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nlexa [21]
3 years ago
5

fictitious element Jz has two common isotopes. Three-fifth of the isotopes have a mass of 329.1 amu, while the rest have a mass

of 334.5 amu. Given this information what is the average atomic mass (in amu) of Jz? Report the answer to 4 significant figures
Chemistry
1 answer:
Alla [95]3 years ago
7 0

The average atomic mass of Jz is 331.3 u.

The average atomic mass of Jz is the <em>weighted average</em> of the atomic masses of its isotopes.

We multiply the atomic mass of each isotope by a number representing its <em>relative importance</em> (i.e., its fractional abundance).  

Thus,

⅗ × 329.1 u = 197.46  u

⅖ × 334.5 u = <u>133.80 u</u>

      TOTAL = 331.3   u

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The vapor pressure of pure acetone (C3H6O) is 216 torr
svetoff [14.1K]

The mole fraction of acetone (C₃H₆O) is 0.333

<h3>Data obtained from the question </h3>
  • Mole of C₃H₆O = 0.1 mole
  • Mole of CHCl₃ = 0.2 mole
  • Mole fraction of C₃H₆O =?

<h3>How to determine the mole fraction </h3>

Mole fraction of a substance can be obtained by using the following formula:

Mole fraction = mole / total mole

With the above formula, we can obtain the mole fraction of C₃H₆O as follow:

  • Mole of C₃H₆O = 0.1 mole
  • Mole of CHCl₃ = 0.2 mole
  • Total mole = 0.1 + 0.2 = 0.3 mole
  • Mole fraction of C₃H₆O =?

Mole fraction = mole / total mole

Mole fraction of C₃H₆O = 0.1 / 0.3

Mole fraction of C₃H₆O = 0.333

Learn more about mole fraction:

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6 0
2 years ago
0.250 moles of NaCl is dissolved in 0.850 L of solution. What is the molar concentration?
Nuetrik [128]

Considering the definition of molarity, the molar concentration is 0.294 \frac{moles}{liter}.

Molarity reflects the concentration of a solution indicating the number of moles of solute that are dissolved in a given volume.

The molarity of a solution is calculated by dividing the moles of the solute by the volume of the solution:

molarity=\frac{amount of moles of solute}{volume}

Molarity is expressed in units \frac{moles}{liter}.

In this case, you know:

  • amount of moles of solute= 0.250 moles
  • volume= 0.850 L

Replacing in the definition of molarity:

molarity=\frac{0.250 moles}{0.850 L}

Solving:

molarity= 0.294 \frac{moles}{liter}

Finally, the molar concentration is 0.294 \frac{moles}{liter}.

Learn more about molarity with this example: brainly.com/question/15406534?referrer=searchResults

6 0
2 years ago
How many grams of lead(II) iodide are produced from 6.000 moles of NaI according to the balanced equation: Pb(NO3)2 + 2 NaI à 2
marta [7]

Answer:

mass PbI₂ formed = 1383 grams

Explanation:

Pb(NO₃)₂ + 2NaI => 2NaNO₃ + PbI₂(s)

6 mol NaI =>  1/2(6 mol) PbI₂ = 3 mol PbI₂ x 461.01 g/mol = 1383.03 grams ≅ 1383 grams (4 sig. figs.)

7 0
2 years ago
A sample of gaseous arsine (AsH3) in a 460 mL flask at 332 Torr and 223 K, is heated to 437 K, at which temperature arsine decom
Taya2010 [7]

Answer:

28/95 = 29.,5 % of Arsine decomposed

Explanation:A sample of gaseous arsine (AsH3) in a 460 mL flask at 332 Torr and 223 K, is heated to 437 K, at which temperature arsine decom- poses to solid arsenic and hydrogen gas. The flask is then cooled to 273 K, at which tem- perature the pressure in the flask is 488 Torr. What percentage of arsine molecules have de- composed?

Answer in units of %.

initial pressure 332 Torr initial volume 0.46 L initial temperature 223K

final pressure 488 Torr final volume 0.46 L final 273 K

Torr is 1/760 atm 332 torr = 0.437 atm 488 Torr =0.642 atm

PV = nRT so n=RT/PV

INITIAL n= 0.082 X 223/(0.437)(0.46) = 91 moles

final n= 0.082 X 273 / (.437)(488) = 105 moles

2AsH3----------> 2As + 3H2

x moles of Arsine decomposed to make 1.5 moles of H2

the final number of moles was

(91 -X)+ 1.5 X = 105 moles

91 + 0.5 X = 105

0.5 X = 14

X =28

CHECK

if 28 moles of Arsine , then the container would have

91 --28 + 1.5(28) = 91 +14 =105 check

so 28/95 = 29.,5 % of Arsine decomposed

Your answer

(quit)

polyalchemVirtuoso

Answer:

Explanation:

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