Answer:
9 liters of CO₂ are produced by this combustion
Explanation:
In order to determine the volume of produced CO₂, we start with the reaction:
C₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(g)
We need, O₂ density to find out the mass, that has reacted.
δ O₂ = O₂ mass / O₂ volume → δ O₂ . O₂ volume = O₂ mass
δ O₂ = 1.429 g /dm₃ (1dm³ = 1L) 1.429 g/L . 15L = 21.4 g of O₂
We convert the mass to moles: 21.4 g . 1mol / 32 g = 0.670 moles
By stoichiometry, 5 moles of O₂ can produce 3 moles of CO₂
Then, 0.670 moles of O₂ will produce (0.670 . 3) /5 = 0.402 moles of dioxide.
We apply Ideal Gases Law for STP, to find out the CO₂ volume
V = (n . R . T) / P → V = (0.402 mol . 0.082 . 273K) / 1 atm = 8.99 L ≅ 9 L
<u>Answer:</u> Antimony ion is the smallest and yttrium ion is the largest.
<u>Explanation:</u>
Atomic radius of an atom is defined as the total distance from the nucleus to the outermost shell of the atom.
An ion is formed when a neutral atom looses or gains electrons.
- When an atom looses electrons, it results in the formation of positive ion known as cation.
- When an atom gains electrons, it results in the formation of negative ion known as anion.
As moving from left to right in a period, more and more electrons get added up in the same shell and the attraction between the last electron and nucleus increases, which results in the shrinkage of size of an atom. Hence, the size of an atom decreases.
The size of the cation is small then their neutral atom because it has less number of electrons while its nuclear charge remains the same. Thus, the nucleus attracts the electron more towards itself and leads to the decrease in size.
We are given three cations: 
Yttrium lies in Period 5, group 3 of the periodic table.
Niobium lie in Period 5, group 5 of the periodic table.
Antimony lies in Period 5, group 15 of the periodic table.
So, the order of atomic radii in increasing order follows:

Hence, antimony ion is the smallest and yttrium ion is the largest.
Answer:
Europeans won land from the Ottoman Empire after WWI
Explanation:
Mass percentages of the the elements in the compound is the only additional information needed to calculate the empirical formula of a compound if the masses of each element of the compound are provided.
<h3>What is Empirical Formula ? </h3>
The Empirical formula is the simplest whole number ratio of atoms present in given compound.
<h3>What is Molecular formula of Compound ?</h3>
Molecular Formula = Empirical Formula × n
n = 
Thus from the above conclusion we can say that Mass percentages of the the elements in the compound is the only additional information needed to calculate the empirical formula of a compound if the masses of each element of the compound are provided.
Learn more about Empirical Formula here: brainly.com/question/1603500
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