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3 years ago

14

A solution with a ph less than 7 is called a(n) ___ and has a higher number of _ than a solution with a ph g

reater than 7.

1 answer:

castortr0y [4]3 years ago

5 0

It will be a BASE; basidity

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Quiz. Formula Weight and % Composition

astraxan [27]

Answer:

#1- the percentage is 2.77%

#2 - 218.1336g

#3- 7.89%

#4- 63.8489g

#5- 136.1406g

#6- 340.3515g

#7- 2.387238

#8-

Explanation:

7 0

4 years ago

A small amount of a solid is added to water. The observation made after fifteen minutes is shown in the figure. Which of these s

natulia [17]

Answer:

B) sand

Explanation:

A) Oil is a wrong choice because it is a liquid not a solid and also if it is oil, it will float over the water surface as a droplets.

B) Sand is the right choice, because sand is a solid and it does not dissolve in water and stabilizes at the bottom.

C) Sugar is a wrong choice, because small amount of sugar will dissolve in water and be a homogeneous solution and does not appear as a particles.

D) Wood ships is also a wrong choice, even it is a solid and does not dissolve in water, but it will float over the water surface.

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4 years ago

What are the reactants in photosynthesis?

Sveta_85 [38]

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6 0

3 years ago

Read 2 more answers

A lack of sewage structures in overcrowded cities may lead to __________.

dangina [55]

A lack of swage structures in overcrowded cities will lead to poor public health, the less structures to hold the mass of peoples sewage will have bacteria in the air causing bacterial infections. <span />

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4 years ago

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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

6 0

3 years ago