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krek1111 [17]
3 years ago
15

Sulfur and fluorine form several different compounds including sulfur hexafluoride and sulfur tetrafluoride. Decomposition of a

sample of sulfur hexafluoride produced 4.44 g of fluorine and 1.25 g of sulfur, while decomposition of a sample of sulfur tetrafluoride produced 4.44 g of fluorine and 1.88 g of sulfur. Calculate the mass of fluorine per gram of sulfur for the sample of sulfur hexafluoride.
Chemistry
1 answer:
svetoff [14.1K]3 years ago
3 0

Answer:

See Explanation

Explanation:

For SF6;

Since;

1.25 g of S corresponds to 4.44g of F

1 g of sulphur corresponds to 1 * 4.44/1.25 = 3.55

For SF4;

Since;

1.88 g of S corresponds to 4.44g of F

1 g of sulphur corresponds to 1 * 4.44/ 1.88  = 2.36

Hence;

Mass of oxygen per gram of sulphur in SF6/Mass of oxygen per gram of sulphur in SF4

=

3.55/2.36 = 1.5

Hence the law of multiple proportion is obeyed here.

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Scientists use a Graduated Cylinder
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A cube has a height of 8 cm and a mass of 457 g. What is its density?
ZanzabumX [31]

Answer:

the answer is b,  0.89

Explanation:

A cube has a height of 8 cm and a mass of 457 g. What is its density?

a. 233,984 g/cm

b.  0.89g/cm3

c. 1.12 g/cm3

Density = mass/volume

the volume of the cube is 8X8X8=512cm3

the mass is 457 gm

the density is 457/512 = 0.889 gm/cm3

the answer is b, 0.89

3 0
2 years ago
Free-energy change, ΔG∘, is related to cell potential, E∘, by the equationΔG∘=−nFE∘where n is the number of moles of electrons t
mart [117]

Answer:

a)\Delta G=372490 J

b)\Delta G=-568614 J

Explanation:

a) The reaction:

Mg(s) +Fe^{2+}(aq) \longrightarrow Mg^{2+}(aq) + Fe(s)

The free-energy expression:

\Delta G=-n*F*E

E=E_{red}-E_{ox]

The element wich is reduced is the Fe and the one that oxidates is the Mg:

-0.44V=E_{red}-(-2.37V)=1.93V

The electrons transfered (n) in this reaction are 2, so:

\Delta G=-2mol*96500 C/mol * 1.93 V

\Delta G=-372490 J

b) If you have values of enthalpy and enthropy you can calculate the free-energy by:

\Delta G=\Delta H - T* \Delta S

with T in Kelvin

\Delta G=-675 kJ*\frac{1000J}{kJ} - 298K*-357 J/K

\Delta G=-568614 J

7 0
2 years ago
I need help please please
inna [77]

Answer:

B should be the answer, and ur low-key valid lol

Explanation:

6 0
3 years ago
A sample of gas contains 3.0L of nitrogen at 320kPa. What volume would be necessary to decrease the pressure at 110kPa
Phoenix [80]

Answer:

V_{2} = 8.92 L

Explanation:

We have the equation for ideal gas expressed as:

PV=nRT

Being:

P = Pressure

V = Volume

n = molar number

R = Universal gas constant

T = Temperature

From the statement of the problem I infer that we are looking to change the volume and the pressure, maintaining the temperature, so I can calculate the right side of the equation with the data of the initial condition of the gas:

P_{1} V_{1} =nRT

320Kpa*0.003m^{3} =nRT

1000L = 1m^{3}

So

nRT= 0.96

Now, as for the final condition:

P_{2}V_{2}=nRT

P_{2} V_{2} =0.96

clearingV_{2}

V_{2} =\frac{0.96}{P_{2} }

V_{2} =0.00872m_{3}

V_{2} = 8.92 L

6 0
3 years ago
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