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3 years ago

What is the density at STP of the gas sulfur hexafluoride, SF6?

Chemistry

2 answers:

svetoff [14.1K]3 years ago

7 0

Answer:

Explanation:

There is a formula for this:

M = DRT/P where M = molar mass. This just derived from PV = nRT where you say n = grams/molar mass. However, just with this formula, we can get D which is density at STP (1 atm and 273K). We find that D = 6.52g/L.

jonny [76]3 years ago

3 0

The third one... 6.52 g/L

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VashaNatasha [74]

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2 years ago

The specific heat of copper is 0.40 joules/ g °c. How much heat is needed in joules to change the temperature of a 55 gram subst

Snezhnost [94]

Answer : The amount of heat needed is, 1188 J

Explanation :

Formula used :

$q=m\times c\times (T_2-T_1)$

where,

q = heat needed = ?

m = mass of copper = 55 g

c = specific heat capacity of copper = $0.40J/g^oC$

$T_1$ = initial temperature = $20.0^oC$

$T_2$ = final temperature = $74.0^oC$

Now put all the given values in the above formula, we get:

$q=55g\times 0.40J/g^oC\times (74.0-20.0)^oC$

$q=1188J$

Thus, the amount of heat needed is, 1188 J

3 0

3 years ago

Consider a voltaic cell where the anode half-reaction is Zn(s) → Zn2+(aq) + 2 e− and the cathode half-reaction is Sn2+(aq) + 2 e

notsponge [240]

Answer: The concentration of $Sn^{2+}$ in the cell is $9.0\times 10^{-3}M$

Explanation:

We are given:

Oxidation half reaction: $Zn(s)\rightarrow Zn^{2+}(aq.)+2e^-$ $E^o_{Zn^{2+}/Zn}=-0.76V$

Reduction half reaction: $Sn^{2+}(aq.)+2e^-\rightarrow Sn(s)$ $E^o_{Sn^{2+}/Sn}=-0.136V$

The substance having highest positive $E^o$ potential will always get reduced and will undergo reduction reaction. Here, fluorine will undergo reduction reaction will get reduced.

Here, tin will undergo reduction reaction and will get reduced.

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

Putting values in above equation, we get:

$E^o_{cell}=-0.136-(-0.76)=0.624V$

To calculate the EMF of the cell, we use the Nernst equation, which is:

$E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[Mn^{2+}]}{[Cu^{2+}]}$

where,

$E_{cell}$ = electrode potential of the cell = 0.660 V

$E^o_{cell}$ = standard electrode potential of the cell = +0.624 V

n = number of electrons exchanged = 2

$[Zn^{2+}]=2.5\times 10^{-3}M$

$[Sn^{2+}]$ = ?

Putting values in above equation, we get:

$0.660=0.624-\frac{0.059}{2}\times \log(\frac{2.5\times 10^{-3}}{[Sn^{2+}})$

$[Sn^{2+}]=9.0\times 10^{-3}M$

Hence, the concentration of $Sn^{2+}$ ions is $9.0\times 10^{-3}M$

3 0

3 years ago

How does one calculate the pH of a solution? O A. The pH is the product of [H+] and [OH-]. O B. The pH is the negative log of [O

blagie [28]

Answer:

I believe it is A

Explanation:

pH = -log[H+] and pOH = -log[OH−].

8 0

2 years ago

How does the Bohr model show you an atom has become an ion?

Step2247 [10]

An early model of the atom was developed in 1913 by Danish scientist Niels Bohr (1885–1962). The Bohr model shows the atom as a central nucleus containing protons and neutrons with the electrons in circular orbitals at specific distances from the nucleus . These orbits form electron shells or energy levels, which are a way of visualizing the number of electrons in the various shells. These energy levels are designated by a number and the symbol "n." For example, 1n represents the first energy level located closest to the nucleus.

6 0

3 years ago