V₁=4.8 l
c₁=2.5 moles/l
c₀=6.0 moles/l
v₀-?
v(H₂O)-?
n(H₂SO₄)=v₁c₁=v₀c₀
v₀=v₁c₁/c₀
v₀=4.8·2.5/6=2.0 l
v(H₂O)=v₁-v₀
v(H₂O)=4,8-2.0=2.8 l
you need to take 2.0 liters of solution (6.0 M H₂SO₄) and 2.8 liters of water
Distillation; either simple or fractional macroscale depending on how far apart the boiling points are.
A.
M = 0.125 M
V = 3.0 L
M=n/V(L) ⇒ n = 0.125 mol/L *3.0L = 0.375 mol
MM of K2SO4
K2 = 2* 39 = 78 g/mol
S= 32 g/mol
O4 = 4*16 = 64 g/mol
MM = 78g/mol + 32g/mol + 64g/mol =174 g/mol
mass = n*MM = 0.375 mol*174 g/mol = 65.25 g
b.
M= 0.015 M
V = 0.55L
n = 0.015 mol/L * 0.55L = 0.00825 mol
MM NaF
Na: 23g/mol
F: 19 g/mol
MM = 23g/mol + 19g/mol = 42g/mol
mass = n * MM = 0.00825 mol * 42g/mol = 0.3465 g
c.
n = M*V(L) = 0.350 mol/L * 0,700L = 0.245 mol
MM = 6*12 g/mol + 12*1g/mol + 6*16g/mol = 180 g/mol
mass = n*MM = 0.245 mol * 180g/mol = 44.1 g
It is cheap, lightweight and extremely malleable.
Answer:
Solute concentration will afect the rate of a chemical reaction, because you must work with molarity
Explanation:
I think that solute mass may be it can affect the rate of reaction, if you have more mass in a solute, you will also have more moles.
If you want to know more, you have to consider temperature in the reaction and the presence of catalysts. They all, affect reactions.