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Inessa05 [86]
3 years ago
9

The equation for density is frequently written as m/v which is mass / volume. Which of the following statements regarding base a

nd derived units is true for density. Options:
A. Density is a base unit that is represented by mass/volume.
B. Density is a value represented by two base units mass/volume.
C. Density is a value represented by two units mass/volume. Mass is a base unit and volume is a derived unit.
D. Density is a value represented by two derived units mass/volume.
Chemistry
1 answer:
ruslelena [56]3 years ago
6 0

The answer for this question is C.

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3. Theoretically how many grams of magnesium is required to produce to 5.0 g of
FinnZ [79.3K]

Answer:

3grams

Explanation:

The reaction for the production of Magnesium dioxide will be

Mg + O2  → MgO

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Consider 2.4 moles of a gas contained in a 4.0 L bulb at a constant temperature of 32°C. This bulb is connected to an evacuated
Luden [163]

Complete Question

The complete question is shown on the first uploaded image

Answer

a

As the valve is opened , the gas will flow into the empty container until the both containers have the same pressure

b

\Delta H = 0\ , \Delta E = 0 , q= 0 , w= 0

c

The driving force for this process is the increase in entropy this is because the movement of the internal energy of the gas into a larger volume, what this does is that it increases the amount of disorder(entropy).

Explanation

In order to obtain the parameter in the part B of the question we are first obtain the initial pressure, using the ideal gas equation  

                      P = \frac{nRT}{V}

                     P = \frac{(2.4mol)(0.0821\frac{1 atm}{K \cdot \ mol} )}{4.0L}

                         P =15 \ atm

The next thing is to obtain the new pressure of the gas , using boyle's law

              P_1V_1 = P_2V_2

                  P_2 = \frac{P_1 V_1}{V_2}

                  P_2 = \frac{(15 \ atm)(4.0L)}{24.0 L}

                  P_2 = 2.5 \ atm    

Since the this process is isothermal , the change in heat is equal to zero

                      i.e  q = 0 J

  The workdone to move  the gas to the other container is zero because the  the pressure at this second container is zero due to the fact that it is a vacuum

    i.e  w = -P_{external} \Delta V

              =-(0 \ atm) (24.0 - 4.0L)

              = 0L \cdot atm

  Since the change in heat is zero and the workdone is zero then the change in internal energy is equal to 0

     This is because the change in internal energy is equal to a summation of change in heat and the workdone

                i.e \Delta E = q + w

                            = 0J

Generally the change in enthalpy is mathematically represented as

                \Delta H = n C_p \Delta T

Since the temperature is zero this means that the change in temperature is zero , substituting this value for change in temperature into the equation for  change in enthalpy

           \Delta H = n C_p (0)

                   = 0J

 

7 0
4 years ago
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