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Alex_Xolod [135]
2 years ago
3

Which of the statements below correctly describes the combustion of glucose, shown below? c6h12o6 + 6o2 → 6co2 + 6h2o?

Chemistry
1 answer:
docker41 [41]2 years ago
3 0
I think I found the complete problem with the given choices. We know that O₂ is the oxidizing agent or the one that's reduced, while glucose is the reducing agent or the one that's oxidized. Following these information, the answer would be letter C, or the third choice.

The oxidation number of H before the reaction is -1, which became 1 after the reaction. The oxidation number increased, thus, H is the reducing agent.

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2 years ago
What is the mass of a solution that has a density of 0.775 g/ml and a volume of 50.0ml?
Margarita [4]
Hey there!

D = m / V

0.775 = m / 50.0

m = 0.775 * 50.0

m = 38.75 g
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Find a part of the article that describes signals that are sent within Diego’s body. Where does the signal come from, and how do
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The article "The Melting Arctic" attempts to win over public opinion by making use of persuasive techniques. One such technique
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The decomposition of copper(II) nitrate on heating is endothermic reaction. 2Cu(NO3)2(s) → 2C10(s) + 4NO2(g) + O2(g) Calculate t
Basile [38]

Answer:

The enthalpy change for the given reaction is 424 kJ.

Explanation:

2Cu(NO_3)_2(s)\rightarrow 2CuO(s) + 4NO_2(g) + O_2(g),\Delta H_{rxn}=?

We have :

Enthalpy changes of formation of following s:

\Delta H_{f,Cu(NO_3)_2}=-302.9 kJ/mol

\Delta H_{f,CuO}=-157.3 kJ/mol

\Delta H_{f,NO_2}= 33.2 kJ/mol

\Delta H_{f,O_2}= 0 kJ/mol (standard state)

\Delta H_{rxn}=\sum [\Delta H_f(product)]-\sum [\Delta H_f(reactant)]

The equation for the enthalpy change of the given reaction is:

\Delta H_{rxn} =

=(2 mol\times \Delta H_{f,CuO}+4\times \Delta H_{f,NO_2}+1 mol\times \Delta H_{f,O_2})-(2mol\times \Delta H_{f,Cu(NO_3)_2})

\Delta H_{rxn}=

(2mol\times (-157.3 kJ/mol)+4\times 33.2 kJ/mol=1 mol\times 0 kJ/mol)-(2 mol\times (-302.9 kJ/mol)

\Delta H_{rxn}=424 kJ

The enthalpy change for the given reaction is 424 kJ.

6 0
2 years ago
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