CuH12O10S i think that’s correct i’m not sure
Earth because it has so much water come on b
It would be the skeletal system. This systems primary purpose it to give body its shape, support, protection, and makes new red blood cells from the bone marrow. Located deep inside bones.
In a solution, when the concentrations of a weak acid and its conjugate base are equal: 1. the -log of the concentration of H+ and the -log of the Ka are equal.
<h3>What are Strong acid strong bases ?</h3>
Any acid that entirely ionises in solutions is considered a strong acid. When positioned, it emits the most hydrogen ions or protons.
- It is a substance with the capacity to take a proton out of an extremely weak acid. In water, they can also entirely separate into its ions. Examples include sodium hydroxide and potassium hydroxide (KOH) (NaOH)
<h3>What are weak acid and weak bases ?</h3>
Acids that don't entirely dissociate in solution are referred to as weak acids. In other words, any acid that is not a strong acid qualifies as a weak acid. The amount of dissociation determines how strong an acid is; the more dissociation, the stronger the acid.
- Weak bases are basic compounds that, when dissolved in solutions, do not entirely separate into their constituent ions.
Learn more about Acids and bases here:
brainly.com/question/27915098
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Answer:
25.0 g of sodium carbonate are present in 220 ml of the solution.
Explanation:
Hi there!
I have found the complete question on the web:
<em>Calculate the volume in milliliters of a 1.07 M sodium carbonate solution that contains 25.0g of sodium carbonate Na2CO3. Be sure your answer has the correct number of significant digits.</em>
First, let's find how many moles of sodium carbonate have a mass of 25.0 g.
The molar mass of Na₂CO₃ is 106 g/mol.
So, if 106 g of sodium carbonate is equal to 1 mol, then 25.0 g will be:
25.0 g · (1 mol / 106 g) = 0.236 mol Na₂CO₃
The solution contains 1.07 mol sodium carbonate in 1 liter.
So, if 1.07 mol sodium carbonate is present in 1 l solution, then, 0.236 mol will be present in:
0.236 mol · (1000 ml / 1.07 mo) = 221 ml (220 ml without any intermediate rounding).
25.0 g of sodium carbonate are present in 220 ml of the solution.