Answer:
The answer to your question is letter A
Explanation:
Abundance Atomic mass
V-50 0.25% 49.947
V-51 99.750% 50.944
Formula
Average atomic mass = [V-50-Abundance x V-50-Atomic mass] +
[ V-51-Abundance x V-51-Atomic mass]
Substitution
Average atomic mass = [ 0.0025 x 49.947] + [0.9975 x 50.944]
= 0.125 + 50.817
Result
= 50.942 amu
N2 + 3H2 = 2NH3
in this question, we are dealing with only NH3 and H2 so we only focus on that
since the ratio of H2 to 2NH3 is 3:2, we say that
3 liters of H2 = 2 liters of 2NH3
3.6 litres of H2 = x liters of 2NH3
We cross multiple to give:
3 × x = 3.6 × 2
3x = 7.2
Divide both sides by 3
x = 7.2 ÷ 3
x = 2.4liters
What are stars made of? Basically, stars are big exploding balls of gas, mostly hydrogen and helium. Our nearest star, the Sun, is so hot that the huge amount of hydrogen is undergoing a constant star-wide nuclear reaction, like in a hydrogen bomb.
In a spiral galaxy like the Milky Way, the stars, gas, and dust are organized into a "bulge," a "disk" containing "spiral arms," and a "halo." Elliptical galaxies have a "bulge-shape" and a "halo," but do not have a "disk.
Hope it helped
Answer:
0.246 moles
Explanation:
Aluminum Sulfate is an ionic equation so we need to write the equation first with proper charges and subscripts:
- Aluminum (Al) has a charge of [3+]. Sulfate (SO4) has a charge of [2-]. The charges need to be swapped to match.
- The ionic equation therefore is: Al_2(SO4)_3. We need 2 Aluminum's to match with the charge of 3 sulfates.
Now that we know the ionic equation, we need to find the molar mass of Al_2(SO4)_3. There's <em><u>2 Aluminum's, each weighing 26.98 g/mol. There are 3 sulfates, each sulfate weighs 96.06 g/mol</u></em>. We need to multiply the number of elements/compounds by their mass to get the total molar mass and then add them up <em>(remember atomic mass = molar mass in g/mol)</em>:
- (2 Al * 26.98 g/mol) + (3 (SO4) * 96.06 g/mol) = 342.15 g/mol of Al_2(SO4)_3
We now know the molar mass of Al_2(SO4)_3 we need to find how many moles of it there are in an 84.2g sample. To do that, we need to do some dimensional analysis:
- 84.2g Al_2(SO4)_3 * ( 1 mol of Al_2(SO4)_3 / 342.15 g/mol of Al_2(SO4)_3 )
The grams of Al_2(SO4)_3 cancel so that we're left with moles of Al_2(SO4)_3 on top:
- 84.2 * ( 1 mol of Al_2(SO4)_3 / 342.15 ) = <u><em>0.246 mol of Al_2(SO4)_3</em></u>
<u><em></em></u>
So in an 84.2g sample of Al_2(SO4)_3 (aka aluminum sulfate), there are 0.246 moles. This is correct because in one mole of Al_2(SO4)_3, there are 342.15g and the sample of 84.2 g of Al_2(SO4)_3 is way smaller than that, so our answer should be less than one mole, which it is.
I hope this helped!
