The vital capacity will be 4600ml.
<h3>What is
vital capacity?</h3>
The highest amount of air a person can inhale following their maximal exhalation is known as their vital capacity. It is equivalent to the total of the inspiratory, tidal, and expiratory reserve volumes. It roughly corresponds to Forced Vital Capacity. A wet or conventional spirometer can assess a person's vital capacity.
Normal people have a 3 to 5-liter vital capacity.
It enables simultaneous inhalation of the greatest possible volume of clean air and exhalation of the greatest possible volume of stale air. So, by increasing gaseous exchange between the body's various tissues, it improves the amount of energy available for bodily function.
VC = TV₊IRV₊ERV
where,
VC = Vital capacity
TV = Tidal volume
IRV = inspiratory reserve volume
ERV = expiratory reserve volume
VC = 500 ₊ 3000 ₊ 1100
VC = 4600ml
Therefore, the vital capacity will be 4600ml.
To know more about vital capacity refer to: brainly.com/question/14877276
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Answer:
K remains the same;
Q < K;
The reaction must run in the forward direction to reestablish the equilibrium;
The concentration of 
will decrease.
Explanation:
In this problem, we're adding an excess of a reactant, chlorine gas, to a system that is already at equilibrium. According to the principle of Le Chatelier, when a system at equilibrium is disturbed, the equilibrium shifts toward the side of the equilibrium that minimizes the disturbance.
Since we'll have an excess of chlorine, the system will try to reduce that excess by shifting the equilibrium to the right. Therefore, the reaction must run in the forward direction to reestablish the equilibrium.
The value of K remains the same, as it's only temperature-dependent, while the value of Q will be lower than K, that is, Q < K, as Q < K is the case when reaction proceeds to the right.
As a result, since is also a reactant, its concentration will decrease.
Answer:
<h2><u>Reason:</u></h2>
Catalyst is used as a very fine powder and larger pieces of iron are not used. This is because the surface area of catalyst needs to be large so that more of the surface is exposed to the substrate and more of the substrate is catalyzed.
<h2><u>Important Info:</u></h2>
=> Larger Pieces of Iron has a smaller surface area than the fine particles.
=> Larger the surface area of catalysts/enzymes , more will be the reaction rate and vice versa.
Hope this helped!
<h2>~AnonymousHelper1807</h2>
<span>54.8 g of MgI2 can be produced.
To solve this, you need to determine the molar mass of each reactant and the product. First, look up the atomic weights of iodine and magnesium
Atomic weight of Iodine = 126.90447
Atomic weight of Magnesium = 24.305
Molar mass of MgI2 = 24.305 + 2 * 126.90447 = 278.11394
Now determine how many moles of Iodine and Magnesium you have
moles of Iodine = 50.0 g / 126.90447 g/mol = 0.393997154 mole
moles of Magnesium = 5.15 / 24.305 g/mol = 0.211890557 mole
Since for every magnesium atom, you need 2 iodine atoms and since the number of moles of available iodine isn't at least 2 times the available moles of magnesium, iodine is the limiting reagent.
So figure out how many moles of magnesium will be consumed by the iodine
0.393997154 mole / 2 = 0.196998577 mole.
This means that you can make 0.196998577 moles of MgI2. Now simply multiply by the previously calculated molar mass of MgI2
0.196998577 mole * 278.11394 g/mole = 54.78805 g
Round the result to the correct number of significant figures.
54.78805 g = 54.8 g</span>
