Go to the spanish site because I don’t think people here would be able to answer
The reaction will shift toward the products.
Adding more reactant will shift the reaction to the product.
A balanced chemical equation for this reaction :
FeS(s)+2HCl(aq)→FeCl2(aq)+ H2S(g)
<h3>Further explanation</h3>
Given
Word equation
Required
Balanced equation
Solution
The chemical equation can be expressed in terms of:
- word equation
- skeleton equation
- balanced equation
Skeleton equation
FeS(s)+ HCl(aq) --> FeCl2 (aq) + H2S (g)
Balanced equation
FeS(s)+2HCl(aq)→FeCl2(aq)+ H2S(g)
Answer:
Cl⁻
Explanation:
Definition of atomic radii
The atomic radius is the distance between center of two bonded atoms.
Trend along period:
As we move from left to right across the periodic table the number of valance electrons in an atom increase.The atomic size tend to decrease in same period of periodic table because the electrons are added with in the same shell. When the electron are added, at the same time protons are also added in the nucleus. The positive charge is going to increase and this charge is greater in effect than the charge of electrons. This effect lead to the greater nuclear attraction. The electrons are pull towards the nucleus and valance shell get closer to the nucleus. As a result of this greater nuclear attraction atomic radius decreases.
Trend along group:
In group by addition of electron atomic radii increase from top to bottom due to increase in atomic number and addition of extra shell.
In this way Cl⁻ will have the largest atomic radii because one extra electron is added and its atomic number is already greater than fluorine.
Answer:
E_a = 103.626 × 10³ KJ/mol
Explanation:
Formula to solve this is given by;
Log(k2/k1) = (E_a/2.303R)((1/T1) - (1/T2))
Where;
k2 is rate constant at second temperature
k1 is rate constant at first temperature
R is universal gas constant
T1 is first temperature
T2 is second temperature
We are given;
k1 = 2.8 × 10^(-3) /s
k2 = 4.8 × 10^(-4) /s
R = 8.314 J/mol.k
T1 = 60°C = 333.15 K
T2 = 45°C = 318.15 K
Thus;
Log((4.8 × 10^(-4))/(2.8 × 10^(-3))) = (E_a/(2.303 × 8.314))((1/333.15) - (1/318.15))
We now have;
-0.76592 = -0.00000739121E_a
E_a = -0.76592/-0.00000739121
E_a = 103.626 × 10³ KJ/mol
