The relation between the volume of the gas and the temperature is established by Charles's law. With a decrease in the temperature, the volume decreases by 45.7 mL. Thus, option c is correct.
<h3>What is Charle's law?</h3>
Charle's law states the direct relation present between the temperature and the volume of the gas. The law is given as:
V₁ ÷ T₁ = V₂ ÷ T₂
Given,
V₁ = 50 mL
T₁ = 303.15 K
T₂ = 277.15 K
Substituting the value the final volume is calculated as:
50 ÷ 303.15 = V₂ ÷ 277.15
V₂ = (50 × 277.15) ÷ 303.15
= 45.71 mL
Therefore, option c. 45.7 mL is the final volume.
Learn more about Charles law here:
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I would say that you should wear a lab coat, safety goggles, and gloves
when the teacher says so - not everything in a lab is dangerous, so
there is no need to always wear these. But when the teacher says you
should - then you should.
Answer:
±0.005 g
Explanation:
The uncertainty depends on whether the measurement was obtained manually or digitally.
1. Manual
The minimum uncertainty is ±0.01 g.
It may be greater, depending on random or personal errors
2. Digital
Most measurements of mass are now made on digital scales.
A digital device must always round off the measurement it displays.
For example, if the display reads 20.00, the measurement must be between 20.005 and 19.995 (±0.005).
If the measured value were 20.006, the display would round up to 20.01.
If the measured value were 19.994, the display would round down to 19.99.
The uncertainty is ±0.005 g.
The scale shown below would display a mass of 20.00 g
We can set up an ICE table for the reaction:
HClO H+ ClO-
Initial 0.0375 0 0
Change -x +x +x
Equilibrium 0.0375-x x x
We calculate [H+] from Ka:
Ka = 3.0x10^-8 = [H+][ClO-]/[HClO] = (x)(x)/(0.0375-x)
Approximating that x is negligible compared to 0.0375 simplifies the equation to
3.0x10^-8 = (x)(x)/0.0375
3.0x10^-8 = x2/0.0375
x2 = (3.0x10^-8)(0.0375) = 1.125x10^-9
x = sqrt(1.125x10^-9) = 0.0000335 = 3.35x10^-5 = [H+]
in which 0.0000335 is indeed negligible compared to 0.0375.
We can now calculate pH:
pH = -log [H+] = - log (3.35 x 10^-5) = 4.47
