Ask question

Ask question

All categories

2 years ago

8

How many moles are in 3.90 x1030 atoms of Zn?

1 answer:

zaharov [31]2 years ago

7 0

Explanation:
The problem basically wants you to find a way to convert between the number of atoms present in the sample and the number of moles they are equivalent to.
To convert between atoms and moles we use something called Avogadro's constant, which basically acts as the definition of a mole.
More specifically, in order to have one mole of an element you need
6.022 x 10^23 atoms of that element. You can thus use this number as a conversion factor to take you from atoms to moles or vice versa.
In your case, you will have

3.90 x 10^ 26 atoms Zn x 1 mole Zn ( Avogrado’s constant) / 6.022 x 10^23 atoms Zn

= 6.5 x 10^8 is the answer

You might be interested in

The pollen grains move in water randomly because of

polet [3.4K]

A the collisions between water particles and polllen grains

6 0

3 years ago

If earth was a hard boiled egg,which part of earth would the egg white represent?

romanna [79]

Answer:

The egg white will represent the outer layer which is the we are located on!

Explanation:

6 0

3 years ago

Suppose that 7.25 x 10^22 atoms of a hypothetical element have a mass of 3.88 g. What would be the molar mass (g/mol) of this el

S_A_V [24]

Answer:

32.23 to 4 significant figures.

Explanation:

The molar mass of the element is the mass of 6.022 * 10^23 atoms (Avogadro's number).

So by proportion it is 6.022 * 10^23 * 3.88 / 7.25 * 10^22

= 32.23 to 4 significant figures.

4 0

3 years ago

A solution contains one or more of the following ions: Ag , Ca2 , and Co2 . Lithium bromide is added to the solution and no prec

koban [17]

Answer:

$Ca^{2+} and Co^{2+}$

Explanation:

Given:

A solution contains one or more of the following ions such as Ag, $Ca_2$ and $Co_2$

Here the Lithium bromide is added to the solution and no precipitate forms

Solution:

Since with LiBr no precipitation takes place therefore Ag+ is absent

Here on adding $Li_2SO_4$ to it precipitation takes place.

Precipitate is as follows,

$Ca^{2+}(aq)+SO_4^2^-(aq)----->CASO_4(s)$

Thus,

$Ca^2^+$ is present

When $Li_3PO_4$ is added again precipitation takes place.

Therefore the reaction is as follows,

$Co^2^+(aq)+PO_4^3^-(aq)------>Co_3(PO_4)_2(s)$

Therefore,

$Ca^{2+} and Co^{2+}$ are present in the solution

3 0

3 years ago

ANSWER FAST PLZZZ!!!!!!!!!!!!!

Anestetic [448]

Answer: D. test tubes, thermometers, etc

I think this is the correct answer. Sorry if it is not.

Explanation:

6 0

3 years ago