Answer is: 1. HClO₃; (chloric acid).
Balance chemical reaction (dissociation):
ClO₃⁻(aq) + H₂O(l) ⇄ HClO₃(aq) + OH⁻(aq).
According
to Bronsted-Lowry theory acid are donor of protons and bases
are acceptors of protons (the hydrogen cation or H⁺).
The chlorate anion (ClO₃⁻) is Bronsted base and it
can accept proton and become conjugate acid HClO₃..
Answer:
See explanation
Explanation:
Full molecular equation;
2NH3(aq) + AgNO3(aq) -------> [Ag(NH3)2]NO3(aq)
Full ionic equation
2NH3(aq) + Ag^+(aq) + NO3^-(aq) --------> [Ag(NH3)2]^+(aq) + NO3^-(aq)
Net ionic equation;
2NH3(aq) + Ag^+(aq) --------> [Ag(NH3)2]^+(aq)
When Silver nitrate is mixed with a solution of aqueous ammonia, a white and cloudy solution was observed.
Answer:
6.
2AI2O3 + heat -> 4AI + 3O2
Reaction type: Combustion
7.
2AI+ 6HCl -> 3H2 + 2AlCl3
Reaction type: Single Replacement
8.
IDK
Reaction type: Double Replacement?
Explanation:
Answer:
V2 = 894.4mL
Explanation:
P1= 124.1, V1= 578mL, P2 = 80.2kPa, V2= ?
Applying Boyle's law
P1V1 = P2V2
Substitute and simplify
124.1*578=80.2*V2
V2= 894.4mL
Answer:
8.625 grams of a 150 g sample of Thorium-234 would be left after 120.5 days
Explanation:
The nuclear half life represents the time taken for the initial amount of sample to reduce into half of its mass.
We have given that the half life of thorium-234 is 24.1 days. Then it takes 24.1 days for a Thorium-234 sample to reduced to half of its initial amount.
Initial amount of Thorium-234 available as per the question is 150 grams
So now we start with 150 grams of Thorium-234





So after 120.5 days the amount of sample that remains is 8.625g
In simpler way , we can use the below formula to find the sample left

Where
is the initial sample amount
n = the number of half-lives that pass in a given period of time.