Answer: 4.08183 or you can round it to 4
Explanation:
I’m pretty sure it’s B since ones charge the neutral while ones is positive, it’s an obvious difference as well/
A. We can calculate the initial concentrations of each by
the formula:
initial concentration ci = initial volume * initial
concentration / total mixture volume
where,
total mixture volume = 10 mL + 20 mL + 10 mL + 10 mL = 50
mL
ci (acetone) = 10 mL * 4.0 M / 50 mL = 0.8 M
ci (H+) = 20 mL * 1.0 M / 50 mL = 0.4 M (note: there is only 1 H+ per
1 HCl)
ci (I2) = 10 mL * 0.0050 M / 50 mL = 0.001 M
B. The rate of reaction is determined to be complete when
all of I2 is consumed. This is signified by complete disappearance of I2 color
in the solution. The rate therefore is:
rate of reaction = 0.001 M / 120 seconds
rate of reaction = 8.33 x 10^-6 M / s
1) T<span>he dissolution of the salt potassium sulfite:
K</span>₂SO₃(aq) → 2K⁺(aq) + SO₃²⁻(aq).
Potassium has +1 charge because it lost one electron to accomplish stabile electron configuration of noble gas argon.
2) From dissolution reaction: n(K⁺) : n(SO₃²⁻) = 2 : 1.
n(K⁺) = 0.700 mol.
0.700 mol : n(SO₃²⁻) = 2 : 1.
n(SO₃²⁻) = 0.700 mol ÷ 2.
n(SO₃²⁻) = 0.350 mol; amount of sulfite anions.
Answer:
0.5mol/L
Explanation:
First, let us calculate the number of mole NaOH = 23 + 16 + 1 = 40g/mol
Mass of NaOH from the question = 30g
Number of mole = Mass /Molar Mass
Number of mole = 30/40 = 0.75mol
Volume = 1.5L
Active mass = mole/Volume
Active mass = 0.75mol/1.5L
Active mass = 0.5mol/L
