1. How isotopes of copper are Cu-63 and Cu-65. How are these two isotopes the same? How are they different?2. What is an isotope?

3. Why do elements in the same group have similar chemical properties?

4.What are valence electrons?

5.Write the symbol for each of the following elements:
The halogen in period three
The alkali metal in period two
The noble gas in period one
The alkaline earth element in period six
Any transition metal in the 5th period
A metaloid in group 14
A nonmetal in group 16

6. What are two differences between a metal and a nonmeta

PLEASE ASAP!!!!!!!!!!!!!!

Answers

Answer 1

Answer:

Answer:

question 6. answer is Metals are considered to be electropositive in nature due to their ability to donate electrons. Whereas non-metals are electronegative as they generally accept electrons

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1. (a) What name is given to the law describing the relationship between volume and pressure at constant temperature? Write a mathematical expression that describes this relationship. (2 marks)(b) Sketch a graph of the relationship described in part (a).

Answers

Explanation:

a)Boyle's law states that pressure is inversely proportional to the volume of the gas at constant temperature.

$Pressure\propto (1)/(Volume)$ (At constant temperature)

The equation given by this law is:

$PV=k$

$P_1V_1=P_2V_2$

where,

$P_1\text{ and }V_1$ are initial pressure and volume respectively.

$P_2\text{ and }V_2$ are final pressure and volume respectively.

b) A graph of the relationship is attached as an image.

a chemist encounters an unknown metal. They drop the metal into a graduated cylinder containing water, and find the volume change is 3.2 mL. If the metal weighs 1.5g, what is the density in g/mL of the metal?

Answers

The density of metal will be "0.469 g/mL".

The given values in the question are:

Mass of metal = 1.5 g
Volume of metal or Volume change = 3.2 mL

Now,

The density of metal will be:

→ $Density = (Mass)/(Volume)$

By substituting the given values, we get

→ $= (1.5)/(3.2)$

→ $=0.469 \ g/mL$

Thus the above is the appropriate answer.

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The density of the metal is 0.47 g/mL

The density of an object is defined as the mass of the object per unit volume of the object.

Density = mass / volume

With the above formula, we can obtain the density of the metal. This is illustrated below:

Volume of metal = change in volume of water = 3.2 mL

Mass of metal = 1.5 g

Density of metal =?

Density = mass / volume

Density = 1.5 / 3.2

Density of metal = 0.47 g/mL

Therefore, the density of the metal is 0.47 g/mL

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In general, atomic radii: A. decrease down a group and remain constant across a period.
B. decrease down a group and increase across a period.
C. increase down a group and increase across a period.
D. increase down a group and remain constant across a period.
E. increase down a group and decrease across a period.

Answers

Answer: The correct answer is Option E.

Explanation:

Atomic radius is defined as the total distance measured from the nucleus of an atom to the outermost shell.

Trend down the group:

Moving from top to bottom, a new shell gets add up around the nucleus and the outermost shell gets far away from the nucleus. Due to this, the distance between the nucleus and outermost shell increases, which results in the increase of atomic radii of the atom.

Trend across the period:

Moving from left to right in a period, more and more electrons gets add up in the same shell. The attraction between the last electron and the nucleus increases. This results in the shrinkage of size of an atom. Thus, decreasing the atomic radii of the atom.

Hence, the correct answer is Option E.

Final answer:

E. increase down a group and decrease across a period. Atomic radii generally increase down a group due to extra electron shells and decrease across a period due to greater nuclear charge.

Explanation:

In general, the correct answer to this question is E: atomic radii increase down a group and decrease across a period on the Periodic Table. The atomic radii increase down a group due to the addition of extra electron shells. Each additional shell means a greater distance between the nucleus and the outermost electrons, which results in a larger atomic radius. On the other hand, as you move across a period from left to right, atomic radii typically decrease. This is due to an increase in positive charge in the nucleus which pulls the electrons closer, thus decreasing the atomic radius.

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Classify these bonds as ionic, polar covalent, or nonpolar covalent p cl k br c c

Answers

Types of Bonds can be predicted by calculating the difference in electronegativity.
If, Electronegativity difference is,

Less than 0.4 then it is Non Polar Covalent

Between 0.4 and 1.7 then it is Polar Covalent

Greater than 1.7 then it is Ionic

For P and Cl,
E.N of Chlorine    = 3.16
E.N of Phosphorous = 2.19
   ________

E.N Difference 0.97 (Polar Covalent)

For K and Br,
E.N of Bromine      = 2.96
E.N of Potassium    = 0.82
   ________

E.N Difference 2.14 (Ionic)

For C and C,
E.N of Carbon      = 2.55
E.N of Carbon     = 2.55
   ________

E.N Difference 0.00 (Non-Polar Covalent)

Could someone pls help me :)

Answers

liquid 1 and 2 have the same color and mass so the answer would be liquid 1 and 2

Explanation:

hope this is helpful

VSEPR theory predicts that an atom with one lone pair and three bonding pairs (such as the N-atom in aniline) will have a tetrahedral electron geometry and a trigonal pyramidal molecular geometry due to steric repulsions between H-atoms and the N-atom lone pair. However, in question 5 you observed that the N-atom in aniline is not perfectly sp3 hybridized (i.e. the hybridization is different from that predicted for a tetrahedral electron geometry). Briefly describe all of the factors that result in the calculated hybridization of the N-atom lone pair

Answers

Answer: The lone pair of electron on nitrogen is accommodated in a 2p orbital hence it interacts with the pi system in aniline.

Explanation:

Aniline is less basic than amines. This is because, the nitrogen atom in aniline is not purely sp3 hybridized. Its actual hybrization state is closer to sp2 because the lone pair on nitrogen is accommodated in a 2p orbital. The nitrogen atim in aniline is planar and its

lonely pair interacts with the pi electron system of aniline. This makes the lone pair unavailable for protonation hence aniline is less basic than amines.

Final answer:

The calculated hybridization of the N-atom lone pair in aniline is affected by electron-electron repulsions, resonance, and steric effects from substituents on the aromatic ring.

Explanation:

The calculated hybridization of the N-atom lone pair in aniline is different from the predicted sp3 hybridization due to a combination of factors:

The presence of a lone pair on the nitrogen atom leads to electron-electron repulsions, causing distortions in the molecule's geometry.
The lone pair on the nitrogen atom can participate in resonance, resulting in delocalization of electrons and a change in hybridization.
The presence of the substituents on the aromatic ring can affect the hybridization of the N-atom lone pair by exerting steric effects.

Overall, these factors contribute to the observed hybridization of the N-atom lone pair in aniline, deviating from the predicted tetrahedral electron geometry.

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