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Elan Coil [88]
3 years ago
11

A gas storage tank has a volume of 3.5 X 105 L when the temperature is 27oC, and the pressure is 101 kPa. What is the new volume

of the tank if the temperature drops to -10oC and the pressure drops to 95 kPa ?
Chemistry
1 answer:
masha68 [24]3 years ago
8 0

Answer: 3.26x10^5L

Explanation:

P1 = 101kPa

P2 = 95kPa

T1 = 27°C = 27 +273 =300k

T2 = —10°C = —10 +273 = 263k

V1 = 3.5 x 10^5 L

V2 =?

P1V1 / T1 = P2V2 /T2

(101 x 3.5 x 10^5)/300 = (95xV2)/263

300x95xV2 = 101x 3.5x10^5x263

V2 = (101x 3.5x10^5x263) /(300x95)

V2 = 3.26x10^5L

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93.0 mL of O2 gas is collected over water at 0.930 atm and 10.0C what would be the volume of this dry gas at STP
Dahasolnce [82]
The  volume  of  the  dry   gas  at     stp  is  calculated as follows

calculate  the  number  on  moles  by use  of   PV =nRT  where  n  is  the number  of  moles

n  is therefore  = Pv/RT
P  = 0.930  atm
R(gas  contant=  0.0821  L.atm/k.mol
V= 93ml  to  liters =  93/1000= 0.093L
T=  10  +  273.15 = 283.15k

n=  (0.930  x0.093)  /(0.0821  x283.15) =  3.  72  x10^-3  moles

At  STp    1  mole  =  22.4L
what about  3.72  x10^-3 moles

by  cross  multiplication

volume =  (3.72  x10^-3)mole  x  22.4L/  1  moles  = 0.083 L   or  83.3 Ml
3 0
3 years ago
Please help! <br> Fe2O3 + C → CO + Fe
Sunny_sXe [5.5K]
Fe2O3 + 3C → 2Fe + 3CO :)
3 0
2 years ago
A certain radioactive isotope decays at a rate of 0.2​% annually. Determine the ​half-life of this​ isotope, to the nearest year
pychu [463]

Answer:

The half-life of the radioactive isotope is 346 years.

Explanation:

The decay rate of the isotope is modelled after the following first-order linear ordinary differential equation:

\frac{dm}{dt} = -\frac{m}{\tau}

Where:

m - Current isotope mass, measured in kilograms.

t - Time, measured in years.

\tau - Time constant, measured in years.

The solution of this differential equation is:

m(t) = m_{o}\cdot e^{-\frac{t}{\tau} }

Where m_{o} is the initial mass of the isotope. It is known that radioactive isotope decays at a yearly rate of 0.2 % annually, then, the following relationship is obtained:

\%e = \frac{m(t)-m(t+1)}{m(t)}\times 100\,\% = 0.2\,\%

1 - \frac{m(t+1)}{m(t)} = 0.002

1 - \frac{m_{o}\cdot e^{-\frac{t+1}{\tau} }}{m_{o}\cdot e^{-\frac{t}{\tau} }}=0.002

1 - e^{-\frac{1}{\tau} } = 0.002

e^{-\frac{1}{\tau} } = 0.998

-\frac{1}{\tau} = \ln 0.998

The time constant associated to the decay is:

\tau = -\frac{1}{\ln 0.998}

\tau \approx 499.500\,years

Finally, the half-life of the isotope as a function of time constant is given by the expression described below:

t_{1/2} = \tau \cdot \ln 2

If \tau \approx 499.500\,years, the half-life of the isotope is:

t_{1/2} = (499.500\,years)\cdot \ln 2

t_{1/2}\approx 346.227\,years

The half-life of the radioactive isotope is 346 years.

6 0
2 years ago
n a semiconductor, the bonding molecular orbitals that contain electrons are referred to as the ________, while the antibonding
Kruka [31]

Answer:

In a semiconductor, the bonding molecular orbitals that contain electrons are referred to as the valence band, while the antibonding orbitals that are completely empty are referred to as the conduction band.

The conduction band occupies a higher energy level than the valence band. The band gap is what separates the two orbitals.

3 0
2 years ago
Ok so the question is Sort the processes based on the type of energy transfer they involve. condensation
Lesechka [4]

Maybe evaporation???????????????

7 0
3 years ago
Read 2 more answers
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