If we know what we are reading, reaction time to the simple reaction task in this project will be less than reaction time to the voluntary action of your quadriceps, however, if we are reading for the first time, reaction time to the simple reaction task in this project will be more than reaction time to the voluntary action of your quadriceps.
<h3><u>
Explanation:</u></h3>
First, we need to know what is the reaction time. Reaction time is the amount of time it takes to respond to a stimulus. Now we will talk about voluntary and involuntary actions.
Voluntary action is an action that is thought of and consciously performed by the person. So we can say the person has full control over the situation. An involuntary action is one which occurs without the conscious choice of an organism. If the involuntary action occurred as a response to some stimulus it will be known as a reflex. If we compare the reaction time of voluntary action of our quadriceps and the simple reaction to the task given in the project, the reaction time could be less only if we have studied about the given question in the given task.
So the lines what we are reading will come to our mind involuntarily, but if we are talking about typing the answer the reaction time would be more as first, we have to form a structured sentence and then type it, so it will take more time.
When 440.23 grams of iron(III) oxide are reacted with hydrogen gas, the amount of iron produced will be 307.66 grams
<h3>Stoichiometric calculation</h3>
From the equation of the reaction:
The mole ratio of iron(III) oxide to produced iron is 1:2.
Mole of 440.23 iron(III) oxide = 440.23/159.69 = 2.76 moles
Equivalent mole of produced iron = 2.76 x 2 = 5.52 moles
Mass of 5.52 moles of iron = 5.52 x 55.8 = 307.66 grams
More on stoichiometric calculations can be found here; brainly.com/question/27287858
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The
reaction is<span>
C</span>₂H₄ +
O₂ → CO₂ + H₂O<span>
To balance the equation, both side have same
number of elements. Here,</span>
In left hand
side has
in right hand side has
4 H atoms
2 H
atoms
2 C atoms 1 C
atom
<span>
2 O atoms 3 O
atoms
First, we have to balance number of C atoms and number of H atoms in both side.
To balance C atoms, '2' should be added before CO</span>₂ and to balance H atoms, '2' should be added
before H₂<span>O.
Then number of oxygen atoms is </span>2 x 2 + 2 = 6 in right hand side. So, 3 should be
added before O₂<span> in left hand side.
After balancing the equation should be,</span>
C₂H₄<span> + 3O</span>₂<span> → 2CO</span>₂<span> + 2H</span>₂O
First, find moles of oxygen gas: (3.01 x10^23 molec.)/(6.02 x10^23) =0.5mol O2
Second, multiply moles by the standard molar volume of a gas at STP:(0.5mol)(22.4L) = 11.2L O2
