Answer:
B. Yes, because the mass of all the products of burning is equal to the mass
of the reactants (wood and oxygen gas).
Explanation:
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<span>H2C2O4(aq) + 2OH- --> C2O4^2- + 2H2O(l)</span>
SrSo4 = Sr(2+) + SO4(2-)
Let’s say that the initial concentration of SrSo4 was 1. ( or we have 1 mole of this reagent).
When The reaction occurs part of SrSo4is dissociated. And we get X mole Sr(2+) and So4(2-).
Ksp=[Sr(2+)]*[SO4(2-)]
X^2=3.2*10^-7
X=5.6*10^-4
Answer:
Ethanamine (also known as ethylamine)
Explanation:
The compound that is requested by the question is ethanamine. Its trivial name is ethylamine.
It is a compound that contained the ethyl moiety (CH3CH2-) as well as the amine moiety (-NH2).
Ethanamine has a structure that can easily be determined by the statements in the question.
The structure of ethanamine is shown in the image attached.
<h3>
Answer:</h3>
8CO₂
<h3>
Explanation:</h3>
We are given;
- Butane is a hydrocarbon in the homologous series known as alkane.
We are required to determine the other product produced in the combustion of butane apart from water.
- We know that the complete combustion of alkane yields carbon dioxide and water.
- Therefore, combustion of butane will yield carbon dioxide and water.
- The balanced equation for the complete combustion of butane will be;
2C₄H₁₀ + 13O₂ → 8CO₂ + 10H₂O
