Answer: The percent ionization of 
in a 0.050 M 
solution is 8.9 %
Explanation:
cM 0 0
So dissociation constant will be:
Give c= concentration = 0.050 M and 
= degree of ionisation = ?
Putting in the values we get:
percent ionisation =
they are all an energy particle
Answer: Mother is bb. Father is Bb
Top left box: Bb
Top Right box: bb
Bottom left box: Bb
Bottom Right box: bb
Hope this helps :D. If you want me to explain my answer just let me know
<span>Ca(NO3)2 + Na2CO3 = CaCO3 + 2NaNO3
Yes a precipitate of Calcium Carbonate is formed since it is insoluble in water.
Mol Wt of Calcium Nitrate is 164. And that of Calcium Carbonate is 100.
One mole of Calcium Nitrate produces one mole of Calcium Carbonate.
i.e. 164 gms will produce 100gms of precipitate
So, 1.74gms of Calcium Carbonate will be obtained from 2.85gms Calcium Nitrate present in the original solution.</span>
Answer:
3.336.
Explanation:
<em>Herein, the no. of millimoles of the acid (HCOOH) is more than that of the base (NaOH).</em>
<em />
So, <em>concentration of excess acid = [(NV)acid - (NV)base]/V total</em> = [(30.0 mL)(0.1 M) - (29.3 mL)(0.1 M)]/(59.3 mL) = <em>1.18 x 10⁻³ M.</em>
<em></em>
<em> For weak acids; [H⁺] = √Ka.C</em> = √(1.8 x 10⁻⁴)(1.18 x 10⁻³ M) = <em>4.61 x 10⁻⁴ M.</em>
∵ pH = - log[H⁺].
<em>∴ pH = - log(4.61 x 10⁻⁴) = 3.336.</em>
