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)
Answer:
The calorimeter constant is = 447 J/°C
Explanation:
The heat absorbed or released (Q) by water can be calculated with the following expression:
Q = c × m × ΔT
where,
c is the specific heat
m is the mass
ΔT is the change in temperature
The water that is initially in the calorimeter (w₁) absorbs heat while the water that is added (w₂) later releases heat. The calorimeter also absorbs heat.
The heat absorbed by the calorimeter (Q) can be calculated with the following expression:
Q = C × ΔT
where,
C is the calorimeter constant
The density of water is 1.00 g/mL so 50.0 mL = 50.0 g. The sum of the heat absorbed and the heat released is equal to zero (conservation of energy).
Qabs + Qrel = 0
Qabs = - Qrel
Qcal + Qw₁ = - Qw₂
Qcal = - (Qw₂ + Qw₁)
Ccal . ΔTcal = - (cw . mw₁ . ΔTw₁ + cw . mw₂ . ΔTw₂)
Ccal . (30.31°C - 22.6°C) = - [(4.184 J/g.°C) × 50.0 g × (30.31°C - 22.6°C) + (4.184 J/g.°C) × 50.0 g × (30.31°C - 54.5°C)]
Ccal = 447 J/°C
Answer:
The final volume will be 5.80 L
Explanation:
Step 1: Data given
Number of moles gas = 0.140 moles
Volume of gas = 2.78 L
Number of moles added = 0.152 moles
Step 2: Calculate the final volume
V1/n1 = V2/n2
⇒ with V1 = the initial volume = 2.78 L
⇒ with n1 = the initial number of moles = 0.140 moles
⇒ with V2 = The new volume = TO BE DETERMINED
⇒ with n2 = the new number of moles = 0.140 + 0.152 = 0.292 moles
2.78/0.140 = V2 /0.292
V2 = 5.80 L
The final volume will be 5.80 L
Answer:
Positron emission
Explanation:
Positron emission involves the conversion of a proton to a neutron. This process increases the mass number of the daughter nucleus by 1 while its atomic number remains the same. The new neutron increases the number of neutrons present in the daughter nucleus hence the process increases the N/P ratio.
A positron is usually ejected in the process together with an anti-neutrino to balance the spins.
