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kykrilka [37]
3 years ago
6

Complete combustion of 1.11 0 g of a gaseous hydrocarbon yields 3.613 g of carbon dioxide and one 1.109 g of water. 80.288 g sam

ple of the hydrocarbon occupies a volume of 131 milliliters at 24.8°C and 753 mm Hg. What is the molecular formula of the hydrocarbon?
Physics
1 answer:
Mamont248 [21]3 years ago
4 0

Answer:

 C4H6

Explanation:

we know that

n = P×V / RT

= (753 ) x (0.131 ) / ((62.36367) x (24.8 + 273.15) ) = 0.005309

molecular weight = (0.288) / (0.005309) = 54.25 g/mol

molecular weight  of CO_2 in terms of C

(3.613 ) / (44.00964 ) x (12.01078) = 0.986033g C

Molecular weight of H_2O in terms of H

(1.109 ) / (18.01532 ) x (2  / 1) x (1.007947) = 0.124096g H

Number of atoms of C

(0.986033) / (1.110) x (54.25) / (12.01078) = 4.012

(0.124096) / (1.110) x (54.25 ) / (1.007947) = 6.017

Round to the nearest whole numbers to find the molecular formula:  C4H6

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B. 14.4 N

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A spinning wheel is slowed down by a brake, giving it a constant angular acceleration of ?5.20 rad/s2. during a 3.80-s time inte
ddd [48]

<span>We can answer this using the rotational version of the kinematic equations:</span><span>
θ = θ₀ + ω₀<span>t + ½αt²     -----> 1</span></span>

ω² = ω₀² + 2αθ            -----> 2

Where:

θ = final angular displacement = 70.4 rad

θ₀ = initial angular displacement = 0

ω₀ = initial angular speed

ω = final angular speed

t = time = 3.80 s

α = angular acceleration = -5.20 rad/s^2

Substituting the values into equation 1:<span>
70.4 = 0 + ω₀(3.80) + ½(-5.20)(3.80)² </span><span>

ω₀ = (70.4 + 37.544) / 3.80 </span><span>

ω₀ = 28.406 rad/s </span><span>


Using equation 2:
ω² = (28.406)² + 2(-5.2)70.4 


ω = 8.65 rad/s 


</span>

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3 years ago
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