I would like all question to be answer show working read the question then follo
ID: 693862 • Letter: I
Question
I would like all question to be answer show working read the question then follow q1 is balance
these are simple question just read the topic by the number each q have a number
L Balance the chenical muction and descrbe which tpe of cheial mactine R ia 2 The ormula far magnedum nitrite is 3 A solution ot0512 M K0 is Sluted fron 1.12 t. to 2.80 L. What is the final concentrotion of Aballonn has an intul.volumeof24S LWhat wil the volume be ifth.temperature is raised from 24'C o35C 5. What is the volume ocrepied by 25 D gaf argon gas at a gressure of L 0S atm and a temperature of 20 0 C? 6. How imany moles of Pe are needed to react with 0.75 moles of Nale acconding to the folowing reaction? 2 How many liters of a 0.225M Na0fHl solution contain 0525 mol of NaOH?Explanation / Answer
1) This is a double replacement precipitation reaction.
BaCl2(aq) + K2SO4(aq) ------> KCl(aq) + BaSO4(s)
balance reation
BaCl2(aq) + K2SO4(aq) ------> 2KCl(aq) + BaSO4(s)
Net ionic: Ba2+(aq) + SO42-(aq) -----> BaSO4(s)
2) Mg(NO3)2
3)initially concentration of KOH solution = 0.512 M and 1.12 L solution
find moles of KOH = molarity x volume
moles of KOH = 0.512 M x 1.12 L = 0.57344 moles
now final volume of KOH solution become = 2.80 L
concentration KOH = moles / volume = (0.57344 moles / 2.80 L) = 0.2048 M
{ note : check value are same or not b/z in fig value not show clear }
4) in this question use ideal gas equation where n, R and P are constant
(V1 / T1) = (V2/T2)
V1 = 2.45 L , T1 = 24 oC , T2 = 35 oC
V2 = 2.45 L x (35 oC / 24 oC)
V2 = 3.573 L
5) use ideal gas equation to find out volume
PV = nRT { P = presure, V = volume , n = number of moles, R = gas constant (0.082 L atm /K mol) , T = tempreture }
find numbers of moles of argon in 25.0 g
moles = mass / molar mass = ( 25.0 g / 39.948 g/mol) = 0.6258 moles
V = (0.6258 moles x 0.082 L atm / k mol x 293 K ) / 1.05 atm
V = 14.32 L
6) first write balance reaction
Fe + 3NaBr ----> FeBr3 + 3Na
1 mole of Fe react with 3 mole of NaBr produce 1 mole of FeBr3 and 3 moles of Na so,
0.75 moles NaBr need = (0.75 / 3 ) = 0.25 moles of Fe
7) molarity = moles / volume
volume = moles / molarity
Volume = 0.525 moles / 0.225 M = 2.33 L