Does it require more energy to "vaporize" water at the boiling point or to melt water at the melting point? Explain.

At the equivalence point of a titration of the [H+] concentration is equal to:

icang [17]

B. At the equivalence point of a titration of the [H+] concentration is equal to 7.

<h3>What is equivalence point of a titration?</h3>

The equivalence point of a titration is a point in titration at which the amount of titrant added is just enough to completely neutralize the analyte solution.

At the equivalence point in an acid-base titration, moles of base equals moles of acid and the solution only contains salt and water.

At the equivalence point, equal amounts of H+ and OH- ions combines as shown below;

H⁺ + OH⁻ → H₂O

The pH of resulting solution is 7.0 (neutral).

Thus, the pH at the equivalence point for this titration will always be 7.0.

Learn more about equivalence point here: brainly.com/question/23502649

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5 0

2 years ago

Material use make car tyres and chewing gum?

Ulleksa [173]

Answer:

Rubber

Make me brainlyist

3 0

3 years ago

What is the Molecular Mass of CuSo4.5H2O

serious [3.7K]

The answer to this question is 159.609 g/mol

8 0

3 years ago

Read 2 more answers

How does the arrangement of particles determine the physical properties of different states of matter?

Usimov [2.4K]

Answer:

See explanation

Explanation:

Matter may exist in three phases; solid, liquid and gas. The state in which matter exists depends on the extent of intermolecular forces operating in the substance.

In solid particles, the molecules that compose the solid are close together because the molecules of a solid do not move from place to place but they continue to vibrate about their fixed position.

For liquids, the molecules that compose a liquid are in random motion but are less energetic than molecules of a gas.

In gases, the molecules are not held together at all. The molecules of a gas have the highest degree of freedom. They move from one point another at a high velocity.

Hence, the order of increasing degree of movement of the particles in different states of matter = solids<liquids< gases.

Solids have well arranged particles, the molecules of a liquid are a little more disorderly than liquid particles while gas particles are the most disorderly of all the states of matter.

6 0

2 years ago

The gas-phase reaction follows an elementary rate law and is to be carried out first in a PFR and then in a separate experiment

astraxan [27]

Answer:

The activation energy is $=8.1\,kcal\,mol^{-1}$

Explanation:

The gas phase reaction is as follows.

$A \rightarrow B+C$

The rate law of the reaction is as follows.

$-r_{A}=kC_{A}$

The reaction is carried out first in the plug flow reactor with feed as pure reactant.

From the given,

Volume "V" = $10dm^{3}$

Temperature "T" = 300 K

Volumetric flow rate of the reaction $v_{o}=5dm^{3}s$

Conversion of the reaction "X" = 0.8

The rate constant of the reaction can be calculate by the following formua.

$V= \frac{v_{0}}{k}[(1+\epsilon )ln(\frac{1}{1-X}-\epsilon X)]$

Rearrange the formula is as follows.

$k= \frac{v_{0}}{V}[(1+\epsilon )ln(\frac{1}{1-X}-\epsilon X)]............(1)$

The feed has Pure A, mole fraction of A in feed $y_{A_{o}}$ is 1.

$\epsilon =y_{A_{o}}\delta$

$\delta$ = change in total number of moles per mole of A reacte.

$=1(2-1)=1$

Substitute the all given values in equation (1)

$k=\frac{5m^{3}/s}{10dm^{3}}[(1+1)ln \frac{1}{1-0.8}-1 \times 0.8] = 1.2s^{-1}$

Therefore, the rate constant in case of the plug flow reacor at 300K is $1.2s^{-1}$

The rate constant in case of the CSTR can be calculated by using the formula.

$\frac{V}{v_{0}}= \frac{X(1+\epsilon X)}{k(1-X)}.............(2)$

The feed has 50% A and 50% inerts.

Hence, the mole fraction of A in feed $y_{A_{o}}$ is 0.5

$\epsilon =y_{A_{o}}\delta$

$\delta$ = change in total number of moles per mole of A reacted.

$=0.5(2-1)=0.5$

Substitute the all values in formula (2)

$\frac{10dm^{3}}{5dm^{3}}=\frac{0.8(1+0.5(0.8))}{k(1-0.8)}=2.8s^{-1}$

Therefore, the rate constant in case of CSTR comes out to be $2.8s^{-1}$

The activation energy of the reaction can be calculated by using formula

$k(T_{2})=k(T_{1})exp[\frac{E}{R}(\frac{1}{T_{1}}-\frac{1}{T_{2}})]$

In the above reaction rate constant at the two different temperatures.

Rearrange the above formula is as follows.

$E= R \times(\frac{T_{1}T_{2}}{T_{1}-T_{2}})ln\frac{k(T_{2})}{k(T_{1})}$

Substitute the all values.

$=1.987cal/molK(\frac{300K \times320K}{320K \times300K})ln \frac{2.8}{1.2}=8.081 \times10^{3}cal\,mol^{-1}$

$=8.1\,kcal\,mol^{-1}$

Therefore, the activation energy is $=8.1\,kcal\,mol^{-1}$

8 0

3 years ago