Answer:
Explanation:
Bronsted Base is an H+ acceptor
No good answer Bronstead base does not accept hydroxide or electrons
The answer is C2O.............
Answer:
869 g Cl₂O
Explanation:
To find the theoretical yield of Cl₂O, you need to (1) convert moles SO₂ to moles Cl₂O (via mole-to-mole ratio from reaction coefficients) and then (2) convert moles Cl₂O to grams Cl₂O (via molar mass). It is important to arrange the conversions/ratios in a way that allows for the cancellation of units (the desired unit should be in the numerator). The final answer should have 3 sig figs to reflect the sig figs of the given amount (10.0 moles).
1 SO₂ (g) + 2 Cl₂ (g) ----> 1 SOCl₂ (g) + 1 Cl₂O (g)
Molar Mass (Cl₂O): 2(35.453 g/mol) + 15.998 g/mol
Molar Mass (Cl₂O): 86.904 g/mol
10.0 moles SO₂ 1 mole Cl₂O 86.904 g
------------------------ x ---------------------- x ------------------ = 869 g Cl₂O
1 mole SO₂ 1 mole
Answer:
- 13.56 g of sodium chloride are theoretically yielded.
- Limiting reactant is copper (II) chloride and excess reactant is sodium nitrate.
- 0.50 g of sodium nitrate remain when the reaction stops.
- 92.9 % is the percent yield.
Explanation:
Hello!
In this case, according to the question, it is possible to set up the following chemical reaction:
Thus, we can first identify the limiting reactant by computing the yielded mass of sodium chloride, NaCl, by each reactant via stoichiometry:
Thus, we infer that copper (II) chloride is the limiting reactant as it yields the fewest grams of sodium chloride product. Moreover the formed grams of this product are 13.56 g. Then, we take 13.56 g of sodium chloride to compute the consumed mass sodium nitrate as it is in excess:
Therefore, the leftover of sodium nitrate is:
Finally, the percent yield is computed via:
Best regards!
Answer :
The basic rules for naming of hydrocarbons are :
First select the longest possible carbon chain.
The longest possible carbon chain should include the carbons of double or triple bonds.
The naming of alkane is done by adding the suffix -ane, alkene by adding the suffix -ene, alkyne by adding the suffix -yne.
The numbering is done in such a way that first carbon of double or triple bond gets the lowest number.
The carbon atoms of the double or triple bond get the preference over the other substituents present in the parent chain.
If two or more similar alkyl groups are present in a compound, the prefixes di-, tri-, tetra- and so on are used to specify the number of times of the alkyl groups in the chain.
