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Svetradugi [14.3K]
2 years ago
15

4NH3+5O2-4NO+6H2Ohow many moles of NH3 must react to produce 5.0 moles of NO?​

Chemistry
1 answer:
tresset_1 [31]2 years ago
7 0

Answer: 5.0 moles

Explanation:

From the equation, we see that for every 4 moles of ammonia consumed, 4 moles of nitrogen monoxide are produced (we can reduce this to moles of ammonia consumed = moles of nitrogen monoxide produced).

This means that the answer is <u>5.0 mol</u>

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Given the molar absorptivity for a species X of 1600 M-1cm-1 at a wavelength of 270 nm, and 400 M-1cm-1 at a wavelength of 540 n
Andre45 [30]

Answer:

Explanation:

From the given information:

At wavelength = 270 nm

\varepsilon x_1 = 1600 \ m^{-1} \ cm^{-1}  \\ \\  \varepsilon y_1 = 200 \ m^{-1} \ cm^{-1}

At 270 nm

Suppose x is said to be the solution for the concentration of x and y to be the solution for the concentration of y;

Then:

\varepsilon x_1  \ l + \varepsilon y_1  \ l= 0.5 \\ \\ A = A_1 + A_2

1600 xl + 200 yl= 0.5

Divide both sides by 200

8xl + yl = \dfrac{0.5}{200}

8x + y = \dfrac{0.5}{200}l

Use l = 1cm (i.e the standard length)

Then;

8x + y = \dfrac{0.5}{200} ---- (1)

<u>For 540 nm:</u>

\varepsilon x_2 x  \ l + \varepsilon y_2 y  \ l= 0.5  \\ \\ 40 xl + 800 yl = 0.5

x + 20 y = \dfrac{0.5}{400 \ l}

since l = 1

x + 20 y = \dfrac{0.5}{400 \ } --- (2)

Equating both (1) and (2) together, we have:

8x + y - 8x - 160 y = \dfrac{0.5}{200} - \dfrac{0.5 \times 8}{400}  \\ \\  \implies - 159 y = \dfrac{0.5}{200} ( 1 - \dfrac{8}{2}) \\ \\  -159 y = \dfrac{-0.5 \times 3}{200}  \\ \\  159 \ y = 0.0075  \\ \\  y = \dfrac{0.0075}{159} \\ \\  y = 0.00004716 \\ \\ y = 4.7 \times 10^{-5 } \ M

3 0
3 years ago
Gaseous methane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 9.6 g of methane is
lianna [129]

Answer:

21.6 g

Explanation:

The reaction that takes place is:

  • CH₄ + 2O₂ → CO₂ + 2H₂O

First we<u> convert the given masses of both reactants into moles</u>, using their <em>respective molar masses</em>:

  • 9.6 g CH₄ ÷ 16 g/mol = 0.6 mol CH₄
  • 64.9 g O₂ ÷ 32 g/mol = 2.03 mol O₂

0.6 moles of CH₄ would react completely with (2 * 0.6) 1.2 moles of O₂. As there are more O₂ moles than required, O₂ is the reactant in excess and CH₄ is the limiting reactant.

Now we <u>calculate how many moles of water are produced</u>, using the <em>number of moles of the limiting reactant</em>:

  • 0.6 mol CH₄ * \frac{2molH_2O}{1molCH_4} = 1.2 mol H₂O

Finally we<u> convert 1.2 moles of water into grams</u>, using its <em>molar mass</em>:

  • 1.2 mol * 18 g/mol = 21.6 g
4 0
3 years ago
What is the value of the specific heat capacity of liquid water in j/mol·°c?
Scilla [17]
<span>he specific heat capacity of liquid water is 4.186 J/gm K.</span>
4 0
3 years ago
Read 2 more answers
A gas cylinder contains only the gases radon, nitrogen, and helium. the radon has a pressure of 222 torr while the nitrogen has
Tema [17]
Dalton's law states that the total pressure of a mixture of non-interacting gases is equal to the sum of their partial pressures.

In this case,
P_{total} = P_{Rn} + P_{Ar} + P_{N_{2}}
771 = 222 + 446 + P_{N_{2}}
P_{N_{2}} = 103 Torr
7 0
3 years ago
In a car crash, large accelerations of the head can lead to severe injuries or even death. A driver can probably survive an acce
Alexxandr [17]

Answer:

Velocity, u = 14.7 m/s

Explanation:

It is given that, a driver can probably survive an acceleration of 50 g that lasts for less than 30 ms, but in a crash with a 50 g acceleration lasting longer than 30 ms, a driver is unlikely to survive.

Let v is the highest speed that the car could have had such that the driver survived. Using a = -50 g and t = 30 ms

Using first equation of kinematics as :

v=u+at

In case of crash the final speed of the driver is, v = 0

0=u+at

-u=at

-u=-50\times 9.8\times 30\times 10^{-3}

u = 14.7 m/s

So, the highest speed that the car could have had such that the driver survived is 14.7 m/s. Hence, this is the required solution.

8 0
3 years ago
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