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Scilla [17]
3 years ago
9

Indicate the number of h2 and n2 molecules needed to yield two molecules of Nh3

Chemistry
1 answer:
kati45 [8]3 years ago
8 0
The balanced chemical reaction and or equation for the famous haber or ammonia process would be :

3H2 + N2 => 2NH3.

You might be interested in
6th grade help me pleaseeeee
Vesnalui [34]

Answer:

the answer would be A. cells

Explanation:

just trust me Im a 8th grader and I did that before

7 0
3 years ago
The pH of a solution prepared by mixing 40.00 mL of 0.10 M NH3 with 50.00 mL of 0.10 M NH4Cl and 30mL of 0.05 M H2SO4 is 5.17. A
liq [111]

Answer:

Following are the answer to this question:

Explanation:

The value of pH solution is =5.17 So, the p^{OH}:

p^{OH}=14-56.17

      =8.823

The volume of the NH_{3} = 40.00 ml  

convert into the liter= 0.040L

The value of the concentrated NH_{3} =0.10 M

The volume of the NH_{4}Cl= 50.00 ml

convert into the liter= 0.050L

 The value of concentrated NH_{4}Cl= 0.10 M

The volume of the H_{2}So_{4}= 30 ml

convert into the liter= 0.030L  

The value of concentrated H_2So_4=0.05 M

Calculating total volume=(0.40+0.050+0.030)

                                       =0.120 L

calculating the new concentrated value of NH_3 = \frac{0.10\times 0.040}{0.120}= 0.33 \ M

calculating the new concentrated value of NH_4Cl= \frac{0.050\times 0.10}{0.120}= 0.04166 \ Mcalculating the new concentrated value of H_2So_4= \frac{0.030\times 0.05}{0.120}= 0.0125 \ M when 1 mol H_2So_4 produced 2 mols H^{+} so, 0.0125 in H_2So_4produced:

=4 \times (2 \times 0.0125) \ mol H^{+}\\\\= 0.025 mol H^{+}

create the ICE table:    

NH_3    \ \ \ \ \ \ \ \     + H^{+}  \ \ \ \ \ \ \longrightarrow NH_4^{+}                    

I (m)       0.033(m)            0.025                       0.04166

C            -0.025                 -0.025                       + 0.025  

E            8.3\times 10^{-3}     0                    0.0667

now calculating pH:

when ph= 8.83:

P^{H}= p^{kb}|+ \log\frac{[NH_4^{+}]}{[NH_3]}\\\\8.83=p^{kb}+\log\frac{0.0667}{8.3 \times 10^{-3}}\\\\p^{kb}=8.83-0.9069\\\\ \ \ \ =7.7231 \\\\\ The P^{kb} \ for \ NH_3 \ is =7.7231\\\\\ The P^{kb} \ for N^{+}H_4=14-7.7231\\\\\ \ \ \ \ \ =6.2769

5 0
3 years ago
What is the volume of 14.0g of nitrogen gas at STP?
lozanna [386]

Answer:

  • <em>The volume of 14.0 g of nitrogen gas at STP is </em><u><em>11.2 liter.</em></u>

Explanation:

STP stands for standard pressure and temperature.

The International Institute of of Pure and Applied Chemistry, IUPAC changed the definition of standard temperature and pressure (STP) in 1982:

  •   Before the change, STP was defined as a temperature of 273.15 K and an absolute pressure of exactly 1 atm (101.325 kPa).

  •    After the change, STP is defined as a temperature of 273.15 K and an absolute pressure of exactly 105 Pa (100 kPa, 1 bar).

Using the ideal gas equation of state, PV = nRT you can calculate the volume of one mole (n = 1)  of gas. With the former definition, the volume of a mol of gas at STP, rounded to 3 significant figures, was 22.4 liter. This is classical well known result.

With the later definition, the volume of a mol of gas at STP is 22.7 liter.

I will use the traditional measure of 22.4 liter per mole of gas.

<u>1) Convert 14.0 g of nitrogen gas to number of moles:</u>

  • n = mass in grams / molar mass
  • Atomic mass of nitrogen: 14.0 g/mol
  • Nitrogen gas is a diatomic molecule, so the molar mass of nitrogen gas = molar mass of N₂ = 14.0 × 2 g/mol = 28.0 g/mol
  • n = 14.0 g / 28.0 g/mol = 0.500 mol

<u>2) Set a proportion to calculate the volume of nitrogen gas:</u>

  • 22.4 liter / mol = x  / 0.500 mol
  • Solve for x: x = 0.500 mol × 22.4 liter / mol = 11.2 liter.

<u>Conclusion:</u> the volume of 14.0 g of nitrogen gas at STP is 11.2 liter.

6 0
4 years ago
815 yd to km (need 3 sig fig)
SpyIntel [72]

<u>Answer:</u> The correct answer is 0.745 km

<u>Explanation:</u>

We are given:

A numerical value of 815 yards

To convert this into kilometers, we use the conversion factor:

1 km = 1093.6 yards

Converting the given value into kilometers, we get:

\Rightarrow 815yd\times \frac{1km}{1093.6yd}=0.745km

Hence, the correct answer is 0.745 km

5 0
3 years ago
How does a sample of hydrogen at 10 °C compare to a sample of hydrogen at 350 K?
Andrej [43]

Answer: -

C. The hydrogen at 10 °C has slower-moving molecules than the sample at 350 K.

Explanation: -

The kinetic energy of gas molecules increase with the increase in the temperature of the gas. With the increase in kinetic energy, the gas molecules also move faster. Thus with the increase of temperature, the speed of the molecules increase.

Temperature of first hydrogen gas sample is 10 °C.

10 °C means 273+10 = 283 K

Thus first sample temperature = 283 K

The second sample temperature of the hydrogen gas  is 350 K.

Thus the temperature is increased.

So both the kinetic energy and speed of molecules is more for the hydrogen gas sample at 350 K.

Thus the hydrogen at 10 °C has slower-moving molecules than the sample at 350 K.

Hence the answer is C.

7 0
3 years ago
Read 2 more answers
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