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3 years ago

The valence of aluminum is +3, and the valence of chlorine is –1. The formula for aluminum chloride is

Chemistry

2 answers:

bogdanovich [222]3 years ago

6 0

AICI3

which is choice c.

Aluminium chloride

Chemical compound

DescriptionAluminium chloride is the main compound of aluminium and chlorine. It is white, but samples are often contaminated with iron(III) chloride, giving it a yellow color. The solid has a low melting and boiling point. Wikipedia

IUPAC ID: aluminium chloride

Molar mass: 133.34 g/mol

Soluble in: Water

Related Lewis acids: Iron(III) chloride; Boron trifluoride

Heat capacity (C): 91.1 J/mol·K

Other names: aluminium(III) chloride; aluminum trichloride

CaHeK987 [17]3 years ago

3 0

The correct option is C.i.e. AlCl3 Given: The valence of aluminum is +3, and the valence of chlorine is –1. Thus, 3 chlorine atoms each of (-1 charge )combine to make the charge balance because of Al (3+). Thus, 1 atom of Aluminum (Al) binds with three atoms of Chlorine (Cl).Thus, the net foromula becomes AlCl3

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The substitution reaction of toluene with Br2 can, in principle, lead to the formation of three isomeric bromotoluene products.

GREYUIT [131]

Answer:

See explanation and image attached

Explanation:

Aromatic hydrocarbons undergo electrophillic substitution. Usually, substituted benzene is more or less reactive to electrophillic substitution compared to unsubstituted benzene.

Substituents on the benzene ring tend to direct the incoming electrophile during electrophillic substititution. The presence of the -CH3 group on toluene directs the incoming Br electrophile to the ortho/para position.

Where the incoming electrphile E is Bromine, we can see that in the ortho/ para product, the electron pushing -CH3 stabilizes the resonance structure formed and increases electron density at the ortho/para position via resonance compared to the meta product as we can see from the image attached. Hence, the ortho and para products predominate over meta products.

Image credit: Chemistry steps

6 0

3 years ago

Consider a mixture of NaCl and NaNO3 that is 31.8% Na by mass. Calculate the percentage by mass of NaCl in the mixture.

lutik1710 [3]

Answer: The percentage by mass of NaCl in the mixture is 38.5 %

Explanation:

To calculate the mass percentage of element in compound, we use the equation:

$\text{Mass percent of element}=\frac{\text{Mass of element}}{\text{Mass of compound}}\times 100$ .......(1)

For NaCl:

Mass of sodium element = 23 g

Mass of NaCl = 58.5 g

Putting values in equation 1, we get:

$\text{Mass percent of sodium element}=\frac{23g}{58.5g}\times 100=39.32\%$

Mass fraction of sodium metal in NaCl = 0.3932

For $NaNO_3$ :

Mass of sodium element = 23 g

Mass of $NaNO_3$ = 85 g

Putting values in equation 1, we get:

$\text{Mass percent of sodium element}=\frac{23g}{85g}\times 100=27.1\%$

Mass fraction of sodium metal in sodium nitrate = 0.271

Let us assume the mass fraction of NaCl in the mixture is 'x'

So, the mass fraction of $NaNO_3$ in the mixture will be '(1-x)'

We are given:

Percent by mass of Na in the mixture = 31.8 %

Mass fraction of Na in the mixture = 0.318

Evaluating the mass fraction of NaCl in the mixture:

$[(x\times 0.3932)+((1-x)\times 0.271)]=0.318$

x = 0.385

Percent by mass of NaCl in the mixture will be = $(0.385\times 100)=38.5\%$

Hence, the percentage by mass of NaCl in the mixture is 38.5 %

4 0

3 years ago

Draw the products obtained from the reaction of 1 equivalent of HBr with 1 equivalent of 2,5-dimethyl-1,3,5-hexatriene. Draw the

fgiga [73]

Answer:

Here's what I get.

Explanation:

According to Markovnikov's rule, the H will add to a terminal carbon, generating three resonance stabilized carbocations.

The Br⁻ ion will add to any of the three carbocations.

There are three possible products:

5-bromo-2,5-dimethylhexa-1,3-triene (1)
3-bromo-2,5-dimethylhexa-1,4-triene (2)
1-bromo-2,5-dimethylhexa-2,4-triene (3)

6 0

3 years ago

Cis-platin is a chlorine-containing chemotherapy agent with the formula pt(nh3)2cl2. What is the mass of one mole of cis-platin?

marishachu [46]

Cis-platin is a chemotherapy agent used to treat and kill cancerous cells in patients. One mole of cis-platin has a mass of 300.06 grams/mol. Thus, option B is correct.

<h3>What is a chemotherapy agent?</h3>

A chemotherapy agent is an alkylating agent that is used to treat cancer and is given to reduce the infection or to relieve the symptoms. Cis-platin (Pt(NH₃)₂Cl₂) is one of the chemotherapy agents that treat lung, ovarian, and neck cancer, etc.

The mass of one mole of Cis-platin is calculated as,

Moles = mass ÷ molar mass

Where,

Moles = 1 mole

The molar mass of (Pt(NH₃)₂Cl₂) is calculated as,

195.06 g/mole + 2(17g/mole) + (35.5)2 = 300.06 grams

Substituting values to calculate mass:

Mass = Molar mass × moles

= 300. 06 × 1

= 300.06 grams/mol

Therefore, option B. 300.06 gm/mol is the mass of Cis-platin.

Learn more about chemotherapy agents here:

brainly.com/question/14260402

#SPJ4

Your question is incomplete, but most probably your full question was, Cis-platin is a chlorine-containing chemotherapy agent with the formula pt(nh3)2cl2. What is the mass of one mole of cis-platin?

488.91 g/mol
300.06 g/mol
492.37 g/mol
283.02 g/mol
2860.5 g/mol

8 0

2 years ago

What is the molality of an aqueous KCl solution with a mole fraction of KCl, XKCl = 0.175? (The molar mass of KCl = 74.55 g/mol

NARA [144]

Answer:

The molality of the KCl solution is 11.8 molal

Explanation:

Step 1: Data given

Mol fracrion KCl = 0.175

Molar mass KCl = 74.55 g/mol

Molar mass H2O = 18.02 g/mol

Step 2: Calculate mol fraction H2O

mol fraction H2O = 1 - 0.175 = 0.825

Step 3: Calulate mass of H2O

Suppose the total moles = 1.0 mol

Mass H2O = moles H2O * molar mass

Mass H2O = 0.825 * 18.02 g/mol

Mass H2O = 14.87 grams = 0.01487 kg

Step 4: Calculate molality

Molality KCl = 0.175 / 0.01487 kg

Molality KCl = 11.8 molal

The molality of the KCl solution is 11.8 molal

5 0

3 years ago