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Eva8 [605]
3 years ago
6

in a torricell barometer a pressure of one atmospheric pressure supports q 760 mm coloumn of mercury.if the original tube contai

ning the mercury is replaced with tybe having twice the diameter of the original with a tube having twice the dimaeter of the original, the height of the mercury___________-
Chemistry
1 answer:
dybincka [34]3 years ago
5 0

Answer:

The answer to the question is

The height of the mercury fluid column remain the same.

Explanation:

The pressure, P in a column of fluid of height, h is given by

P = (Density p)×(height of fluid column h)×(gravity g)

Therefore, when the diameter is doubled we have

Density of the mercury in the tube with twice the diameter = (Mass of mercury)/(volume of mercury) where the volume of mercury = h×pi×(Diameter×2)^2/4 = h×pi×Diameter^2. Therefore the volume increases by a factor of 4 and therefore the mass increases by a factor of 4 which means that the density remains the same hence

P = p×h1×g = p×h2×g Therefore h1 = h2

The height of the fluid column remain the same

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A 100.0 mL sample of 0.300 M NaOH is mixed with a 100.0 mL sample of 0.300 M HNO3 in a coffee cup calorimeter. If both solutions
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Answer:

Qm  = -55.8Kj/mole

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What is the half-life of a pharmaceutical if the initial dose is 500 mg and only 31 mg remains after 6 hours?
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\large \boxed{\text{b. 1.5 h}}

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\begin{array}{rcl}\ln \left (\dfrac{500}{31}\right) & = & k \times 6\\\\\ln 16.1 & = & 6k\\2.78& =& 6k\\k & = & \dfrac{2.78}{6}\\\\& = & 0.463 \text{ h}^{-1}\\\end{array}

2. Calculate the half-life

t_{\frac{1}{2}} = \dfrac{\ln2}{k} = \dfrac{\ln2}{\text{0.463  h}^{-1}} = \textbf{1.5 h}\\\\ \text{The half-life is $\large \boxed{\textbf{1.5 h}}$}

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