All categories

3 years ago

What is the mass of 4 × 1015 atoms of nickel (Ni)? Answer in units of g.

Chemistry

2 answers:

IRISSAK [1]3 years ago

8 0

Answer:

4060

Explanation:

hope this helps

follow me for more answers

AlekseyPX3 years ago

7 0

It would probably be 4060

You might be interested in

Which statement describes the light reaction of photosynthesis?

cupoosta [38]

Answer:

Explanation:

water molecules are combined to make food

5 0

3 years ago

36 g of water vapor takes up how many liters at room temperature and pressure of 293K and 100 K PA

crimeas [40]

100kpa is correct .100 kpa is correct

6 0

4 years ago

Read 2 more answers

How many mole of CO2 are in 58.9g of the compound

vivado [14]

Convert grams to moles and use Avogadro's number 6.022x10tho the 23rd power. Hope this helps

8 0

3 years ago

An atom of helium has a radius of 31. pm and the average orbital speed of the electrons in it is about 4.4x 10 m/s Calculate the

otez555 [7]

The question is incomplete, here is the complete question:

An atom of helium has a radius of 31. pm and the average orbital speed of the electrons in it is about $4.4\times 10^6m/s$ . Calculate the least possible uncertainty in a measurement of the speed of an electron in an atom of helium. Write your answer as a percentage of the average speed, and round it to 2 significant digits

<u>Answer:</u> The percentage of average speed is 41 %

<u>Explanation:</u>

We are given:

Radius of helium atom = 31 pm = $31\times 10^{-12}m$ (Conversion factor: $1m=10^{12}pm$ )

So, diameter of helium atom = $(2\times r)=(2\times 31\times 10^{-12})=64\times 10^{-12}m$

The diameter of the atom will be equal to the uncertainty in position.

The equation representing Heisenberg's uncertainty principle follows:

$\Delta x.\Delta p=\frac{h}{2\pi}$

where,

$\Delta x$ = uncertainty in position = d = $64\times 10^{-12}m$

$\Delta p$ = uncertainty in momentum = $m\Delta v$

m = mass of electron = $9.1095\times 10^{-31}kg$

h = Planck's constant = $6.627\times 10^{-34}kgm^2/s^2$

Putting values in above equation, we get:

$64\times 10^{-12}m\times 9.1095\times 10^{-31}kg\times \Delta v=\frac{6.627\times 10^{-34}kgm^2/s}{2\times 3.14}\\\\\Delta v=\frac{6.627\times 10^{-34}kgm^2/s^2}{2\times 3.14\times 64\times 10^{-12}m\times 9.1095\times 10^{-31}kg}=1.81\times 10^6m/s$