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alukav5142 [94]
3 years ago
10

What is the name of the following compound? A line-angle formula shows a ring with six vertices and a circle inscribed. A Cl ato

m is attached to the first and the fourth vertices. What is the name of the following compound? A line-angle formula shows a ring with six vertices and a circle inscribed. A Cl atom is attached to the first and the fourth vertices.
A. 1,4-dichlorobenzene p-dichlorobenzene
B. phenyldichloride
C. p-dichlorobenzene and 1,4-dichlorobenzene
D. 1,4-dichlorobenzene and phenyldichloride
Chemistry
1 answer:
ozzi3 years ago
4 0

Answer: Option C. p-dichlorobenzene and 1,4-dichlorobenzene.

Explanation:

A line-angle formula with six vertices and a circle inscribed corresponds to the compound known as benzene.

Further, according to the IUPAC standards for naming benzene derivatives, you must first number the position of the substituent. In this case, the substituents (chloros) are located at the positions 1 and 4; also, for the benzene derivatives when they have 2 substituents and the positions are 1 and 4, this configuration is known as <em>para </em>or <em>p </em>configuration.

Additionally, this compound has 2 substituents (chloros) so you have to indicate this number (di).

Therefore, the correct answer is C. p-dichlorobenzene and 1,4-dichlorobenzene.

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Which substance can be decomposed by a chemical change?
11111nata11111 [884]
The answer to this question would be: ammonia
Chemical decomposition is the breakdown of a compound into smaller compound or element. In this question, the Sodium(Na), Argon (Ar) and Sulfur(S) is elements, not compounds.
An element should not breakable since it was the simplest form.
Ammonia(NH3) is a compound so it should be able to be decomposed.
7 0
3 years ago
How much time does it take for a truck to slow down if its acceleration is -15m/s2 and the change in velocity is -30m/s
damaskus [11]

Answer:

2s

Explanation:

Given parameters:

Acceleration of the truck = 15m/s²

change in Velocity  = 30m/s

Unknown:

Time taken for the body to slow down = ?

Solution:

Acceleration is the rate of change of velocity ;

  Acceleration  = \frac{change in velocity}{time}

    time  = \frac{change in velocity }{acceleration}  

Now insert the parameters and solve;

   time  = \frac{30}{15}  = 2s

5 0
3 years ago
Which statements accurately describe parasite-host relationships? Check all that apply.
Fittoniya [83]
I believe it would be:
-parasites live on or on their hosts
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4 0
3 years ago
Read 2 more answers
How many moles of MgCO3 are in 2.57g of MgCO3?
Sveta_85 [38]

Answer:

0.0305 moles of MgCO₃

Explanation:

In order to solve this problem, we first need to calculate the molecular weight of MgCO₃:

  • MgCO₃ MW = Atomic mass Mg + Atomic mass C + (Atomic mass O)*3
  • MgCO₃ MW = 24.3 + 12 + 16*3 = 84.3 g/mol

Finally we <u>divide the mass by the molecular weight</u>, to calculate the <em>number of moles</em>:

  • 2.57 g MgCO₃ ÷ 84.3 g/mol = 0.0305 moles.
7 0
3 years ago
A 57.0 mL sample of a 0.120 M potassium sulfate solution is mixed with 35.5 mL of a 0.118 M lead(II) acetate solution and the fo
Katarina [22]

Answer:

Limiting reagent = lead(II) acetate

Theoretical yield = 1.2704 g

% yield = 78.09 %

Explanation:

Considering:

Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}

Or,

Moles =Molarity \times {Volume\ of\ the\ solution}

Given :

For potassium sulfate :

Molarity = 0.120 M

Volume = 57.0 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 57.0×10⁻³ L

Thus, moles of potassium sulfate:

Moles=0.120 M \times {57.0\times 10^{-3}}\ moles

Moles of potassium sulfate  = 0.00684 moles

For lead(II) acetate :

Molarity = 0.118 M

Volume = 35.5 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 35.5×10⁻³ L

Thus, moles of lead(II) acetate :

Moles=0.118 \times {35.5\times 10^{-3}}\ moles

Moles of lead(II) acetate  = 0.004189 moles

According to the given reaction:

K_2SO_4_{(aq)}+Pb(C_2H_3O_2)_2_{(aq)}\rightarrow 2KC_2H_3O_2_{(s)}+PbSO_4_{(aq)}

1 mole of potassium sulfate react with 1 mole of lead(II) acetate

0.00684 moles potassium sulfate react with 0.00684 mole of lead(II) acetate

Moles of lead(II) acetate = 0.004189 moles

Limiting reagent is the one which is present in small amount. Thus, lead(II) acetate is limiting reagent. ( 0.004189 < 0.00684)

The formation of the product is governed by the limiting reagent. So,

1 mole of lead(II) acetate gives 1 mole of lead(II) sulfate

0.004189 mole of lead(II) acetate gives 0.004189 mole of lead(II) sulfate

Molar mass of lead(II) sulfate = 303.26 g/mol

Mass of lead(II) sulfate = Moles × Molar mass = 0.004189 × 303.26 g = 1.2704 g

Theoretical yield = 1.2704 g

Given experimental yield = 0.992 g

<u>% yield = (Experimental yield / Theoretical yield) × 100 = (0.992/1.2704 g) × 100 = 78.09 %</u>

3 0
3 years ago
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