Answer:
Hydrogen
Explanation:
Just to provide some background, an element is a pure substance consisting of only one type of atom. An atom is the smallest constituent of matter. All elements are comprised of a single type of atom (e.g., gold is composed of gold atoms, helium of helium atoms, phosphorus phosphorus, and so on).
A molecule is a group of two or more atoms. They can be the same atom (homonuclear), such as or different atoms (heteronuclear).
Some examples of homonuclear molecules include:
Hydrogen (H2)
Nitrogen (N2)
Phosphorus (P4)
Some examples of heteronuclear molecules include:
Carbon dioxide (CO2)
Sulfuric acid (H2SO4)
Methane (CH4)
Answer:
C. The lowest-energy electron configuration of an atom has the maximum number of unpaired electrons, all of which have the same spin, in degenerate orbitals.
Explanation:
The Hund's rule is used to place the electrons in the orbitals is it states that:
1. Every orbital in a sublevel is singly occupied before any orbital is doubly occupied;
2. All of the electrons in singly occupied orbitals have the same spin.
So, the electrons first seek to fill the orbitals with the same energy (degenerate orbitals) before paring with electrons in a half-filled orbital. Orbitals doubly occupied have greater energy, so the lowest-energy electron configuration of an atom has the maximum number of unpaired electrons, and for the second statement, they have the same spin.
The other alternatives are correct, but they're not observed by the Hund's rule.
From the stoichiometry of the combustion reaction, we can see that 7.4 L of oxygen is consumed.
<h3>What is combustion?</h3>
Combustion is a reaction in which a substance is burnt in oxygen. The equation of the reaction is; C4H10O(l) + 6O2 (g) → 4CO2 (g) + 5H2O(l)
We can obtain the number of moles of CO2 from;
PV = nRT
n = 1.02 atm * 7.15 L/0.082 atm LK-1mol-1 * (125 + 273) K
n = 7.29 /32.6
n = 0.22 moles
If 6 moles of oxygen produces 4 moles of CO2
x moles of oxygen produces 0.22 moles of CO2
x = 0.33 moles
1 mole of oxygen occupies 22.4 L
0.33 moles of oxygen occupies 0.33 moles * 22.4 L/ 1 mole
= 7.4 L of oxygen
Learn more about stoichiometry: brainly.com/question/13110055
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Answer:
V₂ → 106.6 mL
Explanation:
We apply the Ideal Gases Law to solve the problem. For the two situations:
P . V = n . R . T
Moles are still the same so → P. V / R. T = n
As R is a constant, the formula to solve this is: P . V / T
P₁ . V₁ / T₁ = P₂ .V₂ / T₂ Let's replace data:
(1.20 atm . 73mL) / 112°C = (0.55 atm . V₂) / 75°C
((87.6 mL.atm) / 112°C) . 75°C = 0.55 atm . V₂
58.66 mL.atm = 0.55 atm . V₂
58.66 mL.atm / 0.55 atm = V₂ → 106.6 mL
Answer:
Element Symbol Mass Percent
Cuprum Cu 66.464%
Sulfur S 33.537%
Explanation:
I got this out of my module, sorry if it's wrong but i am pretty sure 97% this is correct!
