The chemical compound's empirical formula is NS.
The chemical compound's molecular formula is N4S4.
<h3>What does a chemical empirical formula look like?</h3>
- The empirical formula of a compound that gives the proportion (ratios) of the elements in the complex but not the precise number or arrangement of atoms is known as an empirical formula.
- This would be the compound's element to whole number ratio with the lowest value.
<h3>What sort of empirical formula would that be?</h3>
- The chemical structure of glucose is C6H12O6. Every mole of carbon and oxygen is accompanied by two moles of hydrogen.
- Glucose has the empirical formula CH2O.
- Ribose has the chemical formula C5H10O5, which can be simplified to the empirical formula CH2O.
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the question you are looking for is
A compound containing only sulfur and nitrogen is 69.6% S by mass; the molar mass is 184 g/mol. What are the empirical and molecular formulas of the compound?
The element that will have the lowest electronegativity is an element with a small number of valence electrons and a large atomic radius.
Electronegativity of an element is the ability or power of that element in a molecule to attract electrons to its Valence electrons. The following are the properties of electronegativity:
- It increases across a period from left to right of the periodic table,
- It decreases down the periodic table groups
- Group 1 elements are the least (lowest) electronegative elements. These elements have the lowest valence electrons with a large atomic radius.
- Group 7 elements are the most electronegative elements.
Atomic radius of elements increase down a group because of a progressive increase in the number of shells occupied by electrons which increases the size. But it decreases across a period because electrons are accommodated within the same shell leading to greater attraction by the protons in the nucleus.
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Answer:
Model
Explanation:
A model of anything is something you make to represent it in it's physical world form
Answer:
23.0733 L
Explanation:
The mass of hydrogen peroxide present in 125 g of 50% of hydrogen peroxide solution:

Mass = 62.5 g
Molar mass of
= 34 g/mol
The formula for the calculation of moles is shown below:
Thus, moles are:

Consider the given reaction as:

2 moles of hydrogen peroxide decomposes to give 1 mole of oxygen gas.
Also,
1 mole of hydrogen peroxide decomposes to give 1/2 mole of oxygen gas.
So,
1.8382 moles of hydrogen peroxide decomposes to give ![\frac {1}{2}\times 1.8382 mole of oxygen gas. Moles of oxygen gas produced = 0.9191 molGiven: Pressure = 746 torr The conversion of P(torr) to P(atm) is shown below: [tex]P(torr)=\frac {1}{760}\times P(atm)](https://tex.z-dn.net/?f=%5Cfrac%20%7B1%7D%7B2%7D%5Ctimes%201.8382%20mole%20of%20oxygen%20gas.%20%3C%2Fp%3E%3Cp%3EMoles%20of%20oxygen%20gas%20produced%20%3D%200.9191%20mol%3C%2Fp%3E%3Cp%3EGiven%3A%20%3C%2Fp%3E%3Cp%3EPressure%20%3D%20746%20torr%0A%3C%2Fp%3E%3Cp%3EThe%20conversion%20of%20P%28torr%29%20to%20P%28atm%29%20is%20shown%20below%3A%0A%3C%2Fp%3E%3Cp%3E%5Btex%5DP%28torr%29%3D%5Cfrac%20%7B1%7D%7B760%7D%5Ctimes%20P%28atm%29)
So,
Pressure = 746 / 760 atm = 0.9816 atm
Temperature = 27 °C
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
So,
T₁ = (27 + 273.15) K = 300.15 K
Using ideal gas equation as:
PV=nRT
where,
P is the pressure
V is the volume
n is the number of moles
T is the temperature
R is Gas constant having value = 0.0821 L.atm/K.mol
Applying the equation as:
0.9816 atm × V = 0.9191 mol × 0.0821 L.atm/K.mol × 300.15 K
<u>⇒V = 23.0733 L</u>