Answer:
0.2 M is the concentration of the resulting solution
Explanation:
M is a unit for concentration which means mole of solute in 1L of solution.
According to the received information, we have 4 g in 500 mL of solution.
In 0.5L we have 4 g of solute
In 1L we have (1 .4)/0.5 = 8 g of solute
Let's convert the mass in mole
Mass / Molar mass = Mole
Molar mass NaOH = 40 g/m
8g / 40 g/m = 0.2M
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Answer:
Explanation:The buffer is a mixture of ammonia (NH3) and ammonium (NH4+). ... Since ammonia (NH3) is a weak base, it will have a pH above 7 and since ammonium (NH4+) is a weak acid, it will have a pH below 7. Ammonia (NH3) is a weak base and ammonium (NH4+) is a weak acid. Ammonia (NH3) is the conjugate base of ammonium (NH4+).
Answer: 2moles
Explanation:
The best way to solve this problem is to write a balance equation for the reaction.
Which is: Fe2 + 2i2______> 2Fei2
With this you can see that iron ii reacting with excess iodine produces two (2) moles of iron ii iodine.
Alternatively.
Using the formular
Moles = reacting mass/molar mass
Reacting mass = 112g
Molar mass of iron ii = 55g/mol
Hence,
Mole = 112g divides 55g/mol
= 2.0mol//