3 Cu + 8 HNO3 = 3 Cu(NO3)2 + 2 NO + 4 H2O i dont know how to explain it but this is the balanced equation basically all the atoms of every element has to be balanced
Explanation:
![[{H}^{ + }] = {10}^{ - pH}](https://tex.z-dn.net/?f=%5B%7BH%7D%5E%7B%20%2B%20%7D%5D%20%20%3D%20%20%7B10%7D%5E%7B%20-%20pH%7D%20)
![= {10}^{ - 11.13}](https://tex.z-dn.net/?f=%20%3D%20%20%7B10%7D%5E%7B%20-%2011.13%7D%20)
![= {10}^{ (- 11 - 0.13)}](https://tex.z-dn.net/?f=%20%3D%20%20%7B10%7D%5E%7B%20%28-%2011%20-%200.13%29%7D%20)
![= {10}^{ - 0.13} \times {10}^{ - 11}](https://tex.z-dn.net/?f=%20%3D%20%20%7B10%7D%5E%7B%20-%200.13%7D%20%20%5Ctimes%20%20%7B10%7D%5E%7B%20-%2011%7D%20)
![= 0.741 \times {10}^{ - 11} \frac{mol}{litre}](https://tex.z-dn.net/?f=%20%3D%200.741%20%5Ctimes%20%20%7B10%7D%5E%7B%20-%2011%7D%20%20%5Cfrac%7Bmol%7D%7Blitre%7D%20)
<h2>HOPE IT HELPS YOU</h2>
Based on <span> Henderson - Hasselbach equation, we have
pH = pka + log </span>
![\frac{[Acid]}{\text{[conjugate base]}}](https://tex.z-dn.net/?f=%20%5Cfrac%7B%5BAcid%5D%7D%7B%5Ctext%7B%5Bconjugate%20%20base%5D%7D%7D%20)
Given: pH = 5.28 and [Acid] = 0.05 m,
∴ pKa = 5.28 - log (0.05) = 6.58
∴ Ka = 2.624 X
![10^{-7}](https://tex.z-dn.net/?f=%2010%5E%7B-7%7D%20)
Thus, ionization constant of boric acid is 2.624 X
The number of protons is same as periodic table number.
In this case 81 Periodic table number element has 81 Protons.
<h3>Answer:</h3>
2.55 × 10²² Na Atoms
<h3>Solution:</h3>
Data Given:
M.Mass of Na = 23 g.mol⁻¹
Mass of Na = 973 mg = 0.973 g
# of Na Atoms = ??
Step 1: Calculate Moles of Na as:
Moles = Mass ÷ M.Mass
Moles = 0.973 g ÷ 23 g.mol⁻¹
Moles = 0.0423 mol
Step 2: Calculate No, of Na Atoms as:
As 1 mole of sodium atoms counts 6.022 × 10²³ and equals exactly to the mass of 23 g. So, we can write,
Moles = No. of Na Atoms ÷ 6.022 × 10²³ Na Atoms.mol⁻¹
Solving for No. of Na Atoms,
No. of Na Atoms = Moles × 6.022 × 10²³ Na Atoms.mol⁻¹
No. of Na Atoms = 0.0423 mol × 6.022 × 10²³ Na Atoms.mol⁻¹
No. of Na Atoms = 2.55 × 10²² Na Atoms
<h3>Conclusion: </h3>
2.55 × 10²² sodium atoms are required to reach a total mass of 973 mg in a substance of pure sodium.