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Tomtit [17]
3 years ago
7

Which observation about methane (ch4) can be explained using the orbital hybridization theory?

Chemistry
1 answer:
MissTica3 years ago
7 0
<span>All 4 C-H bonds are identical.

Hybridisation or hybridization is the idea of blending atomic orbitals into new hybrid orbitals, with various energies, shapes, and so on, than the component atomic orbitals, appropriate for the matching of electrons to make chemical bonds in valence bond theory.
</span>
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Suppose there is 0.63 g of HNO3 per 100 mL of a particular solution. What is the concentration of the HNO3 solution in moles per
Vadim26 [7]

Answer:

There are 0.09996826 moles per liter of the solution.

Explanation:

Molar mass of HNO3: 63.02

Convert grams to moles

0.63 grams/ 63.02= 0.009996826

Convert mL to L and place under moles (mol/L)

100mL=0.1 L

0.009996826/0.1= 0.09996826 mol/L

4 0
3 years ago
What is the answer to 2.5 * 22.5
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Your answer to 2.5*22.56 is 56.25
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Places with _______ altitude will have ______ climates. low; cold high; cold low; freezing high; warm
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Places with high altitudes will have cold climates.
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Honeybee venom is a solution that contains formic acid, which gives the venom an approximate ph of 5.0. which substance is more
Genrish500 [490]
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5 0
4 years ago
Two solutions namely, 500 ml of 0.50 m hcl and 500 ml of 0.50 m naoh at the same temperature of 21.6 are mixed in a constant-pre
weeeeeb [17]

24.6 ℃

<h3>Explanation</h3>

Hydrochloric acid and sodium hydroxide reacts by the following equation:

\text{HCl} \; (aq) + \text{NaOH} \; (aq) \to \text{NaCl} \; (aq) + \text{H}_2\text{O} \; (aq)

which is equivalent to

\text{H}^{+} \; (aq) + \text{OH}^{-} \; (aq) \to \text{H}_2\text{O}\; (l)

The question states that the second equation has an enthalpy, or "heat", of neutralization of -56.2 \; \text{kJ}. Thus the combination of every mole of hydrogen ions and hydroxide ions in solution would produce 56.2 \; \text{kJ} or 56.2 \times 10^{3}\; \text{J} of energy.

500 milliliter of a 0.50 mol per liter "M" solution contains 0.25 moles of the solute. There are thus 0.25 moles of hydrogen ions and hydroxide ions in the two 0.500 milliliter solutions, respectively. They would combine to release 0.25 \times 56.2 \times 10^{3} = 1.405 \times 10^{4} \; \text{J} of energy.

Both the solution and the calorimeter absorb energy released in this neutralization reaction. Their temperature change is dependent on the heat capacity <em>C</em> of the two objects, combined.

The question has given the heat capacity of the calorimeter directly.

The heat capacity (the one without mass in the unit) of water is to be calculated from its mass and <em>specific</em> heat.

The calorimeter contains 1.00 liters or 1.00 \times 10^{3} \; \text{ml} of the 1.0 gram per milliliter solution. Accordingly, it would have a mass of 1.00 \times 10^{3} \; \text{g}.

The solution has a specific heat of 4.184 \; \text{J} \cdot \text{g}^{-1} \cdot \text{K}^{-1}. The solution thus have a heat capacity of 4.184 \times 1.00 \times 10^{3} = 4.184 \times 10^{3} \; \text{J} \cdot\text{K}^{-1}. Note that one degree Kelvins K is equivalent to one degree celsius ℃ in temperature change measurements.

The calorimeter-solution system thus has a heat capacity of 4.634 \times 10^{3} \; \text{J} \cdot \text{K}^{-1}, meaning that its temperature would rise by 1 degree celsius on the absorption of 4.634 × 10³ joules of energy. 1.405 \times 10^{4} \; \text{J} are available from the reaction. Thus, the temperature of the system shall have risen by 3.03 degrees celsius to 24.6 degrees celsius by the end of the reaction.

4 0
3 years ago
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