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tester [92]
3 years ago
11

Is carbon mixed with sand a mixture

Chemistry
1 answer:
Julli [10]3 years ago
6 0
No because carbon is a gas and sand is a solid
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Which polysaccharides are used for STRUCTURE and SUPPORT in cells?
zhenek [66]

Answer:

In models describing the structure of the plant cell wall, the polysaccharide cellulose,which only contains glucose,forms cable-like structures which are associated with interspersed hemicellulose polysaccharides which consist of glucose chains with xylose side groups. The pectin polysaccharides, containing ironic acids, have a more kinked ribbon-like structure and serve to link the cellulose polysaccharides together by providing a gel-like matrix,completing the cell wall structure.

Explanation:

4 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
Which is a chemical property of soda ash?
Anvisha [2.4K]
Dry, white powder, soluble in water to form a slightly basic solution
4 0
4 years ago
Types OF SALTS<br>two examples​
meriva
Sea salt and rock salt, rock salt is gathered from salt deposits found underground, sea salt is obviously from the sea :/

Hope this helps you
8 0
3 years ago
Mass = 100g, volume=20ml, what is density?
Nonamiya [84]

Answer:

<h3>The answer is 5.0 g/mL</h3>

Explanation:

The density of a substance can be found by using the formula

density =  \frac{mass}{volume} \\

From the question

mass = 100 g

volume = 20 mL

So we have

density =  \frac{100}{20}  \\

We have the final answer as

<h3>5.0 g/mL</h3>

Hope this helps you

4 0
3 years ago
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