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aliya0001 [1]
3 years ago
5

How many moles are found in a 350.mL solution of 13.5M HNO3

Chemistry
1 answer:
tiny-mole [99]3 years ago
5 0

Answer: Moles are found in a 350.mL solution of 13.5M HNO_{3} are 4.725 mol.

Explanation:

Given: Volume = 350 mL (1 mL = 0.001 L) = 0.350 L

Molarity = 13.5 M

Molarity is the number of moles of a substance present in a liter of solution.

Therefore, moles of HNO_{3} present in the given solution are calculated as follows.

Molarity = \frac{moles}{Volume (in L)}\\13.5 M = \frac{moles}{0.350 L}\\moles = 4.725 mol

Thus, we can conclude that moles found in a 350.mL solution of 13.5M HNO_{3} are 4.725 mol.

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3 years ago
Which of the following equations is correct for coffee-cup calorimeter?
Bogdan [553]

The equation that is correct for coffee-cup calorimeter is q reaction = -q calorimeter. Details about coffee-cup calorimeter.

<h3>What is a calorimeter?</h3>

A calorimeter is an apparatus for measuring the heat generated or absorbed by either a chemical reaction, change of phase or some other physical change.

A coffee-cup calorimeter is a specific type of calorimeter that involves the absorption of heat of a reaction by water when a reaction occurs.

The enthalpy change of the reaction is equal in magnitude but opposite in sign to the heat flow for the water:

qreaction = -(qwater)

Therefore, the equation that is correct for coffee-cup calorimeter is q reaction = -q calorimeter.

Learn more about coffee-cup calorimeter at: brainly.com/question/27828855

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3 0
2 years ago
What volume (in L) of oxygen will be required to produce 77.4 L of water vapor in the reaction below?
Inga [223]

Answer:

90.3 L

Explanation:

Given data:

Volume of water produced = 77.4 L

Volume of oxygen required = ?

Solution:

Chemical equation:

2C₂H₆ + 7O₂  →  4CO₂ + 6H₂O

It is known that,

1 mole = 22.414 L

There are 7 moles of oxygen = 7×22.414 = 156.9 L

There are 6 moles of water = 6×22.414 = 134.5 L

Now we will compare:

                               H₂O           :              O₂    

                               134.5         :              156.9

                                 77.4         :             156.9/134.5×77.4 =90.3 L

So for the production of 77.4 L water 90.3 L oxygen is required.

8 0
3 years ago
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