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Kazeer [188]
3 years ago
6

2. Use the following combined gas law formula and compute the volume that the butane sample will occupy at STP. (HINT: Convert b

oth temperatures to Kelvin.) Pbutane x Voriginal/Troom = Pstandard x Vfinal/Tstandard
DATA TABLE
Initial mass of butane lighter-21.8 g
Final mass of butane lighter-21.6 g
Volume of butane released-50 mL
Temperature of water-22 oC
Atmospheric pressure-763mm Hg
Chemistry
2 answers:
enyata [817]3 years ago
7 0

Answer: STP stands for standard temperature and pressure and refers to conditions of 273K (0 degree centigrade) at 760mmHg.

Explanation: Temperature, T1= 273 + 22= 295K; Pressure, P1= 763mmHg; Volume, V1= 50mL; T2= 273K; P2= 760mmHg; V2= ?

P1*V1/ T1 = P2*V2/ T2

V2= P1*V1*T2/P2*T1

V2= 763*50*273/760*295

∴ V2= 10414950/224200 = 46.45mL

MrRissso [65]3 years ago
3 0
When we put Boyle's law<span>, Charles' </span>law<span>, and Gay-Lussac's </span>law<span> together, we come up with the </span>combined gas law<span>, which shows that: Pressure is inversely proportional to volume, or higher volume equals lower pressure. It is expressed as:

</span><span>Pbutane x Voriginal/Troom = Pstandard x Vfinal/Tstandard 

It seems that the given values is not complete the pressure of butane is not given and also we cannot calculate for it since there is no more than two data about the water.</span>
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what are we/what has someone did to prevent/lessen the negative side effects of aluminium extraction?
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3 years ago
A chemical reaction in which an uncombined element replaces an element that is part of a compound is called a
san4es73 [151]

Answer:

A chemical reaction in which an uncombined element replaces an element that is part of a compound is called a simple substitution reaction or simple displacement reaction.

Explanation:

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<u><em>A chemical reaction in which an uncombined element replaces an element that is part of a compound is called a simple substitution reaction or simple displacement reaction.</em></u>

3 0
3 years ago
A sample of 211 g of iron (III) bromide is reacted with
Alisiya [41]

FeBr₃ ⇒ limiting reactant

mol NaBr = 1.428

<h3>Further explanation</h3>

Reaction

2FeBr₃ + 3Na₂S → Fe₂S₃ + 6NaBr

Limiting reactant⇒ smaller ratio (mol divide by coefficient reaction)

  • FeBr₃

211 g of Iron (III) bromide(MW=295,56 g/mol), so mol FeBr₃ :

\tt n=\dfrac{mass}{MW}\\\\n=\dfrac{211}{295,56}\\\\n=0.714

  • Na₂S

186 g of Sodium sulfide(MW=78,0452 g/mol), so mol Na₂S :

\tt n=\dfrac{186}{78.0452}=2.38

Coefficient ratio from the equation FeBr₃ :  Na₂S = 2 : 3, so mol ratio :

\tt FeBr_3\div Na_2S=\dfrac{0.714}{2}\div \dfrac{2.38}{3}=0.357\div 0.793

So  FeBr₃ as a limiting reactant(smaller ratio)

mol NaBr based on limiting reactant (FeBr₃) :

\tt \dfrac{6}{3}\times 0.714=1.428

6 0
3 years ago
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trapecia [35]

Answer:

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