Answer:
HClO 7.54
Explanation:
Hypochlorous acid (HClO) is a weakest acid because the pKa value of Hypochlorous acid is very high among the options given in the activity. pKa is a method which is used in order to identify the strength of an acid. The higher the value of pKa of a liquid, lower the strength of an acid while lower the value of pKa of chemical, higher the strength of an acid. In the options, HClO2 is a strong acid due to high lower pKa value.
We can calculate the amount of Oxygen in this volume using the Ideal Gas Equation, PV = nRT, where "P" is the pressure, "V" is the volume, "n" is the number of moles of material, "R" is the gas constant, and "T" is the temperature in Kelvin. To properly answer this problem, all of the information needs to be converted into the proper units. Fortunately, everything except the volume and temperature is in the correct units for the Ideal Gas Equation; the volume can be adjusted by converting mL to L (x1000) and the temperature can be adjusted by adding 273.15 to the current temperature (conversion from Celsius to Kelvin).
Plugging in all the values, we find that:
PV = nRT
(2.7 atm)(0.3 L) = n(0.0821*)(313.15 K)
n = 0.0315 mol Oxygen
Under these conditions, 0.0315 moles of oxygen can be placed in this volume.
Hope this helps!
* - the units are liters times atmospheres divided by moles times Kelvin.
Answer:
physical change. 95%Sure 5% doubtful.
1.03!! Very easy you welcome:)
Answer:
The correct answer is 378 g of water
Explanation:
A concentration of 5.50% by mass means that there are 5.50 grams of solute in 100 grams of solution:
5.50 %m/m = 5.50 g solute/100 g solution
If we have 22.0 g of solute (KCl), we found the total mass of solution as follows:
5.50 g KCl ----- 100 g solution
22.0 g KCl ------ X = 22.0 g KCl x 100 g solution/ 5.50 g KCl = 400 g solution
22.0 g KCl/400 g solution = 5.5 g KCl/100 g solution = 5.50 % m/m
This is the total mass of solution (400 g). The mass of water we have to add is the following:
Mass water = 400.0 g solution - 22.0 g KCl = 378.0 g
So, in order to prepare a 5.50% m/m of solution, we have to weight 22.0 g of KCl and to add 378.0 g of water until reaching 400.0 g of total mass solution.