Finding percent composition is fairly easy. You only need to divide the mass of an element by the total mass of the compound. We can do this one element at a time.
First, let's find the total mass by using the masses of the elements given on the periodic table.
7 x 12.011 (mass of Carbon) = 84.077
5 x 1.008 (mass of Hydrogen) = 5.04
3 x 14.007 (mass of Nitrogen) = 42.021
6 x 15.999 (mass of Oxygen) = 95.994
Add all of those pieces together.
84.077 + 5.04 + 42.021 + 95.994 = 227.132 g/mol is your total. Since we also just found the mass of each individual element, the next step will be very easy.
Carbon: 84.077 / 227.132 = 0.37016 ≈ 37.01 %
Hydrogen: 5.04 / 227.132 = 0.022189 ≈ 2.22 %
Nitrogen: 42.021 / 227.132 = 0.185 ≈ 18.5 %
Oxygen: 95.994 / 227.132 = 0.42263 ≈ 42.26 %
You can check your work by making sure they add up to 100%. The ones I just found add up to 99.99, which is close enough. A small difference (no more than 0.03 in my experience) is just a matter of where you rounded your numbers.
The answer
<span>the molar ratio for the following equation
____C3H8 + ____O2 Imported Asset ____CO2 + ____ H2O
</span>after it has been properly balanced:
__1_C3H8 + ____5O2 Imported Asset ____3CO2 + ____ 4H2O
proof:
number of C =3 (C3H8; 3CO2)
number of H =8 (C3H8 ; 4H2O)
number of O = 10(5x2) or (2x3+4) (5O2;4H2O)
the answer is
<span>Reactants: C3H8 = 1, O2 = 8; Products: CO2 = 3 and H2O = 4</span>
Answer:
0.861 L
Explanation:
We are given pressure, volume, and temperature, so let's apply the Combined Gas Law:
(P₁V₁)/T₁ = (P₂V₂)/T₂
Convert the temperatures to degrees Kelvin.
25.0°C -> 298 K, 100.0°C -> 373 K
Plug in the initial conditions on the left, then the final/new on the right, and solve for the unknown:
(165(2.5))/298 = (600(V₂))/373
V₂ = (165(2.5)(373))/(298(600))
V₂ = 0.861 L
Answer:
About 5 times faster.
Explanation:
Hello,
In this case, since the Arrhenius equation is considered for both the catalyzed reaction (1) and the uncatalized reaction (2), one determines the relationship between them as follows:

By replacing the corresponding values we obtain:

Such result means that the catalyzed reaction is about five times faster than the uncatalyzed reaction.
Best regards.
Answer:
The sun'll likely absorb the radiation if it is close enough (Which it will never be)
Explanation: