constant and r is the distance between both ions, i.e the atomic radii of the ions.
If you look at Coulomb's law, you note that in the force is negative (because q1 is negative while q2 is positive).
In addition to that, the compounds MgO and NaF have similar combined ionic radii, then we can determine the melting point trend from the amount of energy gotten
The melting point of ionic compounds is determined by 1. charge on the ions 2. size of ions. while NaF has smaller charges (+1 and -1), MgO (+2 and -2) has larger charges and greater combined atomic radii. This implies that the compound with greater force would have a higher melting point.
Hence the compound MgO would have a higher melting point than NaF.
<u>Answer:</u> For the given reaction, the value of
is greater than 1
<u>Explanation:</u>
For the given chemical equation:

The expression of
for above equation follows:
![K_c=\frac{[Ca^{2+}]\times [CO_3^{2-}]}{[CaCO_3}]\\\\K_c=[Ca^{2+}]\times [CO_3^{2-}]](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BCa%5E%7B2%2B%7D%5D%5Ctimes%20%5BCO_3%5E%7B2-%7D%5D%7D%7B%5BCaCO_3%7D%5D%5C%5C%5C%5CK_c%3D%5BCa%5E%7B2%2B%7D%5D%5Ctimes%20%5BCO_3%5E%7B2-%7D%5D)
The concentration of pure solids and pure liquids are taken as 1 in equilibrium constant expression
As, the denominator is missing and the numerator is the only part left in the expression. So, the value of
will be greater than 1.
Hence, for the given reaction, the value of
is greater than 1