Since we know that one mole of any gas at STP is equal to 22.4 L we can multiply 135L by the following conversion: 1 mole/22.4L. When you set up the problem it looks like this…: (135L)x 1 mole/22.4L =6.03 moles of oxygen gas The liters cancel out and you are left with moles as your units.
So your answer is then 3.058
Answer:
Passive transport, and Active transport
Explanation:
Do u need an explanation?
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Hope this helped a little!
The number of moles of b2o3 that will be formed is determined as 4 moles.
<h3>
Limiting reagent</h3>
The limiting reagent is the reactant that will be completely used up.
4 b + 3O₂ → 2b₂O₃
from the equation above;
4 b ------------> 2 b₂O₃
2b ------------> b₂O₃
2 : 1
3O₂ -------------> 2b₂O₃
3 : 2
b is the limiting reagent, thus, the amount of b2o3 to be formed is calculated as;
4 b ------------> 2 moles of b2o3
8 moles -------> ?
= (8 x 2)/4
= 4 moles
Thus, the number of moles of b2o3 that will be formed is determined as 4 moles.
Learn more about limiting reactants here: brainly.com/question/14222359
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We can use the ideal gas
equation which is expressed as PV = nRT. At a constant pressure and number of
moles of the gas the ratio T/V is equal to some constant. At another set of
condition of temperature, the constant is still the same. Calculations are as
follows:
T1 / V1 = T2 / V2
V2 = T2 x V1 / T1
V2 = 303.15 x 300 / 333.15
<span>V2 = 272.99 cm³</span>
Answer:
The Prandtl number for this example is 14,553.
Explanation:
The Prandlt number is defined as:

To compute the Prandlt number for this case, is best if we use the same units in every term of the formula.

Now that we have coherent units, we can calculate Pr
