Answer:
0.0827M of H₂SO₃
Explanation:
LiOH reacts with H₂SO₃ to produce water and Li₂SO₃, thus:
2LiOH + H₂SO₃ → 2H₂O + Li₂SO₃
<em>Where 2 moles of lithium hydroxide react with 1 mole of sulfurous acid.</em>
As the chemist requires 22.14mL = 0.02214L of a 0.210M solution to neutralize the acid, moles of LiOH are:
0.02214L × (0.210mol / L) =<em>0.004649 moles of LiOH</em>.
As 2 moles of LiOH react with 1 mole of H₂SO₃, moles of H₂SO₃ are:
0.004649 moles of LiOH ₓ (1 mole H₂SO₃ / 2 mol LiOH) =
<em>0.002325 moles of H₂SO₃</em>
These moles are present in 28.10mL = 0.02810L. Thus, molar concentration of the acid is:
0.002325 moles H₂SO₃ / 0.02810L = <em>0.0827M of H₂SO₃</em>
During the light independent reaction, carbon dioxide is fixed by adding it to a <span>5-carbon compound</span>
Answer:
electroluminescence is a production of light by the flow of electrons, as within certain crystals. An example is at most resataurants with a bright sign that either says open or closed.
Explanation:
Answer:
7,01 g of indole-3-acetic acid
Explanation:
The milimolar concentration (mM) is defined as the ratio of milimoles per Liter of solution. 400mM means 400mmoles / L that is the same of <em>0,4mol / L</em>
100mL are <em>0,1L</em>. Using these values:
0,1L × (0,4mol / L ) = 0,04moles of indole-3-acetic acid.
As the MW of the molecule is 175,2 g/mol:
0,04mol × (175,2g / mol) = <em>7,01 g of indole-3-acetic acid</em>
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Thus, <em>you need 7,01 g of indole-3-acetic acid to generate your solution</em>.
I hope it helps!
If 1000 ml (1 L) of CH₃COOH contain 1.25 mol
let 250 ml of CH₃COOH contain x
⇒ x =
= 0.3125 mol
∴ moles of CH₃COOH in 250ml is 0.3125 mol
Now, Mass = mole × molar mass
= 0.3125 mol × [(12 × 2)+(16 × 2)+(1 × 4)] g/mol
= 18.75 g
∴ Mass of CH₃COOH present in a 250 mL cup of 1.25 mol/L solution of vinegar is <span>18.75 g</span>
