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Veseljchak [2.6K]
2 years ago
14

a compound containing only sulfur and nitrogen is by mass; the molar mass is g/mol. what are the empirical and molecular formula

s of the compound?'
Chemistry
1 answer:
kolezko [41]2 years ago
3 0

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.

learn more about empirical formula here

brainly.com/question/1603500

#SPJ4

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?

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8 0
3 years ago
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Which of the following is NOT considered a branch of Chemistry?
mestny [16]

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Explanation:

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3 years ago
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A 155.0 g piece of copper at 128 oC is dropped into 250.0 g of water at 17.9 oC. (The specific heat of copper is 0.385 J/goC.) C
Mamont248 [21]

Answer:

T_{eq}=23.85^oC

Explanation:

Hello,

In this case, as the copper's heat loss is gained by the water, the following energetic relationship is:

\Delta H_{Cu}=-\Delta H_{H_2O}

Therefore the equilibrium temperature shows up as:

m_{Cu}Cp_{Cu}(T_{Cu}-T{eq}) = m_{H_2O}Cp_{H_2O}(T_{eq}-T_{H_2O})\\\\T_{eq}=\frac{m_{Cu}Cp_{Cu}T_{Cu}-m_{H_2O}Cp_{H_2O}T_{H_2O}}{m_{Cu}Cp_{Cu}-m_{H_2O}Cp_{H_2O}} \\

Thus, by knowing that water's heat capacity is 4.18J/g°C, one obtains:

T_{eq}=\frac{155.0g*0.385\frac{J}{g^oC}*128^oC+250.0g*4.18\frac{J}{g^oC}*17.9^oC}{155.0g*0.385\frac{J}{g^oC}+250.0g*4.18\frac{J}{g^oC}}=23.85^oC

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3 years ago
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