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3 years ago

5

When gasses are heated the average velocity (speed) of the gas particles decreases.

1 answer:

Kitty [74]3 years ago

6 0

I think that would be true!

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1. The forks shown are made of silver (Ag). Some of the silver forks

Anon25 [30]

Answer:

Corrosion

Explanation:

Silver, although known as a nobble metal, is also subject to corrosion process such as having silver tarnish when exposed to sulfur and air.

Tarnishing occurs on the surfaces of some metals such as brass, copper, and silver, which results in a corroded layer. Silver tarnish occurs from the chemical reaction that takes place when silver is exposed to sulfur which results in the formation of black Ag₂S

In order to restore the original silver surface, the silver tarnish (silver sulfide) layer is removed.

We have the statement presented here as follows;

The forks shown are made of silver (Ag). Some of the silver forks shown have lost their luster - they have become tarnished. This is an example of <u>Corrosion.</u>

4 0

3 years ago

Chem help<br>calculate the wavelength of an electromagnetic wave with a frequency of (in pic)

olchik [2.2K]

we have,

wavelenght=c/f

where c= 3×10^8 m/s

f=6.3×10^12 s^-1

so wavelength=(3×10^8)/(6.3×10^12)

=0.476×10^-4 m

4 0

3 years ago

How many millimeters (mm) is the length of a standard table if it is

matrenka [14]

Answer:

1,500 mm

Explanation:

if 1 meter = 1000 mm, 0.5 meters is 500 mm, so 1.50 meters is 1,500 mm

5 0

3 years ago

According to Hund's rule of maximum spin multiplicity, how many singly-occupied orbitals are there in the valence shells of the

leva [86]

Answer:

A) carbon - 2

B) cobalt - 3

C) sulfur - 2

D) fluorine - 1

E) titanium - 2

F) germanium - 2

Explanation:

Hund's rule of maximum multiplicity:-

Firstly, every orbital which is present in the sublevel is singly occupied and then the orbital is doubly occupied.

(A) Carbon.

The electronic configuration is -

$1s^22s^22p^2$

Thus, 2s orbital is fully filled and p orbital can singly filled 3 electrons. Thus, Carbon has 2 singly occupied orbitals.

(B) Cobalt.

The electronic configuration is -

$1s^22s^22p^63s^23p^63d^{7}4s^2$

Thus, 4s orbital is fully filled and d orbital can singly filled 5 electrons. Thus, 4 electrons will be paired in 2 orbitals and 3 orbitals will be singly filled in cobalt.

(C) Sulfur.

The electronic configuration is -

$1s^22s^22p^63s^23p^4$

Thus, 3s orbital is fully filled and p orbital can singly filled 3 electrons. Thus, 2 electrons will be paired in 1 orbital and 2 orbitals will be singly filled in sulfur.

D) fluorine

The electronic configuration is -

$1s^22s^22p^5$

Thus, 2s orbital is fully filled and p orbital can singly filled 3 electrons. Thus, 4 electrons will be paired in 2 orbitals and 1 orbital will be singly filled in fluorine.

E) Titanium

The electronic configuration is -

$1s^22s^22p^63s^23p^63d^{2}4s^2$

Thus, 4s orbital is fully filled and d orbital can singly filled 5 electrons. Thus, 2 orbitals will be singly filled in titanium.

F) Germanium

The electronic configuration is -

$1s^22s^22p^63s^23p^63d^{10}4s^24p^2$

Thus, 4s, 3d orbitals are fully filled and p orbital can singly filled 3 electrons. Thus, Germanium has 2 singly occupied orbitals.

4 0

3 years ago

You have 100 mL of a solution of benzoic acid in water; the amount of benzoic acid in the solution is estimated to be about 0.30

dimaraw [331]

Answer:

0.00370 g

Explanation:

From the given information:

To determine the amount of acid remaining using the formula: $\dfrac{(final \ mass \ of \ solute)_{water}}{(initial \ mass \ of \ solute )_{water}} = (\dfrac{v_2}{v_1+v_2\times k_d})^n$

where;

v_1 = volume of organic solvent = 20-mL

n = numbers of extractions = 4

v_2 = actual volume of water = 100-mL

k_d = distribution coefficient = 10

∴

$\dfrac{(final \ mass \ of \ solute)_{water}}{0.30 \ g} = (\dfrac{100 \ ml}{100 \ ml +20 \ ml \times 10})^4$

$\dfrac{(final \ mass \ of \ solute)_{water}}{0.30 \ g} = (\dfrac{100 \ ml}{100 \ ml +200 \ ml})^4$

$\dfrac{(final \ mass \ of \ solute)_{water}}{0.30 \ g} = (\dfrac{1}{3})^4$

$\dfrac{(final \ mass \ of \ solute)_{water}}{0.30 \ g} = 0.012345$

Thus, the final amount of acid left in the water = 0.012345 * 0.30

= 0.00370 g

3 0

3 years ago