Answer:
11.6mL of the 0.1400M NaOH solution
Explanation:
<em>0.154g of chloroacetic acid diluted to 250mL. Titrated wit 0.1400M NaOH solution.</em>
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The reaction of chloroacetic acid, ClCH₂COOH (Molar mass: 94.5g/mol) with NaOH is:
ClCH₂COOH + NaOH → ClCH₂COO⁻ + Na⁺ + H₂O
<em>Where 1 mole of the acid reacts per mole of the base.</em>
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That means the student will reach equivalence point when moles of chloroacetic acid = moles NaOH.
You will titrate the 0.154g of ClCH₂COOH. In moles (Using molar mass) are:
0.154g ₓ (1mol / 94.5g) = <em>1.63x10⁻³ moles of ClCH₂COOH</em>
To reach equivalence point, the student must add 1.63x10⁻³ moles of NaOH. These moles comes from:
1.63x10⁻³ moles of NaOH ₓ (1L / 0.1400moles NaOH) = 0.0116L of the 0.1400M NaOH =
<h3>11.6mL of the 0.1400M NaOH solution</h3>
The partial pressure of 0.50 Ne gas is 214.71 torr
calculation
the partial pressure of Ne = moles of Ne/total moles x final pressure
find the total moles of the air mixture
that is moles of Ne + moles of K= 0.50 + 1.20 = 1.70 moles
The partial pressure is therefore = 0.50 /1.70 x 730 = 214.71 torr
I would say A but I am not sure
Answer:
9. 8.2x10¹⁴ Hz
Explanation:
Wavelength λ = 366 nm
Formula: Frequency = C / λ
To find the wavelength in m,
1 nm = 1.0x10⁻⁹ m
So, 366 nm = ?
= 3.66x10⁻⁷ m
Frequency = 3.0x10⁸m.s-1 / 3.66x10⁻⁷ m
= 8.2x10¹⁴ s-1
