Here Is A Sentence You Could Possibly Use:
"Hydrogen and Nitrogen are two examples of non-metals."
(You can replace the two given elements with whatever two non-metallic elements you like.)
Hope This Helps!
BTW, You May Want A Second Opinion Just In Case! :)
Answer:
Empirical formula: BH3
Molecular Formula: B2H6
Explanation:
To solve the exercise, we need to know how many boron atoms and how many hydrogen atoms the compound has. We know that of the total weight of the compound, 78.14% correspond to boron and 21.86% to hydrogen. As the weight of the compound is between 27 g and 28 g, using the above percentages we can solve that the compound has between 21.1 g and 21.8 g of boron, and between 5.9 g and 6.1 g of hydrogen:
100% _____ 27 g
78.14% _____ x = 78.14% * 27g / 100% = 21.1 g boron
100% ______27 g
21.86% ______ x = 21.86% * 27g / 100% = 5.9 g hydrogen
100% _____ 28 g
78.14% _____ x = 78.14% * 28g / 100% = 21.8 g boron
100% _____ 28g
21.86% _____ x = 21.86% * 28g / 100% = 6.1 g hydrogen
So, if the atomic weight of boron is 10.8 g, there must be two boron atoms in the compound that sum 21.6 g. The weight of hydrogen is 1 g, so the compound must have six hydrogen atoms.
The molecular formula represents the real amount of atoms that form a compound. Therefore, the molecular formula of the compound is B2H6.
The empirical formula is the minimum expression that represents the proportion of atoms in a compound. For example, ethane has 2 carbon atoms and 6 hydrogen atoms, so its molecular formula is C2H6, however, its empirical formula is CH3. Therefore, the empirical formula of the boron compound is BH3.
Answer:
A I did the exam and I saw A sorry if it is wrong dont have the best of memory
Answer:
The <u>equilibrium constant</u> is:
Explanation:
The correct equation is:
Thus, with the equilibrium concentrations you can calculate the equilibrium constant, Kc.
The equation for the equilibrium constant is:
![k_c=\dfrac{[NH_3]^2}{[N_2]\cdot [H_2]^3}](https://tex.z-dn.net/?f=k_c%3D%5Cdfrac%7B%5BNH_3%5D%5E2%7D%7B%5BN_2%5D%5Ccdot%20%5BH_2%5D%5E3%7D)
Substituting:
