Answer:
pH = 9.25
Explanation:
Reaction of ammonia (NH₃) with hydrochloric acid (HCl) is:
NH₃ + HCl → NH₄⁺ + Cl⁻
As equal volumes of a 0.020M solution of NH₃ and a 0.010M solution of react, molarity of NH₃ after reaction is 0.010M and 0.010M of NH₄⁺.
For the buffer NH₄⁺/NH₃ H-H equation to find pH of solution is:
pH = pKa + log₁₀ [NH₃] / [NH₄⁺]
pH = 9.25 + log₁₀ [0.010M] / [0.010M]
<em>pH = 9.25</em>
<h2> The amount of Hcn formed in grams is = 0.06223 grams</h2><h3> calculation</h3>
write the equation for reaction
that is; KCn + HCl → KCl + HCn
step 1:find the moles of KCn reacted
moles of KCn = mass of KCn/molar mass of KCn( 65.12 g/mol)
= 0.150 g/ 65.12 g/mol =2.303 x10^-3 moles
step 2: use the mole ratio between KCn :HCn which is 1:1 to determine the moles of HCn which is also 2.303 x 10^-3 moles
step 3: find the mass of HCn =moles of HCn x molar mass of HCn
= 2.303 x10^-3 x 27. 02 =0.06223 grams
Sulphur trioxide reacts with water to form a solution of sulphuric acid.
The equation describing this reaction is:
SO3 + H2O .............> H2SO4
Based on the above equation, the formula of the compound formed when sulphur trioxide reacts with water is: H2SO4
The answer to this question would be: 11.5 grams
The molecular mass of hydrogen is 1, oxygen is 16 and carbon is 12. Then the molecular mass of CH4O would be: 12 + 4(1) + 16= 32.
To find the hydrogen mass, you just need to divide the hydrogen with total mass, resulting in the mass ratio of hydrogen to the compound. Then the amount of hydrogen inside 92 gram of CH4O are: 92grams * (4/32)= 11.5 grams
CaCl2 and KCl are both salts which dissociate in water
when dissolved. Assuming that the dissolution of the two salts are 100 percent,
the half reactions are:
<span>CaCl2 ---> Ca2+ + 2 Cl-</span>
KCl ---> K+ + Cl-
Therefore the total Cl- ion concentration would be coming
from both salts. First, we calculate the Cl- from each salt by using stoichiometric
ratio:
Cl- from CaCl2 = (0.2 moles CaCl2/ L) (0.25 L) (2 moles
Cl / 1 mole CaCl2)
Cl- from CaCl2 = 0.1 moles
Cl- from KCl = (0.4 moles KCl/ L) (0.25 L) (1 mole Cl / 1
mole KCl)
Cl- from KCl = 0.1 moles
Therefore the final concentration of Cl- in the solution
mixture is:
Cl- = (0.1 moles + 0.1 moles) / (0.25 L + 0.25 L)
Cl- = 0.2 moles / 0.5 moles
<span>Cl- = 0.4 moles (ANSWER)</span>
