Balanced chemical reaction (dissociation): K₃PO₄(aq) → 3K⁺(aq) + PO₄³⁻(aq).
K₃PO₄ is potassium phosphate, <span>a water-soluble </span>ionic salt.
In water potassium phosphate, ionic compound, dissociates on positive potassium ion (cations) and negative phosphate ions (anions).
Potassium has positive charge (+1), compound has neutral charge.
If you type in the letters (uppercase and lowercase matters) into google it will give you the answers
Answer:
The initial rate of the reaction between substances P and Q was measured in a series of
experiments and the following rate equation was deduced.
Complete the table of data below for the reaction between P and Q
Explanation:
Given rate of the reaction is:
![rate= k[P]^{2} [Q]\\=>[Q]=\frac{rate}{k.[P]^{2} } \\and \\\\\\\ [P]=\sqrt{\frac{rate}{k.[Q]} }](https://tex.z-dn.net/?f=rate%3D%20k%5BP%5D%5E%7B2%7D%20%5BQ%5D%5C%5C%3D%3E%5BQ%5D%3D%5Cfrac%7Brate%7D%7Bk.%5BP%5D%5E%7B2%7D%20%7D%20%5C%5Cand%20%5C%5C%5C%5C%5C%5C%5C%20%5BP%5D%3D%5Csqrt%7B%5Cfrac%7Brate%7D%7Bk.%5BQ%5D%7D%20%7D)
Substitute the given values in this formulae to get the [P], [Q] and rate values.
From the first row,
the value of k can be calulated:
![k=\frac{rate}{[P]^{2}[Q] } \\ =\frac{4.8*10^-3}{(0.2)^{2} 2. (0.30)} \\ =0.4](https://tex.z-dn.net/?f=k%3D%5Cfrac%7Brate%7D%7B%5BP%5D%5E%7B2%7D%5BQ%5D%20%7D%20%5C%5C%20%20%3D%5Cfrac%7B4.8%2A10%5E-3%7D%7B%280.2%29%5E%7B2%7D%202.%20%280.30%29%7D%20%5C%5C%20%3D0.4)
Second row:
2. Rate value:

3.Third row:
![[Q]=\frac{rate}{k.[P]^{2} } \\ =9.6*10^-3 / (0.4 *(0.40)^{2} \\ =0.15mol.dm^{-3}](https://tex.z-dn.net/?f=%5BQ%5D%3D%5Cfrac%7Brate%7D%7Bk.%5BP%5D%5E%7B2%7D%20%7D%20%5C%5C%20%20%20%20%20%3D9.6%2A10%5E-3%20%2F%20%280.4%20%2A%280.40%29%5E%7B2%7D%20%5C%5C%20%20%20%20%3D0.15mol.dm%5E%7B-3%7D)
4. Fourth row:
![[P]=\sqrt{\frac{rate}{k.[Q]} }\\=>[P]=\sqrt{\frac{19.2*10^-3}{0.60*0.4} } \\=>[P]=0.283mol.dm^{-3}](https://tex.z-dn.net/?f=%5BP%5D%3D%5Csqrt%7B%5Cfrac%7Brate%7D%7Bk.%5BQ%5D%7D%20%7D%5C%5C%3D%3E%5BP%5D%3D%5Csqrt%7B%5Cfrac%7B19.2%2A10%5E-3%7D%7B0.60%2A0.4%7D%20%7D%20%5C%5C%3D%3E%5BP%5D%3D0.283mol.dm%5E%7B-3%7D)
Answer:
1. 4-ethyl-1-heptene
2. 6-ethyl-2-octene
3. 1-butyne
Explanation:
The compounds are named according to IUPAC rules.
Compound 1:
- Identify the longest carbon chain. This chain is called the parent chain.
- Identify all of the substituents (groups appending from the parent chain).
- The parent chain is numbered so that the multiple bonds have the lowest numbers (double has the priority over alkyl substituents).
- The longest chain contains 7 carbon atoms, so taken the name hept.
- The double bond between C1 and C2, so take no. 1 and add the suffix ene to hept "1-heptene".
- The ethyl group is the alkyl substituent on position 4.
- So the name is 4-ethyl-1-heptene.
Compound 2:
- Identify the longest carbon chain. This chain is called the parent chain.
- Identify all of the substituents (groups appending from the parent chain).
- The parent chain is numbered so that the multiple bonds have the lowest numbers (double has the priority over alkyl substituents).
- The longest chain contains 8 carbon atoms, so taken the name oct.
- The double bond between C2 and C3, so take no. 2 and add the suffix ene to oct "2-octene".
- The ethyl group is the alkyl substituent on position 6.
- So the name is 6-ethyl-2-octene.
Compound 3:
- Identify the longest carbon chain. This chain is called the parent chain.
- Identify all of the substituents (groups appending from the parent chain), there is no substituents.
- The parent chain is numbered so that the multiple bonds have the lowest numbers (Triple bond here take the lowest number).
- The longest chain contains 4 carbon atoms, so taken the name but.
- The triple bond between C1 and C2, so take no. 1 and add the suffix yne to but "1-butyne".