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makvit [3.9K]
3 years ago
14

In the mid-1900s, most scientists used the Bohr atomic model to study atoms. In the Bohr model, electrons move in set paths arou

nd the atom's nucleus.
In current atomic models, electron clouds are shown around the atom's nucleus. Electron clouds are regions where electrons are likely to be found.

What is a limitation of current atomic models as compared to the Bohr atomic model?

A. Current atomic models do not indicate the position of the nucleus within the atom's structure.

B. Current atomic models do not show how many electrons an atom has.

C. Current atomic models do not show where the electron is but only where the electron is likely to be.
Chemistry
2 answers:
patriot [66]3 years ago
7 0

Answer:

The answer is C.

Explanation:

Novosadov [1.4K]3 years ago
5 0
The answer is c beca it true
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Answer:

We need 13.06 grams of silver nitrate to precipitate out silver bromide in the final solution

Explanation:

<u>Step 1:</u> Data given

Sample 1: The 1.15 M sample  has a volume of 33.O mL

Sample 2: The 0.660 M sample has a volume of 59.0 mL

Molar mass of KBr = 119 g/mol

Molar mass of AgNO3 = 169.87 g/mol

<u>Step 2:</u> Calculate number of moles for both samples

Number of moles = Molarity * Volume

Sample 1:  1.15 M * 33 *10^-3 L = 0.03795 moles

Sample 2: 0.660 M *59*10^-3 L = 0.03894 moles

Total mol KBr = 0.03795 + 0.03894 = 0.07689 moles

<u>Step 3:</u> Calculate total mass

mass = Number of moles * Molar mass

mass = 0.07689 moles * 119 g/moles = 9.15 grams  ( in 55mL)

<u>Step 4</u>: Calculate moles of AgBr

AgNO3 reacts with KBr  

KBr(aq) + AgNO3(aq) → AgBr(s) + KNO3(aq)

1 mole of KBr consumed, needs 1 mole of AgNO3 to produce 1 mole of AgBr and 1 mole of KNO3

So 0.07689 moles of KBr wll need 0.07689 moles of AgNO3

<u>Step 5:</u> Calculate mass of silver nitrate

mass of AgNO3 = Moles of AgNO3 * Molar mass of AgNO3

mass of AgNO3 = 0.07689 moles * 169.87 g/mol = 13.06 grams

We need 13.06 grams of silver nitrate to precipitate out silver bromide in the final solution

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