The equation is: C+O2=>CO2
Since we got 10 molecules of CO2 new balanced equation would be 10C+10O2=>10CO2
from this equation we can see that we have 10 molecules of oxygen, however ,we need to find atoms. There are 2 atoms in the oxygen molecule so we need to multiply 10 by 2 which gives us 20 atoms.
The answer: there are 20 atoms of oxygen
Answer:
Explanation:
You look at the type of atom and their electronegativity difference.
If ΔEN <1.6, covalent; if ΔEN >1.6, ionic
Ar/Xe: Noble gases; no reaction
F/Cs: Non-metal + metal; ΔEN = |3.98 – 0.79| = 3.19; Ionic
N/Br: Two nonmetals; ΔEN = |3.04 - 2.98| = 0.
<h3>
Answer: b) 0.250 mol</h3>
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Work Shown:
Using the periodic table, we see that
- 1 mole of carbon = 12 grams
- 1 mole of oxygen = 16 grams
These are approximations and these values are often found underneath the atomic symbol. For example, the atomic weight listed under carbon is roughly 12.011 grams. I'm rounding to 2 sig figs in those numbers listed above.
So 1 mole of CO2 is approximately 12+2*16 = 44 grams. The 2 is there since we have 2 oxygens attached to the carbon atom.
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Since 1 mole of CO2 is 44 grams, we can use that to convert from grams to moles.
11.0 grams of CO2 = (11.0 grams)*(1 mol/44 g) = (11.0/44) mol = 0.250 mol of CO2
In short,
11.0 grams of CO2 = 0.250 mol of CO2
This is approximate.
We don't need to use any of the information in the table.
Answer: 72L of 30% and 128L of 80%
You can determine the weight of the acid by multiplying the concentration with the volume. Let say v1 is the volume of 30% solution needed and v2 is the volume of 80% solution.
The weight of acid from the used solution should be equal to the product. You can get this equation
final solution= solution1 + solution2
200l * 62%= v1 * 30% + v2*80%
124L= 0.3v1 + 0.8v2
124L- 0.3v1= 0.8v2
v2=155L- 0.375v1
The total volume of both should be 200l. If you use the previous equation, you can calculate:
v1+v2=200L
v1+ (155L- 0.375v1)= 200L
0.625v1= 200L - 155L
v1= 45/ 0.625= 72L
v1+v2=200L
v2= 200L- 72L= 128L
