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deff fn [24]
3 years ago
8

Please help Please Help

Chemistry
1 answer:
Sveta_85 [38]3 years ago
8 0

Answer:

1) false

2) true

3) true

4) false

5) false. I think

6) false

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Calculate the number of moles of C2H6 in 6.29×1023 molecules of C2H6.
Kipish [7]

1.05moles

Explanation:

Given parameters:

Number of molecules of C₂H₆ = 6.29 x 10²³molecules

Unknown:

Number of moles = ?

Solution:

The mole is the amount of substances that contains Avogadro's number of particles i.e 6.02 x 10²³

 To find the number of moles:

  number of moles = \frac{number of particles}{Avogadro's number}

  number of moles = \frac{6.29 x 10^{23} }{6.02 x 10^{23} }

  number of moles = 1.05moles

Learn more:

moles brainly.com/question/1841136

#learnwithBrainly

7 0
3 years ago
When metals form ions, they tend to do so by what
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Losing electrons and forming positive ions
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A 720. cm^3 vessel contains a mixture of Ar and Xe. If the mass of the gas mixture is 2.966 g at 25.0°C and the pressure is 760.
sleet_krkn [62]

Explanation:

The given data is as follows.

      Pressure (P) = 760 torr = 1 atm

      Volume (V) = 720 cm^{3} = 0.720 L

     Temperature (T) = 25^{o}C = (25 + 273) K = 298 K

Using ideal gas equation, we will calculate the number of moles as follows.

                                PV = nRT

   Total atoms present (n) = \frac{PV}{RT}

                                          = 1 \times \frac{0.720 L}{0.0821 \times 298}

                                           = 0.0294 mol

Let us assume that there are x mol of Ar and y mol of Xe.

Hence, total number of moles will be as follows.

               x + y = 0.0294

Also,      40x + 131y = 2.966

             x = 0.0097 mol

              y = (0.0294 - 0.0097)

                = 0.0197 mol

Therefore, mole fraction will be calculated as follows.

Mol fraction of Xe = \frac{y}{(x+y)}

                               = \frac{0.0197}{0.0294}

                              = 0.67

Therefore, the mole fraction of Xe is 0.67.

6 0
3 years ago
In which of the following cases would sound reach each ear out of phase? 
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I hope this is the answer that you are looking for .

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