Answer:
The water will eventually become the same temature (A)
Answer:
At the cathode during the electrolysis of an aqueous solution of magnesium iodide, MgI2 , 2I−(aq) is produced
Explanation:
At cathode, reduction reaction takes place.
The dissociation of MgI2 in aqueous solution is Mg2+(aq) and 2I−(aq)
Here, the Iodine reduces to 2I−(aq) from state of 0 (MgI2) to state of -1 (2I−(aq))
Hence, at the cathode during the electrolysis of an aqueous solution of magnesium iodide, MgI2 , 2I−(aq) is produced
Answer:
Order zero
Explanation:
Let's consider the decomposition of ammonia to nitrogen and hydrogen on a tungsten filament at 800°C.
2 NH₃(g) → N₂(g) + 3 H₂(g)
The generic rate law is:
rate = k × [NH₃]ⁿ
where,
rate: reaction rate
k: rate constant
n: reaction order
When n = 0, we get:
rate = k × [NH₃]⁰ = k
As we can see, when the reaction order with respect to ammonia is zero, the reaction rate is independent of the concentration of ammonia.
given days = 42 days
So, number of half life, n = 42/14.3 = 2.93
So, after 42 days, the mass of sodium phosphate sample left = original mass
x (1/2)^n,
= 157 mg x (1/2)^2.93 = 20.60 g of Na3PO4 after 42 days
Formula mass of Sodium phosphate = 3(23) + 32 + 4(16) amu = 165 amu
165 g of sodium phosphate contains 32 g of Phosphorus
20.60 g ------------------------------? = 20.60 x 32/ 165 =3.99g of P32
