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Kamila [148]
3 years ago
5

What is the reactants start for H2 + O2 --> H2O

Chemistry
2 answers:
zaharov [31]3 years ago
7 0

Answer:

H2 + O2

Explanation:

H2 + O2 --> H2O

The items on the left side is the reactants.  The arrow means yields, and the item the arrow is pointing at is the product.

Alecsey [184]3 years ago
3 0

Answer:

h²0

Explanation:

I'm sorry , id ask g00gle

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Mercury is the only metal that is a liquid at room temperature. When mercury vapor is inhaled, it is readily absorbed by the lun
Lapatulllka [165]

Answer:

P = 0.0166 mm Hg

Explanation:

To solve this question, we need to use the Clausius Clapeyron equation, which is a commonly used expression to calculate vapour pressure at a given temperature. We have the enthalpy of vaporization of the mercury, so, let's write the equation:

Clausius Clapeyron equation:

Ln (P₂ / P₁) = (-ΔHv / R)(1/T₂ - 1/T₁)    (1)

Where:

R: universal constant of gases (8.314 J / K.mol)

P₂: Vapour pressure at 43°C (or 316 K)

P₁: Pressure of mercury at the boiling point (1 atm)

T₂: temperature at 43 °C

T₁: Boiling point of mercury (357 °C or 630 K)

As we are given the boiling point of the mercury, we can safely assume that the pressure at this point is 1 atm, becuase remember that when a sustance boils, is because it's internal pressure has reached the atmospherical pressure of 1 atm. With this clear, all we just need to do is solve for P₂. We are going to do this very slowly so you can understand the process. First let's replace the given data:

Ln (P₂ / 1) = (-59100 J/mol / 8.314 J / K.mol) (1/316 - 1/630)

Ln P₂ = -7108.49 * (3.16x10⁻³ - 1.59x10⁻³)

Ln P₂ = -7108.49 * (1.51x10⁻³)

Ln P₂ = -10.7338

P₂ = 10⁽⁻¹⁰°⁷³³⁸⁾

P₂ = 2.18x10⁻⁵ atm

We can express this value in mm Hg and it will be:

P₂ = 2.18x10⁻⁵ * 760

<h2>P₂ = 0.0166 mm Hg</h2>

Hope this helps

8 0
2 years ago
15 POINTS PLS HELP!!!!!
scZoUnD [109]

Answer: (C) Vaporizing

Explanation:

 Vaporization is the process in which the substance change the state of of liquid into the gas state.

The vaporization process require the largest input of the energy as when the state is in the solid state then, the solid substances contain the strong forces of the attraction and they require high energy to break these strong bonds.

For changing the liquid state into the gases state we require to overcome the surface tension and require enough energy for acquiring the vaporization state.

Therefore, option (C) is correct.

5 0
3 years ago
Use standard enthalpies of formation to calculate Δ H ∘ rxn for each reaction. MISSED THIS? Read Section 7.9; Watch KCV 7.9, IWE
Eva8 [605]

Answer:

Standard Heat of Reaction 1 = -136.2 kJ/mol

Standard Heat of Reaction 2 = -41.166 kJ/mol

Standard Heat of Reaction 3 = -136.07 kJ/mol

Standard Heat of Reaction 4 = 279.448kJ/mol

Explanation:

C₂H₄ (g) + H₂ (g) → C₂H₆ (g)

CO (g) + H₂O (g) → H₂ (g) + CO₂ (g)

3NO₂ (g) + H₂O (l) → 2HNO₃ (aq) + NO (g)

Cr₂O₃ (s) + 3CO (g) → 2Cr (s) + 3CO₂ (g)

The required standard heat of formation for each of the reactants and product above, as obtained from literature is listed below.

C₂H₄ (g), 52.5 kJ/mol

H₂ (g), 0 kJ/mol

C₂H₆ (g), -83.7 kJ/mol

CO (g), -110.525 kJ/mol

H₂O (g), -241.818 kJ/mol

H₂ (g), 0 kJ/mol

CO₂ (g), -393.509 kJ/mol

NO₂ (g), 33.2 kJ/mol

H₂O (l), -285.8 kJ/mol

HNO₃ (aq), -206.28 kJ/mol

NO (g), 90.29 kJ/mol

Cr₂O₃ (s), -1128.4 kJ/mol

CO (g), -110.525 kJ/mol

Cr (s), 0 kJ/mol

CO₂ (g), -393.509 kJ/mol

Note that

ΔH∘(rxn) = ΔH∘(products) - ΔH∘(reactants)

C₂H₄ (g) + H₂ (g) → C₂H₆ (g)

ΔH∘(rxn) = ΔH∘(products) - ΔH∘(reactants)

ΔH∘(products) = (1×-83.7) = -83.7 kJ/mol

ΔH∘(reactants) = (1×52.5) + (1×0) = 52.5 kJ/mol

ΔH∘(rxn) = -83.7 - 52.5 = -136.2 kJ/mol

CO (g) + H₂O (g) → H₂ (g) + CO₂ (g)

ΔH∘(rxn) = ΔH∘(products) - ΔH∘(reactants)

ΔH∘(products) = (1×0) + (1×-393.509) = -393.509 kJ/mol

ΔH∘(reactants) = (1×-110.525) + (1×-241.818) = -352.343 kJ/mol

ΔH∘(rxn) = -393.509 - (-352.343) = -41.166 kJ/mol

3NO₂ (g) + H₂O (l) → 2HNO₃ (aq) + NO (g)

ΔH∘(rxn) = ΔH∘(products) - ΔH∘(reactants)

ΔH∘(products) = (2×-206.28) + (1×90.29) = -322.27 kJ/mol

ΔH∘(reactants) = (3×33.2) + (1×-285.8) = -186.2 kJ/mol

ΔH∘(rxn) = -322.27 - (-186.2) = -136.07 kJ/mol

Cr₂O₃ (s) + 3CO (g) → 2Cr (s) + 3CO₂ (g)

ΔH∘(rxn) = ΔH∘(products) - ΔH∘(reactants)

ΔH∘(products) = (2×0) + (3×-393.509) = -1,180.527 kJ/mol

ΔH∘(reactants) = (1×-1128.4) + (3×-110.525) = -1,459.975 kJ/mol

ΔH∘(rxn) = -1,180.527 - (-1,459.975) = 279.448 kJ/mol

Hope this Helps!!!

4 0
3 years ago
PLLLLLLLZ HELP IS NEEDED ASAP I WOULD REALLY APPRECIATE IT
Lesechka [4]

Answer:

1)Reactants

2)Light

3)An item that can increase reaction rates

4)Reactants must collide with each other

Less molecules lower the chance for collisions

The more collisions there are the higher the reaction rate

5 0
3 years ago
Wind power is:
12345 [234]
Answer) All of the above
7 0
2 years ago
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