Answer:
physical
Explanation:
no chemical reaction is happening
Answer:
According to Avogadro's law, the volume of one mole of any gas at Standard Temperature and Pressure (STP = 273 K and 1 atm) is 22.4 L. Two important Gas Laws are required in order to convert the experimentally determined volume of hydrogen gas to that at STP. 1.
Explanation:
internet keep searching
Answer: 2.4 ml
Solution :
Molar mass of
= 17 g/mole
Given,: 28% w/w of
solution means 28 g of ammonia in 100 g of solution.
Mass of solution = 100 g
Now we have to calculate the volume of solution.
Molarity : It is defined as the number of moles of solute present in one liter of solution.

where,
n = moles of solute 
= volume of solution in liter = 0.11 L
Now put all the given values in the formula of molarity, we get

Using molarity equation:



C
0.70
I hope this is help, I’m so so sorry if I’m incorrect
Answer: 35.4 grams
Explanation:
Molarity of a solution is defined as the number of moles of solute dissolved per Liter of the solution.

where,
Molality = 2.65
n= moles of solute =?
= volume of solution in ml = 445 ml
Putting in the values we get:


Mass of solute in g=
Thus 35.4 grams of
is needed to prepare 445 ml of a 2.65 m solution of
.