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allochka39001 [22]
3 years ago
15

PLEASE HELP ME 20 POINTS!!!!!

Chemistry
1 answer:
julsineya [31]3 years ago
5 0

Answer:

Theoretical yield of the reaction is 121·38 g

The excess reactant is hydrogen

The limiting reactant is nitrogen

Explanation:

By assuming that the reaction between nitrogen and hydrogen taking place in presence of catalyst because at normal conditions the reaction between them will not occur

Number of moles of nitrogen taken are 100÷28 ≈ 3.57

Number of moles of hydrogen taken are 100÷2 = 50

Actually the reaction between nitrogen and hydrogen takes place according to the following equation

<h3>Nx_{2} + 3Hx_{2}  → 2NHx_{3}</h3>

So from the equation for 1  mole of nitrogen and 3 moles of hydrogen we get 2 moles of ammonia

Here in the problem we have approximately 3·57 moles of nitrogen so we require 3×3·57 moles of hydrogen

∴ Number of moles of hydrogen required is 10·71

But we have 50 moles of hydrogen

∴ Excess reagent is hydrogen and limiting reagent is nitrogen

Number of moles of ammonia produced  is 2×3·57 = 7·14

Weight of ammonia is 17 g

∴ Amount of ammonia produced is 17×7·14 = 121·38 g

∴ Theoretical yield of the reaction is 121·38 g

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One method for determining the amount of corn in early Native American diets is the stable isotope ratio analysis (SIRA) techniq
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Answer:

a. i. 8.447 × 10⁻³ T ii.  27.14 cm

b. i. 2.14 cm ii. It is easily detectable.

Explanation:

a.

i. What strength of magnetic field is required?

Since the magnetic force F = Bqv equals the centripetal force F' = mv²/r on the C12 charge, we have

F = F'

Bqv = mv²/r

B = mv/re where B = strength of magnetic field, m = mass of C12 isotope = 1.99 × 10⁻²⁶ kg, v = speed of C 12 isotope = 8.50 km/s = 8.50 × 10³ m/s, q = charge on C 12 isotope = e = electron charge = 1.602 × 10⁻¹⁹ C (since the isotope loses one electron)and r = radius of semicircle = 25.0 cm/2 = 12.5 cm = 12.5 × 10⁻² m

So,

B = mv/rq

B = 1.99 × 10⁻²⁶ kg × 8.50 × 10³ m/s ÷ (12.5 × 10⁻² m × 1.602 × 10⁻¹⁹ C)

B = 16.915 × 10⁻²³ kgm/s ÷ (20.025 × 10⁻²¹ mC)

B = 0.8447 × 10⁻² kg/sC)

B = 8.447 × 10⁻³ T

(ii) What is the diameter of the 13C semicircle?

Since the magnetic force F = Bq'v equals the centripetal force F' = mv²/r' on the C13 charge, we have

F = F'

Bq'v = mv²/r'

r' = mv/Be where r = radius of semicircle, B = strength of magnetic field = 8.447 × 10⁻³ T, m = mass of C12 isotope = 2.16 × 10⁻²⁶ kg, v = speed of C 12 isotope = 8.50 km/s = 8.50 × 10³ m/s, q' = charge on C 13 isotope = e = electron charge = 1.602 × 10⁻¹⁹ C (since the isotope loses one electron) and  = d/2 = 12.5 cm = 12.5 × 10⁻² m

So, r' = mv/Be

r' = 2.16 × 10⁻²⁶ kg × 8.50 × 10³ m/s ÷ (8.447 × 10⁻³ T × 1.602 × 10⁻¹⁹ C)

r' = 18.36 × 10⁻²³ kgm/s ÷ 13.5321 × 10⁻²² TC)

r' = 1.357 × 10⁻¹ kgm/TC)

r' = 0.1357 m

r' = 13.57 cm

Since diameter d' = 2r', d' = 2(13.57 cm) = 27.14 cm

b.

i. What is the separation of the C12 and C13 ions at the detector at the end of the semicircle?

Since the diameter of the C12 isotope is 25.0 cm and that of the C 13 isotope is 27.14 cm, their separation at the end of the semicircle is 27.14 cm - 25.0 cm = 2.14 cm

ii. Is this distance large enough to be easily observed?

This distance of 2.14 cm easily detectable since it is in the centimeter range.

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The Ostwald process for the commercial production of nitric acid from ammonia and oxygen involves the following steps:
bagirrra123 [75]

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)

2 NO (g) + O_2 (g) \longrightarrow 2 NO_2 (g)

3 NO_2 (g) + H_2O (l) \longrightarrow 2 HNO_3 (g) + NO (g)

Combinning the equations:

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

+

(2 NO (g) + O_2 (g) \longrightarrow 2 NO_2 (g))*2

+

(3 NO_2 (g) + H_2O (l) \longrightarrow 2 HNO_3 (g) + NO (g))*4/3

=

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)

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

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

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3 years ago
Sue is building a bike with a metal frame in a warm, sunny area. Which material would be best for Sue to use for the grips on th
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Answer:

Rubber, or another insulator.

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