Neutralization reaction is the reaction between acid and base.
There the neutralization reaction is:
HBr + LiOH ----------> LiBr + H2O.
Hope this helps!
One electron is shared between ethane, H3-C-C-H3
I pretty sure the answer is true but here is also a site to help you. https://www.britannica.com/science/wave-water
Answer: Explanation:
Temperatura. La solubilidad de un soluto en un determinado solvente principalmente depende de la temperatura. Para muchos sólidos disueltos en el agua líquida, la solubilidad aumenta con la temperatura hasta 100 °C, aunque existen casos que presentan un comportamiento inverso.
The volume of the NaOH used is calculated as 14 mL.
<h3>What is stoichiometry?</h3>
The term stoichiometry has to do with the calculation of the amount of substance in a reaction using mass - mole or mass - volume relationship.
Here;
Number of moles of CaCO3 = 0.205 g/100.1 = 0.00205 moles
Number of moles of HCl = 2.00 M * 7/1000 L = 0.014 moles
2 moles of HCl reacts with 1 mole of CaCO3
x moles of HCl reacts with 0.00205 moles of CaCO3
x = 0.00205 moles * 2/1 = 0.0041 moles
Hence HCl is the excess reactant
Amount of excess HCl = 0.014 moles - 0.0041 moles = 0.0099 moles
Concentration of excess HCl reacted = 0.0099 moles/125 * 10^-3 = 0.0792 M
Using;
CAVA/CBVB = NA/NB
CAVANB = CBVBNA
VB = CAVANB/CBNA
VB = 0.0792 M * 10 mL * 1/ 0.058 M
VB = 14 mL
Missing parts;
A 0.205 g sample of caco3 (mr = 100.1 g/mol) is added to a flask along with 7.50 ml of 2.00 m hcl. caco3(aq) + 2hcl(aq) → cacl2(aq) + h2o(l) + co2(g) enough water is then added to make a 125.0 ml solution. a 10.00 ml aliquot of this solution is taken and titrated with 0.058 m naoh. naoh(aq) + hcl(aq) → h2o(l) + nacl(aq) how many ml of naoh are used?
Learn more about stoichiometry: brainly.com/question/9743981
