Can anyone help me please this is due and i need help someone check over my answers

Calculate AH for the reaction:<br> 2N2 (g) + 6H20 (g) → 302 (g) + 4NH3(g)

Ratling [72]

Answer:

Explanation:

4NH₃ (g) + 3O₂ (g) ⇒ 2N₂ (g) + 6H₂ O(1)

Δ $H_r={\sum}{ \Delta }H_f(\text{product})-{\sum}{ \Delta }H_f(\text{reactant})$

ΔH r =(2ΔH f(N 2 )+6ΔH f (H 2 O(l)))−(4ΔH f (NH 3 (g))+3ΔH f (O 2 (g)))

ΔH rex =[2×0+6×(−286)]−[4×(−46)+3×0]=−1716+186

ΔH rex =−1532kJ/mol

Thermodynamics is a branch of physical chemistry that studies heat and its effects and interactions. Governed by the four main laws, thermodynamics plays a huge role in physics and chemistry, and is also responsible for the law of conservation of energy, a fundamental rule in science.

7 0

1 year ago

What is the standard value for T in the common form of the Nernst equation?

LuckyWell [14K]

C.

52° Celsius

Hope this helped!

6 0

3 years ago

If 16.00 g of O₂ reacts with 80.00 g NO, how many the excess reactant are left over? (enter only the value, round to whole numbe

pishuonlain [190]

Answer:

50

Explanation:

We will need a balanced equation with masses, moles, and molar masses of the compounds involved.

1. Gather all the information in one place with molar masses above the formulas and masses below them.

Mᵣ: 30.01 32.00 46.01

2NO + O₂ ⟶ 2NO₂

Mass/g: 80.00 16.00

2. Calculate the moles of each reactant

$\text{moles of NO} = \text{80.00 g NO} \times \dfrac{\text{1 mol NO}}{\text{30.01 g NO}} = \text{2.666 mol NO}\\\\\text{moles of O}_{2} = \text{16.00 g O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{32.00 g O}_{2}} = \text{0.5000 mol O}_{2}$

3. Calculate the moles of NO₂ we can obtain from each reactant

From NO:

The molar ratio is 2 mol NO₂:2 mol NO

$\text{Moles of NO}_{2} = \text{2.333 mol NO} \times \dfrac{\text{2 mol NO}_{2}}{\text{2 mol NO}} = \text{2.333 mol NO}_{2}$

From O₂:

The molar ratio is 2 mol NO₂:1 mol O₂

$\text{Moles of NO}_{2} = \text{0.5000 mol O}_{2}\times \dfrac{\text{2 mol NO}_{2}}{\text{1 mol Cl}_{2}} = \text{1.000 mol NO}_{2}$

4. Identify the limiting and excess reactants

The limiting reactant is O₂ because it gives the smaller amount of NO₂.

The excess reactant is NO.

5. Mass of excess reactant

(a) Moles of NO reacted

The molar ratio is 2 mol NO:1 mol O₂

$\text{Moles reacted} = \text{0.500 mol O}_{2} \times \dfrac{\text{2 mol NO}}{\text{1 mol O}_{2}} = \text{1.000 mol NO}$

(b) Mass of NO reacted

$\text{Mass reacted} = \text{1.000 mol NO} \times \dfrac{\text{30.01 g NO}}{\text{1 mol NO}} = \text{30.01 g NO}$

(c) Mass of NO remaining

Mass remaining = original mass – mass reacted = (80.00 - 30.01) g = 50 g NO

5 0

3 years ago

Part A

Elden [556K]

Answer:

ΔG° = -5.4 kJ/mol

ΔG = 873.2 J/mol = 0.873 kJ /mol

Explanation:

Step 1: Data given

ΔG (NO2) = 51.84 kJ/mol

ΔG (N2O4) = 98.28 kJ/mol

Step 2:

ΔG = ΔG° + RT ln Q

⇒with Q = the reaction quatient

⇒with T = the temperature = 298 K

⇒with R = 8.314 J / mol*K

⇒with ΔG° = ΔG° (N2O4) - 2*ΔG°(NO2 )

⇒ ΔG° = 98.28 kJ/mol - 2* 51.84 kJ/mol

⇒ ΔG° = -5.4 kJ/mol

Part B

ΔG = ΔG° =RT ln Q

⇒with G° = -5.4 kj/mol = -5400 j/mol

⇒ with R = 8.314 J/K*mol

⇒with T = 298 K

⇒with Q = p(N2O4)/ [ p(NO2) ]² = 1.63/0.36² = 12.577

ΔG = -5400 + 8.314 * 298 * ln(12.577)

ΔG = -5400 + 8.314 * 298 * 2.532

ΔG = 873.2 J/mol = 0.873 kJ/mol

3 0

3 years ago

Which statement describes the location of metalloids on the periodic table? The metalloids are located below metals and above no

Talja [164]

The metalloids are Boron, Silicon, Geranium, etc and are found to the right of the metals and the left of the nonmetals. Since that is not an option, the best choice would be: The metalloids are located below nonmetals and above metals within a group.

4 0

3 years ago

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