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Leno4ka [110]
2 years ago
13

An unknown substance visually appears to be a white powder. The substance is analyzed in the laboratory and its melting point is

found to be 264 °C.
What can you predict about the intramolecular forces and intermolecular forces in this substance?
Chemistry
1 answer:
wlad13 [49]2 years ago
5 0

Answer:

See explanation

Explanation:

The substance under analysis must be a polar organic substance of high molecular mass. We must remember that as the magnitude of dispersion forces increases with increase in molecular mass, the melting point increases accordingly. Also polar organic substances have very high melting points due to the presence of polar bonds in the molecule..

Hence the intramolecular forces in the substance are covalent bonds while the intermolecular forces are both dispersion forces and dipole forces.

The substance may likely be 1,2,3,4,6,7,8-Heptachlorodibenzo-p-Dioxin which a white powder and melts at 264 °C.

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The three major types of faults are normal, reverse and syncline. true or false
Novosadov [1.4K]
The three major types of faults are Normal, Reverse and Strike-slip faults. 


Answer: FALSE
6 0
3 years ago
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The atomic symbol Superscript 206 subscript 82 upper P b. represents lead-206 (Pb-206), an isotope that has 82 protons and 124 n
Serggg [28]

Answer:

\left \{ {{y=206} \atop {x=82}}Pb \right.

Explanation:

isotopes are various forms of same elements with different atomic number but different mass number.

Radioactivity is the emission of rays or particles from an atom to produce a new nuclei. There are various forms of radioactive emissions which are

  • Alpha particle emission  \left \{ {{y=4} \atop {x=2}}He \right.
  • Beta particle emission    \left \{ {{y=0} \atop {x=-1}}e \right.
  • gamma radiation             \left \{ {{y=0} \atop {x=0}}γ \right.

in the problem the product formed after radiation was Pb-206. isotopes of lead include Pb-204, Pb-206, Pb-207, Pb-208. they all have atomic number 82. which means the radiation cannot be ∝ or β since both radiations will alter the atomic number of the parent nucleus.

Only gamma radiation with \left \{ {{y=0} \atop {x=0}}γ \right. will produce a Pb-206 of atomic number 82 and mass number 206 , since gamma ray have 0 mass and has 0 atomic number.equation is shown below

\left \{ {{y=206} \atop {x=82}}Pb\right ⇒ \left \{ {{y=206} \atop {x=82}}Pb\right +  \left \{ {{y=0} \atop {x=0}}γ\right.

Thus the atomic symbol is \left \{ {{y=206} \atop {x=82}}Pb\right

8 0
3 years ago
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a 125 g chunk of aluminum at 182 degrees Celsius was added to a bucket filled with 365 g of water at 22.0 degrees Celsius. Ignor
Diano4ka-milaya [45]
<h3>Answer:</h3>

32.98°C

<h3>Explanation:</h3>

We are given the following;

Mass of Aluminium as 125 g

Initial temperature of Aluminium as 182°C

Mass of water as 265 g

Initial temperature of water as 22°C

We are required to calculate the final temperature of the two compounds;

First, we need to know the specific heat capacity of each;

Specific heat capacity of Aluminium is 0.9 J/g°C

Specific heat capacity of water is 4.184 J/g°C

<h3>Step 1: Calculate the Quantity of heat gained by water.</h3>

Assuming the final temperature is X°C

we know, Q = mcΔT

Change in temperature, ΔT = (X-22)°C

therefore;

Q = 365 g × 4.184 J/g°C × (X-22)°C

    = (1527.16X-33,597.52) Joules

<h3>Step 2: Calculate the quantity of heat released by Aluminium </h3>

Using the final temperature, X°C

Change in temperature, ΔT = -(X°- 182°)C (negative because heat was lost)

Therefore;

Q = 125 g × 0.90 J/g°C × (182°-X°)C

  = (20,475- 112.5X) Joules

<h3>Step 3: Calculating the final temperature</h3>

We need to know that the heat released by aluminium is equal to heat absorbed by water.

Therefore;

(20,475- 112.5X) Joules = (1527.16X-33,597.52) Joules

Combining the like terms;

1639.66X = 54072.52

             X = 32.978°C

                = 32.98°C

Therefore, the final temperature of the two compounds will be 32.98°C

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This is true. Water is the solvent in aqueous solutions
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