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77julia77 [94]
3 years ago
14

What is the balanced chemical equation for silicon + oxygen?

Chemistry
1 answer:
ZanzabumX [31]3 years ago
6 0

Answer:

SiO2

Explanation:

Si + O2 → SiO2

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How many atoms are in water is it 24 or 2.<br> Explain
Artemon [7]
There are actually three: One oxygen atom and two hydrogen atoms.
8 0
3 years ago
When optically active (R)-2-methylcyclohexanone is treated with either aqueous base or acid, racemization occurs. Explain.
IrinaK [193]

Answer:

See explanation

Explanation:

Racemization is said to occur when a 1:1 ratio of (+) and (-) enantiomers of a compound are produced in a reaction.

The reaction of optically active (R)-2-methylcyclohexanone with either aqueous base or acid leads to the formation of a planar enol species for reaction with acid and a planar enolate species for reaction with base.

Both reactions involves the formation of achiral species which reverts back to the chiral product with equal chances of the formation of both enantiomers of the product during the process. This leads to racemization of the product in both cases.

3 0
3 years ago
At constant pressure and 25^0C , a sample of gas occupies 4.5lit. at what temperature will the gas occupy 9.0lit.​
olga55 [171]

Answer:

Final temperature = 149 K

Explanation:

Given data:

Initial temperature = 25°C (25+273 = 298 K)

Initial volume = 4.5 L

Final temperature = ?

Final volume = 9.0 L

Solution:

The given problem will be solve through the Charles Law.

According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.

Mathematical expression:

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume  

T₂ = Final temperature

Now we will put the values in formula.

V₁/T₁ = V₂/T₂

T₂ = T₁V₂/V₁  

T₂ = 4.5 L × 298 K / 9.0 L

T₂ = 1341 L K / 9.0L

T₂ = 149 K

4 0
3 years ago
What do you think happens to the matter in a food sample as it is burned?
Alika [10]

Burning and other changes in matter do not destroy matter. The mass of matter is always the same before and after the changes occur. The law of conservation of mass states that matter cannot be created or destroyed

6 0
3 years ago
Consider the following generic chemical equation.
makvit [3.9K]

Answer:

\fbox{ A) A  \:  is  \:  limiting \: reactant }

\fbox{ B) B  \:  is  \:  limiting \: reactant }

\fbox{ C) A  \:  is  \:  limiting \: reactant }

\fbox{ D) A  \:  is  \:  limiting \: reactant }

Explanation:

<em>Given equation:</em>

<em>A+3B \rightarrow \: C</em>

<em>To </em><em>find:</em>

Limiting reactant for corresponding number of moles=?

<em>Solution:</em>

We know that the liming reactant is any atom, ion or molecule which is completely consumed during a reaction and other reactant is still left in reactant vessel.

For the given reaction A+3B→C

for every one mole of A three moles of B are required.

﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌

in A) there is one mole of A and 4 mole of B,

if 1 mole of A will react with 3 moles of B, 1 mole of B will be still there in reaction, the reactant was completely consumed is A which is limiting reactant.

﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌

in B) there is 2 mole of A and 3 mole of B,

if 1 mole of A will react with 3 moles of B, 1 mole of A will be still there in reaction, the reactant was completely consumed is B which is limiting reactant.

﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌

in C) there is 0.5 mole of A and 1.6 mole of B

if one mole of A requires three moles of B to complete the reaction then,

0.5 moles of A will require 1.5 moles of B

if 0.5 mole of A will react with 1.6 moles of B, 0.1 mole of B will be still there in reaction, the reactant was completely consumed is A which is limiting reactant.

﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌

in D) there is 24 mole of A and 72 mole of B

if one mole of A requires three moles of B to complete the reaction then,

24 moles of A will require 72 moles of B.

if 24 mole of A will react with 75 moles of B, 3 mole of B will be left over in reaction, the reactant was completely consumed is A which is limiting reactant.

﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌

\fbox{ A) A  \:  is  \:  limiting \: reactant }

\fbox{ B) B \:  is \:  limiting \:  reactant }

\fbox{ C) A  \:  is  \:  limiting \: reactant }

\fbox{ D) A  \:  is  \:  limiting \: reactant }

<em><u>Thanks for joining brainly community.</u></em>

5 0
2 years ago
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