Protons:
- Have a mass
- Positively charged
- Found inside the nucleus of an atom
Electrons:
- Have a mass. (9.10938188×10−31 kilograms), though this can sometimes be considered negligible due to how small that actually is. Barely factored into atomic mass
- Negatively charged
- Found outside the nucleus in the electron shell
Neutrons:
- Have a mass
- Neutral (no charge)
- Found inside the nucleus of an atom
Atom A:
- 1 proton
- 0 Neutrons
- 1 electron
- Atomic mass of 1
- Atomic number of 1
Atom B:
- 8 Protons
- 10 Neutrons
- 8 electrons
- Atomic mass of 18
- Atomic number of 8
Atomic mass includes the number of protons and neutrons in the nucleus. Atomic number is the number of protons, as this is what defines what type of element the atom is.
... to be called elements<span>. This lesson shows </span>you how to<span> predict the </span>numbers<span> of </span>neutrons, electrons, andprotons<span> of the isotopes they are likely to find in nature. (</span><span>cont.) ... What </span>kind<span> of </span>generalization can you make<span> about how the </span>number<span> of </span>protons<span> and </span>neutrons<span> are </span>related<span> to </span>each other<span> in the </span>elements<span>? Unit 1 • Investigation IV</span>
Answer:
b) Phosphorus acid
Explanation:
To distinguish the type of acid of phosphorus with the oxidation state of +3, we need to be familiar with the chemical formula of each of the compounds:
Orthophosphoric acid H₃PO₄
Phosphorus acid H₃PO₃
Metaphosphoric acid HPO₃
Phyrophosphoric acid H₄P₂O₇
Now that we know the formula of the given compounds, the algebraic sum of all the oxidation numbers of all atoms in a neutral compound is zero:
Only phosphorus acid yielded an oxidation state of +3 for phosphorus in the compound.
H₃PO₃:
we know the oxidation state of H = +1
O = -2
The oxidation state of P is unknown. We can express this as an equation:
3(+1) + P + 3(-2) = 0
3 + P -6 = 0
P-3 = 0
P = +3
Answer:
ρ = 1.08 g/cm³
Explanation:
Step 1: Given data
Mass of the substance (m): 21.112 g
Volume of the substance (V): 19.5 cm³
Step 2: Calculate the density of the substance
The density (ρ) of a substance is equal to its mass divided by its volume.
ρ = m / V
ρ = 21.112 g / 19.5 cm³
ρ = 1.08 g/cm³
The density of the substance is 1.08 g/cm³.
Organic chemistry is all about CARBON!!!!!!
