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IgorLugansk [536]
3 years ago
9

please help asap will give brainiest What is the independent variable in an experiment? A) The variable that remains the same th

roughout the experiment. B) The variable being tested in the experiment. C) The variable being measured in an experiment.
Chemistry
1 answer:
iogann1982 [59]3 years ago
5 0
A because if it remains through the whole expídeme y the same the it doesn’t depend on anything else to change it and I got it correct
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What is the mass in grams of one nitrogen molecule?
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The mass of one Nitrogen molecule is option C

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Synthesize Information In a multistep process, cells can combine the reactants glucose (C6H12O6) and oxygen (O2) to form the pro
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Explanation:

In one of the process, energy is built up from scratch, in the other one, energy is liberated for use by an organism or body.

The first process deals with a metabolic reaction in which energy is liberated:

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In the above process, energy is liberated when glucose combines with oxygen. The waste products are carbon dioxide and water. This process liberates heat energy which can be used to do work.

   In the reverse process:

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Which of the following bonds is most popular?<br> C-C,C-N,C-O,C-H
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Aluminum reacts with oxygen to produce aluminum oxide which can be used as an adsorbent, desiccant or catalyst for organic react
pishuonlain [190]
The balance chemical equation is as follow,

                                        2 Al  +  3 O₂   →   Al₂O₃

Aluminium is the Limiting Reagent,
As,
                                107.92 g Al required  =  96 g of O₂
Then,
                          82.49 g of Al will require  =  X g of O₂
Solving for X,
                                    X  =  (82.49 g × 96 g) ÷ 107.92 g

                                    X  =  73.37 g of O₂
But,
     We are provided with 117.65 g of O₂, So, it is provided in excess and 44.28 g of it will remain unreacted.

Solving for Amount of Al₂O₃ formed,
As,
                        107.92 g of Al produced  =  203.92 g of Al₂O₃
Then,
                      82.49 g of Al will produce  =  X g of Al₂O₃
SOlving for X,
                                          X  =  (82.49 g ×  203.92 g) ÷ 107.92

                                          X  =  155.86 g of Al₂O₃
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