Answer:
The least whole number coefficient for HNO₃ is 6
Explanation:
The chemical equation above is the reaction between calcium orthophosphate and nitric acid.
To balance a chemical equation, we have to consider law of conservation of matter which states that matter can neither be created nor destroyed.
What this law implies is that, whatever we have at the reactant side must be equal to whatever is obtainable at the product side.
The above equation is
Ca₃(PO₄)₂ + HNO₃ → Ca(NO₃)₂ + H₃PO₄
To balance the equation, we'll have to check the number of atoms at each side and possibly balance the equation with the number of moles.
The balanced equation is
Ca₃(PO₄)₂ + 6HNO₃ → 3Ca(NO₃)₂ + 2H₃PO₄
From the balanced equation above, we can see that the number of calcium (Ca), Phosphorus (P), Oxygen(O), Nitrogen(N) and hydrogen (H) are balanced at both sides of the equation.
The least number coefficient for HNO₃ is 6
Answer:
The correct answer is option c.
Explanation:
Formula used to determine an average atomic mass :
Mass of isotope Sb-121 = 120.904 amu
Fractional abundance of Sb-121 = 57.21% = 0.5721
Mass of isotope Sb-123 = 122.904 amu
Fractional abundance of Sb-123 = 42.79% = 0.4279
Average atomic mass of Sb:
Covalent network. <span>A solid that is extremely hard, that has a very high melting point, and that will not conduct electricity either as a solid or when molten is held together by a continuous three-dimensional network of covalent bonds. Examples include diamond, quartz (SiO </span><span>2 </span>), and silicon carbide (SiC). The electrons are constrained in pairs to a region on a line between the centers of pairs of atoms.<span>
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Answer:
Explanation:
Hello there!
In this case, when considering weak acids which have an associated percent dissociation, we first need to set up the ionization reaction and the equilibrium expression:
![HA\rightleftharpoons H^++A^-\\\\Ka=\frac{[H^+][A^-]}{[HA]}](https://tex.z-dn.net/?f=HA%5Crightleftharpoons%20H%5E%2B%2BA%5E-%5C%5C%5C%5CKa%3D%5Cfrac%7B%5BH%5E%2B%5D%5BA%5E-%5D%7D%7B%5BHA%5D%7D)
Now, by introducing x as the reaction extent which also represents the concentration of both H+ and A-, we have:
![Ka=\frac{x^2}{[HA]_0-x} =10^{-4.74}=1.82x10^{-5}](https://tex.z-dn.net/?f=Ka%3D%5Cfrac%7Bx%5E2%7D%7B%5BHA%5D_0-x%7D%20%3D10%5E%7B-4.74%7D%3D1.82x10%5E%7B-5%7D)
Thus, it is possible to find x given the pH as shown below:
So that we can calculate the initial concentration of the acid:
![\frac{(1.82x10^{-5})^2}{[HA]_0-1.82x10^{-5}} =1.82x10^{-5}\\\\\frac{1.82x10^{-5}}{[HA]_0-1.82x10^{-5}} =1\\\\](https://tex.z-dn.net/?f=%5Cfrac%7B%281.82x10%5E%7B-5%7D%29%5E2%7D%7B%5BHA%5D_0-1.82x10%5E%7B-5%7D%7D%20%3D1.82x10%5E%7B-5%7D%5C%5C%5C%5C%5Cfrac%7B1.82x10%5E%7B-5%7D%7D%7B%5BHA%5D_0-1.82x10%5E%7B-5%7D%7D%20%3D1%5C%5C%5C%5C)
![[HA]_0=3.64x10^{-5}M](https://tex.z-dn.net/?f=%5BHA%5D_0%3D3.64x10%5E%7B-5%7DM)
Therefore, the percent dissociation turns out to be:
![\% diss=\frac{x}{[HA]_0}*100\% \\\\\% diss=\frac{1.82x10^{-5}M}{3.64x10^{-5}M}*100\% \\\\\% diss = 50\%](https://tex.z-dn.net/?f=%5C%25%20diss%3D%5Cfrac%7Bx%7D%7B%5BHA%5D_0%7D%2A100%5C%25%20%5C%5C%5C%5C%5C%25%20diss%3D%5Cfrac%7B1.82x10%5E%7B-5%7DM%7D%7B3.64x10%5E%7B-5%7DM%7D%2A100%5C%25%20%5C%5C%5C%5C%5C%25%20diss%20%3D%2050%5C%25)
Best regards!
The correct number of significant figures and digits is 3
