Answer:
It partially ionizes
Explanation:
Higher tier) Weak acids only partially ionise in water. Only a small fraction of their molecules break into hydrogen ions when added to water.
Answer:
12CO2 (g) + 12H2O (l) ⇒ C12H24O12 (s) + 12O2
Explanation:
Start by comparing the moles of carbons on the left to number on the right. The number of moles on both side of the arrow should be the same.
Answer:
B. Ionic Compound
Explanation:
An ionic compound is that compound which contains a positively charged ion called CATION and a negatively charged ion called ANION. The cation loses or transfers electrons to the anion, hence, making the former (cation) positive and the latter (anion) negative.
A polyatomic ion is an ion that contains more than one type of atom e.g OH-, NO3²-, CO3²- etc. A polyatomic ion usually has an overall charge formed from the charges of the individual atoms that makes it up. For example, in OH-, the overall charge is -1.
Since a polyatomic ion can have an overall positive or negative charge, it must enter a reaction with another ion that complements it i.e. a negative polyatomic ion will react with a positive ion to neutralize its charge. Hence, this forms an IONIC COMPOUND. This is why most compounds with polyatomic ions are IONIC COMPOUNDS.
For example, CaCO3 is an ionic compound formed when Ca²+ (cation) reacts with the polyatomic anion: CO3²-
Answer:
Kp = 1.41 x 10⁻⁶
Explanation:
We have the chemical equation:
2 A(g) + 3 B(g)⇌ C(g)
In which A and B are the reactants and C is the product. We calculate first the change in the number of moles of gas (Δn or dn):
dn= (sum moles products - sum moles reactants)
= (moles C - (moles A + moles B))
= (1 - (2+3))
= 1 - 5
= -4
We have also the following data:
Kc = 63.2
T= 81∘C + 273 = 354 K
R = 0.082 L.atm/K.mol (it is a constant)
Thus, we introduce the data in the mathematical expression for the relation between Kp and Kc:
= (0.082 L.atm/K.mol x 354 K)⁻⁴ = 1.41 x 10⁻⁶
Answer:
Dissolve 226 g of KCl in enough water to make 1.5 L of solution
Explanation:
1. Calculate the moles of KCl needed
2. Calculate the mass of KCl
3. Prepare the solution
- Measure out 224 g of KCl.
- Dissolve the KCl in a few hundred millilitres of distilled water.
- Add enough water to make 1.5 L of solution.
Mix thoroughly to get a uniform solution.
