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igomit [66]
3 years ago
14

A sample of an element has a volume of 78.0 mL and a density of 1.85 g/mL. What is the mass in grams of the sample?

Chemistry
1 answer:
abruzzese [7]3 years ago
7 0

Hey there!:

density = 1.85 g/mL

volume = 78.0 mL

Mass = ??

therefore:

D = m / V

1.85 = m / 78.0

m = 1.85 x 78.0

m = 144.3 grams

Hope this helps!

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What are the characteristics of an atom or molecule when in the gas phase
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A 1.25 g gas sample occupies 663 ml at 25∘ c and 1.00 atm. what is the molar mass of the gas?
lakkis [162]

Using ideal gas equation,  

P\times V=n\times R\times T

Here,  

P denotes pressure  

V denotes volume  

n denotes number of moles of gas  

R denotes gas constant  

T denotes temperature  

The values at STP will be:  

P=1 atm  

T=25 C+273 K =298.15K

V=663 ml=0.663L

R=0.0821 atm L mol ⁻¹

Mass of gas given=1.25 g g

Molar mass of gas given=?

Number of moles of gas, n= \frac{Given mass of the gas}{Molar mass of the gas}

Number of moles of gas, n= \frac{1.25}{Molar mass of the gas}

Putting all the values in the above equation,

1\times 0.663=\frac{1.25}{Molar mass of the gas}\times 0.0821\times 298.15

Molar mass of the gas=46.15

3 0
3 years ago
The rate constant of a certain reaction is known to obey the Arrhenius equation, and to have an activation energy Ea = 71.0kJ/mo
aleksandrvk [35]

Explanation:

Relation between temperature and activation energy according to Arrhenius equation is as follows.

            k = A exp^{\frac{-E_{a}}{RT}}

where,   k = rate constant

              A = pre-exponential factor

           E_{a} = activation energy

             R = gas constant

              T = temperature in kelvin

Also,  

         ln (\frac{k_{2}}{k_{1}}) = (\frac{-E_{a}}{R}) \times (\frac{1}{T_{2}} - \frac{1}{T_{2}})

      T_{1} = 244^{o}C = (244 + 273) K = 517.15 K

      T_{2} = 324^{o}C = (597.15 + 273) K = 597.15 K

     k_{1} = 6.7 M^{-1} s^{-1},     k_{2} = ?

         R = 8.314 J/mol K

      E_{a} = 71.0 kJ/mol = 71000 J/mol

Putting the given values into the above formula as follows.

   ln (\frac{k_{2}}{6.7}) = (\frac{-71000}{8.314} \times (\frac{1}{597.15} - \frac{1}{517.15})

                   = 2.2123

        \frac{k_{2}}{6.7} = exp(2.2123)

                    = 9.1364

                k_{2} = 61 M^{-1}s^{-1}

Thus, we can conclude that rate constant of this reaction is 61 M^{-1}s^{-1}.

3 0
4 years ago
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