Answer:
b.) 25 ml of aluminum hydroxide
Explanation:
For this question, we have to assume that we have the <u>same concentration</u> for all the solutions, for example, <u>1 M</u>. Additionally, we have to take into account the <u>ionization reaction</u> for each species:
a)
<u>we have to ions</u>
b)
<u>we have fourth ions</u>
c)
<u>we have two ions</u>
d)
<u>we have one ion</u>
If we have the same volume and the same concentration the variables that will help us to answer the question would be the n<u>umber of ions.</u> If we have <u>more ions we will have more particles dissolved</u>. Therefore the answer would be b) (<u>due to the fourth ions</u>).
I hope it helps
At STP, it is at 273 K and 1.00 atm.
Use PV=nRT to and solve for n (number of mol of N2)
1.00 atm • 4.40L / 0.08206 Latm/molK • 273 K = 0.196 mol N2
There are two mol of N2 for every two mol of NH4NO2 (the number of mol of both are equal)
Find the molar mass of NH4NO2 = 64.1 g
Multiply the number of mol by the molar mass
0.196 mol • 64.1 = 12.6 g NH4NO2
1)sodium chloride/common salt
2)sodium hydroxide
3)sodium carbonate/washing soda
4)sodium bi-carbonate/baking soda
5)calcium hypochlorite/bleaching power
6)hemihydrate calcium sulphate/plaster of Paris
7)calcium sulfate
8)copper sulfate
9)bororn trifluoride
10)potassium nitrate
I could only find ten examples
Using ideal gas equation,
PV=nRT
P=199 kpa
V=4.67 L
T=30+273K= 303K
n= number of moles of gas
R=8.321 L·kPa·K-1·mol-1
So using ideal gas equation,
n=PV/RT
=199*4.67/8.321*303\
=0.37 mol