Each element<span> can usually be classified as a metal or a non-metal based on their ... They are usually </span>dull<span>and therefore show no metallic </span>luster<span> and they do not reflect ... </span>Dull<span>, Brittle solids; Little or no metallic </span>luster<span>; </span>High<span> ionization energies; </span>High<span> ...</span>
<span>Answer: option (1) solubility of the solution increases.
</span><span />
<span>Justification:
</span><span />
<span>The solubility of substances in a given solvent is temperature dependent.
</span><span />
<span>The most common behavior of the solubility of salts in water is that the solubiilty increases as the temperature increase.
</span><span />
<span>To predict with certainty the solubility at different temperatures you need the product solubility constants (Kps), which is a constant of equlibrium of the dissolution of a ionic compound slightly soluble in water, or a chart (usually experimental chart) showing the solubilities at different temperatures.
</span><span />
<span>KClO₃ is a highly soluble in water, so you do not work with Kps.
</span><span />
<span>You need the solubility chart or just assume that it has the normal behavior of the most common salts. You might know from ordinary experience that you can dissolve more sodium chloride (table salt) in water when the water is hot. That is the same with KClO₃.
</span><span>The solubility chart of KlO₃ is almost a straight line (slightly curved upward), with positive slope (ascending from left to right) meaning that the higher the temperature the more the amount of salt that can be dissolved.</span>
Answer:
<u>Mass concentration (g/L) </u><u><em>= 2.49g/L.</em></u>
Explanation:
No. of moles = 
= 
= 0.001245 moles
Concentration of KHP (C1) in litres = n/v
= 
= 0.062 mol/L
We know that:
= 
where c1v1 and c2v2 are the products of concentration and volumes of KHP and NaOH respectively.
Since mole ratio is 1 : 1.
1 mole of NaOH - 40g
0.001245 mole of NaOH = 40 × 0.001245 = 0.0498g
⇒0.0498g of NaOH was used during the titration
<u><em>∴Mass concentration (g/L) = 0.0498g ÷ 0.02L</em></u>
<u><em>= 2.49g/L.</em></u>
Answer:
Percentage dissociated = 0.41%
Explanation:
The chemical equation for the reaction is:
The ICE table is then shown as:
Initial (M) 1.8 0 0
Change (M) - x + x + x
Equilibrium (M) (1.8 -x) x x
![K_a = \frac{[C_3H_6ClCO^-_2][H^+]}{[C_3H_6ClCO_2H]}](https://tex.z-dn.net/?f=K_a%20%20%3D%20%5Cfrac%7B%5BC_3H_6ClCO%5E-_2%5D%5BH%5E%2B%5D%7D%7B%5BC_3H_6ClCO_2H%5D%7D)
where ;
Since the value for 
is infinitesimally small; then 1.8 - x ≅ 1.8
Then;
Dissociated form of 4-chlorobutanoic acid = 
Percentage dissociated = 
Percentage dissociated = 
Percentage dissociated = 0.4096
Percentage dissociated = 0.41% (to two significant digits)
Answer:
the gravity
Explanation:
i hope you help this answer
