What is the mass of 6.3 moles of Sulfur
Answers
Answer:
201.6 grams is the mass of 6.3 moles of sulfur.
Explanation:
Moles of sulfur ,n= 6.3 moles
Atomic mass of sulfur = M = 32 g/mol
Mass of the sulfur = m = ?
201.6 grams is the mass of 6.3 moles of sulfur.
The mass of 6.3 moles of Sulfur is 32×6.3=201.6 grams
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Is this true of false
PLEASE HELPP NO LINKSS
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Answer:
That is true
Explanation:
:D
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Answers
Final answer:
The correct IUPAC chemical names for Copper chloride, Copper bromide, Copper iodide, and Copper hydride are Copper(II) chloride, Copper(II) bromide, Copper(II) iodide, and Copper hydride respectively.
Explanation:
In accordance with International Union of Pure and Applied Chemistry (IUPAC) naming conventions, the chemical names for the given formulas are as follows:
- Copper chloride is referred to as copper(II) chloride in the IUPAC naming system as it contains the copper(II) ion, which has a +2 charge.
- Copper bromide is named copper(II) bromide, reflecting the +2 charge of the copper(II) ion.
- Copper iodide is termed copper(II) iodide to denote the +2 charge of the copper(II) ion.
- Copper hydride is an exception as it is simply named copper hydride in IUPAC nomenclature.
Learn more about IUPAC Naming here:
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2H2 + O2 ⇒⇒ 2H2O
What volume of oxygen at STP is required for the complete combustion of 100.50 mL of C2H2?
201 mL
201.00 mL
251 mL
251.25 mL
Answers
Answer:
251.25 mL of O₂
Solution:
The balance chemical equation is as follow,
2 C₂H₂ + 5 O₂ → 4 CO₂ + 2 H₂O
As we know if the gas is acting ideally then 1 mole of any gas at standard temperature and pressure it will occupy exactly 22.4 L or 22400 mL of volume.
Keeping this in mind according to equation,
44800 mL (2 mol) of C₂H₂ required = 112000 mL (5 mol) of O₂
So,
100.50 mL of C₂H₂ will require = X mL of O₂
Solving for X,
X = (100.5 mL × 112000 mL) ÷ 44800 mL
X = 251.25 mL of O₂
The balanced reaction would be:
C2H2 + 5/2O2 = 2CO2 + H2O
We are given the amount of acetylene in the reaction. This will be the starting point of our calculation. We use the ideal gas equation to find for the number of moles.
n = PV / RT = 1.00(.1005 L) / (0.08206 atm L/mol K ) 273.15 K
n= 4.4837 x 10^-3 mol C2H2
4.4837 x 10^-3 mol C2H2 (5/2 mol O2/ 1 mol C2H2) = 0.0112 mol O2
V = nRT/P = 0.0112 mol O2 x 273.15 K x 0.08206 atm L/mol K / 1 atm
V=0.25125 L or 251.25 mL