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RideAnS [48]
3 years ago
15

A student analyzes soil samples from her backyard. She organizes the samples into three categories, based on their pH. The three

categories are: pH less than 8, pH between 8 and 8.5, pH greater than 8.5. Which type of graph should she use to show the percent of samples in each group?
stem and leaf plot
circle graph
line graph
scatter plot
Chemistry
2 answers:
AveGali [126]3 years ago
8 0

Answer:

The correct answer is circle graph.

Explanation:

The pie chart is also known as a circle graph has the best application in observing the relative size of different sectors. The entire circle being 100 percent, each sector will demonstrate the percent of every category. In the given question, the circle will be differentiated into three different sectors, in which the size is equivalent to the percent of each one.  

There are three categories: pH less than 8, pH in between 8 and 8.5, and pH more than 8.5. The entire circle is 100 percent, that is, the sum of the three categories would be 100 percent.  

Sedaia [141]3 years ago
5 0
I think the best answer is the last option. A scatter plot is the appropriate type of graph for the student to use to show the percent samples per group. This plot is somewhat similar to line graphs. However, they are use for  a specific purpose which is to show the relationship between two parameters. In this case, the correlation between pH and the percent of samples.
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5 0
3 years ago
CH3COOH  CH3COO– + H+
Oxana [17]

(a)

pH = 4.77

; (b)

[

H

3

O

+

]

=

1.00

×

10

-4

l

mol/dm

3

; (c)

[

A

-

]

=

0.16 mol⋅dm

-3

Explanation:

(a) pH of aspirin solution

Let's write the chemical equation as

m

m

m

m

m

m

m

m

l

HA

m

+

m

H

2

O

⇌

H

3

O

+

m

+

m

l

A

-

I/mol⋅dm

-3

:

m

m

0.05

m

m

m

m

m

m

m

m

l

0

m

m

m

m

m

l

l

0

C/mol⋅dm

-3

:

m

m

l

-

x

m

m

m

m

m

m

m

m

+

x

m

l

m

m

m

l

+

x

E/mol⋅dm

-3

:

m

0.05 -

l

x

m

m

m

m

m

m

m

l

x

m

m

x

m

m

m

x

K

a

=

[

H

3

O

+

]

[

A

-

]

[

HA

]

=

x

2

0.05 -

l

x

=

3.27

×

10

-4

Check for negligibility

0.05

3.27

×

10

-4

=

153

<

400

∴

x

is not less than 5 % of the initial concentration of

[

HA

]

.

We cannot ignore it in comparison with 0.05, so we must solve a quadratic.

Then

x

2

0.05

−

x

=

3.27

×

10

-4

x

2

=

3.27

×

10

-4

(

0.05

−

x

)

=

1.635

×

10

-5

−

3.27

×

10

-4

x

x

2

+

3.27

×

10

-4

x

−

1.635

×

10

-5

=

0

x

=

1.68

×

10

-5

[

H

3

O

+

]

=

x

l

mol/L

=

1.68

×

10

-5

l

mol/L

pH

=

-log

[

H

3

O

+

]

=

-log

(

1.68

×

10

-5

)

=

4.77

(b)

[

H

3

O

+

]

at pH 4

[

H

3

O

+

]

=

10

-pH

l

mol/L

=

1.00

×

10

-4

l

mol/L

(c) Concentration of

A

-

in the buffer

We can now use the Henderson-Hasselbalch equation to calculate the

[

A

-

]

.

pH

=

p

K

a

+

log

(

[

A

-

]

[

HA

]

)

4.00

=

−

log

(

3.27

×

10

-4

)

+

log

(

[

A

-

]

0.05

)

=

3.49

+

log

(

[

A

-

]

0.05

)

log

(

[

A

-

]

0.05

)

=

4.00 - 3.49

=

0.51

[

A

-

]

0.05

=

10

0.51

=

3.24

[

A

-

]

=

0.05

×

3.24

=

0.16

The concentration of

A

-

in the buffer is 0.16 mol/L.

hope this helps :)

6 0
2 years ago
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