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astraxan [27]
3 years ago
9

A syringe contains 0.65 moles of he gas that occupy 900.0 ml. what volume (in l) of gas will the syringe hold if 0.35 moles of n

e is added?
Chemistry
2 answers:
V125BC [204]3 years ago
8 0

From the explanation, we are given that:

  • The mole of gas n1 = 0.65 mol
  • The volume it occupied V1 = 900.0 ml
  • Mole of Neon = 0.35 ml
<h2>Further Explanation</h2>

To solve this question, we have to consider the Avogadro’s law, which state that "equal volumes of all gases, at the same temperature and pressure, have the same number of molecules”

Therefore, we have to calculate the total number of moles of the gases.

(0.65 + 0.35) mol

1.00 mol

Also we know that the volume it occupied V2 = unknown

According to the Avogadro’s law, the equation will be

0.65 mol x 900.0ml = 1.00ml x V2

585ml = 1.00 ml x v2

V2 = 585ml

To convert 585ml to litre, then we have to divide the derived value by 1000

Therefore we have

\frac{ 585ml}{1000}

0.6L

Therefore, the volume of gas in liters that syringe will hold if 0.35 moles of neon is added is 0.6L

LEARN MORE:

  • A syringe contains 0.65 moles of he gas that occupy 900.0 ml. what volume (in l) of gas will the syringe hold if 0.35 moles of ne is added?  brainly.com/question/6860167
  • A syringe contains 0.65 moles of he gas that occupy 750.0 ml. brainly.com/question/6442444

KEYWORDS:

  • moles
  • syringe
  • convert
  • litres
  • avogradro's law
Grace [21]3 years ago
5 0
<span>PV=nRT= a universal constant For any condition P1V1/n1T1=R and P2V2/n2T2=R i.e P1V1/n1T1=P2V2/n2T2 Becomes V1/n1=V2/n2 rearranging and solving V2=V1X(n2/n1)= 750 mLx((0.65+0.35)/(0.65))=1200ml=1.2L...2 sig figs</span>
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Explanation:

<u>Step 1: </u>Data given

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For 0.00608 moles of NaNO2 we'll get 0.00608/2 = 0.00304 moles

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Where (E_{bond})_{i}  and (E_{bond})_{j} represents average bond energy in breaking "i" th bond and forming "j" th bond respectively.n_{i} and n_{j} are number of moles of bond break and form respectively.

In this reaction, one mol of C=C, four moles of C-H and one mol of F-F bonds are broken. One mol of C-C bond, four moles of C-H bonds and two moles of C-F bonds are formed

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