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2 years ago

Compare the purple and orange arrows. Does more energy go into breaking bonds, or is more energy released when new bonds form?

Chemistry

1 answer:

Nastasia [14]2 years ago

3 0

Answer:

The breaking of chemical bonds never releases energy to the external environment. Energy is only released when chemical bonds are formed. In general, a chemical reaction involves two steps: 1) the original chemical bonds between the atoms are broken, and 2) new bonds are formed. These two steps are sometimes lumped into one event for simplicity, but they are really two separate events. For instance, when you burn methane (natural gas) in your stove, the methane is reacting with oxygen to form carbon dioxide and water. Chemists often write this as:

CH4 + 2 O2 → CO2 + 2 H2O + energy

Explanation:

here is the answer

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What is the change in enthalpy associated with the combustion of 530 g of methane (CH4)?Report your answer in scientific notatio

Veseljchak [2.6K]

Answer:

$-3.0\times 10^4 kJ$ is the change in enthalpy associated with the combustion of 530 g of methane.

Explanation:

$CH_4+2O_2\rightarrow CO_2+2H_2O,\Delta H_{comb}=-890.8 kJ/mol$

Mass of methane burnt = 530 g

Moles of methane burnt = $\frac{530 g}{16 g/mol}=33.125 mol$

Energy released on combustion of 1 mole of methane = -890.8 kJ/mol

Energy released on combustion of 33.125 moles of methane :

$-890.8 kJ/mol\times 33.125 mol=-29,507.75 kJ=-2.950775\times 10^4 kJ$

$-2.950775\times 10^4 kJ\approx -3.0\times 10^4 kJ$

$-3.0\times 10^4 kJ$ is the change in enthalpy associated with the combustion of 530 g of methane.

7 0

2 years ago

Consider a buffer solution containing CH3COOH and CH3COO-, with an equilibrium represented by: CH3COOH(aq) + H2O (l) ←----→ H3O+

Irina-Kira [14]

Answer:

Here's what I get.

Explanation:

(a) The buffer equilibrium

The equation for the buffer equilibrium is

$\rm CH_{3}COOH(aq) + H$_{2}$O(l) $\, \rightleftharpoons \,$ CH$_{3}$COO$^{-}$(aq) + H$_{3}$O$^{+}$(aq)$

(b) Addition of acid

If you add a strong acid like HNO₃, you are increasing the concentration of hydronium ion.

Per Le Châtelier's Principle, the system will respond in such a way as to decrease the concentration of hydronium ion.

The position of equilibrium will shift to the left.

(c) Addition of base.

If you add a strong base like KOH, The hydroxide ions will react with the hydronium ions to form water.

The concentration of hydronium ions will decrease.

Per Le Châtelier's Principle, the system will respond in such a way as to increase the concentration of hydronium ions.

The position of equilibrium will shift to the right.

5 0

2 years ago

The density of mercury is 13.6 g/cm3 . What volume (in quarts) is occupied by 100. g of Hg? (1 L = 1.06 qt)

SpyIntel [72]

Answer:

0.00077 qt

Explanation:

Density -

Density of a substance is given by the mass of the substance divided by the volume of the substance .

Hence , d = m / V

V = volume

m = mass ,

d = density ,

From the question ,

The mass mercury = 100 g

Density of mercury = 13.6 g/cm³ .

Hence , by using the above formula ,and putting the corresponding values , the volume of mercury is calculated as -

d = m / V

13.6 g/cm³ = 100 g / V

V = 7.35 cm³

1 cm³ = 0.001 L

V = 7.35 * 0.001 L = 0.0073 L

Since ,

1 L = 1.06 qt

V = 0.0073* 1.06 qt = 0.0077 qt

6 0

2 years ago

In May 2016, William Trubridge broke the world record in free diving (diving underwater without the use of supplemental oxygen)

Maru [420]

Answer:

The volume that this same amount of air will occupy in his lungs when he reaches a depth of 124 m is - 0.27 L.

Explanation:

Using Boyle's law

${P_1}\times {V_1}={P_2}\times {V_2}$

Given ,

V₁ = 3.6 L

V₂ = ?

P₁ = 1.0 atm

P₂ = 13.3 atm (From correct source)

Using above equation as:

${P_1}\times {V_1}={P_2}\times {V_2}$

${1.0\ atm}\times {3.6\ L}={13.3\ atm}\times {V_2}$

${V_2}=\frac{{1.0}\times {3.6}}{13.3}\ L$

${V_2}=0.27\ L$

The volume that this same amount of air will occupy in his lungs when he reaches a depth of 124 m is - 0.27 L.

7 0

2 years ago

Compound A, C6H12 reacts with HBr/ROOR to give compound B, C6H13Br. Compound C, C6H14, reacts with bromine and light to produce

Savatey [412]

Answer:

The question involves drawing of structures and showing mechanism in which brainly text editor did not support. I made sure I created a pdf file with both the anwsers and explanations in it. The pdf can be found in the attachment below.

Explanation:

Download pdf

3 0

2 years ago