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Anvisha [2.4K]
3 years ago
12

Which radioisotope is naturally occurring?

Chemistry
1 answer:
Lena [83]3 years ago
5 0

Answer:

I think the answer is 13H

Explanation:

The best known example of a naturally-occurring radioisotope is uranium. All but 0.7 per cent of naturally-occurring uranium is uranium-238; the rest is the less stable, or more radioactive, uranium-235, which has three fewer neutrons in its nucleus.

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A tank with volume of 2.4 cu ft is filled with Methane to a pressure of 1500 psia at 104 degrees F. Determine the molecular weig
Soloha48 [4]

Explanation:

It is known that equation for ideal gas is as follows.

               PV = nRT

The given data is as follows.

     Pressure, P = 1500 psia,     Temperature, T = 104^{o}F = 104 + 460 = 564 R

     Volume, V = 2.4 cubic ft,      R = 10.73 psia ft^{3}/lb mol R

Also, we know that number of moles is equal to mass divided by molar mass of the gas.

                n = \frac{mass}{\text{molar mass}}

            m = n \times W

                = 0.594 \times 16.04

                = 9.54 lb

Hence, molecular weight of the gas is 9.54 lb.

  • We will calculate the density as follows.

                d = \frac{PM}{RT}

                    = \frac{1500 \times 16.04}{10.73 \times 564}

                    = 3.975 lb/ft^{3}

  • Now, calculate the specific gravity of the gas as follows.

  Specific gravity relative to air = \frac{\text{density of methane}}{\text{density of air}}

                         = \frac{3.975 lb/ft^{3}}{0.0765 lb/ft^{3}}

                         = 51.96

6 0
3 years ago
What is often mistakenly regarded as being based on science?
photoshop1234 [79]
Unless you are talking about one specific theory, the answer is pseudoscience.
3 0
3 years ago
What items are true about a block of ice at -10°C as you continue to apply heat
zhenek [66]

1. its temperature will rise continuously until it melts

I don't believe that any of the other answers are correct because it can not stay at a certain temperature if it is melting

5 0
3 years ago
If 1.0 gram of hydrogen reacts with 19.0 grams of fluorine, then what is the percent by mass of fluorine in the compound that is
astraxan [27]
The reaction between hydrogen (H2) and fluorine (F2) is given below,
                                   H2 + F2 ---> 2HF
One mole of both hydrogen and fluorine yields to 2 moles of hydrogen fluoride. This can also be expressed as, 2 grams of hydrogen and 38 grams of fluorine will form 40 grams of hydrogen fluoride. From the given, only 20 grams of HF is formed with 19 g of it being fluorine. Thus, the percentage fluorine of the compound formed is 95%. 
8 0
3 years ago
Read 2 more answers
A frictionless piston cylinder device is subjected to 1.013 bar external pressure. The piston mass is 200 kg, it has an area of
Bad White [126]

Answer:

a) T_{2} = 360.955\,K, P_{2} = 138569.171\,Pa\,(1.386\,bar), b) T_{2} =  347.348\,K, V_{2} = 0.14\,m^{3}

Explanation:

a) The ideal gas is experimenting an isocoric process and the following relationship is used:

\frac{T_{1}}{P_{1}} = \frac{T_{2}}{P_{2}}

Final temperature is cleared from this expression:

Q = n\cdot \bar c_{v}\cdot (T_{2}-T_{1})

T_{2} = T_{1} + \frac{Q}{n\cdot \bar c_{v}}

The number of moles of the ideal gas is:

n = \frac{P_{1}\cdot V_{1}}{R_{u}\cdot T_{1}}

n = \frac{\left(101,325\,Pa + \frac{(200\,kg)\cdot (9.807\,\frac{m}{s^{2}} )}{0.15\,m^{2}} \right)\cdot (0.12\,m^{3})}{(8.314\,\frac{Pa\cdot m^{3}}{mol\cdot K} )\cdot (298\,K)}

n = 5.541\,mol

The final temperature is:

T_{2} = 298\,K +\frac{10,500\,J}{(5.541\,mol)\cdot (30.1\,\frac{J}{mol\cdot K} )}

T_{2} = 360.955\,K

The final pressure is:

P_{2} = \frac{T_{2}}{T_{1}}\cdot P_{1}

P_{2} = \frac{360.955\,K}{298\,K}\cdot \left(101,325\,Pa + \frac{(200\,kg)\cdot (9.807\,\frac{m}{s^{2}} )}{0.15\,m^{2}}\right)

P_{2} = 138569.171\,Pa\,(1.386\,bar)

b) The ideal gas is experimenting an isobaric process and the following relationship is used:

\frac{T_{1}}{V_{1}} = \frac{T_{2}}{V_{2}}

Final temperature is cleared from this expression:

Q = n\cdot \bar c_{p}\cdot (T_{2}-T_{1})

T_{2} = T_{1} + \frac{Q}{n\cdot \bar c_{p}}

T_{2} = 298\,K +\frac{10,500\,J}{(5.541\,mol)\cdot (38.4\,\frac{J}{mol\cdot K} )}

T_{2} =  347.348\,K

The final volume is:

V_{2} = \frac{T_{2}}{T_{1}}\cdot V_{1}

V_{2} = \frac{347.348\,K}{298\,K}\cdot (0.12\,m^{3})

V_{2} = 0.14\,m^{3}

4 0
3 years ago
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