The empirical formula of the given compound is 
.
The correct option is B.
<h3>What is an empirical formula?</h3>
The empirical formula of a chemical compound is the simplest whole-number ratio of atoms contained in the substance.
Given,
1.0 g of S
1.5 g of O
To calculate the empirical formula, we will divide the masses of the elements by their atomic weight.
For sulfur
For oxygen
Now, divide the greater value of mole came by the smaller value
Thus, the empirical formula for the given compound is 1 for S and 3 for O
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brainly.com/question/11588623
Kansas City, Missouri, is often abbreviated as KC (abbreviations often refer to the metro area). It is officially nicknamed the City of Fountains. With over 200 fountains, the city claims to have the second most in the world, just behind Rome.
Ca²⁺ion has a lower ratio of charge to volume.
The charge density of an ion is defined as the ratio of the charge of an ion to its volume. Scandium ion has high charge density than calcium ions. The charge density of an ion is defined as the ratio of the charge of an ion to its volume.
Charge density also depends on the size of the ion and valence electrons. The volume of an ion increases with its size. It is inversely proportional to ion volume and directly proportional to charge magnitude. Scandium and calcium are IV periodic elements with atomic numbers of 21 and 20 respectively. Scandium loses three electrons and has a +3 charge and calcium is a divalent cation. Hence, the Scandium ion has high charge density than the calcium ion.
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Answer:
The enthalpy of the nitrogen-nitrogen bond in N2H4 is 162.6 kJ
Explanation:
For the reaction: N2H4(g)+H2(g)→2NH3(g), the enthalpy change of reaction is
ΔH rxn = 2 ΔHºf NH3 - ΔHºf N2H4
but we also know that the ΔH rxn is calculated by accounting the sum of number of bonds formed and bonds broken as follows:
ΔH rxn = 6H (N-H) + 4 (N-H) + 2H (H-H)
where H is the bond enthalpy .When bonds are broken H is positive, and negative when formed, in the product there are 6 N-H bonds , and in the reactants 4 N-H and 1 H-H bonds).
Consulting an appropiate reference handbook or table the following values are used:
ΔHºf (NH3) = -46 kJ/mol
ΔHºf (N2H4) = 95.94 kJ/mol
(The enthalpy of fomation of hydrogen in its standard state is zero)
H (N-H) = 391 kJ
H (H-H) = 432 kJ
H (N-N) = ?
So plugging our values:
ΔH rxn = 2mol ( -46.0 kJ/mol) - 1mol(95.4 kJ/mol) = -187.40 kJ
-187.40 kJ = 6(-391 kJ) + 4 (391 kJ) + 432 + H(N-N)
-187.40 kJ = -350 kJ + H(N-N)
H(N-N) = 162.6 kJ
Answer:
<u>36 times</u>
Explantaion:
The force exerted between two charged particles is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.
In the case of S and q: The distance of separation is 1 unit
F(Sq) α S*q/1²
In the case of S and p: the distance is 6 units
F(Sp) α S*p/6²
therefore:
F(Sq) = 36 * F(Sp)
