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makvit [3.9K]
2 years ago
8

Propane (c3h8), a common fuel, reacts with oxygen to form carbon dioxide and water according to the equation below. c3h8 5o2 → 3

co2 4h2o how many grams of oxygen are required to produce 37.15 g co2? 37.15 g co2 = g o2
Physics
1 answer:
ahrayia [7]2 years ago
3 0

The amount of oxygen will be 22.29 gm required to produce 37.15 g co₂.

<h3>What is a chemical reaction?</h3>

One or more compounds, known as reactants, are transformed into one or more distinct substances, known as products. Such a process is known as a chemical reaction.

Given reaction;

\rm C_3h_8 +5O_2 }\rightarrow  3CO_2 +4H_2O

The quantity of oxygen is x and the quantity of carbon dioxide is y then the ratio of their composition is found as;

5x=3y

5x=37.15

x=3×37.15/5

x=22.29 gm

Hence the amount of oxygen will be 22.29 gm

To learn more about the chemical reaction refer to the link;

brainly.com/question/22817140

#SPJ1

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A 1-kilogram mass is attached to a spring whose constant is 14 N/m, and the entire system is then submerged in a liquid that imp
ch4aika [34]

Answer:

Part(a): The equation of motion is \bf{x(t) = \dfrac{7}{5}~e^{-2t} - \dfrac{2}{5}~e^{-7t}}.

Part(b): The equation of motion is  \bf{x(t) = -e^{-2t} + \dfrac{11}{5}~e^{-7t}}.

Explanation:

If 'm' be the mass of the object, 'k' be the force constant and '\beta' be the damping constant, then the equation of motion of the particle can be written as

\dfrac{d^{2}x}{dt^{2}} + \dfrac{\beta}{m} \dfrac{dx}{dt} + \dfrac{k}{m}x= 0.........................................(I)

Given m = 1 Kg, k = 14 N~m^{-1}, \beta = 9. Substituting these values in equation (I),

\dfrac{d^{2}x}{dt^{2}} + 9~\dfrac{dx}{dt} + 14~x= 0

Taking a trial solution x(t) = e^{mt}, the auxiliary equation can be written as

m^{2} + 9m + 14 = 0............................................................(II)

and its solutions are m_{1} = -2~and~m_{2} = -7, resulting the general solution

x(t) = C_{1}~e^{-2t} + C_{2}~e^{-7t}....................................................................(III)

The velocity at any instant of time of the mass is

v(t) = -2C_{1}~e^{-2t} _7~C_{2}~e^{-7t}..............................................................(IV)

Part(a):

Given x(t=0) = 1 m,~and~v(t=0) = 0~m~s^{-1}. Substituting these values in equation (III) and (IV),

&& 1 = C_{1} + C_{2}......................(V)\\&and,& 0 = -2C_{1} - 7C_{2}.......................(VI)

Solving equations (V) and (VI), we have

C_{1} = \dfrac{7}{5}~and~C_{2} = \dfrac{-2}{5}

So the equation of motion is

x(t) = \dfrac{7}{5}~e^{-2t} - \dfrac{2}{5}~e^{-7t}

Part(b):

Given x(t=0) = 1 m,~and~v(t=0) = - 12~m~s^{-1}. Substituting these values in equation (III) and (IV),

&& 1 = C_{1} + C_{2}......................(VII)\\&and,& -12 = -2C_{1} -7C_{2}.......................(VIII)

Solving equations (V) and (VI), we have

C_{1} = -1~and~C_{2} = \dfrac{11}{5}

So the equation of motion is

x(t) = -e^{-2t} + \dfrac{11}{5}~e^{-7t}

3 0
3 years ago
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