Answer:
There is a mass of 154 Grams of Carbon Dioxide.
Explanation:
One mole is equal to 6.02 × 10^23 particles.
This means we have 1.05 X 10^24 total particles of Ethane.
Each ethane particle contains 2 carbon atoms.
If every particle of ethane is burned, we will end up with 2.10 x 10^24 molecules of Carbon Dioxide (Particles of Methane x 2, since each Methane particle contains 2 carbon atoms)
Carbon Dioxide has a molar mass of 44.01 g/mol
So if we take our amount of Carbon Dioxide molecules and divide it by 1 mole, ((2.10 x 10^24)/(6.02 x 10^23) = 3.49) we find that we have 3.49 moles of Carbon Dioxide.
Now all we need to do is multiply our moles of carbon dioxide(3.49) by it's molar mass(44.01) while accounting for significant digits.
What you should end up with is 154 Grams of Carbon Dioxide.
Hope this helps (And more importantly I hope I didn't make any errors in my math lol)
As a side note this is all assuming that this takes place at STP conditions.
The missing question is:
<em>What is the percent efficiency of the laser in converting electrical power to light?</em>
The percent efficiency of the laser that consumes 130.0 Watt of electrical power and produces a stream of 2.67 × 10¹⁹ 1017 nm photons per second, is 1.34%.
A particular laser consumes 130.0 Watt (P) of electrical power. The energy input (Ei) in 1 second (t) is:

The laser produced photons with a wavelength (λ) of 1017 nm. We can calculate the energy (E) of each photon using the Planck-Einstein's relation.

where,

The energy of 1 photon is 6.52 × 10⁻²⁰ J. The energy of 2.67 × 10¹⁹ photons (Energy output = Eo) is:

The percent efficiency of the laser is the ratio of the energy output to the energy input, times 100.

The percent efficiency of the laser that consumes 130.0 Watt of electrical power and produces a stream of 2.67 × 10¹⁹ 1017 nm photons per second, is 1.34%.
You can learn more about lasers here: brainly.com/question/4869798
Answer:
ok ............bu am not so sure
Answer:
mass of CO = 210.42 g
mass in three significant figures = 210. g
Explanation:
Given data:
mass of Fe2O3 = 0.400 Kg
mass of CO= ?
Solution:
chemical equation:
Fe2O3 + 3CO → 2Fe + 3CO2
Now we will calculate the molar mass of Fe2O3 and CO.
Molar mass of Fe2O3 = (55.845 × 2) + (16 × 3) = 159.69 g/mol
Molar mass of CO = 12+ 16 = 28 g/mol
now we will convert the kg of Fe2O3 in g.
mass of Fe2O3 = 0.400 kg × 1000 = 400 g
number of moles of Fe2O3 = 400 g/ 159.69 g/mol = 2.505 mol
mass of CO = moles of Fe2O3 × 3( molar mass of CO)
mass of CO = 2.505 mol × 84 g/mol
mass of CO = 210.42 g
mass in three significant figures = 210. g