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3 years ago

Metallic compounds are malleable. Which use of metallic bonds best represents this trait?

1 answer:

8 0

In metallic bonding, the valence electrons are free to move throughout the metal structure. Metallic bonding is the electrostatic attraction between the metal atoms or ions and the delocalized electrons. This is why atoms or layers are allowed to slide past each other, resulting in the characteristic properties of malleability and ductility.

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when particles of matter are in direct contact and movement occurs in the substance due to convection, thermal energy transfers

liq [111]

Thermal energy transfers in a solid state, due to convection, in metalic substances.

This is because the covalent bonds between the atoms are being broken and reformed again while the metal is experiencing stress. covalent bonds store energy.

4 0

3 years ago

(a)-Use Lewis symbol store present there action that occurs between Ca and F atoms. (b)-What is the chemical formula of the most

Charra [1.4K]

Answer:

(a) The Lewis structure is shown in the image below.

(b) $CaF_2$

(c) Calcium loses 2 electrons to 2 atoms of fluorine and these 2 atoms of fluorine accepts each electron to form ionic bond.

(d) Calcium atom loses electrons.

Explanation:

Calcium is the element of second group and forth period. The electronic configuration of Calcium is - 2, 8, 8, 2 or $1s^22s^22p^63s^23p^64s^2$

There are 2 valence electrons of Calcium.

Fluorine is the element of group 17 and second period. The electronic configuration of the element fluorine is - 2, 7 or $1s^22s^22p^5$

There is 1 valence electron of fluorine.

The Lewis structure is drawn in such a way that the octet of each atom is complete.

Thus, calcium loses 2 electrons to 2 atoms of fluorine and these 2 atoms of fluorine accepts each electron to form ionic bond. This is done in order that the octet of the atoms are complete and they become stable.

Thus, the formula of calcium chloride is $CaF_2$ .

(a) The Lewis structure is shown in the image below.

(b) $CaF_2$

(c) Calcium loses 2 electrons to 2 atoms of fluorine and these 2 atoms of fluorine accepts each electron to form ionic bond.

(d) Calcium atom loses electrons.

5 0

2 years ago

HELP PLS !!!! :))Formula unit of Sr and CI Formula unit of AI and S

xenn [34]

Answer : The formula unit of Sr & Cl is $SrCl_2$ and the formula unit of Al & S is $Al_2S_3$

Explanation :

Formula unit : it is defined as the lowest number ratio of ions of an elements in an ionic compound or covalent compound.

For the formula unit of Sr & Cl, two chloride ions $(Cl^-)$ are needed to neutralize the one strontium ion $(Sr^{2+})$ .

For the formula unit of Al & S, three sulfide ions $(S^{2-})$ are needed to neutralize the two aluminium ion $(Al^{3+})$ .

The formula unit of $SrCl_2$ and $Al_2S_3$ are shown below.

4 0

3 years ago

How many representative particles are in one mole of CO2? PLEASE ANSWER ALMOST DUE

igor_vitrenko [27]

Answer:

I hope this helps, sorry if it doesnt. Not really good with chemistry, hope you have a great day/night :}

Explanation:

A mole (mol) is the amount of a substance that contains 6.02 × 10 23 representative particles of that substance.

12.00 g C-12 = 1 mol C-12 atoms = 6.022 × 1023 atoms • The number of particles in 1 mole is called Avogadro's Number (6.0221421 x 1023).

7 0

2 years ago

A 0.4272 g sample of an element contains 2.241 x 10 ^21 atoms . what is the symbol of the element?

grin007 [14]

Answer:

Likely $\rm In$ (indium.)

Explanation:

Number of atoms: $N = 2.241 \times 10^{21}$ .

Dividing, $N$ , the number of atoms by the Avogadro constant, $N_A \approx 6.023 \times 10^{23} \; \rm mol^{-1}$ , would give the number of moles of atoms in this sample:

$\begin{aligned} n &= \frac{N}{N_{A}} \\ &\approx \frac{2.241 \times 10^{21}}{6.023 \times 10^{23}\; \rm mol^{-1}} \approx 3.72 \times 10^{-3}\; \rm mol \end{aligned}$ .

The mass of that many atom is $m = 0.4272\; \rm g$ . Estimate the average mass of one mole of atoms in this sample:

$\begin{aligned}M &= \frac{m}{n} \\ &\approx \frac{0.4272\; \rm g}{3.72 \times 10^{-3}\; \rm mol} \approx 114.82\; \rm g \cdot mol^{-1}\end{aligned}$ .

The average mass of one mole of atoms of an element ( $114.82\; \rm g \cdot mol^{-1}$ in this example) is numerically equal to the average atomic mass of that element. Refer to a modern periodic table and look for the element with average atomic mass $114.82$ . Indium, $\rm In$ , is the closest match.

5 0

3 years ago