Answer:
This question appears incomplete
Explanation:
This question appears incomplete because the data provided only makes it possible to calculate the certainty of the acetic acid content per total volume of the vinegar. Thus, the 4% means for every 100 mL of the vinegar, there is 4 mL of acetic acid present. To calculate the volume of acetic acid in any other volume of vinegar, the formula will be
volume of acetic acid = 4/100 × total volume of vinegar
Polyatomic ions: 
, 
, 
, 
, 
, and 
Monatomic ions: 
, 
, and 
<h3>Monoatomic vs Polyatomic Ions</h3>
In chemistry, monoatomic ions are ions that consist of only a single type of atom. They are usually positive or negatively charged and are otherwise known as simple ions. Examples include , , and
Polyatomic ions, on the other hand, are ions that consist of more than one atom, unlike monoatomic ions. The two or more atoms are covalently bonded and the entire structure behaves like a single chemical entity in reactions. Polyatomic ions are otherwise known as molecular ions.
Examples of polyatomic ions are , , , , , and
Thus, from the diagram:
- Polyatomic ions: , , , , , and
More on ions can be found here: brainly.com/question/14982375
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In order for carbon to be stable and have 8 electrons, it must make 4 total covalent bonds.
In prefer for oxygen to be stable and have 8 electrons, it must make 2 covalent bonds.
So, we can deduce that CO2 looks like this:
O=C=O
This molecule has two double bonds.
Pssst...Can I get a brainliest?
Proton plus neutron is the correct answer. Protons and neutrons have a mass of 1 and electrons have a mass of 0. So in order to find the mass of an atom you need to add the number of protons and the number of neutrons.
Answer:
148.04 kJ/mol
Explanation:
Let's consider the following thermochemical equation.
NO(g) + 1/2 O₂(g) → NO₂(g) ΔH°rxn = -114.14 kJ/mol
We can find the standard enthalpy of formation (ΔH°f) of NO(g) using the following expression.
ΔH°rxn = 1 mol × ΔH°f(NO₂(g)) - 1 mol × ΔH°f(NO(g)) - 1/2 mol × ΔH°f(O₂(g))
ΔH°f(NO(g)) = 1 mol × ΔH°f(NO₂(g)) - ΔH°rxn - 1/2 mol × ΔH°f(O₂(g)) / 1 mol
ΔH°f(NO(g)) = 1 mol × 33.90 kJ/mol - (-114.14 kJ) - 1/2 mol × 0 kJ/mol / 1 mol
ΔH°f(NO(g)) = 148.04 kJ/mol
