<span>This is due to the fact that the air pressure in that certain section of Earth’s atmosphere decreased. As density of gas particles decreases as air pressure decreases. Therefore, density of gas particles and air pressure have a direct relationship. An increase in air pressure would then effect to an increase in gas particles. </span>
Answer:
Gamma
Explanation:
I'm not sure how to do it without calculations but:
E=hv
7*10^7 J/mol=6.626*10^34 Js * v
v=1*10^41
Gamma rays.
More here: https://www.hasd.org/faculty/AndrewSchweitzer/spectroscopy.pdf
Atomic mass Ni = 58.69 a.m.u
58.69 g ----------------- 6.02x10²³ atoms
?? g --------------------- 7.5x10¹⁵ atoms
58.69x (7.5x10¹⁵) / 6.02x10²³
=> 7.31x10⁻⁷ g
Given:
7.20 g sample of Al2(SO4)3
Required:
Mass of oxygen
Solution:
Since you are not given a
chemical reaction, just base your solution to the chemical formula given.
Molar mass of Al2(SO4)3 = 342.15 g/mol
7.20 g Al2(SO4)3 (1 mol/342.15g)(3mol O/2 mol Al)(1 mol O2/1/2 mol
O2)(32g O2/1mol O2) = 4.04 g O2
Ok so this is what we know :
2KClO3 -> 2KCl + 3O2 (Always check if equation is balanced - in this case it is)
4.26moles
So we know that we have 4.26 moles of oxygen (O2). Now lets look at the ratio between KClO3 and O2.
We see that the ratio is 2:3 meaning that we need 2KClO3 in order to produce 3O2.
Therefore divide 4.26 by 3 and then multiply by 2.
4.26/3 = 1.42
1.42 * 2 = 2.84
Now we know that the molarity of KClO3 is 2.84 moles.
Multiply by R.M.M to find how many grams of KClO3 we have.
R.M.M of KClO3
K- 39
Cl- 35.5
3O- 3 * 16 -> 48
---------------------------
<span>122.5
</span>2.84 * 122.5 = 347.9 grams therefore the answer is (a)
348 grams needed of KClO3 to produce 4.26 moles of O2.
Hope this helps :).
