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2 years ago

Which statement about bonding is correct? (1 point)

2 answers:

7 0

The statement correct about bonding is the release of energy in bond formation, and is exothermic. Thus, option C is correct.

The bonds have been defined as the sharing of electrons between atoms for the stabilization and forming compounds. The formation and breaking of the bond results in the absorption and release of the energy.

<h3>Endothermic and Exothermic reaction</h3>

The reaction in which the energy has been released has been termed as exothermic reaction. The reaction in which energy is absorbed is termed as endothermic system.

The bonds are stable, and to break the bond energy has to be provided to the system. The energy will result in the delocalization of electrons and thereby breaking of bonds.

Since energy has been absorbed by the system during breaking of bonds, the reaction has been an endothermic reaction.

The formation of the bond has been releases the energy in the system and is an exothermic reaction. Thus, option C is correct.

Learn more about bonding, here:
brainly.com/question/13559242

MA_775_DIABLO [31]2 years ago

6 0

Answer: Forming bonds releases energy and is exothermic

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What mass of solute is contained in 25.4 ml of a 1.56 m potassium bromide solution?

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Answer is: mass of <span>potassium bromide is 4.71 grams.
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3 0

2 years ago

Read 2 more answers

7.00 of Compound x with molecular formula C3H4 are burned in a constant-pressure calorimeter containing 35.00kg of water at 25c.

beks73 [17]

Answer:

$\Delta H_{f,C_3H_4}=276.8kJ/mol$

Explanation:

Hello!

In this case, since the equation we use to model the heat exchange into the calorimeter and compute the heat of reaction is:

$\Delta H_{rxn} =- m_wC_w\Delta T$

We plug in the mass of water, temperature change and specific heat to obtain:

$\Delta H_{rxn} =- (35000g)(4.184\frac{J}{g\°C} )(2.316\°C)\\\\\Delta H_{rxn}=-339.16kJ$

Now, this enthalpy of reaction corresponds to the combustion of propyne:

$C_3H_4+4O_2\rightarrow 3CO_2+2H_2O$

Whose enthalpy change involves the enthalpies of formation of propyne, carbon dioxide and water, considering that of propyne is the target:

$\Delta H_{rxn}=3\Delta H_{f,CO_2}+2\Delta H_{f,H_2O}-\Delta H_{f,C_3H_4}$

However, the enthalpy of reaction should be expressed in kJ per moles of C3H4, so we divide by the appropriate moles in 7.00 g of this compound:

$\Delta H_{rxn} =-339.16kJ*\frac{1}{7.00g}*\frac{40.06g}{1mol}=-1940.9kJ/mol$

Now, we solve for the enthalpy of formation of C3H4 as shown below:

$\Delta H_{f,C_3H_4}=3\Delta H_{f,CO_2}+2\Delta H_{f,H_2O}-\Delta H_{rxn}$

So we plug in to obtain (enthalpies of formation of CO2 and H2O are found on NIST data base):

$\Delta H_{f,C_3H_4}=3(-393.5kJ/mol)+2(-241.8kJ/mol)-(-1940.9kJ/mol)\\\\\Delta H_{f,C_3H_4}=276.8kJ/mol$

Best regards!

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2 years ago

Which formula can be used to find velocity if kinetic energy and mass are known?

astraxan [27]

Answer:

$v = \sqrt{ \frac{2ke}{m} }$

option D is the correct option.

Here,

If an object of mass 'm' moving with a velocity'v' then,

$kinetic \: energy = \frac{1}{2} m {v}^{2} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: {v}^{2} = \frac{2ke}{m} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: {v}^{2} = \sqrt{ \frac{2ke}{m} }$

hope this helps...

Good luck on your assignment..

5 0

2 years ago