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MatroZZZ [7]
1 year ago
15

Under appropriate conditions, nitrogen and hydrogen undergo a combination reaction to yield ammonia: N2 (g) + 3H2 (g) → 2NH3 (g)

A 9.3-g sample of hydrogen requires ________ g of N2 for a complete reaction.
Chemistry
1 answer:
kherson [118]1 year ago
6 0

Answer:

43 g N₂

Explanation:

To find the mass of N₂ required, you need to (1) convert grams H₂ to moles H₂ (via molar mass), then (2) convert moles H₂ to moles N₂ (via mole-to-mole ratio from equation coefficients), and then (3) convert moles N₂ to grams N₂ (via molar mass). It is important to arrange the conversions in a way that allows for the cancellation of units. The final answer should have 2 sig figs to match the amount of sig figs of the given value.

Molar Mass (H₂): 2(1.008 g/mol)

Molar Mass (H₂): 2.016 g/mol

Molar Mass (N₂): 2(14.009 g/mol)

Molar Mass (N₂): 28.018 g/mol

1 N₂(g) + 3 H₂(g) -----> 2 NH₃(g)

9.3 g H₂           1 mole             1 mole N₂             28.018 g
--------------  x  ----------------  x  --------------------  x  ------------------  =  43 g N₂
                        2.016 g            3 moles H₂            1 mole

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Which of the following is true for balancing equations?
nataly862011 [7]

Answer:

D) There must be equal number of atoms of each elements on both sides of equation.

Explanation:

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According to the law of conservation mass, mass can neither be created nor destroyed in a chemical equation.

Explanation:

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For example:

In given photosynthesis reaction:

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8 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]

<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

3 0
3 years ago
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