Answer:
answer........of the question
Answer:
24.2 mL.
Explanation:
<em>Assuming constant temperature</em>, we can solve this problem using <em>Boyle's law</em>, which states:
Where:
We <u>input the data</u>:
- 0.98 bar * 25 mL = 1.013 bar * V₂
And <u>solve for V₂</u>:
The closest option is the second one: 24.2 mL.
The moles of butane gas and oxygen gas reacted if 2.50 moles of H2O is produced is calculated as below
the equation for reaction
2C4H10 +13 O2 = 8CO2 +10 H2O
the moles of butane (C4H10) reacted calculation
by use of mole ratio between C4H10: H2O which is 2 : 10 the moles of C4H10= 2.50 x2/10 = 0.5 moles of C4H10 reacted
The moles of O2 reacted calculation
by use of mole ratio between O2 : H2O which is 13:10 the moles of O2
= 2.50 x 13/10= 3.25 moles of O2 reacted
<u>The answer is </u><u>9.94 ml.</u>
<h3>
What is density?</h3>
- Density is a word we use to describe how much space an object or substance takes up (its volume) in relation to the amount of matter in that object or substance (its mass).
- Another way to put it is that density is the amount of mass per unit of volume. If an object is heavy and compact, it has a high density.
Given,
The density of ethanol, C2H5OH = 0.789 g/mL


= 7.843 g

Therefore, the answer is 9.94 ml
Learn more about density of ethanol,
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<u>The complete question is -</u>
If the density of ethanol, C2H5OH, is 0.789 g/mL. How many milliliters of ethanol are needed to produce 15.0 g of CO2 according to the following chemical equation?
C2H5OH(l) + 3 O2(g) → 2 CO2(g) + 3 H2O(l)