Answer:
I choose D option because may be it's correct
Answer:
<em>a)</em> <em>1.392 x 10^6 g/cm^3</em>
<em>b) 8.69 x 10^7 lb/ft^3</em>
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Explanation:
mass of the star m = 2.0 x 10^36 kg
radius of the star (assumed to be spherical) r = 7.0 x 10^5 km = 7.0 x 10^8 m
The density of substance ρ = mass/volume
The volume of the star = volume of a sphere = 
==> V = 
= 1.437 x 10^27 m^3
density of the star ρ = (2.0 x 10^36)/(1.437 x 10^27) = 1.392 x 10^9 kg/m^3
in g/cm^3 = (1.392 x 10^9)/1000 = <em>1.392 x 10^6 g/cm^3</em>
in lb/ft^3 = (1.392 x 10^9)/16.018 = <em>8.69 x 10^7 lb/ft^3</em>
Answer:
it's is math that you can answer with your sister and your mother
Answer: hi theree, the answer is c. the processes of endocytosis and exocytosis occur here
hope this helps, have a good day :)
Explanation:
Answer:
93.28%
Explanation:
To solve the percent yield we need to find theoretical yield:
<em>Percent yield = Actual Yield (452.78g) / Theoretical yield * 100</em>
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Theoretical yield is obtained converting the mass of acetylene to moles and using the balanced equation determine the moles of CO₂ produced assuming a 100% of reaction:
<em>Moles acetylene (Molar mass: 26.04g/mol)</em>
143.6g C₂H₂ * (1mol / 26.04g) = 5.515 moles C₂H₂
<em>Moles CO₂:</em>
5.515 moles C₂H₂ * (4 moles CO₂ / 2mol C₂H₂) = 11.029moles CO₂
<em>Mass CO₂ (Molar mass: 44.01g/mol):</em>
11.029moles CO₂ * (44.01g / mol) = 485.39g CO₂ is theoretical yield
Percent yield is:
Percent yield = 452.78g / 485.39g * 100
= 93.28%
