Answer:
[CH₃COOH] = 1.70 M
Explanation:
When we talk about concentration we can determine Molarity
Molarity determines the moles of solute that are contained in 1L of solution.
In this case our solute is the acetic acid.
M = mol/L
M = 0.99 mol /0.58L → 1.70 M
We can also make a rule of three
In 0.58 L of solution we have 0.99 moles of solute
In 1 L of solution we may have (1 . 0.99) / 0.58 = 1.70 moles
Acetic acid is a weak acid, partially dissociated in water.
CH₃COOH + H₂O ⇄ CH₃COO⁻ + H₃O⁺ Ka
Muddy water is a mixture. chex mix is mixture
Answer:
6.2g of NaBr are produced
Explanation:
The reaction of HBr with NaOH occurs as follows:
HBr + NaOH → NaBr + H2O
<em>Where 1 mole of each reactant produce 1 mole of NaBr</em>
To solve this question we need to find the moles of each reactant using their molar mass. With moles we can find limiting reactant and the moles (And mass) of NaBr produced, as follows:
<em>Moles HBr -Molar mass: 80.9119g/mol)-</em>
4.9g * (1mol/80.9119g) = 0.0606 moles HBr
<em>Moles NaOH -Molar mass: 40g/mol-</em>
3.86g * (1mol/40g) = 0.0965 moles NaOH
As the reaction is 1:1 and the moles of HBr < Moles NaOH, the limiting reactant is HBr and moles of NaBr produced are 0.0606 moles.
The mass of NaBr (Molar mass: 102.894g/mol) is:
0.0606 moles * (102.894g/mol) =
<h3>6.2g of NaBr are produced</h3>
Answer: C hope this helps Have a wonderful day :)
Explanation:
