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2 years ago

Chemical bonds contain energy that can be released when they are broken. True False

Chemistry

2 answers:

faust18 [17]2 years ago

7 0

The answer ox Falseeeeeeee......

Ilia_Sergeevich [38]2 years ago

7 0

The statement above is FALSE.

Chemical bonds do not contain energy that can be released when they are broken. In chemical reactions, bonds are typically broken and re-arrange in order to form new products. Energy is normally needed to break existing chemical bonds and energy is released when new chemical bonds are formed. Although energy is involved in breaking and forming of chemical bonds, the chemical bonds themselves are devoid of energy.

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Draw arrows for all of the attractive and repulsive forces between the negatively charged electrons and positively charged proto

Sophie [7]

Answer:

Attraction

(e-) ---> <--- (H+)

Repulsion:

<---(e-) (e-)-->

Neutral:

(e-) (Helium)

Explanation:

Accordingly to coulomb's law:

In the attraction, the hydrogen without an electron has a positive charge and needs to be fulfilled with a negative charge found in an eletron.

In the repulsion, both electrons has the same charge and repulse each other.

In the neutral case, the Helium is highly stable therefore the electron is not attracted.

4 0

2 years ago

50cm3 of 1 mol/dm3 HCl at 30°C was mixed with 50cm3 of 1mol/dm3 NaOH at 30°C in a styrofoam calorimeter. The temperature of the

trapecia [35]

Answer:

-21 kJ·mol⁻¹

Explanation:

Data:

H₃O⁺ + OH⁻ ⟶ 2H₂O

V/mL: 50 50

c/mol·dm⁻³: 1.0 1.0

ΔT = 4.5 °C

C = 4.184 J·°C⁻¹g⁻¹

C_cal = 50 J·°C⁻¹

Calculations:

(a) Moles of acid

$\text{Moles of acid} = \text{0.050 dm}^{3} \times \dfrac{\text{1.0 mol}}{\text{1 dm}^{3}} = \text{0.050 mol}\\\\\text{Moles of base} = \text{0.050 dm}^{3} \times \dfrac{\text{1.0 mol}}{\text{1 dm}^{3}} = \text{0.050 mol}$

So, we have 0.050 mol of reaction

(b) Volume of solution

V = 50 dm³ + 50 dm³ = 100 dm³

(c) Mass of solution

$\text{Mass of solution} = \text{100 dm}^{3} \times \dfrac{\text{1.00 g}}{\text{1 dm}^{3}} = \text{100 g}$

(d) Calorimetry

There are three energy flows in this reaction.

q₁ = heat from reaction

q₂ = heat to warm the water

q₃ = heat to warm the calorimeter

q₁ + q₂ + q₃ = 0

nΔH + mCΔT + C_calΔT = 0

0.050ΔH + 100×4.184×4.5 + 50×4.5 = 0

0.050ΔH + 1883 + 225 = 0

0.050ΔH + 2108 = 0

0.050ΔH = -2108

ΔH = -2108/0.0500

= -42 000 J/mol

= -42 kJ/mol

This is the heat of reaction for the formation of 2 mol of water

The heat of reaction for the formation of mol of water is -21 kJ·mol⁻¹.

5 0

3 years ago

After standardizing a NaOH solution, you use it to titrate an HCl solution known to have a concentration of 0.203 M. You perform

jek_recluse [69]

Answer:

0.203 is the mean of the concentration of the HCl solution

Explanation:

You have 5 concentrations. The most appropiate result is the mean of those results. The mean is a statistical defined as the sum of each result divided by the total amount of results. For the results of the problem, the mean is:

0.210 + 0.204 + 0.201 + 0.202 + 0.197 = 1.014 / 5 =

<h3>0.203 is the mean of the concentration of the HCl solution</h3>

8 0

2 years ago

____ HBr + ____ Mg(OH)2 ---> ____ MgBr2 + ____ H2O

swat32

Hello Camkirkland,
I think that you are trying to balance this equation.
In order to balance a chemical equation, the numbers of atoms of each element must be equal on both sides of the equation.

In this particular equation, the answer would be (2) HBr + (1) Mg(OH)2 ---> (1) MgBr2 + (2) H2O.

Hope this answers your question!

4 0

3 years ago

What role does the wire play in voltaic cell

VashaNatasha [74]

It allows electrons to flow from the anode to the cathode.

8 0

2 years ago