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1 year ago

Explain one way the water cycle affects climate. Use complete sentences.

Chemistry

1 answer:

ladessa [460]1 year ago

4 0

Answer:

The water cycle will affect the climate by causing rain and snow.

Explanation:

if tempatures increase more rain fall will occur because more water evaporates in the air.

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Determine whether each description applies to electrophilic aromatic substitution or nucleophilic aromatic substitution.

Alborosie

Answer:

a. electrophilic aromatic substitution

b. nucleophilic aromatic substitution

c. nucleophilic aromatic substitution

d. electrophilic aromatic substitution

e. nucleophilic aromatic substitution

f. electrophilic aromatic substitution

Explanation:

Electrophilic aromatic substitution is a type of chemical reaction where a hydrogen atom or a functional group that is attached to the aromatic ring is replaced by an electrophile. Electrophilic aromatic substitutions can be classified into five classes: 1-Halogenation: is the replacement of one or more hydrogen (H) atoms in an organic compound by a halogen such as, for example, bromine (bromination), chlorine (chlorination), etc; 2- Nitration: the replacement of H with a nitrate group (NO2); 3-Sulfonation: the replacement of H with a bisulfite (SO3H); 4-Friedel-CraftsAlkylation: the replacement of H with an alkyl group (R), and 5-Friedel-Crafts Acylation: the replacement of H with an acyl group (RCO). For example, the Benzene undergoes electrophilic substitution to produce a wide range of chemical compounds (chlorobenzene, nitrobenzene, benzene sulfonic acid, etc).

A nucleophilic aromatic substitution is a type of chemical reaction where an electron-rich nucleophile displaces a leaving group (for example, a halide on the aromatic ring). There are six types of nucleophilic substitution mechanisms: 1-the SNAr (addition-elimination) mechanism, whose name is due to the Hughes-Ingold symbol ''SN' and a unimolecular mechanism; 2-the SN1 reaction that produces diazonium salts 3-the benzyne mechanism that produce highly reactive species (including benzyne) derived from the aromatic ring by the replacement of two substituents; 4-the free radical SRN1 mechanism where a substituent on the aromatic ring is displaced by a nucleophile with the formation of intermediary free radical species; 5-the ANRORC (Addition of the Nucleophile, Ring Opening, and Ring Closure) mechanism, involved in reactions of metal amide nucleophiles and substituted pyrimidines; and 6-the Vicarious nucleophilic substitution, where a nucleophile displaces an H atom on the aromatic ring but without leaving groups (such as, for example, halogen substituents).

3 0

2 years ago

The mass unit associated with density is usually grams. if the volume (in ml or cm3) is multiplied by the density (g/ml or g/cm3

jok3333 [9.3K]

The weight in grams = 7.93 g

Given volume = 2.00 $in^{3}$

Given density = 0.242 g/ $cm^{3}$

We need to find the Mass(weight) in grams.

To find the weight in grams we need to keep in mind that the volume and density must use the same volume unit for cancellation. So that the volume units will cancel out, leaving only the mass units.

The unit of given volume is $in^{3}$ and unit of volume in density is $cm^{3}$ , so first we need to change the unit of volume from $in^{3}$ to $cm^{3}$ so that the volume units will cancel out, leaving only the mass units.

1 $in^{3}$ = 16.39 $cm^{3}$ (given conversion)

$2 in^{3}\times \frac{(16.39 cm^{3})}{(1 in^{3})}$

$in^{3}$ units get cancel out leaving the $cm^{3}$ unit.

$2 in^{3} = 32.78 cm^{3}$

Mass = Density X Volume.

Density = 0.242 g/ $cm^{3}$ and Volume = 32.78 $cm^{3}$

$Mass =0.242\frac{g}{cm^{3}}\times 32.78 cm^{3}$

Mass = 7.93 grams (g)

3 0

1 year ago

What is the hybridization of the central atom in each of the following? 1. Beryllium chloride 2. Nitrogen dioxide 3. Carbon tetr

Lina20 [59]

Answer :

(1) The hybridization of central atom beryllium in $BeCl_2$ is, $sp$

(2) The hybridization of central atom nitrogen in $NO_2$ is, $sp^2$

(3) The hybridization of central atom carbon in $CCl_4$ is, $sp^3$

(4) The hybridization of central atom xenon in $XeF_4$ is, $sp^3d^2$

Explanation :

Formula used :

$\text{Number of electron pair}=\frac{1}{2}[V+N-C+A]$

where,

V = number of valence electrons present in central atom

N = number of monovalent atoms bonded to central atom

C = charge of cation

A = charge of anion

Now we have to determine the hybridization of the following molecules.

(1) The given molecule is, $BeCl_2$

$\text{Number of electrons}=\frac{1}{2}\times [2+2]=2$

The number of electron pair are 2 that means the hybridization will be $sp$ and the electronic geometry of the molecule will be linear.

(2) The given molecule is, $NO_2$

$\text{Number of electrons}=\frac{1}{2}\times [4]=2$

If the sum of the number of sigma bonds, lone pair of electrons and odd electrons present is equal to three then the hybridization will be, $sp^2$ .

In nitrogen dioxide, there are two sigma bonds and one lone electron pair. So, the hybridization will be, $sp^2$ .

(3) The given molecule is, $CCl_4$

$\text{Number of electrons}=\frac{1}{2}\times [4+4]=4$

The number of electron pair are 4 that means the hybridization will be $sp^3$ and the electronic geometry of the molecule will be tetrahedral.

(4) The given molecule is, $XeF_4$

$\text{Number of electrons}=\frac{1}{2}\times [8+4]=6$

Bond pair electrons = 4

Lone pair electrons = 6 - 4 = 2

The number of electrons are 6 that means the hybridization will be $sp^3d^2$ and the electronic geometry of the molecule will be octahedral.

But as there are four atoms around the central xenon atom, the fifth and sixth position will be occupied by lone pair of electrons. The repulsion between lone and bond pair of electrons is more and hence the molecular geometry will be square planar.

3 0

1 year ago

How much energy is transferred when 30.0 g of water is cooled from 25.0 °C to 12.7 °C.

enot [183]

Answer:

Heat transferred, Q = 1542.42 J

Explanation:

Given that,

Mass of water, m = 30 grams

Initial temperature, $T_i=25^{\circ} C$

Final temperature, $T_f=12.7^{\circ} C$

We need to find the energy transferred. The energy transferred is given by :

$Q=mc\Delta T$

c is specific heat of water, c = 4.18 J/g °C

So,

$Q=30\ g\times 4.18\ J/g-^{\circ} C\times (12.7-25)\ ^{\circ} C\\\\Q=-1542.42\ J$

So, 1542.42 J of energy is transferred.

6 0

1 year ago

1. The specific heat capacity of iron is 0.461 J g–1 K–1 and that of titanium is 0.544 J g–1 K–1. A sample consisting of a mixtu

mezya [45]

Answer:

The answer is 80,1 °C

Explanation:

Let´s start from the mass of the sample and the heat capacities:

First of all, we must calculate an average heat capacity. That's because we have a mixture and it is unknown the heat capacity of the whole sample.

The way we should do this calculation is as follows:

(1) $H_{average}=Mass Fraction_{first component}* H_{first component}+MassFraction_{secondcomponent}*H_{second component}$

For example, the mass fraction of Fe is simply:

(2) $MassFraction_{Fe}=\frac{10g Fe}{10g Fe + 10gTi}=0.5$

If you combine the equations (1) and (2) you have:

(3) $H_{average}=0.5*0.461+0.5*0.544=0.5025\frac{J}{g-K}$

Once calculated the average heat capacity we can solve the problem taking into account the corresponding equation:

(4) $Q=m*H_{average}*(T_2-T_1)$

Remember that:

Q: Heat gained or lost

m: Mass of the sample you want to analize

$H_{average}$ : The value obtained in equation (3)

$T_2$ : Final temperature of the sample

$T_1$ : Initial temperature of the sample

Now we must replace the problem data in equation (4)

Take into account:

Heat gained in a system have a positive value
Heat lost in a system have a negative value

In this problem the sample loses 200 J, for this reason $Q=-200J$

The mass of the whole sample is: 10g of Fe + 10g of Ti = 20g of sample
The temperatures must be in absolute units of temperature (these are: rankine or kelvin)
The initial temperature of the system is 100°C or 373K

Now we are ready to use equation (4):

(5) $-200J=20g*0.5025\frac{J}{g*K} *(T_2-373K)$

It is clear that the unknown in equation (5) is $T_2$

The next step is to calculate $T_2$ . Don't forget the signs; these are important.

Key concept: Since the system is loosing heat, the final temperature of the system ( $T_2$ ) should be lower than the initial temperature ( $T_1$ )

7 0

2 years ago

Read 2 more answers