Answer:
2AlCl3 + Ca3N2 - 2AlN+ 3CaCl2
Hey there!
Molar mass NaCl = 58.44 g/mol
Number of moles
n = mass of solute / molar mass
n = 59.76 / 58.44
n = 1.0225 moles of NaCl
Volume in liters:
270 mL / 1000 => 0.27 L
Therefore:
M = number of moles / volume ( L )
M = 1.0225 / 0.27
= 3.78 M
Hope that helps!
Based on the diagram shown, a numerical setup for calculating the gram-formula mass for reactant 1 would be :
6(1) + 2(12) + 16
Hope this helps
Answer: 5
Explanation: this is because the energy level of the emitted of absorbed photon increases as the number of electron shell decreases, thereby making the inner shell have higher energy than other shells
Answer:
The value of Kc for the reaction is 3.24
Explanation:
A reversible chemical reaction, indicated by a double arrow, occurs in both directions: reagents transforming into products (
direct reaction) and products transforming back into reagents (inverse reaction)
Chemical Equilibrium is the state in which direct and indirect reactions have the same reaction rate. Then taking into account the rate constant of a direct reaction and its inverse the chemical constant Kc is defined.
Being:
aA + bB ⇔ cC + dD
where a, b, c and d are the stoichiometric coefficients, the equilibrium constant with the following equation:
![Kc=\frac{[C]^{c} *[D]^{d} }{[A]^{a} *[B]^{b} }](https://tex.z-dn.net/?f=Kc%3D%5Cfrac%7B%5BC%5D%5E%7Bc%7D%20%2A%5BD%5D%5E%7Bd%7D%20%7D%7B%5BA%5D%5E%7Ba%7D%20%2A%5BB%5D%5E%7Bb%7D%20%7D)
Kc is equal to the multiplication of the concentrations of the products raised to their stoichiometric coefficients divided by the multiplication of the concentrations of the reagents also raised to their stoichiometric coefficients.
Then, in the reaction 3A(g) + 2B(g) ⇔ 2C(g), the constant Kc is:
![Kc=\frac{[C]^{2} }{[A]^{3} *[B]^{2} }](https://tex.z-dn.net/?f=Kc%3D%5Cfrac%7B%5BC%5D%5E%7B2%7D%20%7D%7B%5BA%5D%5E%7B3%7D%20%2A%5BB%5D%5E%7B2%7D%20%7D)
where:
- [A]= 0.855 M
- [B]= 1.23 M
- [C]= 1.75 M
Replacing:

Solving you get:
Kc=3.24
<u><em>The value of Kc for the reaction is 3.24</em></u>