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3 years ago

What does the atomic number of an element represent?

Chemistry

2 answers:

Alexxx [7]3 years ago

8 0

Answer:

The atomic number represents or stands for the distinct identity of a chemical element. It is usually defined as the number or protons present in an atom of an element, which is also equal to the number of electrons.

Explanation:

harkovskaia [24]3 years ago

5 0

Hello!
Whenever you see an atomic number with an element, you know the number of protons in the nucleus of the atom, which determines the chemical of an element and its place on the periodic table.

For example, Hydrogen has only one proton, however, if you were to add say 78 more protons, you'd get gold!

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More than ½ of the world's forests have been lost to deforestation. True False

RideAnS [48]

The answer is false

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4 years ago

Read 2 more answers

Gold is alloyed with other metals to increase its hardness in making jewelery.a) Consider a piece of gold jewelry the weighs 9.8

irinina [24]

Answer:

A). Percentage of gold by mass of the jewellery =

(6.059÷9.85) × 100 = 61.5%

B). Purity of the gold = 0.615 × 24 = 14.76 karat ~ 15 karat gold

Explanation:

Mass of jewellery = 9.85g Volume of jewellery = 0.675cm^3.

Density of gold = 19.3g/cm^3 and density of silver = 10.5g/cm^3

Let the volume of gold in the jewellery be X and the volume of silver in the jewellery be Y

Hence we have

Density = mass/volume or mass = volume × density = for gold = X × 19.3g/cm^3 and for the silver = Y × 10.5g/cm^3

19.3X +10.5Y = 9.85g

Also volume of jewellery is given by

Volume of silver in the jewellery + volume of gold in the jewellery = 0.675cm^3.

X + Y = 0.675cm^3.

Solving the above equations we have

Y = 0.675 - X

Which gives

19.3X + 10.5Y = 9.85g

19.3X + 10.5 × (0.675 - X) =9.85g

19.3X + 7.0875 - 10.5X = 9.85

8.8X + 7.0875 = 9.85

8.8X = 2.7625

or X = 0.3139 cm^3

But Y = 0.675 - X

Hence Y = 0.675 - 0.3139 = 0.3611 cm^3

Mass of gold in the jewellery = volume of gold × Density of gold = 0.3139 × 19.3 = 6.059 g

Also mass of silver = 10.5 × 0.3611 = 3.7913g

A). Percentage of gold by mass of the jewellery =

(6.059÷9.85) × 100 = 61.5%

B). Purity of the gold = 0.615 × 24 = 14.76 karat ~ 15 karat gold

7 0

3 years ago

In the laboratory a student combines 47.8 mL of a 0.321 M aluminum nitrate solution with 21.8 mL of a 0.366 M aluminum iodide so

alex41 [277]

Answer: The final concentration of aluminum cation is 0.335 M.

Explanation:

Given: $V_{1}$ = 47.8 mL (1 mL = 0.001 L) = 0.0478 L

$M_{1}$ = 0.321 M, $V_{2}$ = 21.8 mL = 0.0218 L, $M_{2}$ = 0.366 M

As concentration of a substance is the moles of solute divided by volume of solution.

Hence, concentration of aluminum cation is calculated as follows.

$[Al^{3+}] = \frac{M_{1}V_{1} + M_{2}V_{2}}{V_{1} + V_{2}}$

Substitute the values into above formula as follows.

$[Al^{3+}] = \frac{M_{1}V_{1} + M_{2}V_{2}}{V_{1} + V_{2}}\\= \frac{0.321 M \times 0.0478 L + 0.366 M \times 0.0218 L}{0.0478 L + 0.0218 L}\\= \frac{0.0153438 + 0.0079788}{0.0696}\\= 0.335 M$

Thus, we can conclude that the final concentration of aluminum cation is 0.335 M.

4 0

3 years ago

It is winter where jenna lives. Its been snowing all day, but now the snow has changed to sleet and then to freezing rain. What

Vladimir79 [104]

Hi!

Answer:

The change in temperature.

Explanation:

Snow forms at 0 °C, when water vapor converts directly into solid ice crystals. Sleet forms when raindrops fall through a layer of air colder than 0 °C. This means that from the time it was snowing to the time it started sleeting the air has gotten warmer but one layer of air stayed cold, hence the formation of sleet. Freezing rain is rain that freezes when it hits a cold surface. This means that from the time it was sleeting to the time there was freezing rain the air had completely warmed and is now above 0 °C but the ground and all other surfaces are still cold.

I hope this helps, as this happens all the time where I live! :)

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3 years ago

40POINTS AND BRAINLIEST

mina [271]

Answer:

Explanation:

hope this helps :)

7 0

3 years ago