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marissa
2 months ago
15

Which statement describes the location of metalloids on the periodic table? The metalloids are located below metals and above no

nmetals within a group. The metalloids are located below nonmetals and above metals within a group. The metalloids are located to the right of nonmetals and to the right of metals within a period. The metalloids are located to the left of nonmetals and to the left of metals within a period.
Chemistry
2 answers:
eduard [2.7K]2 months ago
5 0

Answer: The accurate statement is that metalloids are found below nonmetals and above metals in a group.

Explanation:

Metalloids are chemical elements that exhibit characteristics of both metals and nonmetals. The periodic table contains 7 metalloids: Boron (B), Silicon (Si), Germanium (Ge), Arsenic (As), Antimony (Sb), Tellurium (Te), and Polonium (Po).

These elements are situated to the right of metals and to the left of non-metals within a period and they are positioned above metals and below non-metals in a group.

Therefore, the correct statement is that metalloids are located below nonmetals and above metals in a group.

Tems11 [2.7K]2 months ago
4 0
The metalloids include Boron, Silicon, and Germanium, and they are positioned to the right of metals and to the left of nonmetals. Since that is not available as an option, the most accurate description would be: The metalloids are situated below nonmetals and above metals in a group.
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Marianne designs an experiment involving electrically charged objects. She wants to know which objects will be attracted to a ne
castortr0y [3046]

Answer:

The generation of static electricity occurs when two surfaces are rubbed together. This process causes a transfer of electrons, resulting in a build-up of negative charge. For instance, when you shuffle on a carpet, the friction creates multiple contact points which allow electrons to move onto you, thus accumulating a static charge. Touching another individual or object can lead to a sudden discharge, experienced as an electric shock.

In a similar way, rubbing a balloon against your hair generates opposite static charges on both your hair and the balloon. As you gently pull the balloon away from your head, the attraction between these opposite charges can be observed, causing your hair to rise.

Materials

• Balloon

• Woolen item (like a sweater, scarf, or yarn ball)

• Stopwatch

• Wall

• Partner (optional)

Preparation

• Inflate the balloon and secure the end.

• Have your partner ready to time with the stopwatch.

Procedure

• Grip the balloon with minimal hand coverage, such as holding it with just your thumb and index finger, or by its tied neck.

• Rub the balloon on the wool item once, making sure to go in one direction only.

• Press the rubbed side of the balloon against the wall and let go. Is it adhering to the wall? If it's stuck, your partner should start the stopwatch to measure how long it stays there. If it doesn’t stick, continue to the next step.

• Briefly touch the balloon to a metal object. Why is this step necessary?

• Repeat this procedure, but each time increase the number of rubs against the woolly item, ensuring the direction remains the same (do not rub back and forth).

Observations and results

As you increase the number of times you rub the balloon on the woolly material, does the duration of its adhesion to the wall increase?

Wool is an excellent conductor; it easily relinquishes electrons. When you rub wool on a balloon, electrons move from the wool to the surface of the balloon, imparting a negative charge to the rubbed area. Balloons, made from rubber, act as insulators, which means not all areas of the balloon will have a negative charge—only where it was rubbed will have a negative charge, while the rest of the balloon remains neutral.

Once the balloon is sufficiently charged negatively by repeated rubbing, it will adhere to the wall. Though the wall typically has a neutral charge, its internal charges can realign such that a positively charged region can attract the negatively charged balloon. Since the wall is also an insulator, the charge does not dissipate instantly. However, when the balloon is in contact with a metal object, the excess electrons from the balloon flow into the metal quickly, making the balloon lose its attraction and peel away.

HOPE IT HELPS

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6 0
3 months ago
Read 2 more answers
In living organisms, C-14 atoms disintegrate at a rate of 15.3 atoms per minute per gram of carbon. A charcoal sample from an ar
lions [2927]

Answer:

The result is "4,241.17 years"

Explanation:

The disintegration rate for C-14 atoms is indicated in  15.3 \frac{atoms}{min-g}

The dissolution rate of the sample is given by 9.16 \frac{atoms} {min-gram}

The C-14 proportion within the sample can be determined as per = \frac {9.16}{15.3} \\\\ = 0.5987

With a half-life of 5730 years.

Now, we need to compute the number of half-lives (n) that are applicable:

(\frac{1}{2})^n= A\\\\A= fraction of C-14, which is remaining \\\\(\frac{1}{2})^n= 0.5987 \\\\ n \log 2 = - \log 0.5987\\\\

\therefore \\\\ \Rightarrow n= \frac{0.227}{0.3010} \\\\ = 0.740\\

Thus, the age of the sample is represented as = n \times\ half-life

                                                 = 0.740 \times 5730 \ years \\\\=4241.17 \ years\\\\

6 0
3 months ago
Solving applied density problems Mr. Auric Goldfinger, criminal mastermind, intends to smuggle several tons of gold across Inter
KiRa [2933]

Answer:

thickness is 0.29 cm

Explanation:

To create a fake iron ball out of gold, we must ensure that its mass matches that of the iron ball. Therefore, we first find the volume of the iron ball using the provided diameter, applying the formula of 4/3 pi r^3.

Given the diameter d = 6 cm; thus, the radius r = 3 cm (d/2).

We calculate the volume of the iron ball: 4/3 * 3.14 * 3^3 = 113.04 cm^3.

The corresponding mass of the iron ball is the volume multiplied by its density: 113.04 * 5.15 g/cm^3 = 582.156 g.

This value represents the mass for the gold ball; now we determine the volume of the gold ball using its density.

Volume of gold ball = mass of gold ball/density of gold = 582.156 g/19.3 g/cm^3 = 30.1635 cm^3.

So this volume must correspond to a hollow sphere with an outer radius R = 3 cm and an unknown inner radius r.

Volume of the hollow ball can be represented as: 4/3 pi [R^3 - r^3].

Thus, 30.1635 cm^3 = 4/3 pi [3^3 - r^3].

30.1635 * 3/(4 * 3.14) = 27 - r^3.

Simplifying gives 7.2046 = 27 - r^3, resulting in r^3 = 19.7954.

Therefore, r = 2.7051 cm.

This indicates the thickness is the outer radius minus the inner radius: 3 - 2.7051 = 0.2949 cm.

Rounding to two significant figures yields

the thickness = 0.29 cm.

8 0
3 months ago
The table lists the lattice energies of some compounds. Compound Lattice Energy (kJ/mol) LiF –1,036 LiCl –853 NaF –923 KF –821 N
KiRa [2933]

Response: Option A) The lattice energy rises as cations become smaller, as demonstrated by LiF and KF.

Clarification: It has been observed that the lattice energy is largely determined by two primary factors regarding ionic solids:

i) The ionic charges - An increase in the charge of the ions corresponds to an increase in lattice energy.

and

ii) The size or radius of the ions - As the ionic size grows, the lattice energy diminishes accordingly.

Therefore, in this context, the latter factor is evident. Thus, it can be concluded that as cation sizes decrease among ionic solids, the lattice energy increases.

7 0
2 months ago
Read 2 more answers
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