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3 years ago

Which of these expressions are correct variations of the Combined Gas Law? P1V1T2 = P2V2T1 both

Chemistry

2 answers:

Anastasy [175]3 years ago

4 0

Answer:

Explanation:

The combined gas law is also known as general gas law and it is the reduction of the ideal gas equation. In the combined gas law, we assume that n= 1

From PV=nRT

$\frac{PV}{T}$ = R (constant) if n= 1

∴ $\frac{P_{1} V_{1} }{T_{1} }$ = $\frac{P_{2} V_{2} }{T_{2} }$ = $\frac{P_{n} V_{n} }{T_{n} }$

Sidana [21]3 years ago

4 0

Answer: Both!

Explanation:

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3 years ago

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Which of the following is not a characteristic of a solution?What is the concentration of a solution?

vladimir2022 [97]

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2 years ago

PF3Br2 is a nonpolar molecule. Based on this information, determine the F−P−F bond angle, the Br−P−Br bond angle, and the F−P−Br

inna [77]

Answer:

120,180,90.

Explanation:

PF3Br2 has a trigonal bypiramidal geometry. The three atoms of F will be arranged around the P atom in the same plane and each F-P-F bond will have an angle of 120 degrees. There will be only one Br atom above and beneath the P atom so the bond angle for Br-P-Br will be 180 degrees. Finally, the F−P−Br bond angle will be at 90 degrees.

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3 years ago

Peppered moths come in two colors, black and white. What did Kettlewell show, with regard to peppered moth populations and tree

Vera_Pavlovna [14]

A is the answer because the trees turned blackish from the coal so moths that were black could camouflage on those trees easier.

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3 years ago

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Consider the following reaction of the commercial production of SO3 using sulfur dioxide and oxygen: SO2(g) + O2(g) → SO3(g)

kipiarov [429]

Answer:

$m_{SO_3}=18.93gSO_3$

$V_{SO_3}=5.3LSO_3$

Explanation:

Hello,

STP conditions are P=1 atm and T=273.15 K, thus, the reacting moles are:

$n_{SO_2}=\frac{5.3L*1atm}{0.082\frac{atm*L}{mol*K}*273.15K}=0.2366molSO_2\\n_{O_2}=\frac{4.7L*1atm}{0.082\frac{atm*L}{mol*K}*273.15K}=0.2098molO_2$

Now, the balanced chemical reaction turns out into:

$2SO_2(g) + O_2(g) --> 2SO_3(g)$

Thus, the exact moles of oxygen that completely react with 0.2366 moles of sulfur dioxide are (limiting reagent identification):

$0.2366molSO_2*\frac{1molO_2}{2molSO_2}=0.1182molO_2$

Since 0.2098 moles of oxygen are available, we stipulate the oxygen is in excess and the sulfur dioxide is the limiting reagent. In such a way, the yielded grams of sulfur trioxide turn out into:

$m_{SO_3}=0.2366molSO_2*\frac{2molSO_3}{2molSO_2}*\frac{80gSO_3}{1molSO_3} \\m_{SO_3}=18.93gSO_3$

By using the ideal gas equation, one computes the volume as:

$V_{SO_3}=\frac{mRT}{MP}=\frac{18.93g*0.082\frac{atm*L}{mol*K} *273.15K}{80g/mol*1atm}\\V_{SO_3}=5.3LSO_3$

It has sense for volume since the mole ratio is 2/2 between sulfur dioxide and sulfur trioxide.

Best regards.

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3 years ago