What example of scientific thinking does Bohr response best indicate

Chemistry

1 answer:

Ugo [173]4 years ago

4 0

Answer:

The Bohr model shows the atom as a central nucleus containing protons and neutrons with the electrons in circular orbitals at specific distances from the nucleus (Figure 1). These orbits form electron shells or energy levels, which are a way of visualizing the number of electrons in the various shells

You might be interested in

A 25.0 ml sample of an unknown monoprotic acid was titrated with 0.12 m naoh. the student added 31.6 ml of naoh and went past th

Inessa05 [86]

The procedure, which can be used to determine more accurately the concentration of the unknown acid is TO BACK-TITRATE WITH ADDITIONAL HYDROCHLORIC ACID TO NEUTRALIZE THE ADDITIONAL SODIUM HYDROXIDE THAT WAS ADDED.
Monoprotic acids are acids that can donate only one proton per each molecule and they have only one equivalence point. Examples of monoprotic acids are HCI, HNO3 and CH3COOH.
The back titration method is typically used when one needs to determine the concentration of an analyte provided there is a known molar concentration of excess reactants.
From the information given in the question above, we are told that excess NaOH was added. To correct this mistake, the right thing to do is to use additional HCl to carry out back titration, taking note of the quantity of acid that will be needed to neutralize the excess NaOH.

7 0

3 years ago

PLZ HELP ASAP!!!

melomori [17]

Sandy soil because the sand will erode faster then the clay will causeing the well to cave in

6 0

3 years ago

When 1.2383 g of an organic iron compound containing Fe, C, H, and O was burned in O2, 2.3162 g of CO2 and 0.66285 g of H2O were

uysha [10]

Answer:

$\boxed{\text{C$_{15}$H$_{21}$FeO$_{6}$}}$

Explanation:

Let's call the unknown compound X.

1. Calculate the mass of each element in 1.23383 g of X.

(a) Mass of C

$\text{Mass of C} = \text{2.3162 g } \text{CO}_{2}\times \dfrac{\text{12.01 g C}}{\text{44.01 g }\text{CO}_{2}}= \text{0.632 07 g C}$

(b) Mass of H

$\text{Mass of H} = \text{0.66285 g }\text{H$_{2}$O}\times \dfrac{\text{2.016 g H}}{\text{18.02 g } \text{{H$_{2}$O}}} = \text{0.074 157 g H}$

(c)Mass of Fe

(i)In 0.4131g of X

$\text{Mass of Fe} = \text{0.093 33 g Fe$_{2}$O$_{3}$}\times \dfrac{\text{111.69 g Fe}}{\text{159.69 g g Fe$_{2}$O$_{3}$}} = \text{0.065 277 g Fe}$

(ii) In 1.2383 g of X

$\text{Mass of Fe} = \text{0.065277 g Fe}\times \dfrac{1.2383}{0.4131} = \text{0.195 67 g Fe}$

(d)Mass of O

Mass of O = 1.2383 - 0.632 07 - 0.074 157 - 0.195 67 = 0.336 40 g

2. Calculate the moles of each element

$\text{Moles of C = 0.63207 g C}\times\dfrac{\text{1 mol C}}{\text{12.01 g C }} = \text{0.052 629 mol C}\\\\\text{Moles of H = 0.074157 g H} \times \dfrac{\text{1 mol H}}{\text{1.008 g H}} = \text{0.073 568 mol H}\\\\\text{Moles of Fe = 0.195 67 g Fe} \times \dfrac{\text{1 mol Fe}}{\text{55.845 g Fe}} = \text{0.003 5038 mol Fe}\\\\\text{Moles of O = 0.336 40} \times \dfrac{\text{1 mol O}}{\text{16.00 g O}} = \text{0.021 025 mol O}$

3. Calculate the molar ratios

Divide all moles by the smallest number of moles.

$\text{C: } \dfrac{0.052629}{0.003 5038}= 15.021\\\\\text{H: } \dfrac{0.073568}{0.003 5038} = 20.997\\\\\text{Fe: } \dfrac{0.003 5038}{0.003 5038} = 1\\\\\text{O: } \dfrac{0.021025}{0.003 5038} = 6.0006$