Answer:
C = 18.29 g
Explanation:
Given data:
Mass of beryllium needed = ?
Mass of nitrogen = 18.9 g
Solution:
Chemical equation:
3Be + N₂ → Be₃N₂
now we will calculate the number of moles of nitrogen:
Number of moles = mass/molar mass
Number of moles = 18.9 g/ 28 g/mol
Number of moles = 0.675 mol
Now we will compare the moles of nitrogen and Be from balance chemical equation.
N₂ : Be
1 : 3
0.675 : 3/1×0.675 = 2.03 mol
Number of moles of Be needed are 2.03 mol.
Mass of Beryllium:
Mass = number of moles × molar mass
Mass = 2.03 mol × 9.01 g/mol
Mass = 18.29 g
Answer:

Explanation:
1. Calculate the initial moles of acid and base

2. Calculate the moles remaining after the reaction
OH⁻ + H₃O⁺ ⟶ 2H₂O
I/mol: 0.0053 0.005 00
C/mol: -0.00500 -0.005 00
E/mol: 0.0003 0
We have an excess of 0.0003 mol of base.
3. Calculate the concentration of OH⁻
Total volume = 53 mL + 25.0 mL = 78 mL = 0.078 L
![\text{[OH}^{-}] = \dfrac{\text{0.0003 mol}}{\text{0.078 L}} = \textbf{0.0038 mol/L}\\\\\text{The final concentration of OH$^{-}$ is $\large \boxed{\textbf{0.0038 mol/L}}$}](https://tex.z-dn.net/?f=%5Ctext%7B%5BOH%7D%5E%7B-%7D%5D%20%3D%20%5Cdfrac%7B%5Ctext%7B0.0003%20mol%7D%7D%7B%5Ctext%7B0.078%20L%7D%7D%20%3D%20%5Ctextbf%7B0.0038%20mol%2FL%7D%5C%5C%5C%5C%5Ctext%7BThe%20final%20concentration%20of%20OH%24%5E%7B-%7D%24%20is%20%24%5Clarge%20%5Cboxed%7B%5Ctextbf%7B0.0038%20mol%2FL%7D%7D%24%7D)
Answer:
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Explanation: