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KonstantinChe [14]
2 years ago
8

The mass of 1.49 of H2 gas at stp

Chemistry
1 answer:
FinnZ [79.3K]2 years ago
3 0

Answer:

Explanation:

I'll assume 1.49 is liters.

All gases occupy 22.4 L for each mole of gas at STP.  This makles a very useful, and important, conversion factor:

(22.4L/mole gas) for all gases at STP.

Therefore 1.49L of H2 would be:

(1.49L H2)/(22.4L/mole gas) = 0.0665 moles

The molar mass of H2 is 2, so:

(0.0665 moles)*(2g/mole) = 0.133 grams to 3 sig figs

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Gnesinka [82]

Answer:

0.043 M

Explanation:

The reaction that takes place is:

  • Ca(OH)₂ + 2HCl → CaCl₂ + 2H₂O

First we <u>calculate how many HCl moles reacted</u>, using the <em>given concentration and volume required to reach the equivalence point</em>:

  • 0.029 M HCl * 37.3 mL = 1.0817 mmol HCl = 1.0817 mmol H⁺

As 1 mol of H⁺ reacts with 1 mol of OH⁻, in the 25.0 mL of the Ca(OH)₂ sample there are 1.0817 mmoles of OH⁻.

With that in mind we can <u>calculate the hydroxide ion concentration in the original sample solution</u>, using <em>the calculated number of moles and given volume</em>:

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3 years ago
The Ostwald process for the commercial production of nitric acid from ammonia and oxygen involves the following steps:
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Answer:

12 NH_3 (g)+ 21 O_2 (g) + \longrightarrow  14 H_2O (l) + 8 HNO_3 (g) + 4 NO (g)

Explanation:

The steps of the Ostwald process:

4 NH_3 (g) + 5 O_2 (g) \longrightarrow 4 NO (g) + 6 H_2O (g)

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Combinning the equations:

4 NH_3 (g) + 5 O_2 (g) \longrightarrow 4 NO (g) + 6 H_2O (g)

+

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+

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=

4 NH_3 (g)+ 4 NO (g)+ 7 O_2 (g) + 4 NO_2 (g) +4/3 H_2O (l) \longrightarrow 4 NO (g) + 6 H_2O (g) +  4 NO_2(g) + 8/3 HNO_3 (g) + 4/3 NO (g)

Simplifying:

4 NH_3 (g)+ 7 O_2 (g) + \longrightarrow  14/3 H_2O (l) + 8/3 HNO_3 (g) + 4/3 NO (g)

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The overall reaction is endothermic becuase the formation of new chemical bonds requires energy consumption.

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