Answer:
Here's what I get
Explanation:
CH₃CH₂CH₂CH₂CH₂CH₃ — hexane
CH₂=CHCH₂CH₂CH₂CH₃ — hex-1-ene is the preferred IUPAC name (PIN). 1-Hexene is accepted
CH₃C≡CCH₃ — but-2-yne (PIN); 2-butyne is accepted
CH₃CH(CH₃)CH₂CH₂CH₃ — 2-methylpentane
CH₃CH₂CHCICH₂CH₃ — 3-chloropentane
Answer:
![K_a=\frac{[H_3O^+][HCO_3^-]}{[H_2CO_3]}](https://tex.z-dn.net/?f=K_a%3D%5Cfrac%7B%5BH_3O%5E%2B%5D%5BHCO_3%5E-%5D%7D%7B%5BH_2CO_3%5D%7D)
Explanation:
Several rules should be followed to write any equilibrium expression properly. In the context of this problem, we're dealing with an aqueous equilibrium:
- an equilibrium constant is, first of all, a fraction;
- in the numerator of the fraction, we have a product of the concentrations of our products (right-hand side of the equation);
- in the denominator of the fraction, we have a product of the concentrations of our reactants (left-hand side o the equation);
- each concentration should be raised to the power of the coefficient in the balanced chemical equation;
- only aqueous species and gases are included in the equilibrium constant, solids and liquids are omitted.
Following the guidelines, we will omit liquid water and we will include all the other species in the constant. Each coefficient in the balanced equation is '1', so no powers required. Multiply the concentrations of the two products and divide by the concentration of carbonic acid:
![K_a=\frac{[H_3O^+][HCO_3^-]}{[H_2CO_3]}](https://tex.z-dn.net/?f=K_a%3D%5Cfrac%7B%5BH_3O%5E%2B%5D%5BHCO_3%5E-%5D%7D%7B%5BH_2CO_3%5D%7D)
Answer:
take these as notes ig:)
Explanation:
reaction equations to make connections between the reactants and products in reactions. ... pathway: mol of O2 → mol of H2O → grams of H2O. 3.5 mol ... 25 g KClO3 ... How many grams of hydrogen gas are produced from the reaction of 221.
Answer:
Carbon dioxide reacts with limewater (a solution of calcium hydroxide, Ca(OH) 2), to form a white precipitate (appears milky) of calcium carbonate, CaCO 3.
Hope it helps you! :)
Answer:
When sodium chloride dissolves in water to make a saturated solution there is a 2.5 per cent reduction in volume. ... The solubility of salt does not change much with temperature, so there is little profit in using hot water.