Kr 5s2 is the correct noble gas configuration for strontium
Answer:
10.5
Explanation:
well we added together so thats the result
The type of bond between two elements can be predicted by measuring the electronegativity difference between the two atoms. Based on this difference the type of bonds are classified as,
E.N difference between 0.0 to 0.3 = Non Polar bond
E.N difference between 0.4 to 1.0 = Moderate Polar Covalent Bond
E.N difference between 1.1 to 2.0 = Very Polar Covalent Bond
E.N difference greater than 2.0 = Ionic Bond
So,
(E.N of N) 3.04 - (E.N of C) 2.55 = 0.49
As the difference is between 0.4-1.0, So, the bond between C and N is Moderate Polar Covalent Bond (or Polar Covalent Bond).
Answer:
0.0119
Explanation:
There was a part missing. I think this is the whole question:
<em>Before any reaction occurs, the concentration of A in the reaction below is 0.0510 M. What is the equilibrium constant if the concentration of A at equilibrium is 0.0153 M?</em>
A (aq) ⇌ 2B (aq) + C(aq)
<em>Remember to use correct significant figures in your answer. Do not include units in your response.</em>
First, we have to make an ICE Chart, which stands for initial, change and equilibrium. We will call "x" unknown concentrations.
A (aq) ⇌ 2B (aq) + C (aq)
I 0.0510 0 0
C -x +2x +x
E 0.0510-x 2x x
Since the concentration at equilibrium of A is 0.0153 M, we get

We can use the value of x to calculate the concentrations at equilibrium.
![[A]e = 0.0153 M \\[B]e = 2x = 2(0.0357) = 0.0714 M \\[C]e = x = 0.0357 M \\](https://tex.z-dn.net/?f=%5BA%5De%20%3D%200.0153%20M%20%5C%5C%5BB%5De%20%3D%202x%20%3D%202%280.0357%29%20%3D%200.0714%20M%20%5C%5C%5BC%5De%20%3D%20x%20%3D%200.0357%20M%20%5C%5C)
The equilibrium constant, Kc, is the ratio of the equilibrium concentrations of products over the equilibrium concentrations of reactants each raised to the power of their stoichiometric coefficients.
![Kc = \frac{[B]^{2} \times [C]}{[A]} = \frac{0.0714^{2} \times 0.0357}{0.0153} = 0.0119](https://tex.z-dn.net/?f=Kc%20%3D%20%5Cfrac%7B%5BB%5D%5E%7B2%7D%20%20%5Ctimes%20%5BC%5D%7D%7B%5BA%5D%7D%20%3D%20%5Cfrac%7B0.0714%5E%7B2%7D%20%20%5Ctimes%200.0357%7D%7B0.0153%7D%20%3D%200.0119)
The equilibrium constant for this reaction at equilibrium is 0.0119.
You can learn more about equilibrium here: brainly.com/question/4289021
It is important to determine the qcalorimeter before determining the qmetal because the qcalorimeter is a constant and it's value is used to find the value of qmetal.