NaOH will dissociate as Na+ and OH- in the solution.
Answer:
0.6257 M is the molarity of solution that is 5.50 percentage by mass oxalic acid.
Explanation:
Mass percentage of oxalic acid = 5.50%
This means that in 100 grams of solution there are 5.50 grams of oxalic acid.
Mass of solution , m = 100
Volume of the solution = V
Density of the solution = d = 1.024 g/mL

V = 97.66 mL = 0.09766 L
(1 mL = 0.001 L)
Moles of oxalic acid = 

The molarity of the solution :

0.6257 M is the molarity of solution that is 5.50 percentage by mass oxalic acid.
Answer:
The choice of the answer is fourth option that is -61 degrees.
Therefore the temperature drop is -61°Centigrade.
Explanation:
Given:
The temperature in a town started out at 55 degrees
Start temperature = 55°Centigrade. (Initial temperature)
End of the Day = -6°Centigrade. (Final temperature)
To Find:
How far did the temperature drop?
Solution:
We will have,

Substituting the above values in it we get

Therefore the temperature drop is -61°Centigrade.
Answer:
dont know how to remove the answer read it wrong
128 ml is the voume of the balloon if the temperature of the gas increases to 320.0k.
Explanation:
given that:
T1 (initial temperature) = 300K
V1 ( initial volume) = 120ml
T2 (final temperature) = 320 K
V2 (final volume) = ?
Pressure remained constant throughout the process.
From the equation
= 
Since pressure is constant the equation will be:
= 
V2 = 
Putting the values in the above formula:
V2 = 
= 128 ml
128 ml is the volume of the gas if temperature increases from 3OO K to 320k