If 3.00 g of limestone reacted, what mass of calcium chloride would be produced?

Given the three equations below, what is the heat of reaction for the production of glucose, C6H12O6, as described by this equat

UkoKoshka [18]

Answer:

- 1273.02 kJ.

Explanation:

This problem can be solved using Hess's Law.

Hess's Law states that regardless of the multiple stages or steps of a reaction, the total enthalpy change for the reaction is the sum of all changes. This law is a manifestation that enthalpy is a state function.

We should modify the given 3 equations to obtain the proposed reaction:

6C(s) + 6H₂(g) + 3O₂(g) → C₆H₁₂O₆(s),

We should multiply the first equation by (6) and also multiply its ΔH by (6):

6C(s) + 6O₂(g) → 6CO₂(g), ∆H₁ = (6)(–393.51 kJ) = - 2361.06 kJ,

Also, we should multiply the second equation and its ΔH by (6):

6H₂(g) + 3O₂(g) → 6H₂O(l), ∆H₂ = (6)(–285.83 kJ) = - 1714.98 kJ.

Finally, we should reverse the first equation and multiply its ΔH by (- 1):

6CO₂(g) + H₂O(l) → C₆H₁₂O₆(s) + 6O₂(g), ∆H₃ = (-1)(–2803.02 kJ) = 2803.02 kJ.

By summing the three equations, we cam get the proposed reaction:

6C(s) + 6H₂(g) + 3O₂(g) → C₆H₁₂O₆(s),

And to get the heat of reaction for the production of glucose, we can sum the values of the three ∆H:

∆Hrxn = ∆H₁ + ∆H₂ + ∆H₃ = (- 2361.06 kJ) + (- 1714.98 kJ) + (2803.02 kJ) = - 1273.02 kJ.

6 0

3 years ago

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Need help I don’t understand this at all need to show my work and strategy it’s stoichiometry gram to gram please help

laila [671]

Answer:

781×10¯² g of MgCl₂.

Explanation:

The balanced equation for the reaction is given below:

Mg + 2HCl —> MgCl₂ + H₂

Next, we shall determine the mass of HCl that reacted and the mass of MgCl₂ produced from the balanced equation. This is illustrated below:

Molar mass of HCl = 1 + 35.5

= 36.5 g/mol

Mass of HCl from the balanced equation = 2 × 36.5 = 73 g

Molar mass of MgCl₂ = 24 + (35.5×2)

= 24 + 71

= 95 g/mol

Mass of MgCl₂ from the balanced equation = 1 × 95 = 95 g

SUMMARY:

From the balanced equation above,

73 g of HCl reacted to produce 95 g of MgCl₂.

Finally, we shall determine the mass of MgCl₂ produced by the reaction of 6 g of HCl. This can be obtained as follow:

From the balanced equation above,

73 g of HCl reacted to produce 95 g of MgCl₂.

Therefore, 6 g of HCl will react to produce = (6 × 95)/73 = 781×10¯² g of MgCl₂.

Thus, 781×10¯² g of MgCl₂ were obtained from the reaction.

4 0

3 years ago

Tartaric acid, C4H6O6, has the first ionization constant with the value: Ka1 = 9.20 × 10-4. Calculate the value of pKb for the c

nirvana33 [79]

Answer:

pKb = 10.96

Explanation:

Tartaric acid is a dyprotic acid. It reacts to water like this:

H₂Tart + H₂O ⇄ H₃O⁺ + HTart⁻ Ka1

HTart⁻ + H₂O ⇄ H₃O⁺ + Tart⁻² Ka2

When we anaylse the base, we have

Tart⁻² + H₂O ⇄ OH⁻ + HTart⁻ Kb1

HTart⁻ + H₂O ⇄ OH⁻ + H₂Tart Kb2

Remember that Ka1 . Kb2 = Kw, plus pKa1 + pKb2 = 14

Kb2 = Kw / Ka1 → 1×10⁻¹⁴ / 9.20×10⁻⁴ = 1.08×10⁻¹¹

so pKb = - log Kb2 → - log 1.08×10⁻¹¹ = 10.96

7 0

3 years ago

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Pls help The white light of the sun is a mixture of all the colors.

Anettt [7]

false the black light is

7 0

3 years ago

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What are the principal alloy elements of an AISI 4340 steel? How much carbon does it have? Is it hardenable? By what techniques?

amid [387]

Answer:

The principal elements are Iron, Nickel, Chromium

It has 0.37% to 0.43% Carbon

It is hardenable.

It can be hardened by cold working, annealing, quenching.

Explanation:

A)

The chemical composition of AISI 4340 Steel are as follows:

Iron --- 95% to 96%

Nickel --- 1.6% to 2.0%

Chromium --- 0.7% to 0.9%

Manganese --- 0.6% to 0.8%

Carbon --- 0.37% to 0.43%

Molybdenum --- 0.2% to 0.3%

Silicon --- 0.15% to 0.30%

Sulfur --- 0.040%

Phosphorous --- 0.0350%

So, the principal alloy elements from this composition are Iron, Nickel and Chromium

B)

The carbon content in this alloy is 0.37% to 0.43%

C)

Yes, it can be hardened.

D)

It can be Hardened by Cold Working, Annealing or Quenching.

3 0

3 years ago