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2 years ago

Chemistry help!Zoom in to see better!!

Chemistry

1 answer:

inna [77]2 years ago

7 0

Answer:

11.9 g of nitrogen monoxide

Explanation:

We'll begin by calculating the number of mole in 6.75 g of NH₃. This can be obtained as follow:

Mass of NH₃ = 6.75 g

Molar mass of NH₃ = 14 + (3×1)

= 14 + 3

= 17 g/mol

Mole of NH₃ =?

Mole = mass /molar mass

Mole of NH₃ = 6.75 / 17

Mole of NH₃ = 0.397 mole

Next, we shall determine the number of mole of NO produced by the reaction of 0.397 mole of NH₃. This can be obtained as follow:

4NH₃ + 5O₂ —> 4NO + 6H₂O

From the balanced equation above,

4 moles of NH₃ reacted to produce 4 moles of NO.

Therefore, 0.397 mole of NH₃ will also react to produce 0.397 mole of NO.

Finally, we shall determine the mass of 0.397 mole of NO. This can be obtained as follow:

Mole of NO = 0.397 mole

Molar mass of NO = 14 + 16 = 30 g/mol

Mass of NO =?

Mass = mole × molar mass

Mass of NO = 0.397 × 30

Mass of NO = 11.9 g

Thus, the mass of NO produced is 11.9 g

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Calculate the entropy change for the reaction: Fe2O3(s) +3C(s) -> 2Fe(s) + 3CO(g)

Gelneren [198K]

Answer:

ΔS = +541.3Jmol⁻¹K⁻¹

Explanation:

Given parameters:

Standard Entropy of Fe₂O₃ = 90Jmol⁻¹K⁻¹

Standard Entropy of C = 5.7Jmol⁻¹K⁻¹

Standard Entropy of Fe = 27.2Jmol⁻¹K⁻¹

Standard Entropy of CO = 198Jmol⁻¹K⁻¹

To find the entropy change of the reaction, we first write a balanced reaction equation:

Fe₂O₃ + 3C → 2Fe + 3CO

To calculate the entropy change of the reaction we simply use the equation below:

ΔS = ∑S $_{products}$ - ∑S $_{reactants}$

Therefore:

ΔS = [(2x27.2) + (3x198)] - [(90) + (3x5.7)] = 648.4 - 107.1

ΔS = +541.3Jmol⁻¹K⁻¹

5 0

3 years ago

Hydrogen iodide is not produced by the same method is for hydrogen chloride why with reaction

Masja [62]

Using Phosphoric acid will work perfectly for producing Hydrogen halides because its not an Oxidizing agent. ...

Using an ionic chloride and Phosphoric acid

H3PO4 + NaCl ==> HCl + NaH2PO4

H3PO4 + NaI ==> HI + NaH2PO4

H2SO4 + NaCl ==> HCl + NaHSO4

This method(Using H2So4) will work for all hydrogen hydrogen halide except Hydrogen Iodide and Hydrogen Bromide.

The Sulphuric acid won't be useful for producing Hydrogen Iodide because its an OXIDIZING AGENT. Whist producing the Hydrogen Iodide... Some of the Iodide ions are oxidized to Iodine.

2I-² === I2 + 2e-

5 0

3 years ago

using the Bohr model for hydrogen: energy = hc/wavelength = 2.18 x 10^-18 Joules (1/nf2 - 1/ni2) N=15 to n=5

soldier1979 [14.2K]

Answer:

Energy lost is 7.63×10⁻²⁰J

Explanation:

Hello,

I think what the question is requesting is to calculate the energy difference when an excited electron drops from N = 15 to N = 5

E = hc/λ(1/n₂² - 1/n₁²)

n₁ = 15

n₂ = 5

hc/λ = 2.18×10⁻¹⁸J (according to the data)

E = 2.18×10⁻¹⁸ (1/n₂² - 1/n₁²)

E = 2.18×10⁻¹⁸ (1/15² - 1/5²)

E = 2.18×10⁻¹⁸ ×(-0.035)

E = -7.63×10⁻²⁰J

The energy lost is 7.63×10⁻²⁰J

Note : energy is lost / given off when the excited electron jumps from a higher energy level to a lower energy level

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3 years ago

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balu736 [363]

Answer:

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2 years ago

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Kipish [7]

Answer:

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i could help you if you tell me the proper question.

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3 years ago

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Chemistry help!Zoom in to see better!!​

Chemistry help!Zoom in to see better!!