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Rus_ich [418]
3 years ago
5

The pressure of a basketball is 412 mmHg. How many kPa is that equal to?

Chemistry
1 answer:
pychu [463]3 years ago
3 0

Answer:

d. 54.9 kPa

Explanation:

mmHg and Pa are units of pressure used in chemistry principally in the study of gases. 1mmHg is equal to 133.322Pa. 412mmHg are:

412 mmHg * (133.322Pa / 1mmHg) = 54929 Pa

The prefix K (Kilo) represents one thousand of the determined unit.

54929Pa are:

54929Pa * (1KPa / 1000Pa) = 54.9kPa

Right answer is:

<h3>d. 54.9 kPa </h3>

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Solve for b, when c= a/b and a = 5.8 and c= 7.4
vova2212 [387]

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Explanation:

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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]

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

<u>Explanation:</u>

We are given:

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

<u>Reduction half reaction:</u>  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

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3 years ago
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Answer:

There are more nutrients available in estuaries. There are more nutrients available in estuaries.

Explanation:

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