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laiz [17]
3 years ago
9

What volume in kiloliters will a sample of Bay water occupy if it has a volume of 125 ml?

Chemistry
2 answers:
Mamont248 [21]3 years ago
6 0
This question is basically asking us to convert the volume from mL to kL.

We can do this by setting up a multiplication problem and cancelling out units:

\frac{125mL}{1}* \frac{1L}{1000mL} * \frac{1kL}{1000L}  =1.25* 10^{-4}kL

So now we know that the Bay water will occupy 1.25x10^-4 kL.
Sergeu [11.5K]3 years ago
5 0

Answer: 0.125\times 10^{-3}kL.

Explanation:

Volume of the gas is defined as the space occupied by a substance. It is expressed in units like cm^3, m^3 , L and ml.

All these units of volume are inter convertible.

We are given:

Volume of the gas = 125ml

Converting this unit of volume into 'L' by using conversion factor:

1ml=0.001L

125ml=\frac{0.001}{1}\times 125=0.125L

1L=0.001kL

Thus 0.125L=\frac{0.001}{1}\times 0.125=0.125\times 10^{-3}kL

Thus the volume in kilo Liters would be 0.125\times 10^{-3}.

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Determine the molar mass of a compound that has a density of 0.1633 g/L at STP.<br> (show work)
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Answer:

                     M.Mass  =  3.66 g/mol

Data Given:

                  M.Mass  =  M = ??

                  Density  =  d  =  0.1633 g/L

                  Temperature  =  T  =  273.15 K (Standard)

                  Pressure  =  P  =  1 atm (standard)

Solution:

              Let us suppose that the gas is an ideal gas. Therefore, we will apply Ideal Gas equation i.e.

                                             P V = n R T    ---- (1)

Also, we know that;

                       Moles  =  n  =  mass / M.Mass

Or,                                   n  =  m / M

Substituting n in Eq. 1.

                                             P V = m/M R T   --- (2)

Rearranging Eq.2 i.e.

                                             P M = m/V R T   --- (3)

As,

                     Mass / Volume = m/V = Density = d

So, Eq. 3 can be written as,

                                             P M = d R T

Solving for M.Mass i.e.

                                             M = d R T / P

Putting values,

M  =  0.1633 g/L × 0.08205 L.atm.K⁻¹.mol⁻¹ × 273.15 K / 1 atm

M  =  3.66 g/mol

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The collision of the molecules between the hydrogen molecule or H2, and an iodine molecule or I2, provided there would be a sufficient energy is that the system would eventually undergo a chemical change wherein a new chemical compound would be formed from these two molecules.
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The effusion rate of hcl is 43.2 cm/min in a certain effusion apparatus. what is the rate of effusion of ammonia in the same app
Jlenok [28]

The effusion rate is 1.125 cm/sec for ammonia.

How to find effusion rate ?

Effusion rate (r1) HCl = 43.2 cm/min

Molar mass (m2) NH3 =17.04g/mole

Molar mass (m1)  HCl    =36.46g/mole

  • Substitute the molar masses of the gases into Graham's law and solve for the ratio.
  • r1÷r2=√m2÷m1

       firstly convert 43.2 cm/min into cm/sec i.e., 0.72 cm/sec

      Then,

      0.72/r2 =√17.04/36.46

      r2= 1.125 cm/sec

Hence, the rate of diffusion of ammonia is 1.125 times faster than the rate of diffusion of hydrogen chloride.

learn more about effusion here:

brainly.com/question/2097955

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