The given elements put into an equation using their symbols are as follows:
Pb +

=

+ Ag
Since there are 2 Pb on the right side of the equation, you would change the coefficient of Pb on the left side to 2:
2Pb +

=

+ Ag
Since there are 2 Acetate on the right side of the equation, you would change the coefficient of Silver Acetate on the left side to 2:
2Pb +

=

+ Ag
Now there are 2 Silver on the left side, so you change the coefficient of Silver on the right side to 2:
2Pb +

=

+ 2Ag
That is your final equation
The coefficients are 2 + 2 = 1 + 2
Answer: A)
Explanation:
the deposition environment has changed from wind-dominated in the past to stream-dominated today.
The statement is : True
-Hope thi helps.
Answer:
You are mostly correcet but im pretty sure that the nonmetals are brittle because they can break easily
Explanation:
In this reaction 50% of the compound decompose in 10.5 min thus, it is half life of the reaction and denoted by symbol
.
(a) For first order reaction, rate constant and half life time are related to each other as follows:

Thus, rate constant of the reaction is
.
(b) Rate equation for first order reaction is as follows:
![k=\frac{2.303}{t_{1/2}}log\frac{[A_{0}]}{[A_{t}]}](https://tex.z-dn.net/?f=k%3D%5Cfrac%7B2.303%7D%7Bt_%7B1%2F2%7D%7Dlog%5Cfrac%7B%5BA_%7B0%7D%5D%7D%7B%5BA_%7Bt%7D%5D%7D)
now, 75% of the compound is decomposed, if initial concentration
is 100 then concentration at time t
will be 100-75=25.
Putting the values,

On rearranging,

Thus, time required for 75% decomposition is 21 min.