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uranmaximum [27]
3 years ago
15

Classify the reaction: N2 (g) + 3H2 (g) à 2NH3 (g)

Chemistry
1 answer:
Whitepunk [10]3 years ago
4 0
For the reaction N2(g) + 3H2(g) <=> 2NH3(g),
we have two reactants combining to form a single product. This type of reaction is called a Synthesis reaction.

(Quick side note): Because this is a reversible, you could write it in the opposite direction as a decomposition reaction.
You might be interested in
To make a sarurated solution, 36 g of sodium chloride is dissolved in 100 g of water at 293 k . Find its concentration at this t
nexus9112 [7]

Answer:

The answer is "26.47\%"

Explanation:

Sodium chloride  solute mass=36 \ g

Solvent water mass=100\ g

Calculating the solution mass = Solute mass  + solvent mass

                                                  = 36\ g +100\ g\\\\ = 136\ g\\\\

Calculating the percentage of concentration:

= \frac{solute\ mass}{solvent\ mass} \times 100\\\\=\frac{36\ grm}{136\ grm} \times 100\\\\=0.2647 \times 100\\\\=26.47 \%

3 0
3 years ago
For an alloy that consists of 33.5 wt% Pb and 66.5 wt% Sn, what is the composition (a) of Pb (in at%), and (b) of Sn (in at%)? T
Tju [1.3M]

Answer:

Pb: 22.4 at%

Sn: 77.6 at%

Explanation:

It is possible to find at% of Pb and Sn converting mass in moles using molar mass assuming a basis of 100g, thus:

Pb: 33.5g × (1mol / 207.2g) = <em>0.1617mol</em>

Sn: 66.5g × (1mol / 118.7g) = <em>0.5602mol</em>

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Total moles: 0.1617mol + 0.5602mol = 0.7219mol

Composition in at%:

Pb: 0.1617mol / 0.7219mol × 100 = <em>22.4 at%</em>

Sn: 0.5602mol / 0.7219mol × 100 = <em>77.6 at%</em>

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I hope it helps!

5 0
3 years ago
Escriba en termino de moles, de moléculas y de masa las siguientes ecuaciones a. Fe +2HCl ________ FeCl2 + H2 b. CH4 + 2O2 _____
MA_775_DIABLO [31]

Answer:

a.

  • 1 mol de Hierro reacciona con 2 moles de acido clorhidrico para formar 1 mol de cloruro de hierro (II)
  • 6.02×10²³ moleculas de hierro reaccionan con 1.20×10²⁴ moleculas de acido clorhidrico para formar 6.02×10²³ moleculas de cloruro de hierro (II)
  • 55.85 g de hierro reaccionan con 72.9 gramos de acido clorhidrico para formar 126.75 g de cloruro férrico.

b.

  • 6.02×10²³ moleculas de metano reaccionan con 1.20×10²⁴ moleculas de oxigeno para formar 6.02×10²³ moleculas de dioxido de carbono y 1.20×10²⁴ moleculas de agua.
  • 1 mol de metano reacciona con dos moles de oxigeno para generar 1 mol de dióxido de carbono y dos moles de agua en estado de vapor.
  • 16 gramos de metano reaccionan con 64 g de oxigeno para formar 44 gramos de dioxido de carbono y 36 gramos de agua.

c.

  • 3 moles de plata sólida reaccionan con 4 moles de acido nitrico para formar 3 moles de nitrato de plata, 1 mol de monoxido de nitrogeno y 2 moles de agua.
  • 1.80×10²⁴ moleculas de plata reaccionan con 2.41×10²⁴ moleculas de acido nitrico para formar 1.80×10²⁴ moleculas de nitrato de plata, 6.02×10²³ moleculas de monoxido de nitrogeno y 1.20×10²⁴ moleculas de agua.
  • 323.58 g de plata reaccionan con 252 g de acido nitrico para formar 509.58 g de nitrato de plata, 30 g de monoxido de nitrogeno y 36 g de agua.

Explanation:

a. Fe (s)  +2 HCl (aq) → FeCl₂ (aq)

1 mol de Hierro reacciona con 2 moles de acido clorhidrico para formar 1 mol de cloruro de hierro (II)

Calculamos cuanto son dos moles de moleculas sabiendo que:

6.02×10²³ moleculas / 1 mol  . 2 mol = 1.20×10²⁴ moleculas. Entonces

6.02×10²³ moleculas de hierro reaccionan con 1.20×10²⁴ moleculas de acido clorhidrico para formar 6.02×10²³ moleculas de cloruro de hierro (II)

Calculamos las masas molares de cada reactivo y producto

Fe = 55.85 g

HCl = 36.45 g

FeCl₂ = 126.75 g

55.85 g de hierro reaccionan con 72.9 gramos de acido clorhidrico para formar 126.75 g de cloruro férrico.

b. CH₄(g) + 2O₂ (g) → CO₂ (g) + 2H₂O (g)

1 mol de metano reacciona con dos moles de oxigeno para generar 1 mol de dióxido de carbono y dos moles de agua en estado de vapor.

6.02×10²³ moleculas de metano reaccionan con 1.20×10²⁴ moleculas de oxigeno para formar 6.02×10²³ moleculas de dioxido de carbono y 1.20×10²⁴ moleculas de agua.

Calculamos las masas molares:

CH₄ = 16 g

O₂ = 32 g

CO₂ = 44 g

H₂O g = 18 g

16 gramos de metano reaccionan con 64 g de oxigeno para formar 44 gramos de dioxido de carbono y 36 gramos de agua.

c. 3 Ag (s) + 4HNO3 (aq) → 3 AgNO3 (aq) + NO (g) + 2H₂O (aq)

Calculamos cuantos moleculas contienen 3 y 4 moles:

6.02×10²³  . 3 = 1.80×10²⁴ moleculas

6.02×10²³  . 4 = 2.41×10²⁴ moleculas

3 moles de plata sólida reaccionan con 4 moles de acido nitrico para formar 3 moles de nitrato de plata, 1 mol de monoxido de nitrogeno y 2 moles de agua.

1.80×10²⁴ moleculas de plata reaccionan con 2.41×10²⁴ moleculas de acido nitrico para formar 1.80×10²⁴ moleculas de nitrato de plata, 6.02×10²³ moleculas de monoxido de nitrogeno y 1.20×10²⁴ moleculas de agua.

Calcualmos las masas molares:

Ag = 107.86 g

HNO₃ = 63 g

AgNO₃ = 169.86 g

NO = 30 g

H₂O = 18 g

323.58 g de plata reaccionan con 252 g de acido nitrico para formar 509.58 g de nitrato de plata, 30 g de monoxido de nitrogeno y 36 g de agua.

6 0
3 years ago
Place these hydrocarbons in order of decreasing boiling point. Rank from highest to lowest boiling point. 1. Paraffin, C40Hg2 2.
lesya [120]

Answer:

Paraffin > Heptadecane > Hexane > 2,2-dimethylbutane > Propane

Explanation:

It must first be establish that all the molecules listed in the question are alkanes. For alkanes, the intermolecular forces between the molecules of alkanes increases with increasing molecular weight. This is as a result of increase in the surface area of the molecule. Increase in surface area implies a greater degree of dispersion forces.

This is the reason why high boiling points are observed for high molecular weight alkanes.

6 0
3 years ago
What volume (in L) of oxygen will be required to produce 77.4 L of water vapor in the reaction below?
Nostrana [21]

Answer:

For the production of 77.4 L water 90.3 L oxygen is required.

Explanation:

Given data:

Volume of oxygen required = ?

Volume of water produced = 77.4 L

Solution:

Chemical reaction equation:

2C₂H₆ + 7O₂  →  4CO₂ + 6H₂O

1 mole = 22.414 L

There are 6 moles of water = 6×22.414 = 134.5 L

There are 7 moles of oxygen = 7×22.414 = 156.9 L

Now we will compare the litters of water and oxygen:

                              H₂O           :              O₂    

                              134.5         :              156.9

                                77.4         :             156.9/134.5×77.4 =90.3 L

So for the production of 77.4 L water 90.3 L oxygen is required.

8 0
3 years ago
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