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Gelneren [198K]
3 years ago
8

Michelle chooses a 10 kg weight. She sets up a ramp made of smooth metal that makes an angle θ with the floor. She attaches a sp

ring scale to the weight and the top of the ramp in order to hold the weight in place. She records the force from the spring scale, then changes θ and records it again. She repeats this several times.
In this experiment, what is the dependent variable?
A.
the mass of the weight
B.
the angle θ between the ramp and the ground
C.
the amount of force on the spring scale
D.
the material the ramp is made of
Chemistry
1 answer:
Yuki888 [10]3 years ago
5 0
I think it’s gonna be A
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ludmilkaskok [199]

Answer:

\large \boxed{\text{5.9 s}}

Explanation:

Graham’s Law applies to the effusion of gases:

The rate of effusion (r) of a gas is inversely proportional to the square root of its molar mass (M).

r \propto \dfrac{1}{\sqrt{M}}

If you have two gases, the ratio of their rates of effusion is

\dfrac{r_{2}}{r_{1}} = \sqrt{\dfrac{M_{1}}{M_{2}}}

The time for diffusion is inversely proportional to the rate.

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Let CO₂ be Gas 1 and O₂ be Gas 2

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M₁ = 44.01

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Calculation

\begin{array}{rcl}\dfrac{t_{2}}{t_{1}} & = & \sqrt{\dfrac{M_{2}}{M_{1}}}\\\\\dfrac{t_{2}}{\text{5 s}}& = & \sqrt{\dfrac{44.01}{32.00}}\\\\& = & \sqrt{1.375}\\t_{2}& = & \text{5 s}\times 1.173\\& = & \mathbf{5.9 s} \\\end{array}\\\text{It will take $\large \boxed{\textbf{5.9 s}}$ for the carbon dioxide to effuse.}

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Hence, [CO_{2}] will be calculated as follows.

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or,                        = 0.608 \times 10^{-5}

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As,      K_{a} = \frac{[H^{+}]^{2}}{[CO_{2}]}

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So,                      pH = -log (1.64 \times 10^{-6})

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Answer:

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