Its the complete transfer of valence electrons between atoms
5600 why because you take away and then you add then yeah
Answer:
663 g
Explanation:
Step 1: Write the balanced equation
2 LiOH + CO₂ ⇒ H₂O + Li₂CO₃
Step 2: Calculate the moles corresponding to 825 L of CO₂
At standard pressure and temperature, 1 mole of CO₂ has a volume of 22.4 L.
825 L × 1 mol/22.4 L = 36.8 mol
Step 3: Calculate the moles of H₂O formed from 36.8 moles of CO₂
The molar ratio of CO₂ to H₂O is 1:1. The moles of H₂O formed are 1/1 × 36.8 mol = 36.8 mol.
Step 4: Calculate the mass corresponding to 36.8 moles of H₂O
The molar mass of H₂O is 18.02 g/mol.
36.8 mol × 18.02 g/mol = 663 g
Answer:
the forces on it are balanced
Explanation:
To standardize 20.00 mL of 0.495 M H₂SO₄ are used 44.10 mL of 0.450 M NaOH in a neutralization reaction.
We want to standardize a H₂SO₄ solution with NaOH. The neutralization reaction is:
H₂SO₄ + 2 NaOH ⇒ Na₂SO₄ + H₂O
The buret, which contains NaOH, reads initially 0.63 mL, and 44.73 mL at the endpoint. The used volume of NaOH is:

44.10 mL of 0.450 M NaOH are used for the titration. The reacting moles of NaOH are:

The molar ratio of H₂SO₄ to NaOH is 1:2. The moles of H₂SO₄ that react with 0.0198 moles of NaOH are:

0.00990 moles of H₂SO₄ are in 20.00 mL of solution. The molarity of H₂SO₄ is:
![[H_2SO_4] = \frac{0.00990mol}{0.02000} = 0.495 M](https://tex.z-dn.net/?f=%5BH_2SO_4%5D%20%3D%20%5Cfrac%7B0.00990mol%7D%7B0.02000%7D%20%3D%200.495%20M)
To standardize 20.00 mL of 0.495 M H₂SO₄ are used 44.10 mL of 0.450 M NaOH in a neutralization reaction.
Learn more: brainly.com/question/2728613