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2 years ago

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Write the congruence condition in symbolic form for each pair of triangles in pic above. If wrong answer I will report. If corre

ct answer I will mark brainliest.

1 answer:

arsen [322]2 years ago

4 0

Answer: ABC&AMN are congruent by ASA comgruence

angle A=A

sideAN=AB

angle M=C

Explanation:

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Unlike most chemical changes, most physical changes are easily reversed. When Adrian's teacher dissolves some sugar in a beaker

pochemuha

Answer:

By heating the solution

Explanation:

Physical changes and chemical changes are the two types of changes that a substance undergoes. Physical change does not alter the substance's chemical composition, hence, can be easily reversed. There is also no new product formed. This is contrary to the occurrences of a chemical change, which cannot be reversed after a new product has been formed.

A physical change is what occurs when Adrian's teacher dissolves some sugar in a beaker of water to form a sugar solution. This change does not involve any new product formation, hence, can be reversed. The sugar can be derived back from the solution by HEATING THE SOLUTION. The water (solvent) will evaporate and the sugar (solute) will precipitate.

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Use bond energies to calculate the enthalpy of reaction for the combustion of ethane. Average bond energies in kJ/mol C-C 347, C

ivanzaharov [21]

The enthalpy of reaction for the combustion of ethane 2CH₃CH₃ + 7O₂ → 4CO₂ + 6H₂O calculated from the average bond energies of the compounds is -2860 kJ/mol.

The reaction is:

2CH₃CH₃ + 7O₂ → 4CO₂ + 6H₂O (1)

The enthalpy of reaction (1) is given by:

$\Delta H = \Delta H_{r} - \Delta H_{p}$ (2)

Where:

r: is for reactants

p: is for products

The bonds of the compounds of reaction (1) are:

2CH₃CH₃: 2 moles of 6 C-H bonds + 2 moles of 1 C-C bond

7O₂: 7 moles of 1 O=O bond

4CO₂: 4 moles of 2 C=O bonds

6H₂O: 6 moles of 2 H-O bonds

Hence, the enthalpy of reaction (1) is (eq 2):

$\Delta H = \Delta H_{r} - \Delta H_{p}$

$\Delta H = 2*\Delta H_{CH_{3}CH_{3}} + 7\Delta H_{O_{2}} - (4*\Delta H_{CO_{2}} + 6*\Delta H_{H_{2}O})$

$\Delta H = 2*(6*\Delta H_{C-H} + \Delta H_{C-C}) + 7\Delta H_{O=O} - (4*2*\Delta H_{C=O} + 6*2*\Delta H_{H-O})$

$\Delta H = [2*(6*413 + 347) + 7*498 - (4*2*799 + 6*2*467)] kJ/mol$

$\Delta H = -2860 kJ/mol$

Therefore, the enthalpy of reaction for the combustion of ethane is -2860 kJ/mol.

Read more here:

brainly.com/question/11753370?referrer=searchResults

I hope it helps you!

7 0

2 years ago