Chemistry, 09.03.2020 04:55 taterbugwallen7944

A student is examining a bacterium under the microscope. The E. coli bacterial cell has a mass of m = 2.00 fg (where a femtogram, fg, is 10−15g) and is swimming at a velocity of v = 7.00 μm/s , with an uncertainty in the velocity of 2.00 % . E. coli bacterial cells are around 1 μm ( 10−6 m) in length. The student is supposed to observe the bacterium and make a drawing. However, the student, having just learned about the Heisenberg uncertainty principle in physics class, complains that she cannot make the drawing. She claims that the uncertainty of the bacterium's position is greater than the microscope's viewing field, and the bacterium is thus impossible to locate. Part A What is the uncertainty of the position of the bacterium?

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Answer from: meramera50

$1.88\cdot 10^{-10} m$

Explanation:

Heisenberg's uncertainty principle states that it is not possible to know with infinite precision the position and the momentum of an object at the same time. Mathematically, this is written as:

$\Delta x \Delta p \geq \frac{h}{4\pi}$

where:

$\Delta x$ is the uncertainty on the position

$\Delta p$ is the uncertainty on the momentum

$h=6.63\cdot 10^{-34} Js$ is the Planck's constant

Since the momentum can be written as product of mass (m) and velocity:

p=mv

The uncertainty on the momentum can be written as (assuming the mass is known with infinite precision):

$\Delta p = m\Delta v$

Therefore, the previous equation can be rewritten as:

$m\Delta x\Delta v\geq \frac{h}{4\pi}$

In this problem, we have:

$m=2.00 fg = 2.00\cdot 10^{-15} g = 2.00\cdot 10^{-18} kg$ is the mass of the E.Coli

$v=7.00\mu m/s = 7.00\cdot 10^{-6} m/s$ is the E.Coli velocity

The uncertainty on the velocity is 2.00% of this value, so:

$\Delta v = \frac{2}{100}v=0.02\cdot 7.00\cdot 10^{-6}=1.4\cdot 10^{-7} m/s$

Therefore, if we now re-arrange the equation, we can find $\Delta x$ , the minimum uncertainty on the position of the bacterium:

$\Delta x \geq \frac{h}{4\pi m\Delta v}=\frac{6.63\cdot 10^{-34}}{4\pi (2.0\cdot 10^{-18})(1.4\cdot 10^{-7})}=1.88\cdot 10^{-10} m$

Answer from: Quest

Answerbonded to an amino group(nh2);

Answer from: Quest

Note that 2 mol dm⁻³ = 2 mol/(dm³)outline: we're given the concentration of hcl, which is 2 mol dm⁻³ (cubic meters in the denominator can represent volume, so i know this is concentration). we're an amount of calcium carbonate caco₃, which is 5 g. we're presented with a chemical equation which includes caco₃ and hcl. we are told to find the volume of hcl. from what we see so far, this seems to be a stoichiometry problem. we'll have to use the given information about calcium carbonate to find the amount of moles of hcl needed to fully react with calcium carbonate. then we'll have information about the moles of hcl and the concentration of hcl; that will be enough to figure out volume according to the formula concentration = moles / volume.======all the answer choices are not in dm.convert the given concentration so that it uses cm⁻³.we can use conversion factors. we have to cube our conversion factors so that dimensional analysis works.we can convert to meters first, using the fact that 10 dm = 1 m ⇔ 10³ dm³ = 1 m³then we use the fact that 10² cm = 1 m ⇔ (10² cm)³ = (1 m)³ ⇔ 10⁶ cm³ = 1 m³so converting: $\dfrac{\text{2 mol}}{\text{dm}^{3}} \cdot \dfrac{\text{10}^3\text{ dm}^3}{\text{1 m}^3} \cdot \dfrac{\text{1 m}^3}{10^6\text{ cm}^3} \\ \\ = \dfrac{\text{2 mol}}{1} \cdot \dfrac{\text{10}^3}{1} \cdot \dfrac{1}{10^6\text{ cm}^3} \qquad \text{\footnotesize cancelling units} \\ \\ = \dfrac{2\text{ mol}}{10^3 \text{ cm}^3} \\ \\ = 0.002\text{ mol/cm}^3$ now we have to find the moles of hydrochloric acid neededfind the moles of calcium carbonate. we have to find the molar mass of calcium carbonate, caco₃, which can be found by molar mass = (mass ca) + (mass c) + 3(mass o) = 40 + 12 + 3(16) = 100 g/molconvert the given 5 g to moles: mol caco₃ = 5 g caco₃ * 1 mol/(100 g) = 0.05 mol caco₃use stoichiometry (ratios) to figure out the moles of hcl needed. ratio wise, there's 2 hcls for every caco₃, so use dimensional analysis mol hcl = mol caco₃ * (2 mol hcl) / (1 mol caco₃) = 0.05 mol hcl * 2 = 0.1 mol hcl we have 0.1 mol hcl and concentration of hcl (0.002 mol/cm³).we need to find volumesince concentration = moles / volume, rearrange to get volume = moles / concentration volume = 0.1 mol / (0.002 mol/cm³) volume = 50 cm³

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A student is examining a bacterium under the microscope. The E. coli bacterial cell has a mass of m...

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