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3 years ago

Determine the molar mass of CH3COOH (the solute) in a 2.0M aqueous solution of CH3COOH

Chemistry

1 answer:

ser-zykov [4K]3 years ago

4 0

Answer: 120g/mol

Explanation:

Molar mass of 1 mole of CH3COOH is ( 12 +( 1 x 3) + 12 +16 + 16 + 1) = 60g/mol.

Therefore; Molar mass of 2 moles would be; 60 x 2 = 120g/mol.

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Answer:

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An aluminum bar was found to have a mass of 27g. Using water displacement, the volume was measured to be 10 ml. What is the density of the aluminum? Group of answer choices (27 g)/(10 ml) (10 ml )/(2.70 g) (270 g)/(10 ml) (10 ml )/(27 g)

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3 years ago

A 35 L tank of oxygen is at 315 K with an internal pressure of 190 atmospheres. How

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Answer:

600.7 moles

Explanation:

Applying,

PV = nRT................... Equation 1

Where P = Pressure of oxygen, V = Volume of oxygen, n = number of moles, R = molar gas constant, T = Temperature.

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n = PV/RT............... Equation 2

From the question,

Given: P = 190 atm, V = 35 L, T = 135 K

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Is constant (matter that has a composition that is the same everywhere)

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Wat two common uses for Gallium??

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For Na2HPO4:(( (Note that for H3PO4, ka1= 6.9x10-3, ka2 = 6.4x10-8, ka3 = 4.8x10-13 ) a) The active anion is H2PO4- b) The activ

Komok [63]

Answer:

Check the explanation

Explanation:

Answer – Given, $H_3PO_4$ acid and there are three Ka values

$K_{a1}=6.9x10^8, K_{a2} = 6.2X10^8, and K_{a3}=4.8X10^{13}$

The transformation of $H_2PO_4- (aq) to HPO_4^2-(aq)$ is the second dissociation, so we need to use the Ka2 = 6.2x10-8 in the Henderson-Hasselbalch equation.

Mass of KH2PO4 = 22.0 g , mass of Na2HPO4 = 32.0 g , volume = 1.00 L

First we need to calculate moles of each

Moles of KH2PO4 = 22.0 g / 136.08 g.mol-1

= 0.162 moles

Moles of Na2HPO4 = 32.0 g /141.96 g.mol-1

= 0.225 moles

[H2PO4-] = 0.162 moles / 1.00 L = 0.162 M

[HPO42-] = 0.225 moles / 1.00 L = 0.225 M

Now we need to calculate the pKa2

pKa2 = -log Ka

= -log 6.2x10-8

= 7.21

We know Henderson-Hasselbalch equation

pH = pKa + log [conjugate base] / [acid]

pH = 7.21 + log 0.225 / 0.162

= 7.35

The pH of a buffer solution obtained by dissolving 22.0 g of KH2PO4 and 32.0 g of Na2HPO4 in water and then diluting to 1.00 L is 7.35

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3 years ago