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Andreas93 [3]
2 years ago
13

How much 1.5 m sodium hydroxide is necessary to exactly neutralize 20.0 ml of 2.5 m phosphoric acid?

Chemistry
1 answer:
andrew-mc [135]2 years ago
8 0
Answer is: 100 mL of <span>sodium hydroxide.
Chemical reaction: 3NaOH + H</span>₃PO₄ → Na₃PO₄ + 3H₂O.
V(H₃PO₄) = 20.0 mL ÷ 1000 mL/L = 0.02 L.
n(H₃PO₄) = V(H₃PO₄) · c(H₃PO₄).
n(H₃PO₄) = 0.02 L · 2.5 mol/L.
n(H₃PO₄) = 0.05 mol.
From chemical reaction: n(H₃PO₄) : n(NaOH) = 1 : 3.
n(NaOH) = 0.15 mol.
V(NaOH) = n(NaOH) ÷ c(NaOH).
V(NaOH) = 0.15 mol ÷ 1.5 mol/L.
V(NaOH) = 0.1 L · 1000 mL/L = 100 mL.
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Answer:

An octet is formed via ionic bonding when one or more valence electrons are transferred from one atom to another.

An octet is formed via covalent bonding when valence electrons are shared between atoms.

An octet is always formed via ionic bonding

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Zinc metal is added to hydrochloric acid to generate hydrogen gas and is collected over a liquid whose vapor pressure is the sam
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Question:

Zinc metal is added to hydrochloric acid to generate hydrogen gas and is collected over a liquid whose vapor pressure is the same as pure water at 20.0 degrees C (18 torr). The volume of the mixture is 1.7 L and its total pressure is 0.987 atm. Determine the number of moles of hydrogen gas present in the sample.

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Answer:

The correct option is;

E. 0.0681 mol

Explanation:

The equation for the reaction is

Zn + HCl = H₂ + ZnCl₂

Vapor pressure of the liquid = 18 torr = 2399.803 Pa

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Therefore, by Avogadro's law, pressure of the hydrogen gas is given by the following equation

Pressure of H₂ = 100007.775 Pa - 2399.803 Pa = 97607.972 Pa

Volume of H₂ = 1.7 L = 0.0017 m³

Temperature = 20 °C = 293.15 K

Therefore,

n = \frac{PV}{RT} =  \frac{100007.775 \times 0.0017 }{8.3145 \times 293.15} = 0.068078 \ moles

Therefore, the number of moles of hydrogen gas present in the sample is n ≈ 0.0681 moles.

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Answer:

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Each principal level accommodates 2n^2 electrons where n= the principal energy shell.

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<u>Answer:</u> The wavelength of light is 1.094\times 10^{-6}m

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