Answer:
to do day-to day activities easily and make surrounding comfortable for man
Explanation:
as an example , binary numbers help to work computers ,calculators.they help us lot .
scientific methods also need to write large and small numbers in short ways .so, scientists find various patterns .
If 1 mole of helium is 4.002602 then 1.2 mole is about ~ 4.803 ♂️.
Answer:
evaperation
Explanation:
remaining energy is transferred from the earth surface by evaporation
Answer:
Answer 2 i got this correct mark brainliest ~shay
Explanation:
Answer : The empirical formula of a compound is, ![Na_3BO_3](https://tex.z-dn.net/?f=Na_3BO_3)
Solution : Given,
If percentage are given then we are taking total mass is 100 grams.
So, the mass of each element is equal to the percentage given.
Mass of Na = 53.976 g
Mass of B = 8.461 g
Mass of O = [100 - (53.976 + 8.461)] = 37.563 g
Molar mass of Na = 23 g/mole
Molar mass of B = 11 g/mole
Molar mass of O = 16 g/mole
Step 1 : convert given masses into moles.
Moles of Na = ![\frac{\text{ given mass of Na}}{\text{ molar mass of Na}}= \frac{53.976g}{23g/mole}=2.347moles](https://tex.z-dn.net/?f=%5Cfrac%7B%5Ctext%7B%20given%20mass%20of%20Na%7D%7D%7B%5Ctext%7B%20molar%20mass%20of%20Na%7D%7D%3D%20%5Cfrac%7B53.976g%7D%7B23g%2Fmole%7D%3D2.347moles)
Moles of B = ![\frac{\text{ given mass of B}}{\text{ molar mass of B}}= \frac{8.461g}{11g/mole}=0.769moles](https://tex.z-dn.net/?f=%5Cfrac%7B%5Ctext%7B%20given%20mass%20of%20B%7D%7D%7B%5Ctext%7B%20molar%20mass%20of%20B%7D%7D%3D%20%5Cfrac%7B8.461g%7D%7B11g%2Fmole%7D%3D0.769moles)
Moles of O = ![\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{37.563g}{16g/mole}=2.347moles](https://tex.z-dn.net/?f=%5Cfrac%7B%5Ctext%7B%20given%20mass%20of%20O%7D%7D%7B%5Ctext%7B%20molar%20mass%20of%20O%7D%7D%3D%20%5Cfrac%7B37.563g%7D%7B16g%2Fmole%7D%3D2.347moles)
Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.
For Na = ![\frac{2.347}{0.769}=3.05\approx 3](https://tex.z-dn.net/?f=%5Cfrac%7B2.347%7D%7B0.769%7D%3D3.05%5Capprox%203)
For B = ![\frac{0.769}{0.769}=1](https://tex.z-dn.net/?f=%5Cfrac%7B0.769%7D%7B0.769%7D%3D1)
For O = ![\frac{2.347}{0.769}=3.05\approx 3](https://tex.z-dn.net/?f=%5Cfrac%7B2.347%7D%7B0.769%7D%3D3.05%5Capprox%203)
The ratio of Na : B : O = 3 : 1 : 3
The mole ratio of the element is represented by subscripts in empirical formula.
The Empirical formula =
= ![Na_3BO_3](https://tex.z-dn.net/?f=Na_3BO_3)