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Chemistry

1 answer:

kvv77 [185]1 year ago

6 0

The bonds that are present in the pictured compound are C −C and C −H.

A bond is that which connects two atoms together. We know that a compound is composed of bonds. There are a number of bonds in a compound. The bonds are always as many as the atoms that need to be linked in the compound.

We know that butane a shown in the image is an alkane. An alkane has carbon - carbon and hydrogen - carbon bonds. There are no multiple bonds in the compound butane.

Thus, the bonds that are present in the pictured compound are C −C and C −H.

Learn more about chemical bond:brainly.com/question/15444131

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If the oxidation state of H is +1 and O is -2, what is the oxidation state of C in C2H4O?

Sedbober [7]

Answer:

-1

Explanation:

According to this question, the oxidation state/number of H and O in C2H4O is +1 and -2 respectively.

The oxidation state of carbon in the compound can be calculated thus:

Where;

x represents the oxidation number of C

C2H4O = 0 (net charge)

x(2) + 1(4) - 2 = 0

2x + 4 - 2 = 0

2x + 2 = 0

2x = -2

Divide both sides by 2

x = -1

The oxidation number of C in C2H4O is -1.

8 0

2 years ago

Using standard heats of formation, calculate the standard enthalpy change for the following reaction. 2NH3(g) 3N2O(g)4N2(g) 3H2O

trasher [3.6K]

Answer: $\Delta H^{0} = -879.15 kJ/mol$

Explanation: Heats of formation is the amount of heat necessary to create 1 mol of a compound from its molecular constituents. The basic conditions the substance is formed is at standard conditions: 1 atm and 25°C. Each compound has its own heat of formation per mol of compound (kJ/mol), but to an element is assigned a value of zero.

Standard Enthalpy Change is defined as the heat absorbed or released when a reaction takes place. It can be positive or negative, which means reaction is endothermic or exothermic, respectively.

Enthalpy change is calculated as the difference between the sum of heat formation of products and the sum of heat formation of the reactants:

$\Delta H^{0}=\Sigma H^{0}_{f}_{(products)}-\Sigma H^{0}_{f}_{(reactants)}$