The answer I'm pretty sure is a because it can't be d because ionic binding the electrons are affected and it can't be c because that's covalent bonding and it can't be b because they don't swap electrons.
H2SO4 + 2KOH ---> K2SO4 + 2H2O
:)
Answer:
0 M is the silver ion concentration in a solution prepared mixing both the solutions.
Explanation:
![molarity=\frac{\text{Moles of solute}}{\text{Volume of solution (L)}}](https://tex.z-dn.net/?f=molarity%3D%5Cfrac%7B%5Ctext%7BMoles%20of%20solute%7D%7D%7B%5Ctext%7BVolume%20of%20solution%20%28L%29%7D%7D)
Moles of silver nitrate = n
Volume of the solution = 425 mL = 0.425 L (1 mL = 0.001 L)
Molarity of the silver nitrate solution = 0.397 M
![n=0.397 M\times 0.425 L=0.1687 mol](https://tex.z-dn.net/?f=n%3D0.397%20M%5Ctimes%200.425%20L%3D0.1687%20mol)
Moles of sodium phosphate = n'
Volume of the sodium phosphate solution = 427 mL = 0.427 L (1 mL = 0.001 L)
Molarity of the sodium phosphate solution = 0.459 M
![n'=0.459 M\times 0.427 L=0.1960 mol](https://tex.z-dn.net/?f=n%27%3D0.459%20M%5Ctimes%200.427%20L%3D0.1960%20mol)
![3AgNO_3+Na_3PO_4\rightarrow Ag_3PO_4+3NaNO_3](https://tex.z-dn.net/?f=3AgNO_3%2BNa_3PO_4%5Crightarrow%20Ag_3PO_4%2B3NaNO_3)
According to reaction, 3 moles of silver nitrate reacts with 1 mole of sodium phosphate, then 0.1687 moles of silver nitrate will recat with :
of sodium phosphate
This means that only 0.05623 moles of sodium phosphate will react with all the 0.1687 moles of silver nitrate , making silver nitrate limiting reagent and sodium phosphate as an excessive reagent.
So, zero moles of silver nitrate will be left in the solution after mixing of the both solutions and hence zero moles of silver ions will left in the resulting solution.
0 M is the silver ion concentration in a solution prepared mixing both the solutions.
The answer is (D) microscopic. You can remember this, because the name is very close to "microscope," an instrument used to greatly magnify and observe tiny organisms and objects.
Answer:
234.35 °C
Explanation:
Given data:
Volume of balloon = 125000 mL
Moles of oxygen = 3 mol
Pressure = 1 atm
Temperature = ?
Solution:
Formula:
PV = nRT
P = Pressure
V = volume
n = number of moles
R = ideal gas constant
T = temperature
Volume of balloon = 125000 mL × 1 L /1000 mL
Volume of balloon = 125 L
Now we will put the values:
Ideal gas constant = R = 0.0821 atm.L/mol.K
PV = nRT
T = PV/nR
T = 1 atm × 125 L/ 0.0821 atm.L/mol.K × 3 mol
T= 125 /0.2463 /K
T = 507.5 K
K to °C
507.5 K - 273.15 = 234.35 °C