Moles of solute for both a and b are the same = 1 mol
<h3>Further explanation</h3>
Given
a 500 cm³ of solution, of concentration 2 mol/dm³
b 2 litres of solution, of concentration 0.5 mol/dm³
Required
moles of solute
Solution
Molarity shows the number of moles of solute in every 1 liter of solution or mmol in each ml of solution
Can be formulated :

a.
V = 500 cm³ = 0.5 L
M = 2 mol/L
n=moles = M x V
n = 2 mol/L x 0.5 L
n = 1 mol
b.
V = 2 L
M = 0.5 mol/L
n=moles = M x V
n = 0.5 mol/L x 2 L
n = 1 mol
Answer:
A HBr
this is hydrogen acid (HBr) but KNO3 is a crystalline salt,MgCl2 and H2O are neutral
Answer:
0.06 kg
Explanation:
Given data:
Mass of frisbee = ?
Velocity of frisbee = 11 m/s
Energy of frisbee = 3.6 j
Solution:
K.E = 1/2 mv²
Now we will put the values in this formula.
3.6 J = 1/2 m(11m/s)²
3.6 j = 0.5 ×121m²/s² × m
3.6 j = 60.5 m²/s² × m
m = 3.6 j/60.5 m²/s²
m= 0.06 j.m⁻².s²
j = Kg.m².s⁻²
m= 0.06 Kg.m².s⁻² .m⁻².s²
m = 0.06 kg
balanced equation =
3Cu(OH)2 + 2H3PO4 → Cu3(PO4)2 + 6H2O