A! Oxidize. Hope this helps!
Answer:
NH₄Cl (aq) + NaOH (aq) -> NaCl (aq) + H₂O (l) + NH₃ (g)
(NH₄)₂CO₃ (aq) + 2KOH (aq) -> K₂CO₃ (aq) + 2H₂O (l) + 2NH₃(g)
NH₄NO₃ (aq) + LiOH (aq) -> LiNO₃ (aq) + H₂O (l) + NH₃ (g)
Explanation:
The gas liberated when an alkali reacts with an ammonium salt is NH₃ (ammonia), not CO₂.
Answer:
Distance = 10 meters and displacement = 10 meters
D.
Henry's Law is written in equation as:
C = kP
where
C is the concentration
k is the Henry's law constant
P is the partial pressure
This law is applied to soluble gases in liquids. At a certain temperature, there is a specific value of the Henry's Law constant. The C represents the solubility. Hence, we solve for C.
C = (<span>6.26×10</span>⁻⁴ <span>mol/(L⋅atm))*(2.85 atm)
C = 0.0017841 mol/L</span>
1.
V = 200 mL (volume)
c = 3 M = 3 mol/L (concentration)
First we convert mL to L:
200 mL = 0.2 L
Then we calculate the moles using the formula: n = V × c = 0.2 L × 3 mol = 0.6 mol
Finally, we just use the molar mass of CaF2 to calculate the actual mass:
molar mass = 78 g/mol
The formula is: m = n × mm (mass = moles × molar mass)
m = 0.6 mol × 78 g/mol = 46.8 g
2.
For this question the steps are exactly like the first question.
V = 50mL = 0.05 L
c = 12 M = 12 mol/L
n = V × c = 0.05 L × 12 mol/L = 0.6 mol
molar mass (HCl) = 36.5 g/mol
m = n × mm = 0.6 mol × 36.5 g/mol = 21.9 g.
3.
The steps for this question are the opposite way.
m(K2CO3) = 250 g
molar mass = 138 g/mol
n = m ÷ mm = 1.81 mol
c = 2 mol/L
V = n ÷ c = 1.81 mol ÷ 2 mol/L = 0.905 L = 905 mL
