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ivolga24 [154]
3 years ago
7

Determine the percent yield if 155g of Hydrogen should have been produced,

Chemistry
1 answer:
Lyrx [107]3 years ago
7 0

Answer:

Percent yield of hydrogen is 87.096%

Explanation:

Theoretical yield = 155g

Experimental yield = 135g

% yield = [experimental yield / theoretical yield] * 100

% yield = (135 / 155) * 100

% yield = 0.87096 * 100

% yield = 87.096 %

The percentage yield of hydrogen is 87.096%

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cupoosta [38]

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Db

Dubnium/Symbol

Explanation:

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4 0
2 years ago
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Completely reacting 150.0 g of a substance with oxygen releases 395.1 J of energy. How much energy would be released if 450.0 g
MrMuchimi
To solve this problem we just need to use the rule of three:
150g..................395.1J
450g................xJ

x = 450*395.1/150 = 1185,3J

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4 0
3 years ago
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How to calculate from moles to grams?<br> 4.0 moles of Cu(CN)2
andreyandreev [35.5K]

Answer: The mass of given amount of copper (II) cyanide is 462.4 g

Explanation:

To calculate the number of moles, we use the equation:

We are given:

Moles of copper (II) cyanide = 4 moles

Molar mass of copper (II) cyanide = 115.6 g/mol

Putting values in above equation, we get:

Hence, the mass of given amount of copper (II) cyanide is 462.4 g

3 0
2 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 of the following compounds would be considered an electrolyte? a. c6h12o6 b. naoh c. co2 d. agcl
AlekseyPX

Answer: B: NaOH

Explanation:

5 0
10 months ago
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