What are the three types of nuclear radiation?

Describe the stages of volcanic activity

OverLord2011 [107]

Active, Dormant, and Extinct.
Hope this helps.

5 0

3 years ago

Consider a disubstituted aromatic compound. The parent name is benzene and there is a chloro and bromo substituent. Disubstitute

Marina86 [1]

Answer:The correct names for the structure are:

--> 1. 1-bromo-3-chlorobenzene.

--> 3. meta-bromochlorobenzene.

--> 6. m-bromochlorobenzene.

Explanation:

Benzene is the simplest member of the aromatic hydrocarbons. It has a ring structure consisting of six carbon and six hydrogen atoms. This equally means that a benzene can have up to six substituents. One of the chemical properties is that benzene and other members of its series undergo substitution reaction whereby one or more of its six hydrogen atoms is replaced by monoatomic reagents.

Disubstituted benzene consists of two substituents which are described based on either numerical locants or specific words for the three possible forms.

The numerical locant method are used the same naming substitutes of other hydrocarbons. From the question, the numerical locant method was derived through using the following steps:

--> the functional group is benzene

--> there are two substituents which includes bromine( written as bromo) and chlorine ( written as chloro)

--> while placing the number, it's done alphabetically ('1-bromo' comes before '3-chloro') in a clockwise manner. This is to give chorine the lowest locant number.

The second naming method for a disubstituted benzene is the the ortho-, meta-, para- (or their singel letter equivalent) nomenclature method. This is only used for benzene structures.

--> ortho or O : this is used when the substituents are close to each other in the benzene ring.

--> meta or (m) : This is used when the substituents are separated by one carbon in the benzene ring.

--> para or (p): This is used when the substituents are across each other in the benzene ring

From the question, the bromine substituent is separated from the chlorine by one carbon atom, therefore it's meta-bromochlorobenzene or m-bromochlorobenzene.

8 0

3 years ago

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

<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

<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

5 0

3 years ago

Convert 15.2 moles of K to atoms of K.

choli [55]

Answer:

$\boxed {\boxed {\sf 9.15*10^{24} \ atoms \ K}}$

Explanation:

To convert atoms to moles, Avogadro's Number must be used: 6.022*10²³.

This tells us the amount of particles (atoms, molecules, etc.) in 1 mole of a substance. In this case it is the atoms of potassium. We can create a ratio.

$\frac {6.022*10^{23} \ atoms \ K }{ 1 \ mol \ K }$

Multiply by the given number of moles: 15.2

$15.2 \ mol \ K *\frac {6.022*10^{23} \ atoms \ K }{ 1 \ mol \ K }$

The moles of potassium cancel.

$15.2 *\frac {6.022*10^{23} \ atoms \ K }{ 1 }$

The denominator of 1 can be ignored.

$15.2 * {6.022*10^{23} \ atoms \ K }{$

Multiply.

$9.15344*10^{24} \ atoms \ K$

The original measurement of moles has 3 significant figures, so our answer must have the same. For the number we calculated that is the hundredth place. The 3 in the thousandth place tells us to leave 5.

$9.15*10^{24} \ atoms \ K$

In 15.2 moles of potassium, there are <u>9.15*10²⁴ atoms of potassium.</u>

3 0

3 years ago

An ostrich runs north in a straight line for 21 meters. She then turns around and runs 25 meters back toward her nest (south) on

drek231 [11]

The ostrich runs 21 N then goes back by 25 S

The displacement is -4 S

21 - 25 = -4

6 0

3 years ago