100 POINTS & BRAINLIEST !!!! PLEASE HELP NEED HYPOTHESIS

Chemistry

1 answer:

creativ13 [48]4 years ago

4 0

Answer:

The second hypothesis

Explanation:

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Many scientists have contributed to the current accepted atomic model. Which of the following would be a reasonable description

xenn [34]

B. The model gradually changed as new scientific data was gathered.

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4 years ago

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The following reaction occurs in a car’s catalytic converter.

blagie [28]

Answer:The correct answer is ;

The oxidation state of nitrogen in NO changes from +2 to 0, and the oxidation state of carbon in CO changes from +2 to +4 as the reaction proceeds.

Explanation:

$2NO(g)+2CO(g)\rightarrow N_2(g)+2CO_2(g)$

In an oxidation recation addition of oxygen atom takes place or loss of electrons takes place.

In an reduction reaction removal of oxygen atom takes place or gain of electrons takes place.

In the given reaction , the nitrogen atom is present in +2 oxidation state in NO molecule and present in 0 oxidation state in $N_2$ molecule. Hence, nitrogen is getting reduced that is reduction reaction. NO is oxidizing agent

In the given reaction , the carbon atom is present in +2 oxidation state in CO molecule and present in +4 oxidation state in $CO_2$ molecule. Hence ,carbon is getting oxidized that is oxidation reaction. CO is a reducing agent.

8 0

3 years ago

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72.0 grams of water how many miles of sodium with react with it?

Flura [38]

Answer:

$\large \boxed{\text{8.00 mol}}$

Explanation:

We will need a balanced chemical equation with masses, moles, and molar masses.

1. Gather all the information in one place:

Mᵣ: 18.02

2Na + H₂O ⟶ 2NaOH + H₂

m/g: 72.0

2. Moles of H₂O

$\text{Moles of H$_{2}$O} = \text{72.0 g H$_{2}$O} \times \dfrac{\text{1 mol H$_{2}$O}}{\text{18.02 g H$_{2}$O}} = \text{3.996 mol H$_{2}$O}$

3. Moles of Na

The molar ratio is 2 mol Na/1 mol H₂O.

$\text{Moles of Na} = \text{3.996 mol H$_{2}$O} \times \dfrac{\text{2 mol Na}}{\text{1 mol H$_{2}$O}} = \textbf{8.00 mol Na}\\\\\text{The water will react with $\large \boxed{\textbf{ 8.00 mol}}$ of Na}$