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4 years ago

9

HOW CAN YOU TELL THAT A CHEMICAL HABGE HAS OCCURRED. PLZ HELP

1 answer:

denpristay [2]4 years ago

3 0

That is choice A.

The products are different from the reactants.

This may be easy to see - for example when one of the products is precipitated, like when you add BaCL2 to Na2SO4. The BaSO4 formed is insoluble in water and appears as a white precipitate.

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miss Akunina [59]

Answer:

$\boxed{\text{40 mol Al}}$

Explanation:

Al₂O₃ ⟶ 2Al + 3O₂

n/mol: 20

$\text{Moles of Al} = \text{20 mol Al$_{2}$O$_{3}$}\times \dfrac{\text{2 mol Al}}{\text{1 mol Al$_{2}$O$_{3}$}}= \textbf{40 mol Al}\\\text{You can produce }\boxed{\textbf{40 mol Al}}$

8 0

3 years ago

Explain the way ammonia, salt, and water are alike

WARRIOR [948]

All of them are compounds.

3 0

3 years ago

PLEASE HELP!!! WILL MARK BRAINLIEST!

fredd [130]

The longest chain of carbons will contain the highest number of carbon atoms.

Explanation:

question 3 - the longest chain have 7 carbon atoms

question 4 - the longest chain have 8 carbon atoms

question 1 - the longest chain have 5 carbon atoms

You may find in the attached picture the longest chain colored in blue for each of the compounds.

Learn more about:

longest carbon chain

brainly.com/question/4859077

brainly.com/question/3542033

#learnwithBrainly

5 0

3 years ago

GUYS I NEED YOUR HELP. IF ANYONE IS IN K12 CHEM I WILL PAY YOU. I JUST NEED THESE LABS: CHEMICAL REACTIONS LAB HONORS PROJECT I

Lerok [7]

I’m in k12 but only in 6th..

6 0

3 years ago

A 0.250 g sample of hydrocarbon (containing only carbon and hydrogen) undergoes complete combustion to produce 0.845 g of CO2 an

melomori [17]

Answer:

CH

Explanation:

We have to obtain the mass of carbon and hydrogen in CO2 and H2O respectively.

For carbon in CO2;

0.845 * 12/44 = 0.23 g

For hydrogen in H20;

0.173 * 2/18 = 0.019 g

We convert the masses to moles of carbon and hydrogen

For carbon - 0.23/ 12 = 0.019 moles

For hydrogen - 0.019/1 = 0.019 moles

Dividing by 0.019 moles for carbon and hydrogen we have the emperical formula of the compound as CH

5 0

3 years ago