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castortr0y [4]
2 years ago
11

In an experiment, the target bacteria can only live in the solution with 0.7-0.9ml/L NaCl. Your solution has mean 0.8ml/L but st

andard deviation 0.05ml/L NaCl. What is a lower bound of the probability that the bacteria will survive?
Chemistry
1 answer:
Ostrovityanka [42]2 years ago
3 0

Answer:

The probability of survival is at least 75%

Explanation:

Since we do not know the probability distribution of the concentration of the solution, we can not determine the exact probability of survival. Nevertheless, we can use the Chebyshev inequality, that applies to any probability distribution, to calculate the lower bound.

It states that

P( | X - µ | ≤ kσ ) ≥ 1- 1/k²

this means that the probability that X is not further than k standard deviations from the mean (µ) is at least 1- 1/k²  , where k >1

in our case,

k1= (X1 - µ) / σ = (0,9 - 0,8 ) / 0,05 = 2

k2=(X2 - µ) / σ = (0,7- 0,8 ) / 0,05 = -2

therefore k=k1=k2=2 and

P( | X - µ | ≤ 2σ ) ≥ 1- 1/2²= 3/4=75%

consequently, the lower bound of the probability of survival is 75%

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1. For which of these elements would the first ionization energy of the atom be higher than that of the diatomic molecule?
Lyrx [107]

Answer: Option (b) is the correct answer.

Explanation:

The energy necessary to remove an electron from a gaseous atom or ion is known as ionization energy.

This means that smaller is the size of an atom more amount of energy has to be supplied to it in order to remove the valence electron. This is because in small atom or element there will be strong force of attraction between the nucleus and electrons.

So, high amount of energy has to be supplied to remove the valence electrons.

As electronic configuration of helium is 1s^{2}. So, due to completely filled valence shell it is more stable in nature.

As a result, we need to provide very high amount of energy to remove an electron from a helium atom.

Thus, we can conclude that out of the given options helium element would the first ionization energy of the atom be higher than that of the diatomic molecule.

7 0
2 years ago
Sodium thiosulfate (Na2S2O3), photographer’s
vova2212 [387]

3Na2S2O3 + AgBr ------>Na5[Ag(S2O3) 3] +NaBr

from equation 3 mol 1 mol

given x mol 0.10 mol


x= (3*0.10)/1=0.30 mol Na2S2O3


Answer: 0.30 mol Na2S2O3

3 0
2 years ago
Read 2 more answers
You wish to extract an organic compound from an aqueous phase into an organic layer (three to six extractions on a marco scale).
AVprozaik [17]

Answer:

Explanation:

It will be better to use solvents that are lighter than water, because their density has an influence on the miscibility . This will give you a better separation during extraction.

7 0
2 years ago
A gold wire has a diameter of 1.00 mm. What length of this wire contains exactly 1.00 mol of gold? (density of Au = 17.0 g/cm3)
allochka39001 [22]

Answer:

The answer to your question is 7160 cm

Explanation:

Data

diameter = 1 mm

length = ?

amount of gold = 1 mol

density = 17 g/cm³

Process

1.- Get the atomic mass of gold

Atomic mass = 197 g

then, 197g ------------ 1 mol

2.- Calculate the volume of this wire

density = mass/volume

volume = mass/density

volume = 197/17

volume = 5.7 cm³

3.- Calculate the length of the wire

Volume = πr²h

solve for h

h = volume /πr²

radius = 0.05 cm

substitution

h = 5.7/(3.14 x 0.05²)

h = 5.7 / 0.0025

h = 7159.2 cm ≈ 7160 cm

8 0
2 years ago
Read 2 more answers
Isopentyl acetate (C7H14O2), the compound responsible for the scent of bananas, can be produced commercially. Interestingly, bee
lora16 [44]

Answer:

The answer to your question is:

a) 4.64 x 10 ¹⁵ molecules

b) 9.28 x 10 ¹⁵ atoms of O2

Explanation:

MW C7H14O2 = 84 + 14 + 32 = 130 g

a)        130 g of C7H14O2 ---------------- 1 mol of C7H14O2

           1 x 10 ⁻⁶ g              ---------------      x

           x = 7.7 x 10 ⁻⁹ mol

          1 mol of C7H14O2   --------------   6 .023 x 10 ²³ molecules

          7.7 x 10⁻⁹ mol          --------------    x

          x = 4.64 x 10¹⁵ molecules

b)      130 g of C7H14O2   ----------------   1 mol of C7H14O2

         1 x 10⁻⁶  C7H14O2   -----------------     x

         x = 7.7 x 10 ⁻⁹ mol of C7H14O2

        1 mol of C7H14O2    ---------------   2 mol of O2

        7.7 x 10 ⁻⁹                 ----------------   x

         x = 1.54 x 10⁻⁸ mol of O2

       1 mol of O2 -----------------  6.023 x 10 ²³ atoms

       1.54 x 10 ⁻⁸  ----------------   x

        x = 9.28 x 10 ¹⁵ atoms of O2

8 0
2 years ago
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