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3 years ago

What did J.J. Thomson discover about the atom that changed the atomic model previously used?

1 answer:

galina1969 [7]3 years ago

7 0

J.J. Thomson hypothesized and discovered that the atom was not the smallest unit of matter but that instead there were much smaller units. He discovered "sub-atomic particles" which make up atoms. The sub-atomic particle that Thomson discovered was the electron. He discovered this through a process of experiments testing cathode rays.

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Chem help will give brainliest a student makes 255 mL of the solution of hydrochloric acid with a molarity of 0.15 M

SSSSS [86.1K]

The answer is in the photo.

3 0

3 years ago

How many moles of aluminum oxide (Al2O3) can be produced from 12.8 moles of oxygen gas (02)

zhannawk [14.2K]

Answer:

Theoretical Yield

Percent yield

Example stoichiometry problem

How much oxygen can be prepared from 12.25 g KClO3 . (Use molar mass KClO3 = 122.5 g.)

Most stoichiometry problems can be solved using the following steps.

Step 1.

Write and balance the equation for the decomposition of KClO3 with heat (∆). 2KClO3 + ∆ → 2KCl + 3O2

Step 2.

Convert what you have (in this case g KClO3) to moles.

# moles = grams/molar mass = 12.25 g /122.5 = 0.100 mole KClO3.

Step 3.

Using the coefficients in the balanced equation, convert moles of what you have (moles KClO3) to moles of what you want (in this case moles oxygen).

0.100 mol KClO3 x (3 moles O2/2 moles KClO3) = 0.100 x (3/2) = 0.150 mole O2.

Step 4.

Convert moles from step 3 to grams.

moles x molar mass = grams

0.150 mole O2 x (32.0 g O2/mole O2) = 4.80 g O2 produced from 12.25 g KClO3. This is the theoretical yield. If the ACTUAL yield is 4.20 grams, calculate percent yield. Percent yield = (actual yield/theoretical yield) x 100 = (4.20/4.80) x 100 = 87.5% yield

NOTE: In step 1, moles can be obtained other ways; in step 4 moles can be converted to other units.

a. For solutions, M x L = moles (or mL x M = millimoles).

b. For gases, L/22.4 = moles

4 0

3 years ago

A clear liquid in an open container is allowed to evaporate. After three days, a solid is left in the container. Was the clear

Mashutka [201]

The liquid did not chemically bond after 3 days, therefor it is a mixture.

Hope this helps!

7 0

3 years ago

Flourine is more reactive than chlorine

IRISSAK [1]

Justification for your answer

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Chlorine is less reactive than fluorine because the outer electrons in a chlorine atom are further from the nucleus than the outer electrons in a fluorine atom. It is harder for a chlorine atom to gain an electron than it is for a fluorine atom.

There are three things to consider every single time relative reactivity is unknown; atomic radius, shielding, and number of electrons. The reactivity is the halogens ability to gain an electron, so number of electrons already in the atom plays a vital role. Chlorine has more electrons so repels a reacting electron with greater force than fluorine, making it less likely to react.

Fluorine also has fewer electron shells than chlorine, so there are fewer electrons between the positive nucleus and the reacting electron to essentiallly block, or weaken, the electromagnetic attraction. This is shielding. Lastly, fluorine is much smaller molecule than chlorine, and the shorter distance, or radius, between the nucleus and the electron again makes it more likely to attract the electron and react to gain a noble gas configuration.

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7 0

4 years ago

Read 2 more answers

of an unknown protein are dissolved in enough solvent to make 5.00mL of solution. The osmotic pressure of this solution is measu

krok68 [10]

The question is incomplete . The complete question is :

100 mg of an unknown protein are dissolved in enough solvent to make 5.00mL of solution. The osmotic pressure of this solution is measured to be 0.107atm at 25.0°C. Calculate the molar mass of the protein. Round your answer to 3 significant digits.

Answer: The molar mass of the protein is $4.57\times 10^3g/mol$

Explanation:

$\pi =CRT$

$\pi=i\times \frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}\times RT$

where,

$\pi$ = osmotic pressure of the solution = 0.107 atm

i = Van't hoff factor = 1 (for non-electrolytes)

Mass of solute (protein) = 100 mg = 0.1 g (Conversion factor: 1 g = 1000 mg)

Volume of solution = 5.00 mL

R = Gas constant = $0.0821\text{ L.atm }mol^{-1}K^{-1}$

T = temperature of the solution = $25^oC=[273+25]=298K$

Putting values in above equation, we get:

$0.107=1\times \frac{0.1\times 1000}{\text{Molar mass of insulin}\times 5.00}\times 0.0821\text{ Latm }mol^{-1}K^{-1}\times 298K\\\\\text{molar mass of protein}$

$\text{molar mass of protein}=4.57\times 10^3g/mol$

Hence, the molar mass of the protein is $4.57\times 10^3g/mol$

7 0

3 years ago