Answer: AgCl
Explanation:
When the reaction takes place in which the sodium chloride is reacted with silver nitrate then the products formed are sodium nitrate and silver chloride.
NaCl(aq) + AgNO3(aq) ⟶ NaNO3(aq) + AgCl (s)
This reaction is known as precipitation reaction in which the white precipitate of silver chloride is formed.
This precipitate is insoluble and separates out of the solution.
So, the correction answer is AgCl(silver chloride).
Answer:
900g
Explanation:
Given parameters:
Number of moles of Sc = 20moles
Unknown:
Mass of Sc = ?
Solution:
To find the mass of the scandium given, we use the expression below:
Mass = Number of moles x molar mass
Molar mass of Sc = 45g/mol
So;
Mass of Sc = 20 x 45 = 900g
If iodine has a 13.3 hour half life, 3.12 mg of the original sample will still be present after 39.9 hours.
<h3>what is The meaning of half-life?</h3>
The reaction rate must decrease to half its initial value before the process has reached its half life. The split of that first order chemical reaction is a quantity related to the rate constant of the reaction.
Briefing:
Given, the iodine half-life is 13.3 hours. As a result, half-life t1/2 Equals ln 2/ K
k = 0.693 .... t1/2
≈ 0.693 /13.3 hours
= 5.21 *10-2
The rate law sets the flow timetime is set by the rate law, which varies for each emotion. When only one reactant's quantity influences the speed of chemical reaction, this is referred to as a first-order contact. Because of this, it is occasionally referred to as unimolecular process. The rate is as follows for this type of reaction: k= 1/t ln [A]0 / [A] is the rate constant for the first order reaction. 5.21 * 10-2 hours are in 39.9 hours.
By including all the numbers in the initial order reaction's rate constant equation, we obtain [A].
t = 3.12mg
To know more about Half life visit:
brainly.com/question/24710827
#SPJ4
The
chemical reaction is represented as:<span>
2A(g) = B(g) + C(g)
To determine the equilibrium concentration of A, we make use of the equilibrium
constant, Kc, given above. It is expressed as the ratio of the equilibrium
concentrations of the products and the reactants. For this reaction, it is
expressed as:
Kc = [B] [C] / [A]^2
From the problem statement, we are given the following
Kc = 0.035
Volume = 20.0 L
Initial concentrations: [B] = 8.00 mol / 20.0 L = 0.4 M
[C] = 12.00 mol / 20.0 L = 0.6
M
Since the initial reactants are B and C, the reaction is reversed as well as
the Kc.
Kc = [A]^2 / [B][C]
We use the ICE table:
B
C A
I 0.4 0.6
0
C -x -x
+x
------------------------------------------
E 0.4 - x 0.6 - x
x
Kc = x^2 / (0.4-x) (0.6-x) = 0.035
solve for x,
x = 0.07691 = [A]</span>