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kari74 [83]
3 years ago
6

On the basis of bonding, which of the following compounds has the highest boiling point? A. water. . B. glucose. . C. sodium chl

oride. . D. alcohol
Chemistry
2 answers:
katrin2010 [14]3 years ago
5 0

Answer:

C. sodium chloride

Explanation:

Boiling point is temperature at which vapor pressure of the liquid becomes equal to external pressure (atmospheric) surrounding liquid.

On basis of bonding, the compound which has the highest boiling point is sodium chloride as it posses ionic bonding which is comprised of strong electrostatic bonds.

Water and alcohol are liquid and also, glucose has covalent bonding.

The melting points are shown below as:

Alcohol - 78.37 °C.

Glucose - 146 °C.

Water - 100°C.

Sodium chloride - 1413 °C.

<u> Answer - C. sodium chloride</u>

baherus [9]3 years ago
3 0
Alcohol = 78c
sodium chloride = 1413c
water = 100c
<span>Glucose does not have a boiling point firstly it decomposes into carbon and H2O then boils

hope it helps.
</span>


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6. A 25.0-mL sample of 0.125 M pyridine is titrated with 0.100 M HCI. Calculate the pH
Vadim26 [7]

Answer:

a) pH = 9.14

b) pH = 8.98

c) pH = 8.79

Explanation:

In this case we have an acid base titration. We have a weak base in this case the pyridine (C₅H₅N) and a strong acid which is the HCl.

Now, we want the know the pH of the resulting solution when we add the following volume of acid: 0, 10 and 20.

To know this, we first need to know the equivalence point of this titration. This can be known using the following expression:

M₁V₁ = M₂V₂  (1)

Using this expression, we can calculate the volume of acid required to reach the equivalence point. Doing that we have:

M₁V₁ = M₂V₂

V₁ = M₂V₂ / M₁

V₁ = 0.125 * 25 / 0.1 = 31.25 mL

This means that the acid and base will reach the equivalence point at 31.25 mL of acid added. So, the volume of added acid of before, are all below this mark, so we can expect that the pH of this solution will be higher than 7, in other words, still basic.

To know the value of pH, we need to apply the following expression:

pH = 14 - pOH  (2)

the pOH can be calculated using this expression:

pOH = -log[OH⁻]  (3)

The [OH⁻] is a value that can be calculated when the pyridine is dissociated into it's ion. However, as this is a weak acid, the pyridine will not dissociate completely in solution, instead, only a part of it will be dissociated. Now, to know this, we need the Kb value of the pyridine.

The reported Kb value of the pyridine is 1.5x10⁻⁹ so, with this value we will do an ICE chart for each case, and then, calculate the value of the pH.

<u>a) 0 mL of acid added.</u>

In this case, the titration has not begun, so the concentration of the base will not be altered. Now, with the Kb value, let's write an ICE chart to calculate the [OH⁻], the pOH and then the pH:

       C₅H₅N + H₂O <-------> C₅H₅NH⁺ + OH⁻     Kb = 1.5x10⁻⁹

i)       0.125                                0             0

e)        -x                                   +x           +x

c)      0.125-x                              x             x

Writting the Kb expression:

Kb = [C₅H₅NH⁺] [OH⁻] / [C₅H₅N]    replacing the values of the chart:

1.5x10⁻⁹ = x² / 0.125-x --> Kb is really small, so we can assume that x will be very small too, and 0.125-x can be neglected to only 0.125, and then:

1.5x10⁻⁹ = x² / 0.125

1.5x10⁻⁹ * 0.125 = x²

x = [OH⁻] = 1.37x10⁻⁵ M

Now, we can calculate the pOH:

pOH = -log(1.37x10⁻⁵) = 4.86

Finally the pH:

pH = 14 - 4.86

<h2>pH = 9.14</h2>

<u>b) 10 mL of acid added</u>

In this case the titration has begun so the acid starts to react with the base, so we need to know how many moles of the base remains after the volume of added acid:

moles acid = 0.1 * (0.010) = 1x10⁻³ moles

moles base = 0.125 * 0.025 = 3.125x10⁻³

This means that the base is still in higher quantities, and the acid is the limiting reactant here, so the remaining moles will be:

remaining moles of pyridine = 3.125x10⁻³ - 1x10⁻³ = 2.125x10⁻³ moles

The concentration of pyridine in solution:

[C₅H₅N] = 2.125x10⁻³ / (0.025 + 0.010) = 0.0607 M

Now with this concentration, we will do the same procedure of before, with the ICE chart, but replacing this new value of the base, to get the [OH⁻] and then the pH:

        C₅H₅N + H₂O <-------> C₅H₅NH⁺ + OH⁻     Kb = 1.5x10⁻⁹

i)       0.0607                             0             0

e)        -x                                   +x           +x

c)      0.0607-x                           x             x

Writting the Kb expression:

Kb = [C₅H₅NH⁺] [OH⁻] / [C₅H₅N]    replacing the values of the chart:

1.5x10⁻⁹ = x² / 0.0607-x --> 0.0607

1.5x10⁻⁹ = x² / 0.0607

1.5x10⁻⁹ * 0.0607 = x²

x = [OH⁻] = 9.54x10⁻⁶ M

Now, we can calculate the pOH:

pOH = -log(9.54x10⁻⁶) = 5.02

Finally the pH:

pH = 14 - 5.02

<h2>pH = 8.98</h2>

<u>c) 20 mL of acid added:</u>

In this case the titration it's almost reaching the equivalence point and the acid is still reacting with the base, so we need to know how many moles of the base remains after the volume of added acid:

moles acid = 0.1 * (0.020) = 2x10⁻³ moles

moles base = 0.125 * 0.025 = 3.125x10⁻³

This means that the base is still in higher quantities, and the acid is the limiting reactant here, so the remaining moles will be:

remaining moles of pyridine = 3.125x10⁻³ - 2x10⁻³ = 1.125x10⁻³ moles

The concentration of pyridine in solution:

[C₅H₅N] = 1.125x10⁻³ / (0.025 + 0.020) = 0.025 M

Now with this concentration, we will do the same procedure of before, with the ICE chart, but replacing this new value of the base, to get the [OH⁻] and then the pH:

        C₅H₅N + H₂O <-------> C₅H₅NH⁺ + OH⁻     Kb = 1.5x10⁻⁹

i)       0.025                                0             0

e)        -x                                   +x           +x

c)      0.025-x                             x             x

Writting the Kb expression:

Kb = [C₅H₅NH⁺] [OH⁻] / [C₅H₅N]    replacing the values of the chart:

1.5x10⁻⁹ = x² / 0.025-x --> 0.025

1.5x10⁻⁹ = x² / 0.025

1.5x10⁻⁹ * 0.025 = x²

x = [OH⁻] = 6.12x10⁻⁶ M

Now, we can calculate the pOH:

pOH = -log(6.12x10⁻⁶) = 5.21

Finally the pH:

pH = 14 - 5.21

<h2>pH = 8.79</h2>
5 0
3 years ago
a chemist dissolved an 11.9-g sample of koh in 100.0 grams of water in a coffee cup calorimeter. when she did so, the water temp
Snowcat [4.5K]

The heat of solution is -51.8 kJ/mol

<h3>What is the heat of solution?</h3>

We know that in a calorimeter, there is no loss or gain of energy. It is a good example of a closed system.

Number of moles of KOH =  11.9-g/56 g/mol = 0.21 moles

Temperature rise = 26.0 ∘c

Mass of the water = 100.0 grams

Heat capacity =  4.184 j/g⋅°c

Then;

ΔH = mcθ

ΔH = 100g * 4.184 j/g⋅°c * 26.0 ∘c = 10.88 kJ

Heat of solution = -(10.88 kJ/ 0.21 moles) = -51.8 kJ/mol

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4 0
2 years ago
What is the atomic number of the as yet undiscovered element in which the 8s and 8p electron energy levels?
Andreyy89

The atomic number of the undiscovered element is 168

Element 118 will have just filled its 7p orbitals. therefore the predicted element to fill completely up to its 8 p orbital would have to filled a whole set of s, p, d, f and g orbitals

That's another 2 + 6 + 10 +14 + 18 = 50 electrons

To determine the total number of quantum numbers we have to find

Nml × Nms

we have Nml × Nms = ( 2 + 1 ) × 2

8s + 8P + 7d + 6f + 5g = 2 + 6 + 10 + 14 + 18 = 50

The element right below should be

Z = 118 + 50

   = 168

Hence the atomic number of the undiscovered element is 168

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brainly.com/question/14514242

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4 0
1 year ago
Give two examples of how our society uses isotopes
nalin [4]

Answer:

Isotopes – caused by varying numbers of neutrons in an element – have many practical uses in our society. ... In geology and archaeology, radioactive isotopes are used to determine the age of a sample while hydrologists can use isotope signatures to distinguish between different groundwater types.

Explanation:

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how many grams of solid lithium must be added to liquid after in order to obtain 15.0 L of hydrogen gas at STP?
alexdok [17]

Answer:

9.29g

Explanation:

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