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Agata [3.3K]
2 years ago
13

A sample of gas has a volume of 215 cm3 at 23.5 degrees Celsius and 84.6kPa what volume will the gas occupy at stp

Chemistry
2 answers:
miss Akunina [59]2 years ago
5 0

The volume of the gas that occupy at STP is 165. 28 cm^3

calculation

by use of combined gas law that is P1V1/T1=P2V2/T2, where

P1=84.6 kpa

T1=23.5 +273=296.5 K

V1=215 cm^3

At STP T= 273 K and P= 101.325 Kpa

therefore p2 = 101.325 Kpa and T2 = 272 K V2=?

by making V2 the subject of the formula V2 =T2P1V1/P2T1

V2 = 273 K x 84.6 Kpa x 215 cm^3/ 101,.325 Kpa x296.5 K =165.28 cm^3

Fynjy0 [20]2 years ago
4 0

The volume of the gas that occupy at STP is 165. 28 cm^3

calculation

by use of combined gas law that is P1V1/T1=P2V2/T2, where

P1=84.6 kpa

T1=23.5 +273=296.5 K

V1=215 cm^3

At STP T= 273 K and P= 101.325 Kpa

therefore p2 = 101.325 Kpa and T2 = 272 K V2=?

by making V2 the subject of the formula V2 =T2P1V1/P2T1

V2 = 273 K x 84.6 Kpa x 215 cm^3/ 101,.325 Kpa x296.5 K =165.28 cm^3

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Explanation:

From the aqueous acid-base equilibrium we know that

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pOH = 14 - pH

In this problem, the solution has a pH of 3, so:

pOH = 14 - 3 = 11

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How many moles of sodium carbonate are contained by 57.3g of sodium carbonate
Lady_Fox [76]

Answer:

\boxed {\boxed {\sf 0.541 \  mol \ Na_2CO_3}}

Explanation:

We are asked to find how many moles of sodium carbonate are in 57.3 grams of the substance.

Carbonate is CO₃ and has an oxidation number of -2. Sodium is Na and has an oxidation number of +1. There must be 2 moles of sodium so the charge of the sodium balances the charge of the carbonate. The formula is Na₂CO₃.

We will convert grams to moles using the molar mass or the mass of 1 mole of a substance. They are found on the Periodic Table as the atomic masses, but the units are grams per mole instead of atomic mass units. Look up the molar masses of the individual elements.

  • Na:  22.9897693 g/mol
  • C: 12.011 g/mol
  • O: 15.999 g/mol

Remember the formula contains subscripts. There are multiple moles of some elements in 1 mole of the compound. We multiply the element's molar mass by the subscript after it, then add everything together.

  • Na₂ = 22.9897693 * 2= 45.9795386 g/mol
  • O₃ = 15.999 * 3= 47.997 g/mol
  • Na₂CO₃= 45.9795386 + 12.011 + 47.997 =105.9875386 g/mol

We will convert using dimensional analysis. Set up a ratio using the molar mass.

\frac {105.9875386  \ g \ Na_2CO_3}{1 \ mol \ Na_2CO_3}

We are converting 57.3 grams to moles, so we multiply by this value.

57.3 \ g \ Na_2CO_3} *\frac {105.9875386  \ g \ Na_2CO_3}{1 \ mol \ Na_2CO_3}

Flip the ratio so the units of grams of sodium carbonate cancel.

57.3 \ g \ Na_2CO_3} *\frac {1 \ mol \ Na_2CO_3}{105.9875386  \ g \ Na_2CO_3}

57.3 } *\frac {1 \ mol \ Na_2CO_3}{105.9875386 }

\frac {57.3 }{105.9875386 } \ mol \ Na_2CO_3

0.5406295944 \ mol \ Na_2CO_3

The original measurement of moles has 3 significant figures, so our answer must have the same. For the number we found that is the thousandth place. The 6 in the ten-thousandth place to the right tells us to round the 0 up to a 1.

0.541 \  mol \ Na_2CO_3

There are approximately <u>0.541 moles of sodium carbonate</u> in 57.3 grams.

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A scientific reasoning can be defined as the justification which connects the evidence with that of the claim and it rules out all those claims that are not supported by any evidence. It explains the fact that how evidence can support a claim and scientifically acceptable conclusion can be drawn. An argument is always based on one or more claims which can either be true or evidentially proved.

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