Answer:
NH₄Cl ------> NH₃ + HCl
Explanation:
Ammonium Chloride =(NH₄Cl
Ammonia = NH₃
Hydrochloric Acid = HCl
NH₄Cl ------> NH₃ + HCl
In decomposition reaction, the reactant is breaking down into smaller parts. In this case, all of the coefficients are 1. The reaction is already balanced.
Answer: A yellowy-green dense gas with a choking smell. Chlorine kills bacteria – it is a disinfectant. It is used to treat drinking water and swimming pool water. It is also used to make hundreds of consumer products from paper to paints, and from textiles to insecticides.
Explanation: hope this helps u! HAppy Holidays!
The amount of current required to produce 75. 8 g of iron metal from a solution of aqueous iron (iii)chloride in 6. 75 hours is 168.4A.
The amount of Current required to deposit a metal can be find out by using The Law of Equivalence. It states that the number of gram equivalents of each reactant and product is equal in a given reaction.
It can be found using the formula,
m = Z I t
where, m = mass of metal deposited = 75.8g
Z = Equivalent mass / 96500 = 18.6 / 96500 = 0.0001
I is the current passed
t is the time taken = 75hour = 75 × 60 = 4500s
On subsituting in above formula,
75.8 = E I t / F
⇒ 75.8 = 0.0001 × I × 4500
⇒ I = 168.4 Ampere (A)
Hence, amount of current required to deposit a metal is 168.4A.
Learn more about Law of Equivalence here, brainly.com/question/13104984
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Answer:
2.122×10^25atoms
Explanation:
number of moles=mass/molar mass
7.05moles= mass of pyridine/79
reacting mass of pyridine=556.95
C5H5N= (12×5)+(5)+(14)=79
C5=60
to find the mass of carbon in 556.95g of pyridine we take the stoichometric ratio
60[C5] -----> 79[C5H5N]
x[C5] --------> 556.95g[C5H5N]
cross multiply
x=(60×556.95)/79
x=423g of carbon
moles=mass/molar mass
moles of carbon=423/12
moles=35.25moles of carbon
moles=number of particles/Avogadro's constant
35.25=number of particles/6.02×10^23
number of particles=2.122×10^25atoms of carbon
Answer:
Explanation:
We have the reactions:
A: 
B: 
Our <u>target reaction</u> is:
We have 
as a reactive in the target reaction and is present in A reaction but in the products side. So we have to<u> flip reaction A</u>.
A: 
Then if we add reactions A and B we can obtain the target reaction, so:
A:
B: 
For the <u>final Kc value</u>, we have to keep in mind that when we have to <u>add chemical reactions</u> the total Kc value would be the <u>multiplication</u> of the Kc values in the previous reactions.
