Answer is: volume of H₂SO₄ is 42.1 mL.<span>
Chemical reaction: H</span>₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O.<span>
c(H</span>₂SO₄) = 0,4567 M = 0,4567 mol/L.<span>
V(NaOH) = 30 mL </span>÷ 1000 mL/L <span>= 0,03 L.
c(NaOH) = 0,321 M = 0,321 mol/L.
n(NaOH) = c(NaOH) · V(NaOH).
n(NaOH) = 0,321 mol/L · 0,030 L.
n(NaOH) = 0,00963 mol.
From chemical reaction: n(H</span>₂SO₄) : n(NaOH) = 1 : 2.<span>
n(H</span>₂SO₄) = 0,01926 mol.<span>
V(H</span>₂SO₄) = n(H₂SO₄) ÷ c(H₂SO₄).<span>
V(H</span>₂SO₄) = 0,01926 mol ÷ 0,4567 mol/L.<span>
V(H</span>₂SO₄<span>) = 0,0421 L = 42,1 mL.</span>
Answer:
The new pressure exerted by the He, is 1266.6 Torr
Explanation:
A typical problem of gases where the volume is increased and the moles and T° keeps on constant. This is an indirect proportion because when the volume of the flask is increased, pressue decreases because molecules collide to a lesser extent with the walls of the vessel.
P₁ . V₁ = P₂ . V₂
760 Torr . 5L = P₂ . 3L
P₂ = (760 Torr . 5L) / 3L
P₂ = 1266.6 Torr
Answer:
change of color
Explanation:
when a substance undergoes a change of color it is an indication that it was chemical
Lecithin is an emulsifier agent that's composed of 5 smaller molecules: phosphoric acid, choline, glycerol( is the backbone), and two fatty acids.
The fatty acids, which are hydrophobic (afraid of water), makes this substance more similiar to fats and represent the non-polar part of the lecithin.
The phosphate group is the polar portion of the molecule and it's the negatively charged. The choline is positively charged, which readily dissolve in water<span>.
</span><span>Lecithin is a good emulsifier because of these structural features. the hydrophobic contacts with the oil, while the hydrophilic end contacts with the water.</span>
Answer:
See explanation
Explanation:
The noble gas core electron configuration involves writing the inert gas core of an atom followed by the valence electrons. This is shown for the following atoms;
Bismuth;
[Xe]4f14 5d10 6s2 6p3
Chromium;
[Ar]4s1 3d5
Strontium;
[Kr]5s2
Phosphorus;
[Ne]3s2 3p3
2.
Bi
6p- n=6, l= 1, ml= 1, ms= 1/2
Cr
3d- n=3, l=2, ml=2,ms=1/2
Sr
5s- n=5, l=0, ml=0, ms=1/2
P
3p- n=3, l= 1, ml= 1, ms=1/2
3.
a) Tin (Sn) - [Kr] 5s2 4d10 5p2
b) Caesium (Cs)- [Xe] 6s1
c) Copper (Cu)- [Ar] 4s1 3d10
