The numbers in front of the formulas
Answer:
1. (NH₄)₂S(s) -----> NH₄+(aq) + S²-(aq)
2. Al³+ (aq) + PO₄³+ (aq) ----> AlPO₄ (s)
Explanation:
The dissociation of ammonium sulphide, (NH₄)₂S when dissolved in water is given in the equation below:
(NH₄)₂S(s) -----> NH₄+(aq) + S²-(aq)
However very little S²- ions are present in solution due to the very basic nature of the S²- ion (Kb = 1 x 105).
The ammonium ion being a better proton donor than water, donates a proton to sulphide ion to form hydrosulphide ion which exists in equilibrium with aqueous ammonia.
S²- (aq) + NH₄+ (aq) ⇌ SH- (aq) + NH₃ (aq)
Aqueous solutions of ammonium sulfide are smelly due to the release of hydrogen sulfide and ammonia, hence, their use in making stink bombs.
2. The reaction between aluminium nitrate and sodium phosphatein aqueous solution is a double decomposition reaction whish results in the precipitation of insoluble aluminium phosphate. The equation of the reaction is given below :
Al(NO₃)₃ (aq) + Na₃PO₄ (aq) ----> AlPO₄ (s) + 3 NaNO₃ (aq)
The net ionic equation is given below:
Al³+ (aq) + PO₄³+ (aq) ----> AlPO₄ (s)
The osmotic pressure of a solution is a colligative property, which means that it depends on the number of particles of solute in the solution.
Formula: Osmotic pressure = MRT, where M is the molarity of the solution, R is the universal constant of ideal gases and T is the absolute temperature of the solution.
So, the answer is the option .: the osmotic pressure of a solution increases as the number of particles of solute in the solution increases.
Answer:
The average atomic mass of copper is 63.55 amu.
Explanation:
hope this help
Answer:
The molarity of the HCl solution should be 4.04 M
Explanation:
<u>Step 1:</u> Data given
volume of HCl solution = 10.00 mL = 0.01 L
volume of a 1.6 M NaOH solution = 25.24 mL = 0.02524 L
<u>Step 2:</u> The balanced equation
HCl + NaOH → NaCL + H2O
Step 3: Calculate molarity of HCl
n1*C1*V1 = n2*C2*V2
Since the mole ratio for HCl and NaOH is 1:1 we can just write:
C1*V1 =C2*V2
⇒ with C1 : the molarity of HCl = TO BE DETERMINED
⇒ with V1 = the volume og HCl = 10 mL = 0.01 L
⇒ with C2 = The molarity of NaOH = 1.6 M
⇒ with V2 = volume of NaOH = 25.24 mL = 0.02524 L
C1 * 0.01 = 1.6 * 0.02524
C1 = (1.6*0.02524)/0.01
C1 = 4.04M
The molarity of the HCl solution should be 4.04 M
