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4 years ago

Bases produce what in solutions ?

1 answer:

7 0

<span>These particular substances produce hydroxide ions (OH−) in aqueous solutions, and are thus classified as Arrhenius bases. For a substance to be classified as an Arrhenius base, it must produce hydroxide ions in an aqueous solution. In order to do so, Arrhenius believed the base must contain hydroxide in the formula.</span>

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What product is formed when sodium and chlorine undergo a synthesis reaction? sodium chlorite

pochemuha

Sodium chloride is the answer

6 0

3 years ago

Read 2 more answers

If a light bulb is missing or Broken in a parallel circuit,will the other bulb light

umka2103 [35]

Yes, if it’s a parallel circuit the wires are two different wires so it will light because that bulb isn’t connected to the one that went out

5 0

3 years ago

If 2.68 g of benzaldehyde are involved with the mixed aldol condensation reaction, how many moles of benzaldehyde are present? R

Zielflug [23.3K]

Answer:

The number of moles of benzaldehyde = 0.0253 moles

Explanation:

The molecular formula of benzaldehyde is C₇H₆O

Its molecular mass is calculated from the atomic masses of the constituent atoms.

C = 12.0 g: H = 1.0 g; O = 16.0 g

Molecular mass = ( 12 * 7) + (1 * 6) + (16 * 1) = 106.0 g/mol

Number of moles of substance = mass of substance/ molar mass of the substance

mass of benzaldehyde = 2.68; molar mass = 106.0 g/mol

Number of moles of benzaldehyde = 2.68 g/ 106 g/mol = 0.0253 moles

Therefore, the number of moles of benzaldehyde = 0.0253 moles

8 0

3 years ago

IXL Science (PLS HELP)

frez [133]

There are 12 Hydrogens (H) and 12 Oxygens (O) and 6 molecules of Hydrogen Peroxide (H2O2) reacted

6 0

3 years ago

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

6 0

3 years ago